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# Number 58368256848 ### Properties of number 58368256848 Cross Sum: Factorization: 2 * 2 * 2 * 2 * 3 * 3 * 11 * 11 * 29 * 115513 Divisors: Count of divisors: Sum of divisors: Prime number? No Fibonacci number? No Bell Number? No Catalan Number? No Base 3 (Ternary): Base 4 (Quaternary): Base 5 (Quintal): Base 8 (Octal): d9704eb50 Base 32: 1mbg9qqg sin(58368256848) 0.10392580458878 cos(58368256848) 0.99458505274339 tan(58368256848) 0.10449162120637 ln(58368256848) 24.790038031882 lg(58368256848) 10.766176723418 sqrt(58368256848) 241595.23349603 Square(58368256848) 3.4068534074741E+21 ### Number Look Up Look Up 58368256848 (fifty-eight billion three hundred sixty-eight million two hundred fifty-six thousand eight hundred forty-eight) is a great number. The cross sum of 58368256848 is 63. If you factorisate the number 58368256848 you will get these result 2 * 2 * 2 * 2 * 3 * 3 * 11 * 11 * 29 * 115513. The number 58368256848 has 180 divisors ( 1, 2, 3, 4, 6, 8, 9, 11, 12, 16, 18, 22, 24, 29, 33, 36, 44, 48, 58, 66, 72, 87, 88, 99, 116, 121, 132, 144, 174, 176, 198, 232, 242, 261, 264, 319, 348, 363, 396, 464, 484, 522, 528, 638, 696, 726, 792, 957, 968, 1044, 1089, 1276, 1392, 1452, 1584, 1914, 1936, 2088, 2178, 2552, 2871, 2904, 3509, 3828, 4176, 4356, 5104, 5742, 5808, 7018, 7656, 8712, 10527, 11484, 14036, 15312, 17424, 21054, 22968, 28072, 31581, 42108, 45936, 56144, 63162, 84216, 115513, 126324, 168432, 231026, 252648, 346539, 462052, 505296, 693078, 924104, 1039617, 1270643, 1386156, 1848208, 2079234, 2541286, 2772312, 3349877, 3811929, 4158468, 5082572, 5544624, 6699754, 7623858, 8316936, 10049631, 10165144, 11435787, 13399508, 13977073, 15247716, 16633872, 20099262, 20330288, 22871574, 26799016, 27954146, 30148893, 30495432, 36848647, 40198524, 41931219, 45743148, 53598032, 55908292, 60297786, 60990864, 73697294, 80397048, 83862438, 91486296, 110545941, 111816584, 120595572, 125793657, 147394588, 160794096, 167724876, 182972592, 221091882, 223633168, 241191144, 251587314, 294789176, 331637823, 335449752, 405335117, 442183764, 482382288, 503174628, 589578352, 663275646, 670899504, 810670234, 884367528, 1006349256, 1216005351, 1326551292, 1621340468, 1768735056, 2012698512, 2432010702, 2653102584, 3242680936, 3648016053, 4864021404, 5306205168, 6485361872, 7296032106, 9728042808, 14592064212, 19456085616, 29184128424, 58368256848 ) whith a sum of 185743046580. The figure 58368256848 is not a prime number. 58368256848 is not a fibonacci number. 58368256848 is not a Bell Number. The number 58368256848 is not a Catalan Number. The convertion of 58368256848 to base 2 (Binary) is 110110010111000001001110101101010000. The convertion of 58368256848 to base 3 (Ternary) is 12120122210011122221200. The convertion of 58368256848 to base 4 (Quaternary) is 312113001032231100. The convertion of 58368256848 to base 5 (Quintal) is 1424014233204343. The convertion of 58368256848 to base 8 (Octal) is 662701165520. The convertion of 58368256848 to base 16 (Hexadecimal) is d9704eb50. The convertion of 58368256848 to base 32 is 1mbg9qqg. The sine of the figure 58368256848 is 0.10392580458878. The cosine of 58368256848 is 0.99458505274339. The tangent of the number 58368256848 is 0.10449162120637. The square root of 58368256848 is 241595.23349603. If you square 58368256848 you will get the following result 3.4068534074741E+21. The natural logarithm of 58368256848 is 24.790038031882 and the decimal logarithm is 10.766176723418. You should now know that 58368256848 is great figure!

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# What is Equipment Depreciation? (with picture) Jim B. Jim B. Equipment depreciation refers to the process by which equipment used for business purposes loses value over each year of its life span. This is an important concept for business owners to understand, as they are allowed to write off this loss of value each year for tax purposes. The amount of equipment depreciation each year depends on how many years the equipment is scheduled to be used and the method of depreciation used. Most commonly, depreciation is calculated using either the straight-line method or the declining balance method. Many businesses rely on various pieces of equipment to operate on a daily basis, such as the machinery used in factories or the various tools used in construction. Each year that this equipment is in use, it loses a little bit of value, until it eventually reaches a point where it has little or no value to the business due to the wear and tear it has endured. The value lost is known as equipment depreciation, a concept by which businesses are able to offset that loss of value in tax write-offs. Any piece of equipment used for business purposes for a period longer than a year is subject to equipment depreciation. This yearly amount lost is realized on tax returns as the depreciation expense and differs from the accumulated depreciation, which is recognized on a balance sheet as a running amount. For example, a piece of equipment that depreciates each year by \$200 US Dollars (USD) will have a depreciation expense for that amount each year, but the accumulated depreciation will be \$200 USD in the first year, \$400 USD the next year, and so on. The amount of equipment depreciation realized each year depends on the method of depreciation used. In the straight-line method, a piece of equipment depreciates the same amount each year, an amount reached by dividing the cost of the equipment by its life span. For example, if a piece of equipment is worth \$500 USD at purchase and has a life span of five years, it would have a depreciation expense of \$100 USD, or \$500 USD divided by five. Some businesses prefer to expense a piece of equipment most heavily in the year that it is bought, and the declining balance method of equipment depreciation allows this. In this method, a percentage rate of depreciation is applied to the balance of the original cost. Using the example above, if the depreciation rate is 50 percent, then the first year's depreciation expense will be \$500 USD multiplied by 0.5, which yields \$250 USD. The next year the balance of the cost would be down to \$250 USD, or \$500 USD minus \$250 USD, and the 50 percent rate would then be applied to that amount to calculate the depreciation expense for year two.

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Code covered by the BSD License ### Highlights from Numerical Differentiation 5.0 5.0 | 1 rating Rate this file 15 Downloads (last 30 days) File Size: 249 KB File ID: #22807 Version: 1.2 # Numerical Differentiation ### Husam Aldahiyat (view profile) 28 Jan 2009 (Updated ) Performs single dimensional differentiation numerically. File Information Description This is a GUI which performs numerical differentiation of a function over a number of equaly spaced points. Also with it is a code that grants the coefficients used for numerical differentiation. The pictures and example should be more than enough for understanding how to use the file. Example: npoints=3; Order=1; d=datnum(npoints,Order) d= -1.5 2 -0.5 % Forward -0.5 -0 0.5 % Central 0.5 -2 1.5 % Backward % The result is a matrix consisting of coefficients that can be % used to numerically differentiate, like this: x=1; f=inline('cos(x)') h=.1; s = ( d(1,1)*f(x) + d(1,2)*f(x+h) + d(1,3)*f(x+2*h) )/h^Order s = -0.8444 s = ( d(2,1)*f(x-h) + d(2,2)*f(x) + d(2,3)*f(x+h) )/h^Order s = -0.84007 s = ( d(3,1)*f(x-2*h) + d(3,2)*f(x-h) + d(3,3)*f(x) )/h^Order s = -0.84413 % The true answer is s = -0.84147 The code uses the Symbolic Math Toolbox to obtain the true value (in order to calculate the error). If you don't have the Symbolic Math Toolbox then you won't enjoy this benefit (program still works though). Acknowledgements Adaptive Robust Numerical Differentiation inspired this file. MATLAB release MATLAB 7.4 (R2007a) Other requirements Symbolic Math Toolbox (Optional) 07 Dec 2009 Zas Babyk

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How to calculate an exponent A step-by-step walk-through of how to type scientific notation into your calculator and do simple calculations with them. • Fast Expert Tutoring • Solving word questions • Clarify mathematic questions Exponent Calculator Exponent Notation: Note that -4 2 and (-4) 2 result in different answers: -4 2 = -1 * 4 * 4 = -16, while (-4) 2 = (-4) * (-4) = 16. If you enter a negative value for x, such as -4, this calculator assumes (-4) n. When a minus sign occurs with How to Calculate Exponents a fraction ( 1/n) part. So, because m/n = m × (1/n) we can do this: x m/n = x (m × 1/n) = (x m) 1/n = n√xm. The order does not matter, so it also works for m/n = (1/n) × m: x m/n = x (1/n × m) = (x Decide mathematic questions Math is the study of numbers, shapes, and patterns. It is used to solve problems and to understand the world around us. Solve word queries Solving math problems can be a fun and rewarding experience. Track Improvement The track has been improved and is now open for use. User reviews Actually it's really a smart app, even though u have to pay for the premium, you don't really have to because you can always wait for the ads, and know the steps of ur answer, like let's be honest its free, waiting isn't a big deal for me, so I would highly recommend this app, you'll like have to wait 2 to 5 minutes to get ads, but it's worth it because all the answers are correct. Anthony Blanton Great app, solves questions for all grades and can understand a picture, Crazy to me and just a press of a button and solve every question on my homework, yes please. You can download without any confusion. Make some improvement. Guillermo Stinson

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## On the dynamic stability of flexible supersonic vehicles The dynamic stability of a long, slender-bodied vehicle with a flexible fuselage is examined analytically in the supersonic speed regime. The small aspect ratio lifting surfaces are considered to be rigid but dependence of their angles of attack on fuselage flexibility is accounted for. The amplitude of pitching oscillation is restricted to ±10° about the zero-lift line by the nature of the unsteady, supersonic aerodynamic theory used. The stability problem is formulated by a set of non-linear differential equations with the non-linear contributions arising from both the inertia and the aerodynamic forces. The present analysis accounts for non-linear contributions up to third degree in the rigid body angle of attack. The stability of the short period mode is investigated using Routh-Hurwitz criteria and an expression representing a stiffness criterion for dynamic stability is obtained. The analytical development is so presented as to make it easily applicable to a supersonic, flexible vehicle with or without wings, e.g. a supersonic transport or a missile. Moreover to facilitate the evaluation of the effect of flexibility and non-linearities on dynamic stability, four cases are considered separately: a. Rigid body equations of motion, without non-linear terms b. Rigid body equations of motion, with non-linear terms c. Flexible body equations of motion, without non-linear terms d. Flexible body equations of motion, with non-linear terms. A numerical example is presented towards the end which investigates the dynamic stability of a flexible, supersonic transport configuration. The conclusions from the example are: 1. The non-linearities can be safely neglected for rigid aircraft, but not for wingless vehicles. 2. Flexibility affects the stability through the lift and pitching moment and also by introducing two more possible equilibrium points. 3. The amount of work involved in finding a solution is markedly increased by the necessity of solution of more characteristic equations of higher degree. 4. The stiffness criterion can be used to adjust the stiffness distribution to one that can make an unstable configuration stable. The usefulness of the method is two-fold. For a flexible vehicle with known geometric, mass and elastic properties, the method can predict its dynamic stability. This feature is of considerable importance particularly in the design stage. On the other hand, if an aircraft with known geometry, total mass and centre-of-gravity location proves to be dynamically unstable, then the analysis provides a stiffness criterion by which it can be made stable. The analysis involves a considerable amount of computation and hence seems to be particularly suited for solution by a digital computer. / Applied Science, Faculty of / Mechanical Engineering, Department of / Graduate Identifer oai:union.ndltd.org:UBC/oai:circle.library.ubc.ca:2429/38867 Date January 1963 Creators Drummond, Alastair Milne Publisher University of British Columbia Source Sets University of British Columbia Language English Detected Language English Type Text, Thesis/Dissertation Rights For non-commercial purposes only, such as research, private study and education. Additional conditions apply, see Terms of Use https://open.library.ubc.ca/terms_of_use. Page generated in 0.0022 seconds

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Example 15.90.9. Let $k$ be a field and put $R = k[f, T_1, T_2, \ldots ]/(fT_1, fT_2 - T_1, fT_3 - T_2, \ldots ).$ Then $(R, f)$ is not a glueing pair because the map $R[f^\infty ] \to R^\wedge [f^\infty ]$ is not injective as the image of $T_1$ is $f$-divisible in $R^\wedge$. For $R = k[f, T_1, T_2, \ldots ]/(fT_1, f^2T_2, \ldots ),$ the map $R[f^\infty ] \to R^\wedge [f^\infty ]$ is not surjective as the element $T_1 + fT_2 + f^2 T_3 + \ldots$ is not in the image. In particular, by Remark 15.90.8, these are both examples where $R \to R^\wedge$ is not flat. There are also: • 4 comment(s) on Section 15.90: The Beauville-Laszlo theorem In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

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# measuring how much is 6 cups to pounds.... That depends upon whether you're measuring something heavy, like lead pellets, or light, like feathers. Volume measures (cups) don't have equivalent weight (pounds) measurements. If you're measuring a food (apples, peaches, flour, sugar, etc.) for a recipe, you'll need to guess how many pounds you'll need to make six cups. Sometimes food packages give the volume on the label. 1. 👍 0 2. 👎 0 3. 👁 64 ## Similar Questions 1. ### biostatistics among coffee drinkers, men drink a mean of 3.2 cups per day with a standard deviation of 0.8 cups. Assume the number of coffee drinks per day follows a normal distribution. a. What proportion drinks 2 cups per day or more? b. What asked by natasha on November 4, 2011 2. ### Math It is common to see mixed fractions in recipes. A recipe for pizza crust may ask for 1 1/2 cups of flour. You could measure this amount in two ways. You could fill a one-cup measuring cup with flour and a one-half cup measuring asked by Jerald on April 29, 2013 3. ### math If a recipe calls for 2 2\4 of something but I only have 1\4 measuring cups. How many 1\4 cups do I need to use? asked by kentrell on October 2, 2014 4. ### math A recipe calls for 2 2/4 cups of raisins, but Julie only has a 1/4 cup measuring cup. How many 1/4 cups does Julie need to measure out 2 2/4 cups of raisins? My answer is, Julie need 10 cups. asked by ezekiel on November 18, 2014 5. ### math 2 Ben is making a bread that calls for 5 cups of flour. He only has 1/2 cup measuring cup. How many times will Ben need to fill the 1/2 cup measuring cup to get the 5 cups of flour? asked by stefie on May 14, 2014 6. ### Math Hi! You need 2 cups of water, but you only have a liquid measuring cup that measures up to 3 pints. what can you do to get the 2 cups of water that you need? explain and show your work. asked by Jessica L. on January 15, 2018 7. ### Math Ben is making a bread that calls for 5 cups of flour.He only has a cup 1/2-cup measuring cup.how many times will Ben need to fill the 1/2-cup measuring cup to get the 5 cups of flour? 8. ### Math Pen has 1/2 cup and 1/8 cup measuring cups. What are two ways he could measure out 1 3/4 cups of flour? Plz help thanks! asked by Pam on July 5, 2016 9. ### math I solved these can you chack them 1. 24 quarts = 6 gallons? 2. 5 cups = 40 oz? 3. 8 cups = 4 pints? 4. 81000 pounds = 4 1/2 tons? 5. 1 quart = 4 cups? 6. 8 gallons = 32 quarts? 7. 32 quarts = 8 gallons? 8. 13 quarts = 3.25 asked by Dakota on January 9, 2010 10. ### Math Help, I need help on this.. For questions 3-5, use the recipe below. (I'll try and make it..) ----------------------- |Ingredients|Amount| |------------------- |Orange Juice| 4 3/4 cups| |Grape juice| 6 1/2 cups| | Cherry Juice 2 asked by Lexi on November 4, 2014 More Similar Questions

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# Lab_7 - Physics 272Lab Lab 7: Magnetic Field of... This preview shows pages 1–3. Sign up to view the full content. Physics 272Lab Lab 7: Magnetic Field of Current-Carrying Wires Lab 7: Magnetic Field of Current-Carrying Wires OBJECTIVES In this lab you will Measure the deflection of a compass needle due to a magnetic field of a wire Test the relation between current and magnetic field strength Calculate the distance dependence of the magnetic field from a wire You have seen in class how a moving charge will create a microscopic magnetic field. You have also seen how a large number of moving charges (current) will create a macroscopic magnetic field given by: 0 2 0 ˆ μ I Δl ×r ΔB= 4 πε r You can measure this macroscopic field with a simple compass. 1) Warm-Up Problem Problem (1) What is the direction of the magnetic field at the indicated locations inside and outside this current carrying rectangular coil of wire shown? Explain briefly. CHECKPOINT 1: Ask an instructor to check your work for credit. You may proceed while you wait to be checked off 2) Measuring a Magnetic Field with a Compass a) Get out the following items for this lab: 1 short bulb with holder 3 batteries with holder (you may need to temporarily borrow the third battery) 4 alligator clips 1 long wire 1 compass b) Put away everything else (except your lab manual), so your table is neat and clear. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Physics 272Lab Lab 7: Magnetic Field of Current-Carrying Wires c) Make a circuit with two-batteries, a short bulb, the long wire, and the four alligator clips (Fig. 1). Leave one end of one of the alligator clips unconnected (bulb not lit). d) Place your magnetic compass on a flat surface Make sure the compass is not near steel objects, such as the circuit you just made, and the computer. e) Once the compass needle settles, align the north marker with the red end of the compass needle. f) Continue aligning the compass until, at rest, the red end of the needle is aligned with the north marker. g) Stretch your long wire out straight, running north-south. h) Make the last connection so that the bulb glows (the bulb must glow). i) Lift the wire up above the compass. Keeping it stretched out north-south. j) This is the end of the preview. Sign up to access the rest of the document. ## This note was uploaded on 04/23/2011 for the course PHYS 272 taught by Professor K during the Spring '07 term at Purdue University. ### Page1 / 7 Lab_7 - Physics 272Lab Lab 7: Magnetic Field of... This preview shows document pages 1 - 3. Sign up to view the full document. View Full Document Ask a homework question - tutors are online

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# On the regularity of solutions of the boundary value problem without initial condition for Schrödinger systems in domain with conical points ## Abstract In this paper, we deal with the boundary value problems without initial condition for Schrödinger systems in cylinders. We establish several results on the regularity of the solutions. MSC: 35Q41, 35B65, 35D30. ## 1 Introduction The theory of initial boundary value problems for partial differential equations and systems in nonsmooth domains has attracted the attention of many researchers. These problems for Schrödinger systems in domain with conical points are considered in [1], [2], in which authors consider the existence, the uniqueness and the regularity of solutions of the mentioned problems. While problems without initial conditions arise when describing different nonstationary processes in nature under the hypothesis that the initial condition practically has no influence at the present time. These problems are investigated in many works, see [3]–[6] for example, with results only about the well-posedness. In [7], we studied the boundary value problem without initial condition for Schrödinger systems in domain with conical points. By studying the corresponding problem with initial condition $t=h$, then passing to the limits $h\to -\mathrm{\infty }$, we obtained the existence and uniqueness of the solution. In the present paper, we continue considering the mentioned problem and our main aim here is to study the regularity of the solution. There are four sections in this paper. In Section 1, we set the problem and recall an obvious result about the unique existence of solution. Afterwards, in Sections 2 and 3, by a similar method to [1], [2], we give results on the regularity of problems with initial condition $t=h$ for Schrödinger systems in domains with conical points. Then, by letting $h\to -\mathrm{\infty }$, the smoothness of the generalized solution of our problem is obtained in Section 4. ## 2 Setting problem and obvious result Let $\mathrm{\Omega }$ be a bounded domain in ${\mathbb{R}}^{n}$ ($n\ge 2$) with boundary $S=\partial \mathrm{\Omega }$. Assume that $S$ is an infinitely differentiable surface everywhere, except for the coordinate origin, in a neighborhood ${U}_{0}$ satisfying $\mathrm{\Omega }\cap {U}_{0}$ and coincides with the cone $K=\left\{x:x/|x|\in G\right\}$, where $G$ is a smooth domain on the unit sphere ${S}^{n-1}$. For $a, set ${\mathrm{\Omega }}_{a}^{b}=\mathrm{\Omega }×\left(a,b\right)$, ${S}_{a}^{b}=S×\left(a,b\right)$. If $\left(a,b\right)=\mathbb{R}$, we use ${\mathrm{\Omega }}_{\mathbb{R}}$ to refer to ${\mathrm{\Omega }}_{-\mathrm{\infty }}^{\mathrm{\infty }}$ and ${S}_{\mathbb{R}}$ to refer to ${S}_{-\mathrm{\infty }}^{\mathrm{\infty }}$. For each multi-index $\alpha =\left({\alpha }_{1},\dots ,{\alpha }_{n}\right)\in {\mathbb{N}}^{n}$, set $|\alpha |={\alpha }_{1}+\cdots +{\alpha }_{n}$ and ${\partial }^{\alpha }={\partial }_{{x}_{1}}^{{\alpha }_{1}}\cdots {\partial }_{{x}_{n}}^{{\alpha }_{n}}$. Denote $u\left(x,t\right)=\left({u}_{1}\left(x,t\right),\dots ,{u}_{s}\left(x,t\right)\right)$, ${D}^{\alpha }u=\left({D}^{\alpha }{u}_{1},\dots ,{D}^{\alpha }{u}_{s}\right)$, ${|{D}^{\alpha }u|}^{2}={\sum }_{i=1}^{s}{|{D}^{\alpha }{u}_{i}|}^{2}$ and ${u}_{{t}^{j}}=\left(\frac{{\partial }^{j}{u}_{1}}{\partial {t}^{j}},\dots ,\frac{{\partial }^{j}{u}_{s}}{\partial {t}^{j}}\right)$, ${|{u}_{{t}^{j}}|}^{2}={\sum }_{i=1}^{s}{|\frac{{\partial }^{j}{u}_{i}}{\partial {t}^{j}}|}^{2}$. Let us introduce some functional spaces (see [7]) used in this paper. We use ${H}^{k}\left(\mathrm{\Omega }\right)$ to denote the space of $s$-dimensional vector functions defined in $\mathrm{\Omega }$ with the norm ${\parallel u\parallel }_{{H}^{k}\left(\mathrm{\Omega }\right)}={\left(\sum _{|\alpha |=0}^{k}{\int }_{\mathrm{\Omega }}{|{D}^{\alpha }u|}^{2}\phantom{\rule{0.2em}{0ex}}dx\right)}^{\frac{1}{2}}.$ Denote by ${H}^{k,l}\left({\mathrm{\Omega }}_{a}^{b}\right)$ the space consisting of all vector functions $u:{\mathrm{\Omega }}_{a}^{b}⟶{\mathbb{C}}^{s}$ satisfying ${\parallel u\parallel }_{{H}^{k,l}\left({\mathrm{\Omega }}_{a}^{b}\right)}={\left({\int }_{{\mathrm{\Omega }}_{a}^{b}}\left(\sum _{|\alpha |=0}^{k}{|{D}^{\alpha }u|}^{2}+\sum _{j=1}^{l}{|{u}_{{t}^{j}}|}^{2}\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}},$ and ${H}^{k,l}\left(-\gamma ,{\mathrm{\Omega }}_{a}^{b}\right)$ is the space of vector functions with norm ${\parallel u\parallel }_{{H}^{k,l}\left(-\gamma ,{\mathrm{\Omega }}_{a}^{b}\right)}={\left({\int }_{{\mathrm{\Omega }}_{a}^{b}}\left(\sum _{|\alpha |=0}^{k}{|{D}^{\alpha }u|}^{2}+\sum _{j=1}^{l}{|{u}_{{t}^{j}}|}^{2}\right){e}^{-2\gamma t}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}.$ In particular, ${\parallel u\parallel }_{{H}^{k,0}\left(-\gamma ,{\mathrm{\Omega }}_{a}^{b}\right)}={\left(\sum _{|\alpha |=0}^{k}{\int }_{{\mathrm{\Omega }}_{a}^{b}}{|{D}^{\alpha }u|}^{2}{e}^{-2\gamma t}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}.$ Especially, we set ${L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{a}^{b}\right)={H}^{0,0}\left(-\gamma ,{\mathrm{\Omega }}_{a}^{b}\right)$. ${H}_{\beta }^{l,k}\left(-\gamma ,{\mathrm{\Omega }}_{a}^{b}\right)$ is the space of all functions $u\left(x,t\right)$ which have generalized derivatives ${D}^{\alpha }u$, $\frac{{\partial }^{j}u}{\partial {t}^{j}}$, $|\alpha |\le l$, $1\le j\le k$, satisfying ${\parallel u\parallel }_{{H}_{\beta }^{l,k}\left(-\gamma ,{\mathrm{\Omega }}_{a}^{b}\right)}^{2}={\int }_{{\mathrm{\Omega }}_{a}^{b}}\left(\sum _{\alpha =0}^{l}{r}^{2\left(\beta +|\alpha |-l\right)}{|{D}^{\alpha }u|}^{2}+\sum _{j=1}^{k}{|{u}_{{t}^{j}}|}^{2}\right){e}^{-2\gamma t}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt<\mathrm{\infty }.$ ${H}_{\beta }^{l}\left(-\gamma ,{\mathrm{\Omega }}_{a}^{b}\right)$ is the space of all functions $u\left(x,t\right)$ which have generalized derivatives ${D}^{\alpha }{u}_{{t}^{j}}$, $|\alpha |\le l$, $1\le j\le l$, satisfying ${\parallel u\parallel }_{{H}_{\beta }^{l}\left(-\gamma ,{\mathrm{\Omega }}_{a}^{b}\right)}^{2}={\int }_{{\mathrm{\Omega }}_{a}^{b}}\sum _{\alpha +j=0}^{l}{r}^{2\left(\beta +|\alpha |+j-l\right)}{|{D}^{\alpha }{u}_{{t}^{j}}|}^{2}{e}^{-2\gamma t}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt<\mathrm{\infty }.$ Denote by ${\stackrel{\circ }{H}}^{k,l}\left(-\gamma ,{\mathrm{\Omega }}_{a}^{b}\right)$ the completion of infinitely differentiable vector functions vanishing near ${S}_{a}^{b}$ with respect to ${H}^{l,k}\left(-\gamma ,{\mathrm{\Omega }}_{a}^{b}\right)$ norm. Now we introduce a differential operator of order $2m$ $L\left(x,t,D\right)=\sum _{|p|,|q|=0}^{m}{\left(-1\right)}^{|p|}{D}^{p}\left({a}_{pq}\left(x,t\right){D}^{q}\right),$ where ${a}_{pq}$ are $s×s$ matrices with the bounded complex-valued components in ${\overline{\mathrm{\Omega }}}_{\mathbb{R}}$, ${a}_{pq}={a}_{qp}^{\ast }$ (${a}_{qp}^{\ast }$ is the transposed conjugate matrix to ${a}_{pq}$). Set $B\left(t,u,v\right)=\sum _{|p|,|q|=0}^{m}{\int }_{\mathrm{\Omega }}{a}_{pq}{D}^{q}u\overline{{D}^{p}v}\phantom{\rule{0.2em}{0ex}}dx,\phantom{\rule{1em}{0ex}}t\in \mathbb{R}.$ We assume further that the form $B\left(t,\cdot ,\cdot \right)$ satisfies the following condition: there exists a constant ${\mu }_{0}>0$ independent of $t$ and $u$ such that $B\left(t,u,u\right)\ge {\mu }_{0}{\parallel u\left(\cdot ,t\right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}$ (2.1) for all $u\in {\stackrel{\circ }{H}}^{m}\left(\mathrm{\Omega }\right)$ and a.e. $t\in \mathbb{R}$, where ${\stackrel{\circ }{H}}^{m}\left(\mathrm{\Omega }\right)$ is the closure in ${H}^{m}\left(\mathrm{\Omega }\right)$ of infinitely differentiable complex $s$-dimensional vector functions with compact support in $\mathrm{\Omega }$. Now we consider the following problem in the cylinder ${\mathrm{\Omega }}_{\mathbb{R}}$: (2.2) $\frac{{\partial }^{j}u}{\partial {\nu }^{j}}{|}_{{S}_{\mathbb{R}}}=0,\phantom{\rule{1em}{0ex}}j=0,\dots ,m-1,$ (2.3) where $\nu$ is the unit vector of outer normal to the surrounding surface ${S}_{\mathbb{R}}$. Let $f\in {L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$, a complex vector-valued function $u\in {\stackrel{\circ }{H}}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$ is called a generalized solution of problem (2.2)-(2.3) if and only if, for any $T>0$, the equality ${\left(-1\right)}^{m-1}i{\int }_{-\mathrm{\infty }}^{T}B\left(t,u,\eta \right)\phantom{\rule{0.2em}{0ex}}dt+{\int }_{{\mathrm{\Omega }}_{-\mathrm{\infty }}^{T}}u\overline{{\eta }_{t}}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt={\int }_{{\mathrm{\Omega }}_{-\mathrm{\infty }}^{T}}f\overline{\eta }\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt$ (2.4) holds for all $\eta \in {\stackrel{\circ }{H}}^{m,1}\left(\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$, $\eta \left(x,t\right)=0$ with $t\ge T$. For any $h\in \mathbb{R}$, we also consider the following problem with initial condition corresponding to problem (2.2)-(2.3): (2.5) $v{|}_{t=h}=0,\phantom{\rule{1em}{0ex}}x\in \mathrm{\Omega },$ (2.6) $\frac{{\partial }^{j}v}{\partial {\nu }^{j}}{|}_{{S}_{h}^{\mathrm{\infty }}}=0,\phantom{\rule{1em}{0ex}}j=0,\dots ,m-1.$ (2.7) A function $v\in {\stackrel{\circ }{H}}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)$ is a generalized solution of problem (2.5)-(2.7) if and only if, for any $T>0$, we have ${\left(-1\right)}^{m-1}i{\int }_{h}^{T}B\left(t,v,\eta \right)\phantom{\rule{0.2em}{0ex}}dt+{\int }_{{\mathrm{\Omega }}_{h}^{T}}v\overline{{\eta }_{t}}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt={\int }_{{\mathrm{\Omega }}_{h}^{T}}f\overline{\eta }\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt$ (2.8) for all $\eta \in {\stackrel{\circ }{H}}^{m,1}\left(\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)$, $\eta \left(x,t\right)=0$ for all $t\ge T$. We recall the results about the solvability of problem (2.2)-(2.3) proved in [7]. ### Theorem 2.1 Assume that: 1. (i) $sup\left\{|\frac{\partial {a}_{pq}}{\partial t}|:\left(x,t\right)\in {\mathrm{\Omega }}_{\mathbb{R}},0\le |p|,|q|\le m\right\}=\mu <\mathrm{\infty }$; 2. (ii) $|\frac{\partial {a}_{pq}}{\partial t}|\le {\mu }_{1}\cdot {e}^{2\gamma t}$, for all $\left(x,t\right)\in {\mathrm{\Omega }}_{\mathbb{R}}$, $0\le |p|,|q|\le m$; 3. (iii) $f,{f}_{t}\in {L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$, where$\mu$, ${\mu }_{1}$are positive constants. Then for all$\gamma >{\gamma }_{0}=\frac{{m}^{\star }\mu }{2{\mu }_{0}}$, ${m}^{\star }={\sum }_{|\alpha |\le m}1$, there exists a uniquely generalized solution$u\in {\stackrel{\circ }{H}}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$of problem (2.2)-(2.3) satisfying ${\parallel u\parallel }_{{\stackrel{\circ }{H}}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)}^{2}\le C\left[{\parallel f\parallel }_{{L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)}^{2}+{\parallel {f}_{t}\parallel }_{{L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)}^{2}\right].$ ## 3 The regularity with respect to time variable of the problem with initial condition The regularity of problem (2.2)-(2.3) is implied by analogous property for problem (2.5)-(2.7). In [1], the regularity with respect to time variable of problem (2.5)-(2.7) is studied in the case $h=0$ and ${f}_{{t}^{k}}\in {L}^{\mathrm{\infty }}\left(0,\mathrm{\infty };{L}_{2}\left(\mathrm{\Omega }\right)\right)$. Now, by a similar method, we consider the problem in the case $h\in \mathbb{R}$ and ${f}_{{t}^{k}}\in {L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)$. Moreover, we also show that the constants $C$ in all prior estimates do not depend on $h$. ### Theorem 3.1 Assume that $l\in \mathbb{N}$ and there exists $\mu >0$ such that 1. (i) $sup\left\{|\frac{\partial {a}_{pq}}{\partial t}|:\left(x,t\right)\in {\mathrm{\Omega }}_{\mathbb{R}},0\le |p|,|q|\le m,\right\}=\mu <\mathrm{\infty }$, $|\frac{{\partial }^{k}{a}_{pq}}{\partial {t}^{k}}|\le {\mu }_{2}$, ${\mu }_{2}>0$, for $2\le k\le l+1$; 2. (ii) ${f}_{{t}^{k}}\in {L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)$, for $k\le l+1$, ${f}_{{t}^{k}}\left(x,h\right)=0$ for all $x\in \mathrm{\Omega }$, $0\le k\le l$. Then for all$\gamma >\left(2l+1\right){\gamma }_{0}$, the solution$v\in {\stackrel{\circ }{H}}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$of problem (2.5)-(2.7) has derivatives with respect to$t$up to order$l$satisfying${v}_{{t}^{k}}\in {H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)$, $\mathrm{\forall }k=\overline{0,l}$, such that ${\parallel {v}_{{t}^{k}}\parallel }_{{H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\sum _{{m}_{1}=0}^{k+1}{\parallel {f}_{{t}^{{m}_{1}}}\parallel }_{{L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2},$ (3.1) where the constant$C$does not depend on$v$, $f$, $h$. ### Proof Suppose that ${\left\{{\phi }_{k}\right\}}_{k=1}^{\mathrm{\infty }}$ is an orthogonal basis of ${H}^{m}\left(\mathrm{\Omega }\right)$ which is orthonormal in ${L}_{2}\left(\mathrm{\Omega }\right)$. For any $N\in \mathbb{N}$, the approximating solutions can be found in the form ${v}^{N}\left(x,t\right)={\sum }_{k=0}^{N}{C}_{k}^{N}\left(t\right){\phi }_{k}\left(x\right)$ where ${\left({C}_{k}^{N}\left(t\right)\right)}_{k=1}^{N}$ is the solution of the ordinary differential system ${\left(-1\right)}^{m-1}iB\left(t,{v}^{N},{\phi }_{k}\right)-{\int }_{\mathrm{\Omega }}{v}_{t}^{N}\overline{{\phi }_{k}}\phantom{\rule{0.2em}{0ex}}dx={\int }_{\mathrm{\Omega }}f\overline{{\phi }_{k}}\phantom{\rule{0.2em}{0ex}}dx,$ (3.2) ${C}_{k}^{N}\left(h\right)=0,\phantom{\rule{1em}{0ex}}k=1,\dots ,N.$ (3.3) From the assumptions it follows that the coefficients ${C}_{k}^{N}\left(t\right)$, defined uniquely by (3.2), have derivatives up to order l + 1 and ${v}^{N}\left(x,h\right)=0$. It is very easy to check that ${D}^{p}{v}^{N}\left(x,h\right)=0$, $\mathrm{\forall }0\le |p|\le m$, $x\in \mathrm{\Omega }$. Step 1: Prove that ${D}^{p}{v}_{{t}^{k}}^{N}\left(x,h\right)=0$ for all $0\le k\le l$, $0\le |p|\le m$, $x\in \mathrm{\Omega }$. Taking the derivative $\left(k-1\right)$ times of both sides of (3.2) with respect to $t$, then multiplying both sides by $\frac{{d}^{k}}{d{t}^{k}}\overline{\left({C}_{k}^{N}\left(t\right)\right)}$ and taking the sum with respect to $k$ from 1 to $N$, we arrive at $\begin{array}{r}-i{\int }_{\mathrm{\Omega }}{|{v}_{{t}^{k}}^{N}|}^{2}\phantom{\rule{0.2em}{0ex}}dx+{\left(-1\right)}^{m}\sum _{|p|,|q|=0}^{m}{\int }_{\mathrm{\Omega }}{a}_{pq}{D}^{q}{v}_{{t}^{k-1}}^{N}\overline{{D}^{p}{v}_{{t}^{k}}^{N}}\phantom{\rule{0.2em}{0ex}}dx\\ \phantom{\rule{1em}{0ex}}=i{\int }_{\mathrm{\Omega }}{f}_{{t}^{k-1}}\overline{{v}_{{t}^{k}}^{N}}\phantom{\rule{0.2em}{0ex}}dx+{\left(-1\right)}^{m-1}\sum _{|p|,|q|=0}^{m}\sum _{s=0}^{k-2}\left(\genfrac{}{}{0}{}{k-1}{s}\right){\int }_{\mathrm{\Omega }}\frac{{\partial }^{k-s-1}{a}_{pq}}{\partial {t}^{k-s-1}}{D}^{q}{v}_{{t}^{s}}^{N}\overline{{D}^{p}{v}_{{t}^{k-s-1}}^{N}}\phantom{\rule{0.2em}{0ex}}dx.\end{array}$ (3.4) Because of the fact that ${f}_{{t}^{k}}\left(x,h\right)=0$ for all $x\in \mathrm{\Omega }$, $0\le k\le l-1$, using induction with respect to $k$ from (3.4) we obtain ${D}^{p}{v}_{{t}^{k}}^{N}\left(x,h\right)=0$ for all $0\le k\le l$, $0\le |p|\le m$, $x\in \mathrm{\Omega }$. Step 2: A priori estimate for ${v}_{{t}^{k}}^{N}$. For $k=l+1$, adding (3.4) to its complex conjugate, integrating with respect to $t$ from $h$ to $\tau$, and then integrating by parts, we get $\begin{array}{r}{\left(-1\right)}^{m}B\left(\tau ,{v}_{{t}^{l}}^{N}\left(x,\tau \right),{v}_{{t}^{l}}^{N}\left(x,\tau \right)\right)\\ \phantom{\rule{1em}{0ex}}={\left(-1\right)}^{m}\sum _{|p|,|q|=0}^{m}{\int }_{{\mathrm{\Omega }}_{h}^{\tau }}\frac{\partial {a}_{pq}}{\partial t}{D}^{q}{v}_{{t}^{l}}^{N}\overline{{D}^{p}{v}_{{t}^{l}}^{N}}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{2em}{0ex}}+{\left(-1\right)}^{m}2Re\sum _{|p|,|q|=0}^{m}l\cdot {\int }_{{\mathrm{\Omega }}_{h}^{\tau }}\frac{\partial {a}_{pq}}{\partial t}{D}^{q}{v}_{{t}^{l}}^{N}\overline{{D}^{p}{v}_{{t}^{l}}^{N}}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{2em}{0ex}}+{\left(-1\right)}^{m}2Re\sum _{|p|,|q|=0}^{m}\sum _{s=0}^{l-1}\left(\genfrac{}{}{0}{}{l}{s}\right){\int }_{{\mathrm{\Omega }}_{h}^{\tau }}\frac{{\partial }^{l-s+1}{a}_{pq}}{\partial {t}^{l-s+1}}{D}^{q}{v}_{{t}^{s}}^{N}\overline{{D}^{p}{v}_{{t}^{l}}^{N}}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{2em}{0ex}}+{\left(-1\right)}^{m}2Re\sum _{|p|,|q|=0}^{m}\sum _{s=0}^{l-2}\left(\genfrac{}{}{0}{}{l}{s}\right){\int }_{{\mathrm{\Omega }}_{h}^{\tau }}\frac{{\partial }^{l-s}{a}_{pq}}{\partial {t}^{l-s}}{D}^{q}{v}_{{t}^{s+1}}^{N}\overline{{D}^{p}{v}_{{t}^{l}}^{N}}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{2em}{0ex}}-{\left(-1\right)}^{m}2Re\sum _{|p|,|q|=0}^{m}\sum _{s=0}^{l-1}\left(\genfrac{}{}{0}{}{l}{s}\right){\int }_{\mathrm{\Omega }}\frac{{\partial }^{l-s}{a}_{pq}}{\partial {t}^{l-s}}\left(x,\tau \right){D}^{q}{v}_{{t}^{s}}^{N}\left(x,\tau \right)\overline{{D}^{p}{v}_{{t}^{l}}^{N}\left(x,\tau \right)}\phantom{\rule{0.2em}{0ex}}dx\\ \phantom{\rule{2em}{0ex}}+2Im{\int }_{\mathrm{\Omega }}{f}_{{t}^{l}}\left(x,\tau \right)\overline{{v}_{{t}^{l}}^{N}\left(x,\tau \right)}\phantom{\rule{0.2em}{0ex}}dx-2Im{\int }_{{\mathrm{\Omega }}_{h}^{\tau }}{f}_{{t}^{l+1}}\overline{{v}_{{t}^{l}}^{N}}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt.\end{array}$ (3.5) We denote by $I$, $II$, $III$, $IV$, $V$, and $VI$ the terms from the first, second, third, fourth, fifth, sixth, and seventh, respectively, of the right-hand sides of (3.5). We will give estimations for these terms. Using Cauchy’s inequality, for any ${ϵ}_{1}>0$, we have $\begin{array}{c}I\le \frac{\mu }{2}\sum _{|p|,|q|=0}^{m}{\int }_{{\mathrm{\Omega }}_{h}^{\tau }}\left({|{D}^{q}{v}_{{t}^{l}}^{N}|}^{2}+{|{D}^{p}{v}_{{t}^{l}}^{N}|}^{2}\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt=\mu {m}^{\star }{\int }_{h}^{\tau }{\parallel {v}_{{t}^{l}}^{N}\left(\cdot ,t\right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt,\hfill \\ II\le 2l\mu {m}^{\star }{\int }_{h}^{\tau }{\parallel {v}_{{t}^{l}}^{N}\left(\cdot ,t\right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt,\hfill \\ III\le {\mu }_{2}{m}^{\star }\left({2}^{l}-1\right){ϵ}_{1}{\int }_{h}^{\tau }{\parallel {v}_{{t}^{l}}^{N}\left(\cdot ,t\right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt+{C}_{1}\sum _{s=0}^{l-1}{\int }_{h}^{\tau }{\parallel {v}_{{t}^{s}}^{N}\left(\cdot ,t\right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt,\hfill \\ IV\le {\mu }_{2}{m}^{\star }\left({2}^{l}-l-1\right){ϵ}_{1}{\int }_{h}^{\tau }{\parallel {v}_{{t}^{l}}^{N}\left(\cdot ,t\right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt+{C}_{1}\sum _{s=0}^{l-1}{\int }_{h}^{\tau }{\parallel {v}_{{t}^{s}}^{N}\left(\cdot ,t\right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt,\hfill \\ V\le {\mu }_{2}{m}^{\star }\left({2}^{l}-1\right){ϵ}_{1}{\parallel {v}_{{t}^{s}}^{N}\left(\cdot ,\tau \right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}+{C}_{1}\sum _{s=0}^{l-1}{\parallel {v}_{{t}^{s}}^{N}\left(\cdot ,\tau \right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2},\hfill \\ \begin{array}{rl}VI\le & {C}_{1}\left({\parallel {f}_{{t}^{l}}\left(\cdot ,\tau \right)\parallel }_{{L}_{2}\left(\mathrm{\Omega }\right)}^{2}+{\int }_{h}^{\tau }{\parallel {f}_{{t}^{l+1}}\left(\cdot ,t\right)\parallel }_{{L}_{2}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt\right)\\ +{ϵ}_{1}\left({\parallel {v}_{{t}^{l}}^{N}\left(\cdot ,\tau \right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}+{\int }_{h}^{\tau }{\parallel {v}_{{t}^{l}}^{N}\left(\cdot ,t\right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt\right),\end{array}\hfill \end{array}$ where ${C}_{1}=max\left\{\frac{\mu {m}^{\star }M}{{ϵ}_{1}},\frac{1}{{ϵ}_{1}}\right\}$, $M={max}_{s=0,\dots ,l-1}\left(\genfrac{}{}{0}{}{l}{s}\right)$. Denote $ϵ=\left(\left({2}^{l+1}-2-l\right)\mu {m}^{\star }+1\right){ϵ}_{1}>0$ for $l>0$, then for all $0<ϵ<{\mu }_{0}$, using above estimations and (3.5), we obtain $\begin{array}{rl}{\parallel {v}_{{t}^{l}}^{N}\left(\cdot ,\tau \right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}\le & \frac{\left(2l+1\right)\mu {m}^{\star }+ϵ}{{\mu }_{0}-ϵ}\\ :=& {\alpha }_{l}{\int }_{h}^{\tau }{\parallel {v}_{{t}^{l}}^{N}\left(\cdot ,t\right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt+C\cdot \sum _{k=0}^{l-1}{\parallel {v}_{{t}^{k}}^{N}\left(\cdot ,\tau \right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}\\ +C\cdot \sum _{k=0}^{l-1}{\int }_{h}^{\tau }{\parallel {v}_{{t}^{k}}^{N}\left(\cdot ,t\right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt\\ +C\cdot {\parallel {f}_{{t}^{l}}\left(\cdot ,\tau \right)\parallel }_{{L}_{2}\left(\mathrm{\Omega }\right)}^{2}+C\cdot {\int }_{h}^{\tau }{\parallel {f}_{{t}^{l+1}}\left(\cdot ,t\right)\parallel }_{{L}_{2}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt,\end{array}$ (3.6) where $C$ is a positive constant. On the other hand, in [6] one already has the following estimate: $\begin{array}{rl}{\parallel {v}^{N}\left(\cdot ,\tau \right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}\le & 2\alpha {\int }_{h}^{\tau }{e}^{2\alpha \left(\tau -t\right)}\left({\parallel f\left(\cdot ,t\right)\parallel }_{{L}_{2}\left(\mathrm{\Omega }\right)}^{2}+{\int }_{h}^{\tau }{\parallel {f}_{s}\left(\cdot ,s\right)\parallel }_{{L}_{2}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}ds\right)\phantom{\rule{0.2em}{0ex}}dt\\ +C\left({\parallel f\left(\cdot ,\tau \right)\parallel }_{{L}_{2}\left(\mathrm{\Omega }\right)}^{2}+{\int }_{h}^{\tau }{\parallel {f}_{t}\left(\cdot ,t\right)\parallel }_{{L}_{2}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt\right).\end{array}$ (3.7) Multiplying both sides of (3.7) by ${e}^{-2\gamma \tau }$, then integrating with respect to $\tau$ from $h$ to $\mathrm{\infty }$ we obtain ${\parallel {v}^{N}\parallel }_{{H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\left({\parallel f\parallel }_{{L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}+{\parallel {f}_{t}\parallel }_{{L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}\right),$ (3.8) where the constant $C$ does not depend on $h$. Now using (3.8) and (3.6) for $l=1$ yields $\begin{array}{r}{\parallel {v}_{t}^{N}\parallel }_{{H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}\\ \phantom{\rule{1em}{0ex}}\le C\left({\parallel {f}_{t}\parallel }_{{L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}+{\parallel {f}_{{t}^{2}}\parallel }_{{L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}+{\parallel {v}^{N}\parallel }_{{H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}\right)\\ \phantom{\rule{2em}{0ex}}+C\cdot {\int }_{h}^{\mathrm{\infty }}{e}^{-2\gamma \tau }{\int }_{h}^{\tau }{e}^{{\alpha }_{1}\left(\tau -t\right)}\left({\parallel {v}^{N}\left(\cdot ,t\right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}+{\parallel {f}_{t}\left(\cdot ,t\right)\parallel }_{{L}_{2}\left(\mathrm{\Omega }\right)}^{2}\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}d\tau \\ \phantom{\rule{2em}{0ex}}+C\cdot {\int }_{h}^{\mathrm{\infty }}{e}^{-2\gamma \tau }{\int }_{h}^{\tau }{e}^{{\alpha }_{1}\left(\tau -t\right)}{\int }_{h}^{t}\left({\parallel {v}^{N}\left(\cdot ,s\right)\parallel }_{{H}^{m}\left(\mathrm{\Omega }\right)}^{2}+{\parallel {f}_{{t}^{2}}\left(\cdot ,s\right)\parallel }_{{L}_{2}\left(\mathrm{\Omega }\right)}^{2}\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}d\tau .\end{array}$ (3.9) Remember that $\underset{{\mu }_{0}>ϵ>0}{inf}2{\alpha }_{1}=\underset{{\mu }_{0}>ϵ>0}{inf}\frac{3\mu {m}^{\star }+ϵ}{{\mu }_{0}-ϵ}=6{\gamma }_{0}<2\gamma .$ So we can choose $ϵ$ small enough such that ${\alpha }_{1}<\gamma$. By (3.9) and using the Fubini theorem we obtain ${\parallel {v}_{t}^{N}\parallel }_{{H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\left(\sum _{k=0}^{2}{\parallel {f}_{{t}^{k}}\parallel }_{{L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}+{\parallel {v}^{N}\parallel }_{{H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}\right),$ which implies ${\parallel {v}_{t}^{N}\parallel }_{{H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\sum _{k=0}^{2}{\parallel {f}_{{t}^{k}}\parallel }_{{L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}.$ (3.10) Repeating the above arguments, from (3.6) one can prove for any $l$, if ${f}_{{t}^{k}}\in {L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)$, $k=\overline{0,l+1}$, that ${v}^{N}$ has derivative with respect to $t$ up to the $l$th and for any $k=\overline{0,l}$ ${\parallel {v}_{{t}^{k}}^{N}\parallel }_{{H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\sum _{{m}_{1}=0}^{k+1}{\parallel {f}_{{t}^{{m}_{1}}}\parallel }_{{L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}.$ (3.11) The inequality (3.11) implies that $\left\{{v}_{{t}^{k}}^{N}\right\}$ is uniformly bounded in ${H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)$. By a standard weakly convergence argument, we can conclude that the sequence ${\left\{{v}_{{t}^{k}}^{N}\right\}}_{N=1}^{\mathrm{\infty }}$ possesses a subsequence convergent to a vector function ${v}_{{t}^{k}}$ in ${H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)$. Moreover, it follows from (3.11) that (3.1) holds. □ ## 4 Further results on the regularity of solution of problem with initial condition Assume that $\omega$ is a local coordinate system on ${S}^{n-1}$. Moreover, assume that the principal part of the operator $L\left(x,t,D\right)$ at origin 0 can be written in the form ${L}_{0}\left(0,t,D\right)={r}^{-2m}Q\left(\omega ,t,r{D}_{r},{D}_{\omega }\right),\phantom{\rule{1em}{0ex}}{D}_{r}=\frac{i\partial }{\partial r},$ where $Q$ is a linear operator with smooth coefficients. Consider the following spectral problem: $Q\left(\omega ,t,\lambda ,{D}_{\omega }\right)v\left(\omega \right)=0,\phantom{\rule{1em}{0ex}}\omega \in G;$ (4.1) ${D}_{\omega }^{j}v\left(\omega \right)=0,\phantom{\rule{1em}{0ex}}\omega \in \partial G,j=0,\dots ,m-1.$ (4.2) Proceeding similarly to Lemma 2.1 in [1], the following lemma holds. ### Lemma 4.1 Let$f,{f}_{t},{f}_{tt}\in {L}_{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$, $\gamma >\left(4m+1\right){\gamma }_{0}$and$f\left(x,h\right)={f}_{t}\left(x,h\right)=0$. If$v\left(x,t\right)$is a generalized solution of problem (2.5)-(2.7) in the space${\stackrel{\circ }{H}}^{m,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$such that$u\equiv 0$whenever$|x|>R$, $R=\mathit{\text{const}}.$, then$v\in {H}_{m}^{2m,1}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$and the following estimate holds: ${\parallel v\parallel }_{{H}_{m}^{2m,1}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\left[{\parallel f\parallel }_{{L}_{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}+{\parallel {f}_{t}\parallel }_{{L}_{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}+{\parallel {f}_{tt}\parallel }_{{L}_{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}\right],$ (4.3) where the constant$C$is independent of$v$, $f$, $h$. ### Lemma 4.2 Let$v\left(x,t\right)$be a generalized solution of problem (2.5)-(2.7), $\gamma >\left(4m+1\right){\gamma }_{0}$and let${f}_{{t}^{k}}\in {L}_{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$for$k\le 2m+1$, ${f}_{{t}^{k}}\left(x,h\right)=0$for$k\le 2m$. In addition suppose that the strip $m-\frac{n}{2}\le Im\lambda \le 2m-\frac{n}{2}$ does not contain points of the spectrum of problem (4.1)-(4.2) for every$t\in \left[h,\mathrm{\infty }\right)$. Then$v\in {H}_{0}^{2m}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$and the following estimate holds: ${\parallel v\parallel }_{{H}_{0}^{2m}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\sum _{k=0}^{2m+1}{\parallel {f}_{{t}^{k}}\parallel }_{{L}_{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2},$ (4.4) where the constant$C$is independent of$v$, $f$, $h$. ### Proof First, we prove that ${\parallel {v}_{{t}^{s}}\parallel }_{{H}_{0}^{2m,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\sum _{k=0}^{2m+1}{\parallel {f}_{{t}^{k}}\parallel }_{{L}_{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2},$ (4.5) where $C$ does not depend on $h$, $s\le 2m$. Doing the same in [1], Proposition 2.1] we have ${\parallel v\parallel }_{{H}_{0}^{2m}\left(K\right)}^{2}\le C\left[{\parallel f\parallel }_{{L}_{2}\left(K\right)}^{2}+{\parallel {v}_{t}\parallel }_{{L}_{2}\left(K\right)}^{2}+{\parallel v\parallel }_{{H}_{m}^{2m}\left(K\right)}^{2}\right],$ for a.e. $t\in \left[h,\mathrm{\infty }\right)$, where $C=\text{constant}$. Multiplying both sides of this inequality by ${e}^{-2\gamma t}$, then integrating with respect to $t$ from $h$ to $\mathrm{\infty }$, we get ${\parallel v\parallel }_{{H}_{0}^{2m,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\left[{\parallel f\parallel }_{{L}_{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}+{\parallel {v}_{t}\parallel }_{{L}_{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}+{\parallel v\parallel }_{{H}_{0}^{2m,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}\right].$ By (3.11) ${\parallel {v}_{t}\parallel }_{{L}_{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\sum _{k=0}^{2}{\parallel {f}_{{t}^{k}}\parallel }_{{L}_{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2},$ and by Lemma 4.1 ${\parallel v\parallel }_{{H}_{0}^{2m,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}\le {\parallel v\parallel }_{{H}_{m}^{2m,1}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\sum _{k=0}^{2}{\parallel {f}_{{t}^{k}}\parallel }_{{L}_{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2},$ from which it follows that ${\parallel v\parallel }_{{H}_{0}^{2m,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\sum _{k=0}^{2}{\parallel {f}_{{t}^{k}}\parallel }_{{L}_{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}.$ That means (4.5) is proved for $s=0$. Now assume that (4.5) is true for $s-1$. By differentiating the systems (2.5) $s$ times with respect to $t$ and by putting $w={v}_{{t}^{s}}$, we obtain ${\left(-1\right)}^{m-1}Lw=-i\left({w}_{t}+{f}_{{t}^{s}}\right)+{\left(-1\right)}^{m}\sum _{k=1}^{s}\left(\genfrac{}{}{0}{}{s}{k}\right){L}_{{t}^{k}}{v}_{{t}^{s-k}},$ where ${L}_{{t}^{k}}={\sum }_{|p|,|q|=0}^{m}{D}^{p}\left(\frac{{\partial }^{k}{a}_{pq}}{\partial {t}^{k}}{D}^{q}\right)$. From the inductive hypothesis and repeating the arguments of the proof in the case $s=0$, the inequality (4.5) holds for $s$. Since ${\parallel v\parallel }_{{H}_{0}^{2m}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\left(\sum _{s=0}^{2m-1}{\parallel {v}_{{t}^{s}}\parallel }_{{H}_{0}^{2m,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}+{\parallel {v}_{{t}^{2m}}\parallel }_{{H}_{0}^{0,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}\right),$ from (4.5) and Theorem 4.1, (4.4) is true. The lemma is proved. □ ### Theorem 4.1 Let l be a nonnegative integer. Assume that$v\left(x,t\right)$is a weak solution in the space${H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)$with$\gamma >\left(2\left(2m+l\right)+1\right){\gamma }_{0}$of problem (2.5)-(2.7) and${f}_{{t}^{k}}\in {H}_{0}^{l,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)$, for$k\le 2m+l+1$, ${f}_{{t}^{k}}\left(x,h\right)=0$for$k\le 2m+l$. In addition, suppose that in the strip $m-\frac{n}{2}\le Im\lambda \le 2m+l-\frac{n}{2}$ there is no point from the spectrum of (4.1)-(4.2). Then we have$v\in {H}_{0}^{2m+l}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)$and the following estimate holds: ${\parallel v\parallel }_{{H}_{0}^{2m+l}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\sum _{k=0}^{2m+l+1}{\parallel {f}_{{t}^{k}}\parallel }_{{H}_{0}^{l,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}.$ (4.6) ### Proof 1. 1. Firstly, we study the case $u\equiv 0$ outside ${U}_{0}$. We use the induction by $l$. For $l=0$, this theorem is proved by Lemma 4.2 with noting that ${H}_{0}^{0,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)={L}_{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$. Assume that the theorem’s assertion holds up to $l-1$, we need to prove that this holds up to $l$. Firstly we will prove the following inequality: ${\parallel {v}_{{t}^{j}}\parallel }_{{H}_{0}^{2m+l-j}}^{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)\le C\sum _{k=0}^{2m+l+1}{\parallel {f}_{{t}^{k}}\parallel }_{{H}_{0}^{l,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}$ (4.7) for all $j=l,l-1,\dots ,0$, where the constant $C$ is independent of $h$. Since ${f}_{{t}^{k}}\in {H}_{0}^{l}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$, ${f}_{{t}^{k}}\in {L}_{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$. So using similar arguments in the proof of Theorem 4.1 we get ${v}_{{t}^{l}}\in {H}^{2m,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$. By Lemma 4.2, one obtains ${v}_{{t}^{l}}\in {H}_{0}^{2m}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$. This means that (4.7)holds for $j=l$. Assume that (4.7) holds for $j=l,l-1,\dots ,s+1$. By the assumption of the induction of $l,v\in {H}_{0}^{2m+l-1}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$, put $w={v}_{{t}^{s}}$ then $w\in {H}_{0}^{2m+l-s-1}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$. Differentiating (2.5) $s$-times with respect to $t$, we have $Lw=F=-i\left({w}_{t}+{f}_{{t}^{s}}\right)+\sum _{p=1}^{s}\left(\genfrac{}{}{0}{}{s}{p}\right){L}_{{t}^{p}}{v}_{{t}^{s-p}},$ (4.8) where ${L}_{{t}^{k}}={\sum }_{|p|,|q|=0}^{m}{\left(-1\right)}^{|p|}{D}^{p}\left(\frac{{\partial }^{k}{a}_{pq}}{\partial {t}^{k}}{D}^{q}\right)$. Following the assumption of the induction of $j$ and the hypothesis of the function $f$ one has ${w}_{t}\in {H}_{0}^{2m+l-s-1}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$, ${f}_{{t}^{s}}\in {H}_{0}^{l-s}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$. It follows that $F\in {H}_{0}^{l-s}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)\subset {H}_{-1}^{l-s-1}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$. Because the strip $2m+l-s-1-\frac{n}{2}\le Im\lambda \le 2m+l-s-\frac{n}{2}$ does not contain spectral points of problem (4.1)-(4.2), then using results in [8], one gets $w\in {H}_{-1}^{2m+l-s-1}\left(K\right)$, a.e. $t\in \left(h,\mathrm{\infty }\right)$. So $w\in {H}_{-1}^{2m+l-s-1,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$. Note that $F\in {H}_{0}^{l-s}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$, then by using similar method to the one used in Lemma 2.2 in [1] we have $w\in {H}_{-1}^{2m+l-s-1,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)$ and the following estimate holds: ${\parallel w\parallel }_{{H}_{-1}^{2m+l-s-1,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\left[{\parallel F\parallel }_{{H}_{0}^{l-s,0}}^{2}+{\parallel w\parallel }_{{H}_{-1}^{l-s+1,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}\right],$ (4.9) where $C$ is a constant independent of $h$. So using the inductive assumption and (4.9), we obtain ${\parallel {v}_{{t}^{j}}\parallel }_{{H}_{0}^{2m+l-s}}^{2}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)\le C\sum _{k=0}^{2m+l+1}{\parallel {f}_{{t}^{k}}\parallel }_{{H}_{0}^{l,0}\left(-\gamma ,{K}_{h}^{\mathrm{\infty }}\right)}^{2}.$ (4.10) It means that (4.7) is proved and our theorem is completed by fixing $j=0$ in (4.7). 2. The general case: Consider a function ${\phi }_{0}\in {C}_{0}^{\mathrm{\infty }}\left({U}_{0}\right)$ such that ${\phi }_{0}\equiv 1$ in some neighborhood of 0. Definite ${v}_{0}={\phi }_{0}v$, which satisfies the system ${\left(-1\right)}^{m-1}iL\left(x,t,D\right){v}_{0}-{\left({v}_{0}\right)}_{t}={\phi }_{0}f+{L}^{\prime }\left(x,t,D\right)v,$ where ${L}^{\prime }\left(x,t,D\right)$ is a linear operator having order less than $2m$ and the coefficients of this operator is equal to 0 outside ${U}_{0}$. So the first case of this theorem implies that ${\parallel {v}_{0}\parallel }_{{H}_{0}^{2m+l}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\sum _{k=0}^{2m+l+1}{\parallel {f}_{{t}^{k}}\parallel }_{{H}_{0}^{l,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}.$ (4.11) Write $v={v}_{0}+{v}_{1}$ where ${v}_{1}=\left(1-{\phi }_{0}\right)v$. The function ${v}_{1}$ is equal to 0 in ${U}_{0}$, so we can apply the famous results on the smoothness of solutions of Schrödinger systems in a smooth domain to this function and obtain ${\parallel {v}_{1}\parallel }_{{H}_{0}^{2m+l}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\sum _{k=0}^{2m+l+1}{\parallel {f}_{{t}^{k}}\parallel }_{{H}_{0}^{l,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}.$ (4.12) From inequalities (4.11) and (4.12) it follows that ${\parallel v\parallel }_{{H}_{0}^{2m+l}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}\le C\sum _{k=0}^{2m+l+1}{\parallel {f}_{{t}^{k}}\parallel }_{{H}_{0}^{l,0}\left(-\gamma ,{\mathrm{\Omega }}_{h}^{\mathrm{\infty }}\right)}^{2}.$ The theorem is proved. □ ## 5 The regularity of solution of problem (2.2)-(2.3) The generalized solution of problem (2.2)-(2.3) can be approximated by a sequence of solutions of problems with initial condition (2.5)-(2.7). It is well known that there is a smooth function $\chi \left(t\right)$ which is equal to 1 on $\left[1,\mathrm{\infty }\right)$, is equal to 0 on $\left(-\mathrm{\infty },0\right]$, and assumes value in $\left[0,1\right]$ on $\left[0;1\right]$ (see [9], Th. 5.5] for more details). Moreover, we can suppose that all derivatives of $\chi \left(t\right)$ are bounded. Let $h\in \left(-\mathrm{\infty },0\right]$ be an integer. We set ${f}^{h}\left(x,t\right)=\chi \left(t-h\right)f\left(x,t\right)$ and we consider a generalized solution ${u}^{h}$ and ${u}^{j}$ of problems (2.5)-(2.7) in cylinders ${\mathrm{\Omega }}_{h}^{\mathrm{\infty }}$ and ${\mathrm{\Omega }}_{j}^{\mathrm{\infty }}$ with $f\left(x,t\right)$ replaced by ${f}^{h}\left(x,t\right)$ and ${f}^{j}\left(x,t\right)$, respectively. If $h>j$, ${u}^{h}$ can be defined in ${\stackrel{\circ }{H}}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{j}^{\mathrm{\infty }}\right)$ with ${u}^{h}\left(x,t\right)=0$, $\mathrm{\forall }j\le t\le h$, then we put ${u}^{jh}={u}^{j}\left(x,t\right)-{u}^{h}\left(x,t\right)$, and ${u}^{jh}$ is the generalized solution of problem (2.5)-(2.7) in which $f\left(x,t\right)$ is replaced by ${f}^{jh}={f}^{j}\left(x,t\right)-{f}^{h}\left(x,t\right)$. According to (4.6), ${\parallel {u}_{{t}^{k}}^{jh}\parallel }_{{H}_{0}^{2m+l}\left(-\gamma ,{\mathrm{\Omega }}_{j}^{\mathrm{\infty }}\right)}^{2}={\parallel {u}_{{t}^{k}}^{jh}\parallel }_{{H}_{0}^{2m+l}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)}^{2}\le C\sum _{{m}_{1}=0}^{2m+l+1}{\parallel {f}_{{t}^{{m}_{1}}}^{jh}\parallel }_{{H}_{0}^{l,0}\left(-\gamma ,{\mathrm{\Omega }}_{j}^{\mathrm{\infty }}\right)}^{2}.$ Because $\begin{array}{rl}{\parallel {f}^{jh}\parallel }_{{H}^{l,0}\left(-\gamma ,{\mathrm{\Omega }}_{j}^{\mathrm{\infty }}\right)}^{2}& ={\int }_{j}^{h+1}{e}^{-2\gamma t}{\parallel {f}^{h}-{f}^{j}\parallel }_{{H}_{0}^{l}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt\\ ={\int }_{j}^{h+1}{e}^{-2\gamma t}|\chi \left(t-h\right)-\chi \left(t-j\right)|{\parallel f\parallel }_{{H}_{0}^{l}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt\\ \le 2{\int }_{j}^{h+1}{e}^{-2\gamma t}{\parallel f\parallel }_{{H}_{0}^{l}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt.\end{array}$ Since $f\in {H}_{0}^{l,0}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$, $lim{\int }_{j}^{h+1}{e}^{-2\gamma t}{\parallel f\parallel }_{{H}_{0}^{l}\left(\mathrm{\Omega }\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt=0$ when $h,j\to -\mathrm{\infty }$. So $lim{\parallel {f}^{jh}\parallel }_{{H}_{0}^{l,0}\left(-\gamma ,{\mathrm{\Omega }}_{j}^{\mathrm{\infty }}\right)}^{2}=0$ when $h,j\to -\mathrm{\infty }$. Repeating this argument, we discover $lim{\parallel {f}_{{t}^{{m}_{1}}}^{jh}\parallel }_{{H}_{0}^{l,0}\left(-\gamma ,{\mathrm{\Omega }}_{j}^{\mathrm{\infty }}\right)}^{2}=0$ when $h,j\to -\mathrm{\infty }$, for all ${m}_{1}=0,\dots ,k+1$. It follows that ${\left\{{u}_{{t}^{k}}^{h}\right\}}_{h=0}^{-\mathrm{\infty }}$ is a Cauchy sequence and ${u}_{{t}^{k}}^{h}$ is convergent to ${u}_{{t}^{k}}$ in ${H}_{0}^{2m+l}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$. Moreover, ${H}_{0}^{2m+l}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$ is continuously embedded in ${H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$, since ${\parallel u\parallel }_{{H}_{0}^{2m+l}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)}^{2}\le \sum _{|\alpha |=0}^{2m+l}{\int }_{{\mathrm{\Omega }}_{\mathbb{R}}}{e}^{-2\gamma t}{r}^{2\left(|\alpha |-2m-l\right)}{|{D}^{\alpha }u|}^{2}\le C{\parallel u\parallel }_{{H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)}^{2}.$ So it is very easy to verify that $u\in {H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$ is a generalized solution of problem (2.2)-(2.3); see [7]. We obtain the following main results. ### Theorem 5.1 Assume that $l\in \mathbb{N}$ and there exists $\mu >0$ such that 1. (i) $sup\left\{|\frac{\partial {a}_{pq}}{\partial t}|:\left(x,t\right)\in {\mathrm{\Omega }}_{\mathbb{R}},0\le |p|,|q|\le m,\right\}=\mu <\mathrm{\infty }$, $|\frac{{\partial }^{k}{a}_{pq}}{\partial {t}^{k}}|\le {\mu }_{2}$, ${\mu }_{2}>0$, for $2\le k\le l+1$; 2. (ii) $|\frac{\partial {a}_{pq}}{\partial t}|\le {\mu }_{1}\cdot {e}^{2\gamma t}$, for all $\left(x,t\right)\in {\mathrm{\Omega }}_{\mathbb{R}}$, $0\le |p|,|q|\le m$; 3. (iii) ${f}_{{t}^{k}}\in {L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$, for all $x\in \mathrm{\Omega }$, $0\le k\le l+1$. Then for all$\gamma >\left(2l+1\right){\gamma }_{0}$, the solution$u\in {\stackrel{\circ }{H}}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$of problem (2.2)-(2.3) has derivatives with respect to$t$up to order$l$satisfying${u}_{{t}^{k}}\in {H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$, $\mathrm{\forall }k=\overline{0,l}$, such that ${\parallel {u}_{{t}^{k}}\parallel }_{{H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)}^{2}\le C\sum _{{m}_{1}=0}^{k+1}{\parallel {f}_{{t}^{{m}_{1}}}\parallel }_{{L}_{2}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)}^{2},$ where the constant$C$does not depend on$u$, $f$. ### Theorem 5.2 Let l be a nonnegative integer. Assume that$u\left(x,t\right)$is a weak solution in the space${H}^{m,0}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$with$\gamma >\left(2\left(2m+l\right)+1\right){\gamma }_{0}$of the problem (2.2)-(2.3) and${f}_{{t}^{k}}\in {H}_{0}^{l,0}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$, for$k\le 2m+l+1$. In addition, suppose that in the strip $m-\frac{n}{2}\le Im\lambda \le 2m+l-\frac{n}{2}$ there is no point from the spectrum of (4.1)-(4.3). Then we have$u\in {H}_{0}^{2m+l}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)$and the following estimate holds: ${\parallel u\parallel }_{{H}_{0}^{2m+l}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)}^{2}\le C\sum _{k=0}^{2m+l+1}{\parallel {f}_{{t}^{k}}\parallel }_{{H}_{0}^{l,0}\left(-\gamma ,{\mathrm{\Omega }}_{\mathbb{R}}\right)}^{2}.$ ## References 1. Hung NM, Anh CT: On the smoothness of solutions of the first initial boundary value problem for Schrödinger systems in domain with conical points. Vietnam J. Math. 2005, 33(2):135-147. 2. Hung NM, Son NTK: On the regularity of solution of the second initial boundary value problem for Schrodinger systems in domains with conical points. Taiwan. J. Math. 2009, 13(6):1885-1907. 3. Bokalo NM: Dynamical problems without initial conditions for elliptic-parabolic equations in spatial unbounded domains. Electron. J. Differ. Equ. 2010., 2010: 4. Bokalo NM: Problem without initial conditions for some classes of nonlinear parabolic equations. J. Sov. Math. 1990, 51: 2291-2322. 10.1007/BF01094990 5. Bokalo M, Dmytryshyn Y: Problems without initial conditions for degenerate implicit evolution equations. Electron. J. Differ. Equ. 2008., 2008: 6. Moiseev EI, Vafodorova GO: Problems without initial conditions for some differential equations. Differ. Equ. 2002, 38(8):1162-1165. 10.1023/A:1021676322884 7. Hung NM, Lien NT: On the solvability of the boundary value problem without initial condition for Schrödinger systems in infinite cylinders. Bound. Value Probl. 2013., 2013: 10.1186/1687-2770-2013-156 8. Maz’ya VG, Plamenevskii BA: On the coefficients in the asymptotic of solutions of the elliptic boundary problem in domains with conical points. Transl. Am. Math. Soc. 1984, 123: 57-88. Translated from: Math. Nachr. 76, 29-60 (1977) ## Acknowledgements We would like to thank the reviewers for a careful reading and valuable comments on the original manuscript. ## Author information Authors ### Corresponding author Correspondence to Nguyen Thi Lien. ### Competing interests The authors declare that they have no competing interests. ### Authors’ contributions All authors read and approved the final manuscript. ## Authors’ original submitted files for images Below are the links to the authors’ original submitted files for images. ## Rights and permissions Reprints and permissions Lien, N.T., Hung, N.M. On the regularity of solutions of the boundary value problem without initial condition for Schrödinger systems in domain with conical points. Bound Value Probl 2014, 181 (2014). https://doi.org/10.1186/s13661-014-0181-8

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## Advent Calendar - December 12, 2022 Monday, Dec 12, 2022| Tags: Perl ### |   Day 11   |   Day 12   |   Day 13 The gift is presented by `W. Luis Mochan`. Today he is talking about his solution to “The Weekly Challenge - 152”. This is re-produced for Advent Calendar 2022 from the original post by him. ## Perl Weekly Challenge 152 My solutions (task 1 and task 2) to The Weekly Challenge - 152. ## Task 1: Triangle Sum Path You are given a triangle array. Write a script to find the minimum sum path from top to bottom. ### Example 1 ``````Input: \$triangle = [ [1], [5,3], [2,3,4], [7,1,0,2], [6,4,5,2,8] ] 1 5 3 2 3 4 7 1 0 2 6 4 5 2 8 Output: 8 Minimum Sum Path = 1 + 3 + 2 + 0 + 2 => 8 `````` ### Example 2 ``````Input: \$triangle = [ [5], [2,3], [4,1,5], [0,1,2,3], [7,2,4,1,9] ] 5 2 3 4 1 5 0 1 2 3 7 2 4 1 9 Output: 9 Minimum Sum Path = 5 + 2 + 1 + 0 + 1 => 9 `````` From the examples it is clear that any path down the triangle is valid, so I just have to choose the minimum values for each row and sum them. This yields a very compact oneliner: ``````perl -MList::Util=min,sum0 -d -E ' say "In: \$_, Out: ", sum0 map {min @\$_} @{eval \$_} for @ARGV; ' `````` ### Results: ``````In: [[1], [5,3], [2,3,4], [7,1,0,2],[6,4,5,2,8]], Out: 8 In: [[5],[2,3],[4,1,5],[0,1,2,3],[7,2,4,1,9]], Out: 9 `````` I assume the input is a list of strings, each describing a correct multidimensional `Perl` array describing the triangle, which by the way, does not even need a triangular shape, so that I can eval it. The result is an array of references to the rows of the given triangle. I convert each of them to an array, find their minimum value and sum over the rows to obtain the output. I use the `min` and `sum0` subroutines from `List::Util`. A full version is ``````#!/usr/bin/env perl # Perl weekly challenge 152 # Task 1: Minimum sum path # use v5.12; use warnings; use List::Util qw(min sum0 all); use Try::Tiny; die "Usage: ./ch-1.pl T1 [T2]...\n" . "where Ti are triangles of the form '[[T00],[T10,T11],[T20,T21,T22]...'" unless @ARGV; for my \$triangle_string (@ARGV){ try { my \$triangle=eval \$triangle_string; my @rows=@\$triangle; # Seems unnecesary, but test shape die "Wrong row-size in \$triangle_string" unless all{\$_+1==@{\$rows[\$_]}}(0..@rows-1); my @minima=map {min @\$_} @rows; my \$sum=sum0 @minima; say "Input: \$triangle_string\nOutput: \$sum\nPath values: ", join "-", @minima; } catch { say \$_; } } `````` ### Example: ``````./ch-1.pl '[[1], [5,3], [2,3,4], [7,1,0,2],[6,4,5,2,8]]' '[[5],[2,3],[4,1,5],[0,1,2,3],[7,2,4,1,9]]' `````` ### Results: ``````Input: [[1], [5,3], [2,3,4], [7,1,0,2],[6,4,5,2,8]] Output: 8 Path values: 1-3-2-0-2 Input: [[5],[2,3],[4,1,5],[0,1,2,3],[7,2,4,1,9]] Output: 9 Path values: 5-2-1-0-1 `````` ### Examples with errors and extreme cases: ``````./ch-1.pl '[[1], [5,3], [2,4], [7,1,0,2],[6,4,5,2,8]]'\ '[5],[2,3],[4,1,5],[0,1,2,3],[7,2,4,1,9]' '[[1]]' '[[]]' `````` ### Results: ``````Wrong row-size in [[1], [5,3], [2,4], [7,1,0,2],[6,4,5,2,8]] at ./ch-1.pl line 19. Can't use string ("7") as an ARRAY ref while "strict refs" in use at ./ch-1.pl line 19. Input: [[1]] Output: 1 Path values: 1 Wrong row-size in [[]] at ./ch-1.pl line 19. `````` You are given coordinates bottom-left and top-right corner of two rectangles in a 2D plane. Write a script to find the total area covered by the two rectangles. ### Example 1 ``````Input: Rectangle 1 => (-1,0), (2,2) Rectangle 2 => (0,-1), (4,4) Output: 22 `````` ### Example 2 ``````Input: Rectangle 1 => (-3,-1), (1,3) Rectangle 2 => (-1,-3), (2,2) Output: 25 `````` This problem may be messy, as there are many possibilities for the relative positions of the different vertices. The problem may be simplified by dividing rectangles that have a partial overlap. This allows a generalization of the problem to that of the area of N rectangles. I obtain the coordinates of each rectangle, four at a time, from @ARGV. I build a list of non-overlapping rectangles. If a rectangle doesn’t overlap the area of any previously added rectangle, I add it. If it does, I split it in two and try to add the fragments instead. After a sequence of splitting horizontally and vertically the pending rectangles, I either end up with some rectangles that do not overlap any other rectangle, in which case I must add them to the list, or with rectangles that are fully contained in one of the previous rectangles, in which case I simply throw them away. The area is the sum of the areas of the non-overlapping rectangles. ``````#!/usr/bin/env perl # Perl weekly challenge 152 # use v5.12; use warnings; use List::Util qw(min sum0 all); die "Usage: ./ch-2.pl L1 B1 R1 T1 L2 B2 R2 T2 ..." . "where L's B's R's and T's denote left, bottom, right and top coordinates" unless @ARGV; my @non_ol; my @input; my @pending; while(@ARGV){ die "# coordinates must be multiple of four" unless @ARGV>=3; my (\$L,\$B,\$R,\$T)=splice @ARGV, 0, 4; (\$L, \$R)=(\$R, \$L) if \$L>\$R; # reorder if necessary (\$B, \$T)=(\$T, \$B) if \$B>\$T; push @input, {left=>\$L, bottom=>\$B, right=>\$R, top=>\$T}; } my @pending=@input; while(@pending){ my \$rectangle=shift @pending; foreach(@non_ol){ my @fragments=divide(\$rectangle, \$_); next ADD1 if @fragments==1; # rectangle contained in some other piece push(@pending, @fragments), next ADD1 if @fragments>1; } push @non_ol, \$rectangle; } say "Input: "; say "\tRectangle \$_: (\$input[\$_]->{left},\$input[\$_]->{bottom}), ", "(\$input[\$_]->{right},\$input[\$_]->{top})" for(0..@input-1); say "Area: ", sum0 map {(\$_->{right}-\$_->{left})*(\$_->{top}-\$_->{bottom})} @non_ol; say "Non-overlapping subregions: "; say "\tRectangle \$_: (\$non_ol[\$_]->{left},\$non_ol[\$_]->{bottom}), ", "(\$non_ol[\$_]->{right},\$non_ol[\$_]->{top})" for(0..@non_ol-1); sub divide { my (\$p, \$q)=@_; return if \$p->{left}>=\$q->{right} # no overlap or \$p->{right}<=\$q->{left} or \$p->{bottom}>=\$q->{top} or \$p->{top}<=\$q->{bottom}; return ({left=>\$p->{left}, bottom=>\$p->{bottom}, right=>\$q->{left}, top=>\$p->{top}}, {left=>\$q->{left}, bottom=>\$p->{bottom}, right=>\$p->{right}, top=>\$p->{top}}) if \$p->{left}<\$q->{left}; # split left return ({left=>\$p->{left}, bottom=>\$p->{bottom}, right=>\$q->{right}, top=>\$p->{top}}, {left=>\$q->{right}, bottom=>\$p->{bottom}, right=>\$p->{right}, top=>\$p->{top}}) if \$p->{right}>\$q->{right}; # split right return ({left=>\$p->{left}, bottom=>\$p->{bottom}, right=>\$p->{right}, top=>\$q->{bottom}}, {left=>\$p->{left}, bottom=>\$q->{bottom}, right=>\$p->{right}, top=>\$p->{top}}) if \$p->{bottom}<\$q->{bottom}; # split bottom return ({left=>\$p->{left}, bottom=>\$p->{bottom}, right=>\$p->{right}, top=>\$q->{top}}, {left=>\$p->{left}, bottom=>\$q->{top}, right=>\$p->{right}, top=>\$p->{top}}) if \$p->{top}>\$q->{top}; # split top return \$p; # \$p contained in \$q } `````` ### Examples ``````./ch-2.pl -1 0 2 2 0 -1 4 4 ./ch-2.pl -3 -1 1 3 -1 -3 2 2 `````` ### Results: ``````Input: Rectangle 0: (-1,0), (2,2) Rectangle 1: (0,-1), (4,4) Area: 22 Non-overlapping subregions: Rectangle 0: (-1,0), (2,2) Rectangle 1: (2,-1), (4,4) Rectangle 2: (0,-1), (2,0) Rectangle 3: (0,2), (2,4) Input: Rectangle 0: (-3,-1), (1,3) Rectangle 1: (-1,-3), (2,2) Area: 25 Non-overlapping subregions: Rectangle 0: (-3,-1), (1,3) Rectangle 1: (1,-3), (2,2) Rectangle 2: (-1,-3), (1,-1) `````` ### Example with several overlapping squares along a diagonal: ``````./ch-2.pl 0 0 2 2 1 1 3 3 2 2 4 4 3 3 5 5 `````` ### Results: ``````Input: Rectangle 0: (0,0), (2,2) Rectangle 1: (1,1), (3,3) Rectangle 2: (2,2), (4,4) Rectangle 3: (3,3), (5,5) Area: 13 Non-overlapping subregions: Rectangle 0: (0,0), (2,2) Rectangle 1: (2,2), (4,4) Rectangle 2: (4,3), (5,5) Rectangle 3: (1,2), (2,3) Rectangle 4: (2,1), (3,2) Rectangle 5: (3,4), (4,5) `````` ### Example with overlapping crosses: ``````./ch-2.pl -1 -5 1 5 -2 -4 2 4 -3 -3 3 3 -4 -2 4 2 -5 -1 5 1 `````` ### Results: ``````Input: Rectangle 0: (-1,-5), (1,5) Rectangle 1: (-2,-4), (2,4) Rectangle 2: (-3,-3), (3,3) Rectangle 3: (-4,-2), (4,2) Rectangle 4: (-5,-1), (5,1) Area: 60 Non-overlapping subregions: Rectangle 0: (-1,-5), (1,5) Rectangle 1: (-2,-4), (-1,4) Rectangle 2: (1,-4), (2,4) Rectangle 3: (-3,-3), (-2,3) Rectangle 4: (2,-3), (3,3) Rectangle 5: (-4,-2), (-3,2) Rectangle 6: (3,-2), (4,2) Rectangle 7: (-5,-1), (-4,1) Rectangle 8: (4,-1), (5,1) `````` In the problem description there are only two rectangles, and their intersection is either empty or some rectangle. These allows a simplified solution, where we just add the area of each rectangle and subtract the area of the intersection (or 0, if empty). This suggest a one-liner (5-liner): ``````perl -MList::Util=min,max -E '(\$p,\$q)=map {[splice @ARGV, 0, 4]} (1,2); say a(\$p)+a(\$q)-i(\$p,\$q);sub a{my (\$x0,\$y0,\$x1,\$y1)=@{\$_[0]};(\$x1-\$x0)*(\$y1-\$y0)} sub i{my (\$x0,\$y0,\$x1,\$y1,\$x2,\$y2,\$x3,\$y3)=map {@{\$_}} @_; return 0 if \$x0>=\$x3 || \$x1<=\$x2 || \$y0>=\$y3 || \$y1<=\$y2; a([max(\$x0,\$x2),max(\$y0,\$y2),min(\$x1,\$x3),min(\$y1,\$y3)])} ' -- -1 0 2 2 0 -1 4 4 perl -MList::Util=min,max -E '(\$p,\$q)=map {[splice @ARGV, 0, 4]} (1,2); say a(\$p)+a(\$q)-i(\$p,\$q);sub a{my (\$x0,\$y0,\$x1,\$y1)=@{\$_[0]};(\$x1-\$x0)*(\$y1-\$y0)} sub i{my (\$x0,\$y0,\$x1,\$y1,\$x2,\$y2,\$x3,\$y3)=map {@{\$_}} @_; return 0 if \$x0>=\$x3 || \$x1<=\$x2 || \$y0>=\$y3 || \$y1<=\$y2; a([max(\$x0,\$x2),max(\$y0,\$y2),min(\$x1,\$x3),min(\$y1,\$y3)])} ' -- -3 -1 1 3 -1 -3 2 2 `````` ### Results: ``````22 25 `````` Here the subroutine a calculates the area of a rectangle and i the area of the intersection of two rectangles. The intersection between `A` and `B` is empty if `AL > BR`, or if `AR < BL`, or… where `L` denotes the leftmost and `R` rightmost coordinate. A similar consideration may be done for the vertical coordinates. I change the strict inequalities to `>=` and `<=`, as overlaping lines don’t contribute to the area. In case of overlap, the overlapping rectangle lies to the right of `AL` and `BL`, to the left of `AR` and `BR` and so on. This allows the calculation and subtraction of the area of the intersection. If you have any suggestion then please do share with us perlweeklychallenge@yahoo.com. ## SO WHAT DO YOU THINK ? If you have any suggestions or ideas then please do share with us.

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Distance between Narrabri (NAA) and Moranbah (MOV) Flight distance from Narrabri to Moranbah (Narrabri Airport – Moranbah Airport) is 579 miles / 932 kilometers / 503 nautical miles. Estimated flight time is 1 hour 35 minutes. Driving distance from Narrabri (NAA) to Moranbah (MOV) is 718 miles / 1156 kilometers and travel time by car is about 13 hours 49 minutes. Map of flight path and driving directions from Narrabri to Moranbah. Shortest flight path between Narrabri Airport (NAA) and Moranbah Airport (MOV). How far is Moranbah from Narrabri? There are several ways to calculate distances between Narrabri and Moranbah. Here are two common methods: Vincenty's formula (applied above) • 579.002 miles • 931.814 kilometers • 503.139 nautical miles Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth. Haversine formula • 580.999 miles • 935.027 kilometers • 504.874 nautical miles The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points). Airport information A Narrabri Airport City: Narrabri Country: Australia IATA Code: NAA ICAO Code: YNBR Coordinates: 30°19′9″S, 149°49′37″E B Moranbah Airport City: Moranbah Country: Australia IATA Code: MOV ICAO Code: YMRB Coordinates: 22°3′28″S, 148°4′37″E Time difference and current local times The time difference between Narrabri and Moranbah is 1 hour. Moranbah is 1 hour behind Narrabri. AEDT AEST Carbon dioxide emissions Estimated CO2 emissions per passenger is 110 kg (242 pounds). Frequent Flyer Miles Calculator Narrabri (NAA) → Moranbah (MOV). Distance: 579 Elite level bonus: 0 Booking class bonus: 0 In total Total frequent flyer miles: 579 Round trip?

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## Duotrigintatrecentillitresvigintiducentillion A Duotrigintatrecentillitresvigintiducentillion (1 Duotrigintatrecentillitresvigintiducentillion) is 10 to the power of 996672 (10^996672). This is a stupendously astronomical number! ## How many zeros in a Duotrigintatrecentillitresvigintiducentillion? There are 996,672 zeros in a Duotrigintatrecentillitresvigintiducentillion. ## What's before Duotrigintatrecentillitresvigintiducentillion? A Duotrigintatrecentilliduovigintiducentillion is smaller than a Duotrigintatrecentillitresvigintiducentillion. ## What's after Duotrigintatrecentillitresvigintiducentillion? A Duotrigintatrecentilliquattuorvigintiducentillion is larger than a Duotrigintatrecentillitresvigintiducentillion. ## Duotrigintatrecentillitresvigintiducentillionaire A Duotrigintatrecentillitresvigintiducentillionaire is someone whos assets, net worth or wealth is 1 or more Duotrigintatrecentillitresvigintiducentillion. It is unlikely anyone will ever be a true Duotrigintatrecentillitresvigintiducentillionaire. If you want to be a Duotrigintatrecentillitresvigintiducentillionaire, play Tap Tales! ## Is Duotrigintatrecentillitresvigintiducentillion the largest number? Duotrigintatrecentillitresvigintiducentillion is not the largest number. Infinity best describes the largest possible number - if there even is one! We cannot comprehend what the largest number actually is. ## Duotrigintatrecentillitresvigintiducentillion written out Duotrigintatrecentillitresvigintiducentillion is written out as: 1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,00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# calculating iv drip rate • ### Drug Dosage IV Rates CalculationsGeorge Brown 2020-5-13 · Calculation of Intravenous Drip Rates In these types of calculations for a given volume time period and drop factor (gtts/mL) the required IV flow rate in drops per minute (gtts/min) is calculated. Note Since a fraction of a drop is not possible to give to a patient it is usual to round • ### How to Calculate IV Drip RatesMedicNerd 2016-9-11 · How to Calculate IV Drip Rates. By Michael Stewart on September 11 2016 in Training Video. This tutorial shows you the easiest way to calculate an IV drip rate. Using this 3 step formula if you use the correct units in each step you will get the right answer every time. Remember • ### How to Calculate IV Drip RatesMedicNerd 2016-9-11 · How to Calculate IV Drip Rates. By Michael Stewart on September 11 2016 in Training Video. This tutorial shows you the easiest way to calculate an IV drip rate. Using this 3 step formula if you use the correct units in each step you will get the right answer every time. Remember • ### How to Calculate Drip Rate. A drip is a small individual drop of a liquid. Drip rate is how fast the liquid drips per unit time. Knowledge on drip rate is mostly used in hospitals to determine the intravenous (IV) tubing to be used. Formula to calculate drip rate. To calculate drip rate we divided the • ### Calculating Infusion Rates in units/h 2015-11-10 · Calculating Infusion Rates in units/h Given the Infusion Rate (mL/h) If an IV fluid contains an additive and you need to calculate the rate of infusion of the additive in units/h given a rate of infusion in mL/h of the IV fluid you can do so by following the two steps shown in the yellow box below Step 1 • ### Manual IV CalculatorWatch Count Drip Rate Setting Description This calculator determines the drip rate for a manually regulated IV. Flow rate is the rate a liquid volume is delivered over a period of time. Manually regulated IVs are also known gravity tubing IVs. Notes IV stands for intravenous. IV tubing comes in 10 15 and 20 gtt/mL for macrodrop and 60 gtt/mL for microdrop factors. • ### Intravenous Infusion Calculations 2018-11-8 · Drip Rates — is when the infusion volume is calculated into drops. The formula for the Drip Rate Drip Rate = Volume (mL) Time (h). Example 1 A patient is ordered to receive 1000 mL of intravenous fluids to run over 8 hours. Cal-culate the drip rate. Solution Volume = 1000 mL. Time = 8 hours. Substituting in the formula gives Drip Rate = 1000 mL 8 h • ### Intravenous Calculations PDFIntravenous Calculations Formula for Calculating IV RATES drops per minute Calculate the drip rate in drops per minute for each of the following blood transfusions. Give each answer to the nearest whole number. Rate dpm = Total volume (mL) x drip factor (drops/ml) Time (mins) A post-operative adult male is to be given 1 unit of autologous blood in 4 hours. • ### Intravenous Infusion Calculations 2018-11-8 · Intravenous Infusion Calculations Drip Rates — is when the infusion volume is calculated into drops. The formula for the Drip Rate Drip Rate = Volume (mL) Time (h). Example 1 A patient is ordered to receive 1000 mL of intravenous fluids to run over 8 hours. • ### How do you calculate drip per minute AskingLot 2020-5-28 · To calculate the drops per minute the drop factor is needed. The formula for calculating the IV flow rate (drip rate) is total volume (in mL) divided by time (in min) multiplied by the drop factor (in gtts/mL) which equals the IV flow rate in gtts/min. Click to see full answer. • ### Calculating Basic IV Drip RatesNursing School of Success 2019-4-24 · There are usually at least 2 conversions you need for basic drip rates the IV tubing size and time. This scenario gives you an IV tubing size of 15gtts/mL. The second conversion you need to use is 1 hr/60 min. Write your conversion factors in the middle of your paper. Step 4 Solve the problem. • ### No Infusion Pump You Can Calculate IV Drip Rates 2011-6-28 · nervous about calculating an IV drip rate". Rather than calculate the drip rate she consulted the pharmacist who calculated for her. When she returned to work the next day an incident report had been filed because the patient had received half the required dose of dopamine. There is no • ### IV Maintenance Fluids Calculator Example of a IV fluid calculation. These are the two methods for calculating pediatric maintenance fluid rates applied in the case of a child weighing 26 kg. 1) Daily volume formula (100 mL for each of the first 10 kg) (50 mL for each kg between 11 and 20) (20 mL for each additional kg past 20 kg) = 1 000 mL 500 mL 120 mL = 1 620 mL. • ### IV Flow Rate Calculations 2020-4-20 · Intravenous (IV) flow rate is the volume of a solution or drug administered in a given amount of time for an IV preparation. Units mL/h and drops/min (gtt/min). Drop (drip) factor refers to the size of the IV tubing and is stated on the IV administration package. It is calibrated in number of drops/mL. • ### IV Drip Rate Calculator IV Drip rate = ( Volume to be given in ml Drop factor in gtts/min)/Time in minutes. -In case the time is specified in hours then the flow rate will be IV Drip rate = ( Volume to be given in ml Drop factor in gtts/min)/ (Time in hours 60) • ### 3 Steps to Calculate IV Drip Rates an infographic 2019-1-7 · 3 Steps to calculate IV Drip Rates. Step1. The Flow Rate The Flow Rate is simply how many millilitres per hour is the fluid supposed to me infusing. It is unique to every question. In order to calculate this divide the volume to be infused in millilitres by the time in hours. • ### Calculating IV Drips FINALe-LeaRN 2015-12-10 · IV fluids with or without additives are medications. It is essential that Many nurses today work in facilities where electronic devices automatically calculate drip rate factors and deliver the amount of fluid/medication needed as scheduled. The nurse uses the • ### Calculating Infusion Rates in units/h 2015-11-10 · Calculating Infusion Rates in units/h Given the Infusion Rate (mL/h) If an IV fluid contains an additive and you need to calculate the rate of infusion of the additive in units/h given a rate of infusion in mL/h of the IV fluid you can do so by following the two steps shown in the yellow box below Step 1 • ### Intravenous Calculations Basicmedical Key 2017-2-11 · CALCULATING IV FLOW RATES IN gtt/min. When an electronic infusion device is not used the nurse manually regulates the IV rate. To manually regulate an IV the nurse must determine the rate in gtt/min. IV flow rates in gtt/min are determined by the type of IV administration tubing. The drop size is regulated by the size of the tubing. • ### Appendix 2 CalculationsWiley Online Library The drip rate must be worked out. Different giving sets admin-ister different amounts of drops per ml. • Paediatric giving set = 60 drops/ml • General giving sets = 20 drops/ml So the amount required Total amount in 24 hours ÷ 24 = hourly rate hourly rate ÷ 60 = minute rate minute rate giving set drip factor = number of drops per min • ### IV Drip Rate Calculation Formula Nursing Review Video 2021-6-25 · The formula for calculating the IV flow rate (drip rate) is total volume (in mL) divided by time (in min) multiplied by the drop factor (in gtts/mL) which equals the IV flow rate in gtts/min. Let s try an example. The provider has ordered 1 000 mL Lactated Ringers to infuse over 8 hours. • ### How to Calculate Drip Rate. A drip is a small individual drop of a liquid. Drip rate is how fast the liquid drips per unit time. Knowledge on drip rate is mostly used in hospitals to determine the intravenous (IV) tubing to be used. Formula to calculate drip rate. To calculate drip rate we divided the volume in milliliters by time in hours. • ### Drip Rate Calculations WorksheetsTeacher Worksheets Drip Rate Calculations. Showing top 8 worksheets in the categoryDrip Rate Calculations. Some of the worksheets displayed are Work 1 Drug dosage iv rates calculations Iv and drug calculations for busy paramedics Calculating iv drips accurately Drug calculations Pharmacology drug dosage read only Study guide with sample questions dosage calculation Healthcare math calculating iv flow • ### Fluid Therapy CalculatorSmall animal 2014-6-27 · Small Animal Fluid Therapy Calculator. Body Weight (kg) Animal type. / Small Dog Medium Dog Large Dog. Maintenance rate (ml/kg/day) Maintenance requirement (ml) over 24 hours. Multiplication factor of the maintenance (M) rate. M1 M2 M3. Fluid requirement (ml) over 24 hours. • ### Manual IV CalculatorWatch Count Drip Rate Setting Description This calculator determines the drip rate for a manually regulated IV. Flow rate is the rate a liquid volume is delivered over a period of time. Manually regulated IVs are also known gravity tubing IVs. Notes IV stands for intravenous. IV tubing comes in 10 15 and 20 gtt/mL for macrodrop and 60 gtt/mL for microdrop factors. • ### IV Flow Rate Calculations 2020-4-20 · Drop (drip) factor refers to the size of the IV tubing and is stated on the IV administration package. It is calibrated in number of drops/mL. There are two main types of drips Macrodrip (10 gtt/mL 15 gtt/mL and 20 gtt/mL) and Microdrip (60 gtt/mL). There • ### Dosage Calculations IV Drip FactorSimple Nursing Calculating intravenous drip rates (gtt/min) would involve these three main elements Total volume Drip factor Time So the formula usually goes as Total volume x drip factor ÷ time = flow rate (gtt/min) Smart Recall. We ve established an easy way for you to recall the formula listed above. • ### Calculating IV Drips FINALe-LeaRN 2015-12-10 · Many nurses today work in facilities where electronic devices automatically calculate drip rate factors and deliver the amount of fluid/medication needed as scheduled. The nurse uses the device s built-in computer entering the amount of fluid and the time it is to run the infusion machine then calculates the rate of infusion. • ### How do you manually calculate drip rate The formula for calculating the IV flow rate (drip rate) is total volume (in mL) divided by time (in min) multiplied by the drop factor (in gtts/mL) which equals the IV flow rate in gtts/min. Let s try an example. The provider has ordered 1 000 mL Lactated Ringers to infuse over 8 hours. • ### Dosage Calculations IV Drip FactorSimple Nursing Calculating intravenous drip rates (gtt/min) would involve these three main elements Total volume Drip factor Time So the formula usually goes as Total volume x drip factor ÷ time = flow rate (gtt/min) Smart Recall. We ve established an easy way for you to recall the formula listed above. • ### IV Flow Rate Calculations 2020-4-20 · Intravenous (IV) flow rate is the volume of a solution or drug administered in a given amount of time for an IV preparation. Units mL/h and drops/min (gtt/min). Drop (drip) factor refers to the size of the IV tubing and is stated on the IV administration package. It is calibrated in number of drops/mL. • ### IV Maintenance Fluids Calculator Example of a IV fluid calculation. These are the two methods for calculating pediatric maintenance fluid rates applied in the case of a child weighing 26 kg. 1) Daily volume formula (100 mL for each of the first 10 kg) (50 mL for each kg between 11 and 20) (20 mL for each additional kg past 20 kg) = 1 000 mL 500 mL 120 mL = 1 620 mL. • ### Intravenous Therapy Dose and Flow Rate Calculation 2020-5-12 · • Concentration of a drug (C) the amount of medication diluted in a volume of IV solution (e.g. 400 mg dopamine/250 ml) (Box 2) • Flow rate the speed at which the IV fluid infuses expressed as volume over time (e.g. 20 ml/hr) • Drop factor the number of drops in the IV drip chamber that is equivalent to 1 ml • ### Worksheet #1 2018-8-19 · a) Calculate the drip rate using a macrodrip (10 drops/mL) set. b) What is the total volume of IV solution infused over a 24-hour period 2. Order Administer 3 L of D5 1 2 NS over 24 hours a) Calculate the hourly rate. b) Calculate the drip rate using a microdrip administration set. 3. You have 1 L of IV fluid infusing at 60 mL/hr. • ### Worksheet #1 2018-8-19 · a) Calculate the drip rate using a macrodrip (10 drops/mL) set. b) What is the total volume of IV solution infused over a 24-hour period 2. Order Administer 3 L of D5 1 2 NS over 24 hours a) Calculate the hourly rate. b) Calculate the drip rate using a microdrip administration set. 3. You have 1 L of IV fluid infusing at 60 mL/hr. • ### Calculating Infusion Rates in units/h 2015-11-10 · Calculating Infusion Rates in units/h Given the Infusion Rate (mL/h) If an IV fluid contains an additive and you need to calculate the rate of infusion of the additive in units/h given a rate of infusion in mL/h of the IV fluid you can do so by following the two steps shown in the yellow box below Step 1 • ### How do you manually calculate drip rate The formula for calculating the IV flow rate (drip rate) is total volume (in mL) divided by time (in min) multiplied by the drop factor (in gtts/mL) which equals the IV flow rate in gtts/min. Let s try an example. The provider has ordered 1 000 mL Lactated Ringers to infuse over 8 hours. • ### Drug Dosage IV Rates CalculationsGeorge Brown 2020-5-13 · Calculation of Intravenous Drip Rates In these types of calculations for a given volume time period and drop factor (gtts/mL) the required IV flow rate in drops per minute (gtts/min) is calculated. Note Since a fraction of a drop is not possible to give to a patient it is usual to round • ### Calculating IV Drips FINALe-LeaRN 2015-12-10 · IV fluids with or without additives are medications. It is essential that Many nurses today work in facilities where electronic devices automatically calculate drip rate factors and deliver the amount of fluid/medication needed as scheduled. The nurse uses the • ### Dosage Calculations IV Drip FactorSimple Nursing Calculating intravenous drip rates (gtt/min) would involve these three main elements Total volume Drip factor Time So the formula usually goes as Total volume x drip factor ÷ time = flow rate (gtt/min) Smart Recall. We ve established an easy way for you to recall the formula listed above.

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# Problem Solving Using Hypothetical data Question 1 The attached file contains hypothetical data for working this problem. Goodman Corporation’s and Landry Incorporated’s stock prices and dividends, along with the Market Index, are shown in the file. Stock prices are reported for December 31 of each year, and dividends reflect those paid during the year. The market data are adjusted to include dividends. 1. Use the data given to calculate annual returns for Goodman, Landry, and the Market Index, and then calculate average returns over the five-year period.  (Hint: Remember, returns are calculated by subtracting the beginning price from the ending price to get the capital gain or loss, adding the dividend to the capital gain or loss, and dividing the result by the beginning price. Assume that dividends are already included in the index.   Also, you cannot calculate the rate of return for 2015 because you do not have 2014 data.) 2. Calculate the standard deviation of the returns for Goodman, Landry, and the Market Index.  (Hint:  Use the sample standard deviation formula given in the chapter, which corresponds to the STDEV function in Excel.) 3. On a stand-alone basis which corporation is the least risky? 4. Construct a scatter diagram graph that shows Goodman’s and Landry’ returns on the vertical axis and the Market Index’s returns on the horizontal axis. 5. Estimate Goodman’s and Landry’s betas as the slopes of regression lines with stock returns on the vertical axis (y-axis) and market return on the horizontal axis (x-axis).  (Hint: use Excel’s SLOPE function.)  Are these betas consistent with your graph? 6. The risk-free rate on long-term Treasury bonds is 8.04%.  Assume that the market risk premium is 6%.  What is the expected return on the market?  Now use the SML equation to calculate the two companies’ required returns. 7. If you formed a portfolio that consisted of 60% Goodman stock and 40% Landry stock, what would be its beta and its required return? 8. Suppose an investor wants to include Goodman Industries’ stock in his or her portfolio.  Stocks A, B, and C are currently in the portfolio, and their betas are 0.769, 0.985, and 1.423, respectively.  Calculate the new portfolio’s required return if it consists of 30% of Goodman, 20% of Stock A, 30% of Stock B, and 20% of Stock C. ## Get Professional Assignment Help Cheaply Are you busy and do not have time to handle your assignment? Are you scared that your paper will not make the grade? Do you have responsibilities that may hinder you from turning in your assignment on time? Are you tired and can barely handle your assignment? Are your grades inconsistent? Whichever your reason may is, it is valid! You can get professional academic help from our service at affordable rates. 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We have a team of professional academic writers who can handle all your assignments. Our essay writers are graduates with diplomas, bachelor's, masters, Ph.D., and doctorate degrees in various subjects. The minimum requirement to be an essay writer with our essay writing service is to have a college diploma. When assigning your order, we match the paper subject with the area of specialization of the writer. ## Why Choose Our Academic Writing Service? • Plagiarism free papers • Timely delivery • Skilled, Experienced Native English Writers • Ability to tackle bulk assignments • Reasonable prices ## How It Works #### 1.      Place an order You fill all the paper instructions in the order form. Make sure you include all the helpful materials so that our academic writers can deliver the perfect paper. It will also help to eliminate unnecessary revisions. #### 2.      Pay for the order Proceed to pay for the paper so that it can be assigned to one of our expert academic writers. 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# Properties Label 8.0.2067245.1 Degree $8$ Signature $[0, 4]$ Discriminant $2067245$ Root discriminant $6.16$ Ramified primes $5, 643$ Class number $1$ Class group trivial Galois group $C_2 \wr S_4$ (as 8T44) # Related objects Show commands for: SageMath / Pari/GP / Magma ## Normalizeddefining polynomial sage: x = polygen(QQ); K.<a> = NumberField(x^8 - x^7 + x^6 - x^4 + x^2 - x + 1) gp: K = bnfinit(x^8 - x^7 + x^6 - x^4 + x^2 - x + 1, 1) magma: R<x> := PolynomialRing(Rationals()); K<a> := NumberField(R![1, -1, 1, 0, -1, 0, 1, -1, 1]); $$x^{8} - x^{7} + x^{6} - x^{4} + x^{2} - x + 1$$ sage: K.defining_polynomial() gp: K.pol magma: DefiningPolynomial(K); ## Invariants Degree: $8$ sage: K.degree()  gp: poldegree(K.pol)  magma: Degree(K); Signature: $[0, 4]$ sage: K.signature()  gp: K.sign  magma: Signature(K); Discriminant: $$2067245$$$$\medspace = 5\cdot 643^{2}$$ sage: K.disc()  gp: K.disc  magma: Discriminant(Integers(K)); Root discriminant: $6.16$ sage: (K.disc().abs())^(1./K.degree())  gp: abs(K.disc)^(1/poldegree(K.pol))  magma: Abs(Discriminant(Integers(K)))^(1/Degree(K)); Ramified primes: $5, 643$ sage: K.disc().support()  gp: factor(abs(K.disc))[,1]~  magma: PrimeDivisors(Discriminant(Integers(K))); $|\Aut(K/\Q)|$: $2$ This field is not Galois over $\Q$. This is not a CM field. ## Integral basis (with respect to field generator $$a$$) $1$, $a$, $a^{2}$, $a^{3}$, $a^{4}$, $a^{5}$, $a^{6}$, $a^{7}$ sage: K.integral_basis() gp: K.zk magma: IntegralBasis(K); ## Class group and class number Trivial group, which has order $1$ sage: K.class_group().invariants() gp: K.clgp magma: ClassGroup(K); ## Unit group sage: UK = K.unit_group() magma: UK, f := UnitGroup(K); Rank: $3$ sage: UK.rank()  gp: K.fu  magma: UnitRank(K); Torsion generator: $$-1$$ (order $2$) sage: UK.torsion_generator()  gp: K.tu[2]  magma: K!f(TU.1) where TU,f is TorsionUnitGroup(K); Fundamental units: $$a^{7} - a^{6} + a^{5} - a^{3} + a - 1$$,  $$a^{6} - a^{5} - a^{2} + 1$$,  $$a^{4} + a - 1$$ sage: UK.fundamental_units()  gp: K.fu  magma: [K!f(g): g in Generators(UK)]; Regulator: $$0.453534943932$$ sage: K.regulator()  gp: K.reg  magma: Regulator(K); ## Class number formula $\displaystyle\lim_{s\to 1} (s-1)\zeta_K(s) \approx\frac{2^{0}\cdot(2\pi)^{4}\cdot 0.453534943932 \cdot 1}{2\sqrt{2067245}}\approx 0.2458126658969$ ## Galois group $C_2^3:S_4.C_2$ (as 8T44): sage: K.galois_group(type='pari') gp: polgalois(K.pol) magma: GaloisGroup(K); A solvable group of order 384 The 20 conjugacy class representatives for $C_2 \wr S_4$ Character table for $C_2 \wr S_4$ ## Intermediate fields Fields in the database are given up to isomorphism. Isomorphic intermediate fields are shown with their multiplicities. ## Sibling fields Degree 8 siblings: data not computed Degree 16 siblings: data not computed Degree 24 siblings: data not computed Degree 32 siblings: data not computed ## Frobenius cycle types $p$ $2$ $3$ $5$ $7$ $11$ $13$ $17$ $19$ $23$ $29$ $31$ $37$ $41$ $43$ $47$ $53$ $59$ Cycle type ${\href{/LocalNumberField/2.8.0.1}{8} }$ ${\href{/LocalNumberField/3.8.0.1}{8} }$ R ${\href{/LocalNumberField/7.3.0.1}{3} }^{2}{,}\,{\href{/LocalNumberField/7.2.0.1}{2} }$ ${\href{/LocalNumberField/11.4.0.1}{4} }^{2}$ ${\href{/LocalNumberField/13.8.0.1}{8} }$ ${\href{/LocalNumberField/17.8.0.1}{8} }$ ${\href{/LocalNumberField/19.4.0.1}{4} }^{2}$ ${\href{/LocalNumberField/23.6.0.1}{6} }{,}\,{\href{/LocalNumberField/23.1.0.1}{1} }^{2}$ ${\href{/LocalNumberField/29.3.0.1}{3} }^{2}{,}\,{\href{/LocalNumberField/29.1.0.1}{1} }^{2}$ ${\href{/LocalNumberField/31.6.0.1}{6} }{,}\,{\href{/LocalNumberField/31.2.0.1}{2} }$ ${\href{/LocalNumberField/37.2.0.1}{2} }^{3}{,}\,{\href{/LocalNumberField/37.1.0.1}{1} }^{2}$ ${\href{/LocalNumberField/41.4.0.1}{4} }^{2}$ ${\href{/LocalNumberField/43.2.0.1}{2} }^{3}{,}\,{\href{/LocalNumberField/43.1.0.1}{1} }^{2}$ ${\href{/LocalNumberField/47.4.0.1}{4} }{,}\,{\href{/LocalNumberField/47.1.0.1}{1} }^{4}$ ${\href{/LocalNumberField/53.3.0.1}{3} }^{2}{,}\,{\href{/LocalNumberField/53.2.0.1}{2} }$ ${\href{/LocalNumberField/59.4.0.1}{4} }^{2}$ In the table, R denotes a ramified prime. Cycle lengths which are repeated in a cycle type are indicated by exponents. sage: p = 7; # to obtain a list of $[e_i,f_i]$ for the factorization of the ideal $p\mathcal{O}_K$: sage: [(e, pr.norm().valuation(p)) for pr,e in K.factor(p)] gp: p = 7; \\ to obtain a list of $[e_i,f_i]$ for the factorization of the ideal $p\mathcal{O}_K$: gp: idealfactors = idealprimedec(K, p); \\ get the data gp: vector(length(idealfactors), j, [idealfactors[j][3], idealfactors[j][4]]) magma: p := 7; // to obtain a list of $[e_i,f_i]$ for the factorization of the ideal $p\mathcal{O}_K$: magma: idealfactors := Factorization(p*Integers(K)); // get the data magma: [<primefactor[2], Valuation(Norm(primefactor[1]), p)> : primefactor in idealfactors]; ## Local algebras for ramified primes $p$LabelPolynomial $e$ $f$ $c$ Galois group Slope content $5$5.2.1.1$x^{2} - 5$$2$$1$$1$$C_2$$[\ ]_{2} 5.2.0.1x^{2} - x + 2$$1$$2$$0$$C_2$$[\ ]^{2}$ 5.4.0.1$x^{4} + x^{2} - 2 x + 2$$1$$4$$0$$C_4$$[\ ]^{4}$ 643Data not computed ## Artin representations Label Dimension Conductor Artin stem field $G$ Ind $\chi(c)$ * 1.1.1t1.a.a$1$ $1$ $$\Q$$ $C_1$ $1$ $1$ 1.643.2t1.a.a$1$ $643$ $$\Q(\sqrt{-643})$$ $C_2$ (as 2T1) $1$ $-1$ 1.3215.2t1.a.a$1$ $5 \cdot 643$ $$\Q(\sqrt{-3215})$$ $C_2$ (as 2T1) $1$ $-1$ 1.5.2t1.a.a$1$ $5$ $$\Q(\sqrt{5})$$ $C_2$ (as 2T1) $1$ $1$ 2.643.3t2.a.a$2$ $643$ 3.1.643.1 $S_3$ (as 3T2) $1$ $0$ 2.16075.6t3.b.a$2$ $5^{2} \cdot 643$ 6.0.33230963375.1 $D_{6}$ (as 6T3) $1$ $0$ * 3.643.4t5.a.a$3$ $643$ 4.2.643.1 $S_4$ (as 4T5) $1$ $1$ 3.413449.6t8.a.a$3$ $643^{2}$ 4.2.643.1 $S_4$ (as 4T5) $1$ $-1$ 3.80375.6t11.a.a$3$ $5^{3} \cdot 643$ 6.2.51681125.1 $S_4\times C_2$ (as 6T11) $1$ $1$ 3.51681125.6t11.a.a$3$ $5^{3} \cdot 643^{2}$ 6.2.51681125.1 $S_4\times C_2$ (as 6T11) $1$ $-1$ * 4.3215.8t44.b.a$4$ $5 \cdot 643$ 8.0.2067245.1 $C_2 \wr S_4$ (as 8T44) $1$ $-2$ 4.80375.8t44.b.a$4$ $5^{3} \cdot 643$ 8.0.2067245.1 $C_2 \wr S_4$ (as 8T44) $1$ $-2$ 4.1329238535.8t44.b.a$4$ $5 \cdot 643^{3}$ 8.0.2067245.1 $C_2 \wr S_4$ (as 8T44) $1$ $2$ 4.33230963375.8t44.b.a$4$ $5^{3} \cdot 643^{3}$ 8.0.2067245.1 $C_2 \wr S_4$ (as 8T44) $1$ $2$ 6.51681125.8t41.b.a$6$ $5^{3} \cdot 643^{2}$ 8.4.258405625.1 $V_4^2:(S_3\times C_2)$ (as 8T41) $1$ $2$ 6.213...125.12t108.b.a$6$ $5^{3} \cdot 643^{4}$ 8.4.258405625.1 $V_4^2:(S_3\times C_2)$ (as 8T41) $1$ $-2$ 6.33230963375.8t41.b.a$6$ $5^{3} \cdot 643^{3}$ 8.4.258405625.1 $V_4^2:(S_3\times C_2)$ (as 8T41) $1$ $0$ 6.33230963375.12t108.b.a$6$ $5^{3} \cdot 643^{3}$ 8.4.258405625.1 $V_4^2:(S_3\times C_2)$ (as 8T41) $1$ $0$ 8.267...625.24t708.b.a$8$ $5^{6} \cdot 643^{4}$ 8.0.2067245.1 $C_2 \wr S_4$ (as 8T44) $1$ $0$ 8.427...025.24t708.b.a$8$ $5^{2} \cdot 643^{4}$ 8.0.2067245.1 $C_2 \wr S_4$ (as 8T44) $1$ $0$ Data is given for all irreducible representations of the Galois group for the Galois closure of this field. Those marked with * are summands in the permutation representation coming from this field. Representations which appear with multiplicity greater than one are indicated by exponents on the *.

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###### Exercise7 Write the fraction as a decimal number. Do not round your answers. 1. $${{\frac{3}{5}}}$$ = 2. $${{\frac{7}{16}}}$$ = in-context

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Question and answer true or false The Pacific campaign was largely conducted by the navy and the marines. The Pacific campaign was largely conducted by the navy and the marines. TRUE. Question Asked 6/26/2012 11:40:24 PM Updated 8/23/2016 2:10:03 AM 1 Answer/Comment Flagged by Wallet.ro [8/23/2016 1:49:32 AM], Unflagged by jeifunk [8/23/2016 2:10:01 AM], Edited by jeifunk [8/23/2016 2:10:03 AM] Get an answer New answers Rating 3 The Pacific campaign was largely conducted by the navy and the marines. TRUE. Added 8/23/2016 1:49:31 AM This answer has been confirmed as correct and helpful. Confirmed by jeifunk [8/23/2016 2:10:05 AM] Comments There are no comments. Add an answer or comment Log in or sign up first. 27,199,081 questions answered * Get answers from Weegy and a team of really smart live experts. Popular Conversations What is Amendments III? Weegy: The First Amendment prohibits the making of any law respecting an establishment of religion, impeding the free ... 8/13/2018 7:06:29 PM| 3 Answers 15. What is 26% as a fraction in simplest form? Weegy: 22% as a fraction in the simplest form is 11/50. 8/15/2018 9:52:26 AM| 3 Answers Veto Weegy: A veto is the power to unilaterally stop an official action, especially the enactment of legislation. ... 8/13/2018 7:41:47 PM| 2 Answers A paragraph with two topics is called what 8/13/2018 7:53:35 PM| 2 Answers 19% of 400 8/8/2018 11:52:34 AM| 2 Answers _______ are found only in animal cells and not in plant cells.      ... Weegy: Plant cells have chloroplasts and cell walls, which are not present in animal cells. User: Which of the ... 8/8/2018 3:17:30 PM| 2 Answers Add and simplify: 9/16 + 1/2 = Weegy: 9/16 + 1/2 = 9/16 + 8/16 = 17/16 = 1 1/16. User: 7 7/8-3 1/4 is? 8/8/2018 8:30:12 PM| 2 Answers S L Points 141 [Total 141] Ratings 0 Comments 111 Invitations 3 Online S L P P Points 109 [Total 563] Ratings 0 Comments 109 Invitations 0 Offline S L R P R P R Points 108 [Total 568] Ratings 0 Comments 8 Invitations 10 Offline S R L R P R P R R R Points 55 [Total 877] Ratings 0 Comments 5 Invitations 5 Offline S Points 45 [Total 45] Ratings 0 Comments 15 Invitations 3 Offline S Points 16 [Total 33] Ratings 0 Comments 16 Invitations 0 Offline S Points 12 [Total 12] Ratings 0 Comments 12 Invitations 0 Offline S Points 10 [Total 40] Ratings 1 Comments 0 Invitations 0 Offline S Points 2 [Total 2] Ratings 0 Comments 2 Invitations 0 Offline S Points 1 [Total 1] Ratings 0 Comments 1 Invitations 0 Offline * Excludes moderators and previous winners (Include) Home | Contact | Blog | About | Terms | Privacy | © Purple Inc.

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# GeometryProcessing/Spring2009/Schedule/Surface Generation Extraction1 (Difference between revisions) Revision as of 22:37, 21 January 2009 (view source)Loos (Talk | contribs) (→Questions)← Older edit Revision as of 22:37, 21 January 2009 (view source)Loos (Talk | contribs) (→Questions)Newer edit → Line 5: Line 5: === Questions === === Questions === - What steps would take place when using Marching Cubes to extract a surface from an implicit representation? + ''What steps would take place when using Marching Cubes to extract a surface from an implicit representation?'' (I'm not sure I understood the question, so feel free to tell me I didn't get it) If you already have a volumetric dataset stored as a 3D set of voxels in space, then you can just run Marching Cubes directly on that representation, one voxel being one cube. If what you have is an evaluation formula for each point in space, then you can first evaluate the function in a fixed volumetric grid, giving you the voxels and then running Marching Cubes. (I'm not sure I understood the question, so feel free to tell me I didn't get it) If you already have a volumetric dataset stored as a 3D set of voxels in space, then you can just run Marching Cubes directly on that representation, one voxel being one cube. If what you have is an evaluation formula for each point in space, then you can first evaluate the function in a fixed volumetric grid, giving you the voxels and then running Marching Cubes. ## Revision as of 22:37, 21 January 2009 • John Meier I liked the structure of Tuesday's lecture: 1) the build-up to the topic of marching cubes from marching tets, and 2) examination of the aspects of marching cubes that have been "solved", like how to prevent ambiguous surface extraction by automatically generating the intersection lookup table, and optimizing runtime with structures likes octrees or span spaces. The biggest mental hurdle for me among the lecture topics was the reason for building an octree of an implicit surface from the bottom up as a preprocessing step, which Carlos cleared up well (once the structure is built, it can be queried for any constant). The analogy to sorting an array before searching for elements was intuitive and appreciated. ### Questions What steps would take place when using Marching Cubes to extract a surface from an implicit representation? (I'm not sure I understood the question, so feel free to tell me I didn't get it) If you already have a volumetric dataset stored as a 3D set of voxels in space, then you can just run Marching Cubes directly on that representation, one voxel being one cube. If what you have is an evaluation formula for each point in space, then you can first evaluate the function in a fixed volumetric grid, giving you the voxels and then running Marching Cubes. Guidance Field Not sure if this is something we've talked about or are going to talk about, but for an advancing front method to generate iso-surfaces, what's a guidance field? I would assume that it's some sort of function that guides the creation of new triangles, but how? And what's it look like?

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# trig Evaluate each value as a trigonometric function of an angle in Quadrant 1. #1 cos (27pi/8) 1. 👍 0 2. 👎 0 3. 👁 282 1. 27 pi/8 = 3 pi + 3 pi/8 that is 180 degrees + 67.5 degrees that is in quadrant 3 where cos is negative so or -cos 67.5 degrees 1. 👍 0 2. 👎 0 ## Similar Questions 1. ### Math Write the first trigonometric function in terms of the second for θ in the given quadrant. tan(θ), cos(θ); θ in Quadrant IV I believe to start this is 1/cos^2*theta. Am I write? If not, can you please help me?! 2. ### Math If sin(x) = 1/3 and sec(y) = 29/21 , where x and y lie between 0 and π/2, evaluate the expression using trigonometric identities. (Enter an exact answer.) cos(2y) 3. ### trig The terminal side of θ lies on a given line in the specified quadrant. Find the values of the six trigonometric functions of θ by finding a point on the line. Line: (55/48)x Quadrant:III sin(θ)= cos(θ)= tan(θ)= csc(θ)= 4. ### Trig Evaluate the trigonometric function using its period as an aid. cos 5pi how would i go about finding the period and what is a period? 1. ### Math Evaluate the two trigonometric functions using its period as an aid. 1. cos (9pi/2) = ? Ans: 0 2. cos (-9pi/2) = ? Ans: 0 I think they are both 0 because cos(t) = x. 2. ### math Find the values of the trigonometric functions of t from the given information. tan(t) =1/9 terminal point of t is in Quadrant III sin(t)= cos(t)= csc(t)= sec(t)= cot(t)= 3. ### Math Use a calculator to evaluate the trigonometric function. Round to four decimal places. Make sure that your calculator is in the correct angle mode. 1) csc 2pi/5 1.3213? 2) sec 0.7 1.3075? 1. ### math given that csc(theta) =-3 and cos(theta)>0, find the remaining 5 trigonometric functions of theta. i have my triangle graphed in the fourth quadrant but I'm getting confused because of the negatives... 2. ### math Sketch a right triangle corresponding to the trigonometric function of the acute angle θ. Use the Pythagorean Theorem to determine the third side and then find the other five trigonometric functions of θ. cot(θ) = 3 sin(θ)=

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Entertainment News Check the I Had 5 Dollars My Mom Gave Me 10 Riddle, Answer Logically explained Written by Examine the I Had 5 {Dollars} My Mother Gave Me 10 Riddle, Reply Logically defined Why is I Had 5 {Dollars} My Mother Gave Me 10 Riddle trending on social media? Amid the COVID-19, many individuals will probably be on the lookout for methods to maintain them energetic and busy. Youngsters and adults are attempting to resolve puzzles to place their minds to the check and maintain busy in the course of the lockdown. These puzzles, riddles, and challenges have develop into viral very quickly as folks have been in search of totally different and enjoyable methods to attach. Such mind-challenging riddles are making rounds on WhatsApp teams and on Social Media. From Bollywood film riddles, Basic Information inquiries to enjoyable challenges, all the pieces has been trending on social media. Lately folks have been sharing the I Had 5 {Dollars} My Mother Gave Me 10 Riddle with their family and friends. This I Had 5 {Dollars} My Mother Gave Me 10 Riddle has garnered many responses and totally different solutions because it has been designed to examine your mathematical and logical considering abilities. Right here is the I Had 5 {Dollars} My Mother Gave Me 10 Riddle so that you can remedy. Attempt to remedy the riddle and in addition Problem your family and friends with this difficult riddle. What’s ‘I Had 5 {Dollars} My Mother Gave Me 10’ Riddle? The I Had 5 {Dollars} My Mother Gave Me 10 Riddle reads as follows: “I had 5 {dollars}, my mother gave me 10 {dollars}, whereas my dad gave me 30 {dollars}. My aunt and uncle gave me 100 {dollars}. I had one other 5 {dollars}. How a lot cash did I actually have?” What’s the reply to I Had 5 {Dollars} My Mother Gave Me 10 Riddle? The proper reply to the I Had 5 {Dollars} My Mother Gave Me 10 Riddle is “10” Clarification: Right here is the reason for the riddle. While you check out the riddle, in the primary a part of the query, the riddle says that the particular person has 5 {dollars} with him. And within the third a part of the riddle, it says that the particular person has one other 5 {dollars}. So in whole, leaving what the mom, father, and aunt & uncle gave, the particular person has 10 {dollars}. #Examine #{Dollars} #Mother #Gave #Riddle #Reply #Logically #defined

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SOPHIA will be down Wednesday, December 7, 2016 at 4:30 AM (CST) for a little housecleaning. We are sorry if this causes you any trouble. The outage should last less than 5 minutes. ### Free Educational Resources • > ACCT434 Final Exam + # ACCT434 Final Exam ##### Rating: (0) Author: Christine Farr ##### Description: http://www.devrygenius.com/product/acct434-final-exam 1. (TCO 1) A significant limitation of activity-based costing is the (Points : 5) 2. (TCO 1) Ireland Company produces a special spray nozzle. The budgeted indirect total cost of inserting the spray nozzle is \$180,000. The budgeted number of nozzles to be inserted is 80,000. What is the budgeted indirect cost allocation rate for this activity? (Points : 5) 3. (TCO 2)Fixed overhead costs include (Points : 5) 4. (TCO 2) Information pertaining to Brenton Corporation's sales revenue ispresented in the following table: February March April Cash Sales \$160,000 \$150,000 \$120,000 Credit Sales 300,000 400,000 280,000 Total Sales \$460,000 \$550,000 \$400,000 Management estimates that 5% of credit sales are not collectible. Of the credit sales that are collectible, 75% are collected in the month of sale and the remainder in the month following the sale. Cost of purchases of inventory each month are 80% of the next month's projected total sales.All purchases of inventory are on account; 50% are paid in the month of purchase, and the remainder is paid in the month following the purchase. Brenton's budgeted total cash receipts in April are (Points : 5) 5. (TCO 2)Budgeting provides all of the following except (Points : 5) 6. (TCO 3) The cost function y = 1,000 + 5X (Points : 5) 7. (TCO 3) Which cost estimation method uses a formal mathematical method to develop cost functions based on past data? (Points : 5) 8. (TCO 4)Sunk costs (Points : 5) 9. (TCO 5) Through put contribution equals revenues minus (Points : 5) 10. (TCO 5) Producing more nonbottleneck output (Points : 5) 11. (TCO 6) What type of cost is the result of an event that results in more than one product or service simultaneously? (Points : 5) 12. (TCO 6) Which of the following is a disadvantage of the physical-measuremethod of allocating joint costs? (Points : 5) 13. (TCO 7) Life-cyclecosting is the name given to (Points : 5) 14. (TCO 7) Each month, Haddon Company has \$300,000 total manufacturingcosts (20% fixed) and \$125,000 distribution and marketing costs (75% fixed). Haddon's monthly sales are \$500,000. The markup percentage on variable costs to arrive at the existing (target) selling price is (Points : 5) 15. (TCO 8) The costs used in cost-based transfer prices (Points : 5) (more) ##### Try a College Course Free Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to over 2,000 colleges and universities.* No credit card required 25 Sophia partners guarantee credit transfer. 221 Institutions have accepted or given pre-approval for credit transfer. * The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 20 of Sophia’s online courses. More than 2,000 colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs. Tutorial

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Related posts of 1 1 Points Lines And Planes Worksheet Answers. 1 day ago Geometry plane and simple worksheet answers - To discover the image more evidently in this article you are able to click on the preferred image to look at the photo in its original dimensions or in full. Back To 1 1 Points Lines And Planes Worksheet Answers. Explanation of Geometry Points Lines Line Segments and Rays. 1 1 Points Lines and Planes Worksheet Answers with Parallel and Perpendicular Lines Geometry Proving Lines Para. So they are coplanar. Worksheet 9 Geometry Points Lines Line Segments and Rays. 25122020 Points lines and planes worksheets answers prentice hall gold. Name three points that are collinear. Source: promotiontablecovers.blogspot.com Line n intersects line m at R but does not intersect line l. Related posts of 1 1 Points Lines And Planes Worksheet Answers. Although these words are not formally defined it is important to have general agreement on what each word means. KITES Kite making has become an art form using numerous. Different planes is a line. Source: promotiontablecovers.blogspot.com They color each one accordingly and end up with a design that can be checked quickly for accuracy but cannot be easily predicted by students. Plane R contains lines and which intersect at point P. We tried to locate some good of 1 1 Points Lines and Planes Worksheet Answers with Parallel and Perpendicular Lines Geometry Proving Lines Para image to suit your needs. Making your individual worksheets is easy and it lets you comprise just the right fabric that you want to ensure your scholars can learn and commit to memory. Interesting descriptive charts multiple choice questions and word problems are included in these pdf worksheets. Source: promotiontablecovers.blogspot.com Worksheet 9 Geometry Points Lines Line Segments and Rays. EXAMPLE 1 Name points lines and planes. 22022018 Version 1 Worksheet - Students determine whether each statement is always true. So they are coplanar. Lesson 1-2 Measure segments and determine accuracy of. Source: pinterest.com Plane R contains lines and which intersect at point P. Give two other names for and for plane Z. Diy 10 Creative Points Lines and Planes Worksheet Answer Key Worksheets are definitely the spine to pupils getting to know and grasping principles taught through the teacher. Worksheet September 06 2018. For instance points H E and G do not lie on the same line. Source: pinterest.com Although these words are not formally defined it is important to have general agreement on what each word means. Diy 10 Creative Points Lines and Planes Worksheet Answer Key Worksheets are definitely the spine to pupils getting to know and grasping principles taught through the teacher. Some of the worksheets for this concept are Unit 1 tools of geometry reasoning and proof Ms work 132 153 geometry 06 Geometry chapter 1 notes practice work Geometry unit 1 workbook Chapter 4 lesson1 0 points line segments lines and rays Geometry plane and simple answer. Point A point has no dimension. This ensemble of printable worksheets for grade 8 and high school contains exercises to identify and draw the points lines and planes. Source: promotiontablecovers.blogspot.com Lesson 1-2 Measure segments and determine accuracy of. They color each one accordingly and end up with a design that can be checked quickly for accuracy but cannot be easily predicted by students. 1 1 Points Lines and Planes Worksheet Answers with Parallel and Perpendicular Lines Geometry Proving Lines Para. Points lines and planes are the basic building blocks used in geometry. Related posts of 1 1 Points Lines And Planes Worksheet Answers. Source: promotiontablecovers.blogspot.com Add point C on plane R. 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Worksheet 9 Geometry Points Lines Line Segments and Rays. Source: promotiontablecovers.blogspot.com Some of the worksheets for this concept are Name answer key Chapter 4 lesson1 0 points line segments lines and rays P points lines angles and parallel lines p Identify points lines and planes Points lines and planes 1 Lines and angles Points lines planes angles. Points D E and F lie on the same line. Making your individual worksheets is easy and it lets you comprise just the right fabric that you want to ensure your scholars can learn and commit to memory. E At which line do the planes L and T intersect. Interesting descriptive charts multiple choice questions and word problems are included in these pdf worksheets. Source: pinterest.com Point A point has no dimension. Name four points that are coplanar. 1 day ago Geometry plane and simple worksheet answers - To discover the image more evidently in this article you are able to click on the preferred image to look at the photo in its original dimensions or in full. They can Points A G and B lie in a plane but point E does not lie in 3. The intersection of two The intersection of two. Source: promotiontablecovers.blogspot.com Name four points that are coplanar. Points lines and planes are the basic building blocks used in geometry. For instance points H E and G do not lie on the same line. Lesson 1-2 Measure segments and determine accuracy of. A Name the planes. Source: pinterest.com Handphone Tablet Desktop Original Size This process continues until the first letter combination is found after which the student can move onto the next letter and so on and so forth. Lines l and m intersect at point P. Some of the worksheets for this concept are Unit 1 tools of geometry reasoning and proof Ms work 132 153 geometry 06 Geometry chapter 1 notes practice work Geometry unit 1 workbook Chapter 4 lesson1 0 points line segments lines and rays Geometry plane and simple answer. So they are coplanar. It is usually represented by a small dot and named by a capital letter. Source: pinterest.com E At which line do the planes L and T intersect. 22022018 Version 1 Worksheet - Students determine whether each statement is always true. The intersection of two The intersection of two. Worksheet 9 Geometry Points Lines Line Segments and Rays. Our free printable worksheets focus on developing knowledge and skill of these fundamentals of geometry. Source: pinterest.com Although these words are not formally defined it is important to have general agreement on what each word means. So they are coplanar. Explanation of Geometry Points Lines Line Segments and Rays. Lines l m and n are coplanar and lie in plane Q. Back To 1 1 Points Lines And Planes Worksheet Answers. This site is an open community for users to share their favorite wallpapers on the internet, all images or pictures in this website are for personal wallpaper use only, it is stricly prohibited to use this wallpaper for commercial purposes, if you are the author and find this image is shared without your permission, please kindly raise a DMCA report to Us. If you find this site convienient, please support us by sharing this posts to your own social media accounts like Facebook, Instagram and so on or you can also save this blog page with the title lively points lines and planes worksheets answers by using Ctrl + D for devices a laptop with a Windows operating system or Command + D for laptops with an Apple operating system. If you use a smartphone, you can also use the drawer menu of the browser you are using. 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Difficulty level In this tutorial, users will learn how to create a trial-averaged BOLD response and store it in a matrix in MATLAB. Difficulty level: Intermediate Duration: 20:12 Speaker: : Mike X. Cohen This tutorial teaches users how to create animations of BOLD responses over time, to allow researchers and clinicians to visualize time-course activity patterns. Difficulty level: Intermediate Duration: 12:52 Speaker: : Mike X. Cohen This tutorial demonstrates how to use MATLAB to create event-related BOLD time courses from fMRI datasets. Difficulty level: Intermediate Duration: 13:39 Speaker: : Mike X. Cohen In this tutorial, users learn how to compute and visualize a t-test on experimental condition differences. Difficulty level: Intermediate Duration: 17:54 Speaker: : Mike X. Cohen This lesson introduces various methods in MATLAB useful for dealing with data generated by calcium imaging. Difficulty level: Intermediate Duration: 5:02 Speaker: : Mike X. Cohen This tutorial demonstrates how to use MATLAB to generate and visualize animations of calcium fluctuations over time. Difficulty level: Intermediate Duration: 15:01 Speaker: : Mike X. Cohen This tutorial instructs users how to use MATLAB to programmatically convert data from cells to a matrix. Difficulty level: Intermediate Duration: 5:15 Speaker: : Mike X. Cohen In this tutorial, users will learn how to identify and remove background noise, or "blur", an important step in isolating cell bodies from image data. Difficulty level: Intermediate Duration: 17:08 Speaker: : Mike X. Cohen This lesson teaches users how MATLAB can be used to apply image processing techniques to identify cell bodies based on contiguity. Difficulty level: Intermediate Duration: 11:23 Speaker: : Mike X. Cohen This tutorial demonstrates how to extract the time course of calcium activity from each clusters of neuron somata, and store the data in a MATLAB matrix. Difficulty level: Intermediate Duration: 22:41 Speaker: : Mike X. Cohen This lesson demonstrates how to use MATLAB to implement a multivariate dimension reduction method, PCA, on time series data. Difficulty level: Intermediate Duration: 17:19 Speaker: : Mike X. Cohen This is a tutorial introducing participants to the basics of RNA-sequencing data and how to analyze its features using Seurat. Difficulty level: Intermediate Duration: 1:19:17 Speaker: : Sonny Chen This tutorial introduces pipelines and methods to compute brain connectomes from fMRI data. With corresponding code and repositories, participants can follow along and learn how to programmatically preprocess, curate, and analyze functional and structural brain data to produce connectivity matrices. Difficulty level: Intermediate Duration: 1:39:04 This is a tutorial on designing a Bayesian inference model to map belief trajectories, with emphasis on gaining familiarity with Hierarchical Gaussian Filters (HGFs). This lesson corresponds to slides 65-90 of the PDF below. Difficulty level: Intermediate Duration: 1:15:04 Speaker: : Daniel Hauke This lesson introduces the practical exercises which accompany the previous lessons on animal and human connectomes in the brain and nervous system. Difficulty level: Intermediate Duration: 4:10 Speaker: : Dan Goodman This tutorial provides instruction on how to simulate brain tumors with TVB (reproducing publication: Marinazzo et al. 2020 Neuroimage). This tutorial comprises a didactic video, jupyter notebooks, and full data set for the construction of virtual brains from patients and health controls. Difficulty level: Intermediate Duration: 10:01 This lecture and tutorial focuses on measuring human functional brain networks, as well as how to account for inherent variability within those networks. Difficulty level: Intermediate Duration: 50:44 Speaker: : Caterina Gratton In this lesson, you will learn about the Python project Nipype, an open-source, community-developed initiative under the umbrella of NiPy. Nipype provides a uniform interface to existing neuroimaging software and facilitates interaction between these packages within a single workflow. Difficulty level: Intermediate Duration: 1:25:05 Speaker: : Satrajit Ghosh

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# Pages 70-74 Seven Reasons for Standards-Based Grading Date conversion 20.05.2016 Size 42.05 Kb. October 2008 | Volume 66 | Number 2 Expecting Excellence    Pages 70-74 Patricia L. Scriffiny If your grading system doesn't guide students toward excellence, it's time for something completely different. Each week brings some new idea that teachers are supposed to implement, while still preparing lessons, grading papers, and keeping their classrooms in some semblance of order. Amid all these challenges, a call to change grading policies can seem particularly unrealistic. My school, Montrose High School, is located in a small but rapidly growing rural community in southwestern Colorado. We serve a community that is primarily white but that has a significant Latino population. After spending the last three years implementing standards-based grading in my high school math classroom, I have discovered seven solid reasons for replacing point-based grades with a standards-based system. Reason 1: Grades Should Have Meaning Each letter grade that a student earns at the high school level is connected to a graduation credit, and many classes reflect only one step in a sequence of learning. So what does each grade indicate to students, parents, and teachers of later courses in the sequence? When I first considered this question, I realized I had no answers. When I was pressed to describe the qualitative difference between an A, B, C, D, or F, my answers were vague. So, I developed a much more focused idea of what I want my grades to mean: • An A means the student has completed proficient work on all course objectives and advanced work on some objectives. • A B means the student has completed proficient work on all course objectives. • A C means the student has completed proficient work on the most important objectives, although not on all objectives. The student can continue to the next course. • A D means the student has completed proficient work on at least one-half of the course objectives but is missing some important objectives and is at significant risk of failing the next course in the sequence. The student should repeat the course if it is a prerequisite for another course. • An F means the student has completed proficient work on fewer than one-half of the course objectives and cannot successfully complete the next course in sequence. Reason 2: We Need to Challenge the Status Quo Many notions I had at the beginning of my career about grading didn't stand up to real scrutiny. The thorny issue of homework is one example of how the status quo needed to change. I once thought it was essential to award points to students simply for completing homework. I didn't believe students would do homework unless it was graded. And yet, in my classroom, students who were clearly learning sometimes earned low grades because of missing work. Conversely, some students actually learned very little but were good at “playing school.” Despite dismal test scores, these students earned decent grades by turning in homework and doing extra credit. They would often go on to struggle in later courses, while their parents watched and worried. Over the past three years, I have radically changed how I formally assess homework—I don't. Of course, it is essential for students to do homework that is tied closely to learning objectives and for students to see those connections (Marzano, Pickering, & Pollock, 2001). Systematic and extensive feedback on assignments sends students the message that they can and should do homework as practice. A typical homework assignment for my students consists of a small collection of problems, each of which is linked to a learning objective. At first, I make those connections for my students, but eventually they make them on their own. When I assign homework, I discuss with my students where and how it applies to their assessments. My goal is to get students to constantly ask themselves, “Do I know this? Can I do this?” To my surprise, my homework completion rates have remained steady over the past three years. Some students don't do all of the homework that I assign, but they know that they are accountable for mastering the standard connected to it. Of course, not every student who needs to practice always does so, but I am amazed and encouraged that students ask me for extra practice fairly regularly. Reason 3: We Can Control Grading Practices One of the biggest sources of frustration in schools today is the sense that we are at the mercy of factors we teachers cannot control. We cannot control student socioeconomic levels, school funding, our salaries, our teaching assignments, increasing class sizes, difficult parents, or a host of other important issues. However, we can control how we assess students. When I approached my principal and district officials with the idea of using an experimental grading system, I received support and encouragement from all of them. In addition, a number of colleagues have been intrigued and want to make standards-based grading work in their classrooms. If a teacher must use a point system to satisfy an administrative mandate or to use a particular grade book, that teacher can still use a standards-based system. The crucial idea is to use a system that is not based on the inappropriate use of averages. The system must not allow students to mask their level of understanding with their attendance, their level of effort, or other peripheral issues. I have found that avoiding point values that might appear in a traditional percentage-based system is helpful because parents and students can get confused if they see numbers that look like what they've seen in the past but refer to a different scale. Teachers who have to assign points can avoid this confusion by using completely different numbers. A point value in the range of 1 to 10, for example, would not have the strong associations of a point value of 85, and thus would not be as easily misinterpreted. Reason 4: Standards-Based Grading Reduces Meaningless Paperwork Since I adopted standards-based grading, my load of meaningless paperwork has been drastically reduced, which provides time for more important considerations. Standards-based grading enables me to get the most from every piece of paper students turn in. Writing feedback only on selected homework problems saves my time when marking papers while still giving me a sense of where students are in their learning. These homework assignments and other formative assessments help me judge the progress of the group as a whole before deciding how to proceed. I don't assess student mastery of any objective until I am confident that a reasonable number of students will score proficiently, and that makes each assessment mean much more. Students who are still struggling after a significant portion of the class has demonstrated mastery can retest individually. The bottom line is that when I review any set of papers, I walk away knowing a great deal more about what my students know than I ever did before. Reason 5: It Helps Teachers Adjust Instruction Imagine two different grade books for the same set of students, as shown in Figure 1. Which one of the two better illustrates what students know and what they still need to learn? Figure 1. Comparing Traditional and Standards-Based Grade Books Traditional Grade Book Name Homework Average Quiz 1 Chapter 1 Test John 90 65 70 Bill 50 75 78 Susan 110 50 62 Felicia 10 90 85 Amanda 95 100 90 Standards-Based Grade Book Name Objective 1: Write an alternate ending for a story Objective 2: Identify the elements of a story Objective 3: Compare and contrast two stories John Partially proficient Proficient Partially proficient Bill Proficient Proficient Partially proficient Susan Partially proficient Partially proficient Partially proficient Felicia Advanced Proficient Proficient Amanda Partially proficient Advanced Proficient The standards-based grade book gives a wealth of information to help the teacher adjust instruction. Note that two objectives (1 and 3) may require more class instruction. The notations for Objective 2, on the other hand, suggest that the class only needs practice and one student needs some reteaching. Students can also see much more information about their learning. In the traditional grade book, Amanda would assume she is in great shape, but standards-based grading reveals that she has not mastered a crucial concept. Gifted and talented students can be truly challenged in a standards-based classroom because if they show early mastery of fundamental skills and concepts, they can then concentrate on more challenging work that is at higher levels of Bloom's taxonomy or that seeks connections among objectives. Students who struggle can continue to retest and use alternate assessments until they show proficiency, and they are not penalized for needing extended time. I guide students with special needs to modify their work and, if needed, develop different ways of demonstrating that they've met their proficiency goals. Their working styles can be easily accommodated in this system because modified assignments and assessments require no special adjustments in the grade book. The grade book simply shows where they are in meeting the standards, without reference to how they are demonstrating their learning or what modifications needed to be made. Reason 6: It Teaches What Quality Looks Like In the adult world, everything is a performance assessment. If adults on the job make poor decisions or cannot determine the quality of their own work, the results are generally undesirable. Quality matters, and the ability to measure the quality of one's own work is a learned skill. So how can we teach this essential skill? One way to teach quality is to demand it. We must create an environment where standards can and must be met and where students are not permitted to submit substandard work without being asked to revise. If we base our grades on standards rather than attendance, behavior, or extra credit (which often has nothing to do with course objectives), we can actually help students grapple with the idea of quality and walk away with a higher degree of self-sufficiency. We can and should report information about student performance in areas like attendance and effort, but we can report it separately from academic achievement (O'Connor, 2007; Tomlinson & McTighe, 2006). Reason 7: It's a Launchpad to Other Reforms When I began using standards-based grading, I quickly discovered that I needed to reexamine my curriculum. Each class needed a clear and concise set of standards with precise levels of mastery. This prompted a number of discussions with other teachers in my department, and each year we continue to adapt our objectives. No one can use standards-based grading without clear standards. In addition to improving curriculum, I have found new ways to use formative assessments and intervention strategies. My work with special education students and English language learners in particular goes much more smoothly because all the modification needed is already built into what I do. I have also been able to work much more effectively with parents by giving them better information. How do students respond to this style of grading? Of course, their reactions vary. It takes time, discussion, and reflection for students to understand their rights and responsibilities in such a system, and teachers must be patient as students and parents adjust. Many students have expressed increased satisfaction with having a larger degree of control over their grades, although some students do not like the revisions they are required to do. Some struggle to overcome test anxiety and need access to alternate assessments. As for parents, many of them simply want opportunities for their children to succeed, so they are grateful for the revision and retesting. Each year, parents ask thoughtful questions, with some noting that this method of grading is more similar to evaluation in the workplace. These seven reasons to change to standards-based grading are merely a starting point. High school teachers need to hold their own practices up to scrutiny and decide whether those practices are worth keeping. By doing so, we unleash a force for change that we can control, with our students and parents as partners. References Marzano, R., Pickering, D., & Pollock, J. (2001). Classroom instruction that works. Alexandria, VA: ASCD. O'Connor, K. (2007). A repair kit for grading: 15 fixes for broken grades. Portland, OR: Educational Testing Service. Tomlinson, C., & McTighe, J. (2006). Integrating differentiated instruction and understanding by design. Alexandria, VA: ASCD. Patricia L. Scriffiny is a math teacher at Montrose High School in Montrose, Colorado; pscriffiny@mcsd.k12.co.us.

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# What is 21? Triple-stumper from Tuesday, 2-1: In Science: “It’s the largest number in the Fibonacci sequence that’s also a day in a month.” I think I knew this at one point, but I couldn’t remember what the sequence was, and I wanted to!  It turns out to be quite simple: You simply start with 1, and add the two numbers just prior in the sequence to get the next number.  Okay, so 1.  1+0=1.  1+1=2.  2+1=3.  3+2=5.  5+3=8.  8+5=13.  And 13+8=21.  The next number in the sequence, 13+21=33, cannot be a day in a month.  So 21 is the correct question. According to An Incomplete Education by Judy Jones and William Wilson, Fibonacci came up with the sequence when he was trying to determine “how many pairs of rabbits can be produced from a single pair of rabbits, if every month every pair begets a new pair, which from the second month on is itself productive, assuming that none of the rabbits die…and each pair consists of a male and a female.” I remembered that there was some other significance to the sequence, and Jones and Wilson go on to explain that the sequence is found repeatedly in nature as well as art and architecture.  I did already remember my favorite example, the pineapple, as well as the “Golden Rectangle.”  Here we go: The pineapple:  This animation, from a website created by Jill Britton, shows it best:          There are three different lines of “bumps” going different directions that pass through the same bump (in the center), and each line contains a number from the Fibonacci sequence (5, 8, and 13).  That ratio is consistent among pineapples. Now for the golden rectangle: First, the golden section is a line divided into two parts where the larger part is to the smaller part what the whole part is to the larger part.  That ratio is 1 : 1.618, the ratio between any two adjacent numbers in the Fibonacci sequence after three.  The golden rectangle has the same relationship with its length and width.  (That is, if you were to take the two parts from a golden section, each part would make the length and width.)  The facade of the Parthenon is a golden rectangle.  So is the human body, in multiple places.  An example: The distance from a person’s bellybutton to their feet is 1 unit, and their whole height is then the 1.618 in the ratio. (I hope this has not been too confusing!)

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## DFT Bin Value Formulas for Pure Real Tones April 17, 20151 comment Introduction This is an article to hopefully give a better understanding to the Discrete Fourier Transform (DFT) by deriving an analytical formula for the DFT of pure real tones. The formula is used to explain the well known properties of the DFT. A sample program is included, with its output, to numerically demonstrate the veracity of the formula. This article builds on the ideas developed in my previous two blog articles: ## DFT Graphical Interpretation: Centroids of Weighted Roots of Unity April 10, 20151 comment Introduction This is an article to hopefully give a better understanding to the Discrete Fourier Transform (DFT) by framing it in a graphical interpretation. The bin calculation formula is shown to be the equivalent of finding the center of mass, or centroid, of a set of points. Various examples are graphed to illustrate the well known properties of DFT bin values. This treatment will only consider real valued signals. Complex valued signals can be analyzed in a similar manner with... ## The Exponential Nature of the Complex Unit Circle Introduction This is an article to hopefully give an understanding to Euler's magnificent equation: $$e^{i\theta} = cos( \theta ) + i \cdot sin( \theta )$$ This equation is usually proved using the Taylor series expansion for the given functions, but this approach fails to give an understanding to the equation and the ramification for the behavior of complex numbers. Instead an intuitive approach is taken that culminates in a graphical understanding of the equation. Complex... ## Computing Translated Frequencies in Digitizing and Downsampling Analog Bandpass Signals October 31, 20131 comment In digital signal processing (DSP) we're all familiar with the processes of bandpass sampling an analog bandpass signal and downsampling a digital bandpass signal. The overall spectral behavior of those operations are well-documented. However, mathematical expressions for computing the translated frequency of individual spectral components, after bandpass sampling or downsampling, are not available in the standard DSP textbooks. The following three sections explain how to compute the... ## A Quadrature Signals Tutorial: Complex, But Not Complicated Introduction Quadrature signals are based on the notion of complex numbers and perhaps no other topic causes more heartache for newcomers to DSP than these numbers and their strange terminology of j operator, complex, imaginary, real, and orthogonal. If you're a little unsure of the physical meaning of complex numbers and the j = √-1 operator, don't feel bad because you're in good company. Why even Karl Gauss, one the world's greatest mathematicians, called the j-operator the "shadow of... ## A Fixed-Point Introduction by Example Introduction The finite-word representation of fractional numbers is known as fixed-point.  Fixed-point is an interpretation of a 2's compliment number usually signed but not limited to sign representation.  It extends our finite-word length from a finite set of integers to a finite set of rational real numbers [1].  A fixed-point representation of a number consists of integer and fractional components.  The bit length is defined... ## A Recipe for a Basic Trigonometry Table October 4, 2022 Introduction This is an article that is give a better understanding to the Discrete Fourier Transform (DFT) by showing how to build a Sine and Cosine table from scratch. Along the way a recursive method is developed as a tone generator for a pure tone complex signal with an amplitude of one. Then a simpler multiplicative one. Each with drift correction factors. By setting the initial values to zero and one degrees and letting it run to build 45 values, the entire set of values needed... ## Exact Near Instantaneous Frequency Formulas Best at Peaks (Part 2) Introduction This is an article that is a continuation of a digression from trying to give a better understanding of the Discrete Fourier Transform (DFT). It is recommended that my previous article "Exact Near Instantaneous Frequency Formulas Best at Peaks (Part 1)"[1] be read first as many sections of this article are directly dependent upon it. A second family of formulas for calculating the frequency of a single pure tone in a short interval in the time domain is presented. It... ## An Alternative Form of the Pure Real Tone DFT Bin Value Formula December 17, 2017 Introduction This is an article to hopefully give a better understanding of the Discrete Fourier Transform (DFT) by deriving alternative exact formulas for the bin values of a real tone in a DFT. The derivation of the source equations can be found in my earlier blog article titled "DFT Bin Value Formulas for Pure Real Tones"[1]. The new form is slighty more complicated and calculation intensive, but it is more computationally accurate in the vicinity of near integer frequencies. This... ## GPS - some terminology! Hi! For my first post, I will share some information about GPS - Global Positioning System. I will delve one step deeper than a basic explanation of how a GPS system works and introduce some terminology. GPS, like we all know is the system useful for identifying one's position, velocity, & time using signals from satellites (referred to as SV or space vehicle in literature). It uses the principle of trilateration  (not triangulation which is misused frequently) for... ## New Video: Parametric Oscillations January 4, 2017 I just posted this last night.  It's kinda off-topic from the mission of the channel, but I realized that it had been months since I'd posted a video, and having an excuse to build on helped keep me on track. ## Pentagon Construction Using Complex Numbers October 13, 2023 A method for constructing a pentagon using a straight edge and a compass is deduced from the complex values of the Fifth Roots of Unity. Analytic values for the points are also derived.

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# In Defense Of FOIL I do sometimes read online forums of educators, particularly math educators, since it’s fun to have somewhere to talk shop, and the topics of conversation are constant enough you don’t have to spend much time getting the flavor of a particular group before participating. If you suppose the students are lazy, the administrators meddling, the community unsupportive, and the public irrationally terrified of mathematics you’ve covered most forum threads. I had no luck holding forth my view on one particular topic, though, so I’ll try fighting again here where I can easily squelch the opposition. The argument, a subset of students-are-lazy (as they don’t wish to understand mathematics), was about a mnemonic technique called FOIL. It’s a tool to help people multiply binomials. Binomials are the sum (or difference) of two quantities, for example, (a + 2) or (b + 5). Here a and b are numbers whose value I don’t care about; I don’t care about the 2 or 5 either, but by picking specific values I avoid having too much abstraction in my paragraph. The product of (a + 2) with (b + 5) is the sum of all the pairs made by multiplying one term in the first binomial by one term in the second. There are four such pairs: a times b, and a times 5, and 2 times b, and 2 times 5. And therefore the product (a + 2) * (b + 5) will be a*b + a*5 + 2*b + 2*5. That would usually be cleaned up by writing 5*a instead of a*5, and by writing 10 instead of 2*5, so the sum would become a*b + 5*a + 2*b + 10. FOIL is a way of making sure one has covered all the pairs. The letters stand for First, Outer, Inner, Last, and they mean: take the product of the First terms in each binomial, a and b; and those of the Outer terms, a and 5; and those of the Inner terms, 2 and b; and those of the Last terms, 2 and 5. Here is my distinguished colleague’s objection to FOIL: Nobody needs it. This is true. You need to find all the pairs made of one term in the first binomial and one term in the second (so that a times 2 and b times 5 are right out), but there’s no need for an extra tool for that. More, if one has, say, a trinomial — three terms added together, such as (a + π [pi] + 2) — then FOIL breaks down altogether. There should be terms representing π [pi] * b and π [pi] * 5 which are neither First, Outer, Inner, nor Last, and no one has found a mnemonic which helps matters any. There are similar problems if one wants to multiply two trinomials, or something with a monomial (a single term, such as b alone), or more complicated terms. And it does nothing if one needs to multiply three or more sets of binomials. Better, the argument goes, to teach students that this is all a consequence of the distributive law and not bother with a tool only good for multiplying two binomials. I agree with the complaints that a tool to help people correctly multiply two binomials does nothing to help them multiply non-two non-binomials. And I agree that if one is comfortable with the reasoning behind why FOIL works where it does, then one does not need it. But I disagree that’s a reason not to teach it at all. Here’s why. For one, FOIL does help people do a particular type of problem correctly. That shouldn’t be overlooked. There are many reasons to study mathematics, and the beauty of the field is a powerful and romantic reason. But one at least as potent must be that it is a practical field, letting one do things one wishes to do, and if one wishes to do something, one surely wishes to do it correctly. Simple tools which help that goal should not be lightly thrown away. There is a bias in mathematical thinking: tools which work for only a small set of problems, like FOIL, look like clutter compared to general tools of broad applicability. I’m sympathetic to that bias. But there is value in tools which work for a very specific problem. An Allen wrench is a much less general tool than an adjustable screwdriver would be, but it can be much better for assembling flat-pack furniture. It’s simple and obvious to use, and I can use it without my father complaining I’ve put everything back in his toolbox wrong. Also, I feel that it is harder to multiply two trinomials together than it is to multiply two binomials. Likely you agree. But the idea that one can use the distributive law to do both kinds of multiplications is a general application of the distributive law, the idea that a times (b + 5) must necessarily be equal to a times b plus a times 5, and that (a + 2) times b must be equal to a times b plus 2 times b. There are some general ideas that people can understand at a glance. Usually, in my experience, these are ideas close to ones they have already seen. More often, one needs practice, on multiple examples which demonstrate the thing being studied without being too complicated. And what are the simplest cases of using the distributive law? … Well, a single number times a binomial, yes, but after that we get to binomials multiplied together. I don’t see how one can reasonably expect a student to get comfortable with the general principle of the distributive law without exercising on the simple case of multiplying binomials together. And that exercise has to work out the problems correctly, at least most of the time. (It can be instructive to get a problem brilliantly wrong; at least, it can enlighten the student who goes on to learn how things went wrong, and makes needed relief for the person doing the grading.) FOIL is a tool to get that exercise done correctly. Not all students will need it, and not all will need it for long; but I’m hesitant to throw away a simple and useful tool just because a more powerful yet more abstract tool exists. ## Author: Joseph Nebus I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him. ## 8 thoughts on “In Defense Of FOIL” 1. Geoffrey Brent says: I still use FOIL occasionally; as you say, while it doesn’t apply directly to trinomials etc, it’s a useful habit for binomials. Which then leaves me with more time to get the trinomials right. Even when multiplying longer expressions together, I do it as sigma-over-i(sigma-over-j(i*j)), and FOIL is a special case of that approach, so it’s not like I’m really changing styles. Just generalising the approach. Like 1. You know, this has reminded me: one of my mathematics teachers did point out how to use FOIL for trinomials (and higher terms), by rewriting a product of trinomials (a + (c + 2)) * (b + (d + 5)), and using a parenthetical expression and so recursion for the second part there. Of course nothing was said about “recursion”, as that would have been terrifying, but we did come out with the process being tediously long but never hard. Like 2. I agree. I’m reminded of an electronics teacher who taught me a mnemonic for calculating power from current and resistance: “Twinkle, twinkle, little star / power equals I squared R.” Now, it’s fairly easy to derive P = I^2 * R from Ohm’s law, but this is a particularly common case and it’s nice to have a way to remember it specifically. FOIL strikes me as a similar sort of mental tool. Like 1. It seems along those lines to me. Ohm’s Law reminds me of a mnemonic tool that literally uses pie charts, too: you drawa circle divided as in a medieval T-O map, with “E” in the upper half, “I” in the lower left quadrant, and “R” in the lower right quadrant. Take out the quantity you want; what’s left is the operation to get that … so “E” is “I R”; “I” is “E / R”, and “R” is “E / I”. And, of course, you get the same with P-I-E. Back in middle school I learned a weird magic square-based scheme for factoring polynomials that I’ve never seen anyone else talk about ever. I should make that an entry someday, so it’s not lost to the ages. Like 3. “If you suppose the students are lazy, the administrators meddling, the community unsupportive, and the public irrationally terrified of mathematics you’ve covered most forum threads.” Mutatis mutandis, this is what philosophy profession forums look like. And there’s a reason: all these things are substantially true. Like 1. Sssh! I’m just starting my first class in years next week. I need to hold on to my wild optimism about the experience. Like 4. So, is the objection that FOIL is bad because it doesn’t solve the general case of polynomial multiplication? Isn’t that the same as saying that the Pythagorean Theorem is bad because it doesn’t solve the unknown side length problem for all triangles? Similarly that the Power Rule is bad because it won’t solve all problems of derivation? In the interest of disclosure, I am not a real Mathematician since I really only have a Bachelor’s in Applied. Thinking about it, I’m not sure which I’d rather not do by hand: figure out the value of cosine for an angle other than the ones you memorized the answer for or compute the square root of a Real number. Both to some arbitrary level of precision. Like 1. The objection as I understand it is more that since the distributive law is a more general and powerful approach, that it’s a waste of limited class time and student attention to focus on that when the teacher could go to the distributive law instead. I feel this assumes the general case is easier to learn than I believe it to be, but now I’m sorry I didn’t think of the Pythagorean Theorem/Law of Cosines example to use in the first past. I’d rather do the square root by hand, but that’s because I actually learned how to do that. I’ve seen the rules about working out cosines for arbitrary angles and done a couple of cases, but none enough to say I actually know how to do it. I was enchanted by the use of cosine angle addition formulas as a way of simplifying multiplication, though, and it shows how desperate the need for calculators was that that was ever seen as a good idea. Like This site uses Akismet to reduce spam. Learn how your comment data is processed.

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## Static Electricity Facts Did you know that there were different types of electricity? Well there are, and one of them is called static electricity. We’re going to give you all the cool facts about static electricity and you’ll be one smart kid! You can also try out our experiment at the end of the article, both for fun and to learn more. Amazingly we see static electricity every single day; it even builds up on us. Wow! Have you ever rubbed your feet on the carpet and then zapped something when you touch it? Well that’s exactly what static electricity is. If you’ve had a bad hair day and it sticks straight up, well this means that your hair has been charged. Maybe you don’t want to go to school like that! Then there’s another irritating one, when pants or skirts stick to your legs, and they just keep on irritating you, no matter what you do. There you go…that’s how we see static electricity every day. ## What is Static Electricity? Static electricity is the build-up of an electrical charge on the surface of an object. The reason that it’s actually called static electricity is because the charges stay in one area for some time and don’t flow or move to a different area. Makes sense, doesn’t it? Atoms are made up of neutrons, protons, and electrons. The electrons spin around on the outside. A static charge happens when two surfaces touch each other and the electrons move from one object to another. One of the objects will have a positive charge and the other a negative charge. If you rub an object quickly, like a balloon, or your feet on the carpet, these will build-up a rather large charge. ### What Causes Static Electricity? Static electricity is created when positive and negative charges aren’t balanced. Items with different charges (positive and negative) will attract each other, while items with similar charges (positive and positive) will push away from each other. It’s kind of like a magnet! So you’ve gone flying down a slide and your hair stands straight up. This is because of the friction effect of sliding. It has caused a positive charge on each piece of hair. As each hair has the same charge, they all try and push away from each other, causing that funny straight hair to stand right up! The same happens with your skin. If it’s charged with static electricity, and then you touch something metal, your skin will get rid of the static electricity when you touch it. Pretty cool what it can do. ### How is Static Electricity Used? Static electricity has quite a few uses, here are some: It is used in printers and photocopiers where static electric charges attract the ink, or toner, to the paper. Other uses include paint sprayers, air filters, and dust removal. ### Can Static Electricity Cause Damage? Some electronic chips, like you find in computers are very sensitive to static electricity. There are special bags that they are stored in. People who work with these types of electronics wear special straps that keep them ‘grounded’ so they won’t build up a charge and ruin the electronic components. ### Cool Facts About Static Electricity A spark of static electricity can measure thousands of volts, but has very little current and only lasts for a short while. It has small amounts of power or energy. Lightning is also static electricity, and it is powerful and dangerous. Even though lightning is really dangerous, about 70% of people who are struck by lightning survive. Wow! Temperatures in a lightning bolt can hit 50,000°F or 27,760°C. Wow, that’s seriously hot! ### Words you Need to Know Atoms the smallest particle of a chemical element that can exist Neutrons – a particle of about the same mass as a proton but without an electric charge Protonsa stable particle with a positive electric charge equal in size to that of an electron Electronsa stable particle with a charge of negative electricity, found in all atoms and it acts as the primary carrier of electricity in solids Frictionthis is the resistance that one surface or object meets when moving over another Volts a volt is the unit of electric potential difference, or the size of the force that sends electrons through a circuit #### Static Electricity Experiment Now, test what you have learned by trying our static electricity experiment! You will need a balloon, a cotton towel, a scissors and a plastic bag (make sure an adult is present). Want to learn more about fascinating topics such as friction, acceleration, energy, light, heat and much more? Then head on over to our Physics section! Physics

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Arctan question 1. Jan 21, 2010 nhrock3 $$-\frac{\pi}{4}=arctan(\frac{\frac{-2L}{3}}{R})\\$$ why it equals $$\frac{\frac{-2L}{3}}{R}=-1$$ ? Last edited: Jan 21, 2010 2. Jan 21, 2010 dacruick $-\frac{\pi}{4}=arctan(\frac{\frac{-2L}{3}}{R})\\$ why it equals $\frac{\frac{-2L}{3}}{R}=-1$ 3. Jan 21, 2010 Staff: Mentor Take the tangent of both sides.

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February 26, 2024 ##### Author : Bill Are you struggling with the cartesian equation? Are you wondering where you can find the right guidance? Well, your search officially ends here, as we will be discussing it in detail. Now, as you know, this equation is one of the important parts of mathematics. But what exactly is it? And How can you understand it uncomplicatedly? And, if you start working on your assignments without any clarity, you cannot expect yourself to do well with the presentation. To make the whole process easier, we will be guiding you through the process and help you understand how you can get the deserving grades. In layman's terms, cartesian equations are like a rule that helps you understand the relationships between two variables. They are usually represented as letters ‘X’ and ‘Y’. Let’s talk with an example. Think about a simple graph with paper with vertical and horizontal lines creating a grid. Now, all points on the grid possess an X coordinate on the horizontal line and a y one on the vertical axis. Here, the equation allows you to figure out how to figure out the coordinates. Like, you have a straight line, and the equation goes: An expression such as 'y equals two times x plus 3' shows that for any value of x, determining the corresponding y consists of multiplying x by two and adding 3. A Cartesian equation serves as a useful tool to articulate various shapes and relationships on a Cartesian grid, facilitating the comprehension and manipulation of mathematical concepts. Once you start to understand Cartesian equations, you can begin to understand its dimensions. If needed, seeking assistance from experts for assignments is always an option. Below are some of the types: ## Types of Cartesian Equations • Linear equations • Other Polynomial Equation • Non-polynomial Equations ## Common Uses of Cartesian Equation Although you may not hear a lot about it, you can use this in mathematics for different industries and fields. To make this easier, you should learn to use a Cartesian equation calculator. It is a free tool available on the web which displays the Cartesian coordinates for the given sets. Scroll down to explore more! ### Graphing Relationships With this equation, you can graphically represent the relationship between two variables or more than one variable on a coordinate plane. It is used in math to visualize functions and data. ### Describing Shapes and Figures Here, you can use geometric shapes like lines, circles, and parabolas. Furthermore, it can help you in defining the characteristics and boundaries of all these shapes. ### Physics and Engineering The majority of physicists and engineers use cartesian equation to analyze the model of physical phenomena. Things like the motion of trajectories, objects and electrical circuits can all be described with the use of these mathematical equations. ### Economics and Finance Financial analysts and economists use equations to cast relationships between variables. Moreover, you can analyze all the trends, understand economic systems and make predictions. ### Geometry and Trigonometry How do you solve all the problems related to distance, geometric constructions, and angles? Simply with the equations. To dig deeper into detail, you can consider getting math assignment help online and learn from subject matter experts. ## Exploring Effective Writing Tips For Theoretical Part Apart from dealing with the equations, you have to explain yourself together by mentioning proper definitions. No matter how well you have managed to solve the equations, you will not get proper grades without exceptional presentation. To make this easier, we have got some of the tips mentioned below; read on to know! ### 1. Clarity and Simplicity Try to break the whole equation into parts while explaining. Use steps, bullet points and headings for proper transition. Refrain from incorporating too much technical jargon into your assignments. This way, you can increase the creativity of your work. ### 2. Keep things Relatable When you discuss in detail the Cartesian equation, it becomes exhausting to find the right terms. If you express yourself without the right terms, your research and knowledge will be questionable. Therefore, it would be best to mention the definition of tough terms in brackets. It will also keep things relatable. ### 3. Logical Flow How do you feel about reading a write-up without any connectivity? It can be confusing, right? To avoid this, you need to maintain a flow in your work. For this, you need to plan the structure of your work before writing and then approach things step-wise. Get assistance from a cheap assignment writing service for a better understanding of the concepts. ### 4. Use of Visuals As you cannot afford to bore your readers with too many paragraphs, you should make things interesting with visuals. Be it infographics, graphs, charts or tables, make the most of all the things. Remember not to overuse the visuals, and do not use too many loud colours for your visuals. Just keep things plain and professional. ### 5. Application Emphasis How will you show the relevance of the Cartesian equation? For this, you can try adding real-world examples. Emphasize some of the past successful examples to prove your point. In case you cannot find examples, talk to the assignment help experts and find suitable examples. ## Get More Clarity on Concepts with My Assignment Services! Your university life can become a rollercoaster when you pick advanced math as your major. Even though sometimes it can be complicated, it starts to become interesting to have a grasp on the basics. When you hear that you will stop making mistakes to master the concepts, it might not be true. After mastering the concepts, the quality of your mistakes will be enhanced. Also, you will learn to deal with your mistakes effectively. To make things more effective, notice the signs in your schedules that are indicating you to take help. My Assignment Services are here assignment writers Australia who can add charm to all your assignments. Starting from complicated mathematical problems to lengthy theoretical assignments, we can help you with all the things. Moreover, they can offer you with proper feedback to work on your skills. Wondering which subjects we should cover? We have got it all! Be it nursing, physics, law or marketing; we have it! That, too, at affordable prices! Make sure to convey your requirements well to get the grades you deserve. So, what’s the hesitation now? Let's start working and complete the first draft of your assignment! ### Related Study Materials Loved reading this Blog? Share your valuable thoughts in the comment section. #### Bill Bill is an academic expert in the fields of law, nursing, business, and management. His diligence in editing and writing assignments solutions has been applauded by students from around the globe; who swear by his eclectic writing style and subject matter expertise in Law and Nursing Studies. He is full time associated with My Assignment Services as a Senior Academic Writer and loves binge-watching on anything sci-fi. ## Get Quote in 5 Minutes* Country Applicable Time Zone is AEST [Sydney, NSW] (GMT+11) + • 1,212,718Orders • 4.9/5Rating • 5,063Experts Trending now H P P H I H B M H W B A

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### Re: How to get last day of 2 months ago by Corion (Pope) on Mar 11, 2011 at 16:05 UTC ( #892685=note: print w/replies, xml ) Need Help?? in reply to How to get last day of 2 months ago The key to finding the "last day of a month" is to realize that the "last day" of a month is always one day before the "first day" of the next month. As the date of the first day of a month is easy to construct, you just have to construct the first day of the current month, then go to the last day of the month before. Then extract the year and month of that month, and construct its first day from it. Repeat until you have skipped enough months. Personally, I like to use Time::Local and localtime, together with POSIX::strftime to manipulate and create date strings. The more heavyweight and somewhat different approach is to use DateTime, but as I often have to do date calculations, and not always in Perl, the strftime approach has worked more often for me than the DateTime approach. Replies are listed 'Best First'. Re^2: How to get last day of 2 months ago by dirtdog (Scribe) on Mar 11, 2011 at 16:49 UTC Thank you all for the replies. I did figure out what i was doing wrong..I had \$t[3] \$t[4]-1 ```use POSIX 'strftime'; my @t = (localtime(time)); my \$lastday = POSIX::mktime(0,0,0,0,\$t[4],\$t[5],0,0,-1); my \$lastday_prev = POSIX::mktime(0,0,0,0,\$t[4]-1,\$t[5],0,0,-1); my @ld_prev = localtime(\$lastday_prev); my @ld = localtime(\$lastday); my \$monthly_date_beg = strftime('%m/' . \$ld_prev[3] . '/%Y', \$t[0],\$t[ +1], \$t[2], \$t[3], \$t[4]-2, \$t[5]); my \$monthly_date_end = strftime('%m/' . \$ld[3] . '/%Y', \$t[0],\$t[1], \$ +t[2], \$t[3], \$t[4]-1, \$t[5]); print "beg = \$monthly_date_beg\n"; print "end = \$monthly_date_end\n"; Create A New User Node Status? node history Node Type: note [id://892685] help Chatterbox? and all is quiet... How do I use this? | Other CB clients Other Users? Others wandering the Monastery: (8) As of 2017-07-20 16:47 GMT Sections? Information? Find Nodes? Leftovers? Voting Booth? I came, I saw, I ... Results (308 votes). Check out past polls.

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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A182244 Sum of all parts of the shell model of partitions of A135010 with n regions. 9 1, 4, 9, 11, 20, 23, 35, 37, 43, 46, 66, 69, 76, 80, 105, 107, 113, 116, 129, 134, 138, 176, 179, 186, 190, 204, 207, 216, 221, 270, 272, 278, 281, 294, 299, 303, 326, 330, 340, 346, 351, 420, 423, 430, 434, 448, 451, 460, 465, 492, 497, 501, 516, 523, 529, 616 (list; graph; refs; listen; history; text; internal format) OFFSET 1,2 LINKS Omar E. Pol, Illustration of the seven regions of 5 FORMULA a(A000041(k)) = A066186(k), k >= 1. EXAMPLE The first four regions of the shell model of partitions are [1],[2, 1],[3, 1, 1],[2], so a(4) = (1)+(2+1)+(3+1+1)+(2) = 11. Written as a triangle begins: 1; 4; 9; 11,  20; 23,  35; 37,  43, 46, 66; 69,  76, 80,105; 107,113,116,129,134,138,176; 179,186,190,204,207,216,221,270; 272,278,281,294,299,303,326,330,340,346,351,420; 423,430,434,448,451,460,465,492,497,501,516,523,529,616; ... From Omar E. Pol, Aug 08 2013: (Start) Illustration of initial terms: .                                                _ _ _ _ _ .                                      _ _ _    |_ _ _    | .                            _ _ _ _  |_ _ _|_  |_ _ _|_  | .                    _ _    |_ _    | |_ _    | |_ _    | | .            _ _ _  |_ _|_  |_ _|_  | |_ _|_  | |_ _|_  | | .      _ _  |_ _  | |_ _  | |_ _  | | |_ _  | | |_ _  | | | .  _  |_  | |_  | | |_  | | |_  | | | |_  | | | |_  | | | | . |_| |_|_| |_|_|_| |_|_|_| |_|_|_|_| |_|_|_|_| |_|_|_|_|_| . .  1    4      9       11       20        23        35 . .                                          _ _ _ _ _ _ .                             _ _ _       |_ _ _      | .                _ _ _ _     |_ _ _|_     |_ _ _|_    | .   _ _         |_ _    |    |_ _    |    |_ _    |   | .  |_ _|_ _ _   |_ _|_ _|_   |_ _|_ _|_   |_ _|_ _|_  | .  |_ _ _    |  |_ _ _    |  |_ _ _    |  |_ _ _    | | .  |_ _ _|_  |  |_ _ _|_  |  |_ _ _|_  |  |_ _ _|_  | | .  |_ _    | |  |_ _    | |  |_ _    | |  |_ _    | | | .  |_ _|_  | |  |_ _|_  | |  |_ _|_  | |  |_ _|_  | | | .  |_ _  | | |  |_ _  | | |  |_ _  | | |  |_ _  | | | | .  |_  | | | |  |_  | | | |  |_  | | | |  |_  | | | | | .  |_|_|_|_|_|  |_|_|_|_|_|  |_|_|_|_|_|  |_|_|_|_|_|_| . .       37           43           46           66 (End) MATHEMATICA lex[n_]:=DeleteCases[Sort@PadRight[Reverse /@ IntegerPartitions@n], x_ /; x==0, 2]; A186412 = {}; l = {}; For[j = 1, j <= 56, j++,   mx = Max@lex[j][[j]]; AppendTo[l, mx];   For[i = j, i > 0, i--, If[l[[i]] > mx, Break[]]];   AppendTo[A186412, Total@Take[Reverse[First /@ lex[mx]], j - i]];   ]; Accumulate@A186412  (* Robert Price, Jul 25 2020 *) CROSSREFS Partial sums of A186412. Row j has length A187219(j). Right border gives A066186. Cf. A000041, A135010, A138121, A141285, A182244, A182181, A182276, A186114, A206437, A210990. Sequence in context: A329897 A312845 A312846 * A312847 A141365 A179055 Adjacent sequences:  A182241 A182242 A182243 * A182245 A182246 A182247 KEYWORD nonn,tabf AUTHOR Omar E. Pol, Apr 23 2012 STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified September 18 11:20 EDT 2021. Contains 347518 sequences. (Running on oeis4.)

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# Number 116832 Number 116,832 spell 🔊, write in words: one hundred and sixteen thousand, eight hundred and thirty-two . Ordinal number 116832th is said 🔊 and write: one hundred and sixteen thousand, eight hundred and thirty-second. Color #116832. The meaning of number 116832 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 116832. What is 116832 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 116832. ## What is 116,832 in other units The decimal (Arabic) number 116832 converted to a Roman number is (C)(X)(V)MDCCCXXXII. Roman and decimal number conversions. #### Weight conversion 116832 kilograms (kg) = 257567.8 pounds (lbs) 116832 pounds (lbs) = 52994.6 kilograms (kg) #### Length conversion 116832 kilometers (km) equals to 72597 miles (mi). 116832 miles (mi) equals to 188023 kilometers (km). 116832 meters (m) equals to 383303 feet (ft). 116832 feet (ft) equals 35611 meters (m). 116832 centimeters (cm) equals to 45996.9 inches (in). 116832 inches (in) equals to 296753.3 centimeters (cm). #### Temperature conversion 116832° Fahrenheit (°F) equals to 64888.9° Celsius (°C) 116832° Celsius (°C) equals to 210329.6° Fahrenheit (°F) #### Time conversion (hours, minutes, seconds, days, weeks) 116832 seconds equals to 1 day, 8 hours, 27 minutes, 12 seconds 116832 minutes equals to 2 months, 3 weeks, 4 days, 3 hours, 12 minutes ### Codes and images of the number 116832 Number 116832 morse code: .---- .---- -.... ---.. ...-- ..--- Sign language for number 116832: Number 116832 in braille: QR code Bar code, type 39 Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ... ## Mathematics of no. 116832 ### Multiplications #### Multiplication table of 116832 116832 multiplied by two equals 233664 (116832 x 2 = 233664). 116832 multiplied by three equals 350496 (116832 x 3 = 350496). 116832 multiplied by four equals 467328 (116832 x 4 = 467328). 116832 multiplied by five equals 584160 (116832 x 5 = 584160). 116832 multiplied by six equals 700992 (116832 x 6 = 700992). 116832 multiplied by seven equals 817824 (116832 x 7 = 817824). 116832 multiplied by eight equals 934656 (116832 x 8 = 934656). 116832 multiplied by nine equals 1051488 (116832 x 9 = 1051488). show multiplications by 6, 7, 8, 9 ... ### Fractions: decimal fraction and common fraction #### Fraction table of 116832 Half of 116832 is 58416 (116832 / 2 = 58416). One third of 116832 is 38944 (116832 / 3 = 38944). One quarter of 116832 is 29208 (116832 / 4 = 29208). One fifth of 116832 is 23366,4 (116832 / 5 = 23366,4 = 23366 2/5). One sixth of 116832 is 19472 (116832 / 6 = 19472). One seventh of 116832 is 16690,2857 (116832 / 7 = 16690,2857 = 16690 2/7). One eighth of 116832 is 14604 (116832 / 8 = 14604). One ninth of 116832 is 12981,3333 (116832 / 9 = 12981,3333 = 12981 1/3). show fractions by 6, 7, 8, 9 ... ### Calculator 116832 #### Is Prime? The number 116832 is not a prime number. The closest prime numbers are 116827, 116833. #### Factorization and factors (dividers) The prime factors of 116832 are 2 * 2 * 2 * 2 * 2 * 3 * 1217 The factors of 116832 are 1 , 2 , 3 , 4 , 6 , 8 , 12 , 16 , 24 , 32 , 48 , 96 , 1217 , 2434 , 3651 , 4868 , 7302 , 9736 , 14604 , 19472 , 116832 show more factors ... Total factors 24. Sum of factors 306936 (190104). #### Powers The second power of 1168322 is 13.649.716.224. The third power of 1168323 is 1.594.723.645.882.368. #### Roots The square root √116832 is 341,806963. The cube root of 3116832 is 48,886311. #### Logarithms The natural logarithm of No. ln 116832 = loge 116832 = 11,668492. The logarithm to base 10 of No. log10 116832 = 5,067562. The Napierian logarithm of No. log1/e 116832 = -11,668492. ### Trigonometric functions The cosine of 116832 is -0,771759. The sine of 116832 is 0,635916. The tangent of 116832 is -0,823983. ### Properties of the number 116832 Is a Friedman number: No Is a Fibonacci number: No Is a Bell number: No Is a palindromic number: No Is a pentagonal number: No Is a perfect number: No ## Number 116832 in Computer Science Code typeCode value PIN 116832 It's recommendable to use 116832 as a password or PIN. 116832 Number of bytes114.1KB CSS Color #116832 hexadecimal to red, green and blue (RGB) (17, 104, 50) Unix timeUnix time 116832 is equal to Friday Jan. 2, 1970, 8:27:12 a.m. GMT IPv4, IPv6Number 116832 internet address in dotted format v4 0.1.200.96, v6 ::1:c860 116832 Decimal = 11100100001100000 Binary 116832 Decimal = 12221021010 Ternary 116832 Decimal = 344140 Octal 116832 Decimal = 1C860 Hexadecimal (0x1c860 hex) 116832 BASE64MTE2ODMy 116832 MD55693fee6b53257464869d70bde87ffb3 116832 SHA1170627e709db0b855eda6076c695d236d0eb0e1c 116832 SHA224fb30095137b6815dc25fec46b37eee059b91616cbff34e6dbfec7f8c 116832 SHA256810e019423903177a3a0933f3a1bda964df8c07ef4476f4e28659d61388cc3fa More SHA codes related to the number 116832 ... If you know something interesting about the 116832 number that you did not find on this page, do not hesitate to write us here. ## Numerology 116832 ### Character frequency in number 116832 Character (importance) frequency for numerology. Character: Frequency: 1 2 6 1 8 1 3 1 2 1 ### Classical numerology According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 116832, the numbers 1+1+6+8+3+2 = 2+1 = 3 are added and the meaning of the number 3 is sought. ## Interesting facts about the number 116832 ### Asteroids • (116832) 2004 FR32 is asteroid number 116832. It was discovered by Spacewatch from Kitt peak on 3/16/2004. ## № 116,832 in other languages How to say or write the number one hundred and sixteen thousand, eight hundred and thirty-two in Spanish, German, French and other languages. The character used as the thousands separator. Spanish: 🔊 (número 116.832) ciento dieciseis mil ochocientos treinta y dos German: 🔊 (Anzahl 116.832) einhundertsechzehntausendachthundertzweiunddreißig French: 🔊 (nombre 116 832) cent seize mille huit cent trente-deux Portuguese: 🔊 (número 116 832) cento e dezesseis mil, oitocentos e trinta e dois Chinese: 🔊 (数 116 832) 十一万六千八百三十二 Arabian: 🔊 (عدد 116,832) مائة و ستة عشر ألفاً و ثمانمائة و اثنان و ثلاثون Czech: 🔊 (číslo 116 832) sto šestnáct tisíc osmset třicet dva Korean: 🔊 (번호 116,832) 십일만 육천팔백삼십이 Danish: 🔊 (nummer 116 832) ethundrede og sekstentusindottehundrede og toogtredive Dutch: 🔊 (nummer 116 832) honderdzestienduizendachthonderdtweeëndertig Japanese: 🔊 (数 116,832) 十一万六千八百三十二 Indonesian: 🔊 (jumlah 116.832) seratus enam belas ribu delapan ratus tiga puluh dua Italian: 🔊 (numero 116 832) centosedicimilaottocentotrentadue Norwegian: 🔊 (nummer 116 832) en hundre og seksten tusen, åtte hundre og tretti-to Polish: 🔊 (liczba 116 832) sto szesnaście tysięcy osiemset trzydzieści dwa Russian: 🔊 (номер 116 832) сто шестнадцать тысяч восемьсот тридцать два Turkish: 🔊 (numara 116,832) yüzonaltıbinsekizyüzotuziki Thai: 🔊 (จำนวน 116 832) หนึ่งแสนหนึ่งหมื่นหกพันแปดร้อยสามสิบสอง Ukrainian: 🔊 (номер 116 832) сто шiстнадцять тисяч вiсiмсот тридцять двi Vietnamese: 🔊 (con số 116.832) một trăm mười sáu nghìn tám trăm ba mươi hai Other languages ... ## News to email Privacy Policy. ## Comment If you know something interesting about the number 116832 or any natural number (positive integer) please write us here or on facebook.

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https://www.meritnation.com/cbse-class-12-commerce/math/rd-sharma-xii-vol-2-2019/straight-line-in-space/textbook-solutions/59_1_3515_24597_28.9_93838

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Rd Sharma XII Vol 2 2019 Solutions for Class 12 Commerce Math Chapter 9 Straight Line In Space are provided here with simple step-by-step explanations. These solutions for Straight Line In Space are extremely popular among Class 12 Commerce students for Math Straight Line In Space Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 2 2019 Book of Class 12 Commerce Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 2 2019 Solutions. All Rd Sharma XII Vol 2 2019 Solutions for class Class 12 Commerce Math are prepared by experts and are 100% accurate. #### Question 8: Find the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are $\frac{x-3}{2}=\frac{y+1}{7}=\frac{z-2}{-3}.$ #### Answer: We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is . Here, Vector equation of the required line is #### Question 9: The cartesian equations of a line are $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}.$ Find a vector equation for the line. #### Answer: The cartesian equation of the given line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$. It can be re-written as $\frac{x-5}{3}=\frac{y-\left(-4\right)}{7}=\frac{z-6}{2}$ Thus, the given line passes through the point having position vector $\stackrel{\to }{a}=5\stackrel{^}{i}-4\stackrel{^}{j}+6\stackrel{^}{k}$ and is parallel to the vector $\stackrel{\to }{b}=3\stackrel{^}{i}+7\stackrel{^}{j}+2\stackrel{^}{k}$. We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is . Vector equation of the required line is #### Question 10: Find the cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are $\frac{x-3}{1}=\frac{y-1}{2}=\frac{z+1}{-2}$. Also, reduce the equation obtained in vector form. #### Answer: We know that the cartesian equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{m}$ is $\frac{x-{x}_{1}}{a}=\frac{y-{y}_{2}}{b}=\frac{z-{z}_{3}}{c}$. Here, Here, Cartesian equation of the required line is $\frac{x-1}{1}=\frac{y-\left(-1\right)}{2}=\frac{z-2}{-2}\phantom{\rule{0ex}{0ex}}⇒\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{-2}\phantom{\rule{0ex}{0ex}}$ We know that the cartesian equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{m}$ is $\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{m}$. Here, the line is passing through the point and its direction ratios are proportional to 1, 2, $-$2. Vector equation of the required line is $\stackrel{\to }{r}=\left(\stackrel{^}{i}-\stackrel{^}{j}+2\stackrel{^}{k}\right)+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}-2\stackrel{^}{k}\right)$ #### Question 11: Find the direction cosines of the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}.$ Also, reduce it to vector form. #### Answer: The cartesian equation of the given line is $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$ It can be re-written as $\frac{x-4}{-2}=\frac{y-0}{6}=\frac{z-1}{-3}$ This shows that the given line passes through the point and its direction ratios are proportional to . So, its direction cosines are Thus, the given line passes through the point having position vector $\stackrel{\to }{a}=4\stackrel{^}{i}+\stackrel{^}{k}$ and is parallel to the vector $\stackrel{\to }{b}=-2\stackrel{^}{i}+6\stackrel{^}{j}-3\stackrel{^}{k}$. We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is . Here, Vector equation of the required line is #### Question 12: The cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form. #### Answer: The cartesian equation of the given line is It can be re-written as $\frac{x-b}{a}=\frac{y-0}{1}=\frac{z-d}{c}$ Thus, the given line passes through the point  and its direction ratios are proportional to a, 1, c. It is also parallel to the vector $\stackrel{\to }{b}=a\stackrel{^}{i}+\stackrel{^}{j}+c\stackrel{^}{k}$. We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is . Vector equation of the required line is #### Question 13: Find the vector equation of a line passing through the point with position vector $\stackrel{^}{i}-2\stackrel{^}{j}-3\stackrel{^}{k}$and parallel to the line joining the points with position vectors Also, find the cartesian equivalent of this equation. #### Answer: We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is . Here, Vector equation of the required line is Reducing (1) to cartesian form, we get #### Question 14: Find the points on the line $\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$ at a distance of 5 units from the point P (1, 3, 3). #### Answer: The coordinates of any point on the line $\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$ are given by Let the coordinates of the desired point be . The distance between this point and (1, 3, 3) is 5 units. Substituting the values of $\lambda$ in (1), we get the coordinates of the desired point as ($-$2, $-$1, 3) and (4, 3, 7). #### Question 15: Show that the points whose position vectors are are collinear. #### Answer: Let the given points be P, Q and R and let their position vectors be Vector equation of line passing through P and Q is If points P, Q and R are collinear, then point R must satisfy (1). $7\stackrel{^}{i}+9\stackrel{^}{k}=\left(-2\stackrel{^}{i}+3\stackrel{^}{j}\right)+\lambda \left(3\stackrel{^}{i}-\stackrel{^}{j}+3\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}$ Comparing the coefficients of , we get $\lambda$=3 These three equations are consistent, i.e. they give the same value of $\lambda$. Hence, the given three points are collinear. Disclaimer: The question given in the book has a minor error. The third vectors should be $7\stackrel{^}{i}+9\stackrel{^}{k}$. The solution here is created accordingly. #### Question 16: Find the cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line $\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}.$ #### Answer: We have $\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}$ It can be re-written as $\frac{x+2}{-1}=\frac{y+3}{7}=\frac{z-3}{\left(\frac{3}{2}\right)}\phantom{\rule{0ex}{0ex}}=\frac{x+2}{-2}=\frac{y+3}{14}=\frac{z-3}{3}$ This shows that the given line passes through the point and its direction ratios are proportional to $-$2, 14, 3. Thus, the parallel vector is $\stackrel{\to }{b}=-2\stackrel{^}{i}+14\stackrel{^}{j}+3\stackrel{^}{k}$. We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is . Here, . Vector equation of the required line is Reducing (1) to cartesian form, we get #### Question 17: The cartesian equation of a line are 3x + 1 = 6y − 2 = 1 − z. Find the fixed point through which it passes, its direction ratios and also its vector equation. #### Answer: The cartesian equation of the given line is 3x + 1 = 6y − 2 = 1 − z It can be re-written as $\frac{x+\frac{1}{3}}{\frac{1}{3}}=\frac{y-\frac{1}{3}}{\frac{1}{6}}=\frac{z-1}{-1}\phantom{\rule{0ex}{0ex}}=\frac{x-\left(-\frac{1}{3}\right)}{2}=\frac{y-\frac{1}{3}}{1}=\frac{z-1}{-6}$ Thus, the given line passes through the point $\left(-\frac{1}{3},\frac{1}{3},1\right)$ and its direction ratios are proportional to 2, 1, $-$6. It is parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}+\stackrel{^}{j}-6\stackrel{^}{k}$. We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is . Vector equation of the required line is #### Question 18: Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z. #### Answer: The equation of the line 5x − 25 = 14 − 7y = 35z can be re-written as Since the required line is parallel to the given line, so the direction ratios of the required line are proportional to . The vector equation of the required line passing through the point (1, 2, $-$1) and having direction ratios proportional to is $\stackrel{\to }{r}=\left(\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}\right)+\lambda \left(7\stackrel{^}{i}-5\stackrel{^}{j}+\stackrel{^}{k}\right)$. #### Question 1: Show that the three lines with direction cosines are mutually perpendicular. #### Answer: The direction cosines of the three lines are Hence, the given lines are perpendicular to each other. #### Question 2: Show that the line through the points (1, −1, 2) and (3, 4, −2) is perpendicular to the through the points (0, 3, 2) and (3, 5, 6). #### Answer: Suppose vector $\stackrel{\to }{a}$ is passing through the points (1, $-$1, 2) and (3, 4,$-$2) and $\stackrel{\to }{b}$ is passing through the points (0, 3, 2) and (3, 5, 6). Then, Now, $\stackrel{\to }{a}.\stackrel{\to }{b}=\left(2\stackrel{^}{i}+5\stackrel{^}{j}-4\stackrel{⏜}{k}\right).\left(3\stackrel{^}{i}+2\stackrel{^}{j}+4\stackrel{⏜}{k}\right)=0$ Hence, the given lines are perpendicular to each other. #### Question 3: Show that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (−1, −2, 1) and, (1, 2, 5). #### Answer: Equations of lines passing through the points are given by $\frac{x-{x}_{1}}{{x}_{2}-{x}_{1}}=\frac{y-{y}_{1}}{{y}_{2}-{y}_{1}}=\frac{z-{z}_{1}}{{z}_{2}-{z}_{1}}$ So, the equation of a line passing through (4, 7, 8) and (2, 3, 4) is $\frac{x-4}{2-4}=\frac{y-7}{3-7}=\frac{z-8}{4-8}\phantom{\rule{0ex}{0ex}}⇒\frac{x-4}{-2}=\frac{y-7}{-4}=\frac{z-8}{-4}$ Also, the equation of the line passing through the points ($-$1, $-$2, 1) and (1, 2, 5) is $\frac{x+1}{1+1}=\frac{y+2}{2+2}=\frac{z-1}{5-1}\phantom{\rule{0ex}{0ex}}⇒\frac{x+1}{2}=\frac{y+2}{4}=\frac{z-1}{4}$ We know that two lines are parallel if We observe $\frac{-2}{2}=\frac{-4}{4}=\frac{-4}{4}=-1\phantom{\rule{0ex}{0ex}}$ Hence, the given lines are parallel to each other. #### Question 4: Find the cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}.$ #### Answer: We know that the cartesian equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is $\frac{x-{x}_{1}}{a}=\frac{y-{y}_{2}}{b}=\frac{z-{z}_{3}}{c}$. Here, The cartesian equation of the required line is $\frac{x-\left(-2\right)}{3}=\frac{y-4}{5}=\frac{z-\left(-5\right)}{6}\phantom{\rule{0ex}{0ex}}=\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\phantom{\rule{0ex}{0ex}}$ #### Question 5: Show that the lines are perpendicular to each other. #### Answer: We have These equations can be re-written as Now, Hence, the given two lines are perpendicular to each other. #### Question 6: Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1) and (4, 3, −1). #### Answer: The direction ratios of the line joining the origin to the point (2, 1, 1) are 2, 1, 1. Let $\stackrel{\to }{{b}_{1}}=2\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}$ The direction ratios of the line joining the points (3, 5, $-$1) and (4, 3, $-$1) are 1, $-$2, 0. Let $\stackrel{\to }{{b}_{2}}=\stackrel{^}{i}-2\stackrel{^}{j}+0\stackrel{^}{k}$ Now, Hence, the two lines joining the given points are perpendicular to each other. #### Question 7: Find the equation of a line parallel to x-axis and passing through the origin. #### Answer: The direction ratios of the line parallel to x-axis are proportional to 1, 0, 0. Equation of the line passing through the origin and parallel to x-axis is $\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}\phantom{\rule{0ex}{0ex}}=\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$ #### Question 8: Find the angle between the following pairs of lines: (i) (ii) (iii) #### Answer: (i) Let be vectors parallel to the given lines. Now, If $\theta$ is the angle between the given lines, then (ii) Let be vectors parallel to the given lines. Now, If $\theta$ is the angle between the given lines, then (iii) Let be vectors parallel to the given lines. Now, If $\theta$ is the angle between the given lines, then #### Question 9: Find the angle between the following pairs of lines: (i) (ii) (iii) (iv) (v) (vi) #### Answer: (i) Let be vectors parallel to the given lines. If $\theta$ is the angle between the given lines, then (ii) Let be vectors parallel to the given lines. Now, If $\theta$ is the angle between the given lines, then (iii) The equations of the given lines can be re-written as Let be vectors parallel to the given lines. Now, If $\theta$ is the angle between the given lines, then (iv) The equations of the given lines can be re-written as Let be vectors parallel to the given lines. Now, If $\theta$ is the angle between the given lines, then (v ) The equations of the given lines can be re-written as Let be vectors parallel to the given lines. Now, $\stackrel{\to }{{b}_{1}}=\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{{b}_{2}}=3\stackrel{^}{i}+4\stackrel{^}{j}+5\stackrel{^}{k}$ If $\theta$ is the angle between the given lines, then Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book. (vi) The equations of the given lines can be re-written as Let be vectors parallel to the given lines. Now, If $\theta$ is the angle between the given lines, then #### Question 10: Find the angle between the pairs of lines with direction ratios proportional to (i) 5, −12, 13 and −3, 4, 5 (ii) 2, 2, 1 and 4, 1, 8 (iii) 1, 2, −2 and −2, 2, 1 (iv) a, b, c and bc, ca, ab. #### Answer: (i) 5, −12, 13 and −3, 4, 5 (ii) 2, 2, 1 and 4, 1, 8 (iii) 1, 2, −2 and −2, 2, 1 (iv) a, b, c and bc, ca, ab. #### Question 11: Find the angle between two lines, one of which has direction ratios 2, 2, 1 while the  other one is obtained by joining the points (3, 1, 4) and (7, 2, 12). #### Answer: The direction ratios of the line joining the points (3, 1, 4) and (7, 2, 12) are proportional to 4, 1, 8. Let be vectors parallel to the lines having direction ratios proportional to 2, 2, 1 and 4, 1, 8. Now, If $\theta$ is the angle between the given lines, then #### Question 12: Find the equation of the line passing through the point (1, 2, −4) and parallel to the line $\frac{x-3}{4}=\frac{y-5}{2}=\frac{z+1}{3}.$ #### Answer: The direction ratios of the line parallel to line $\frac{x-3}{4}=\frac{y-5}{2}=\frac{z+1}{3}$ are proportional to 4, 2, 3. Equation of the required line passing through the point (1, 2, $-$4) having direction ratios proportional to 4, 2, 3 is $\frac{x-1}{4}=\frac{y-2}{2}=\frac{z-\left(-4\right)}{3}\phantom{\rule{0ex}{0ex}}=\frac{x-1}{4}=\frac{y-2}{2}=\frac{z+4}{3}$ #### Question 13: Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line $\frac{2x-1}{4}=\frac{3y+5}{2}=\frac{2-z}{3}.$ #### Answer: The equation of line $\frac{2x-1}{4}=\frac{3y+5}{2}=\frac{2-z}{3}$ can be re-written as $\frac{x-\frac{1}{2}}{2}=\frac{y+\frac{5}{3}}{\frac{2}{3}}=\frac{z-2}{-3}$ The direction ratios of the line parallel to line $\frac{2x-1}{4}=\frac{3y+5}{2}=\frac{2-z}{3}$ are proportional to 2, $\frac{2}{3}$, $-$3. Equation of the required line passing through the point ($-$1, 2, 1) having direction ratios proportional to 2, $\frac{2}{3}$, $-$3 is $\frac{x-\left(-1\right)}{2}=\frac{y-2}{\frac{2}{3}}=\frac{z-1}{-3}\phantom{\rule{0ex}{0ex}}=\frac{x+1}{2}=\frac{y-2}{\frac{2}{3}}=\frac{z-1}{-3}$ #### Question 14: Find the equation of the line passing through the point (2, −1, 3) and parallel to the line $\stackrel{\to }{r}=\left(\stackrel{^}{i}-2\stackrel{^}{j}+\stackrel{^}{k}\right)+\mathrm{\lambda }\left(2\stackrel{^}{i}+3\stackrel{^}{j}-5\stackrel{^}{k}\right).$ #### Answer: The given line is parallel to the vector $2\stackrel{^}{i}+3\stackrel{^}{j}-5\stackrel{^}{k}$ and the required line is parallel to the given line. So, the required line is parallel to the vector $2\stackrel{^}{i}+3\stackrel{^}{j}-5\stackrel{^}{k}$. Hence, the equation of the required line passing through the point (2,$-$1, 3) and parallel to the vector $2\stackrel{^}{i}+3\stackrel{^}{j}-5\stackrel{^}{k}$ is $\stackrel{\to }{r}=\left(2\stackrel{^}{i}-\stackrel{^}{j}+3\stackrel{^}{k}\right)+\mathrm{\lambda }\left(2\stackrel{^}{i}+3\stackrel{^}{j}-5\stackrel{^}{k}\right)$. #### Question 15: Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines #### Answer: Let: Since the required line is perpendicular to the lines parallel to the vectors , it is parallel to the vector $\stackrel{\to }{b}=\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}$. Now, Thus, the direction ratios of the required line are proportional to 2, $-$7, 4. The equation of the required line passing through the point (2, 1, 3) and having direction ratios proportional to 2, $-$7, 4 is $\frac{x-2}{2}=\frac{y-1}{-7}=\frac{z-3}{4}$. #### Question 16: Find the equation of the line passing through the point $\stackrel{^}{i}+\stackrel{^}{j}-3\stackrel{^}{k}$ and perpendicular to the lines #### Answer: The required line is perpendicular to the lines parallel to the vectors . So, the required line is parallel to the vector $\stackrel{\to }{b}=\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}$. Now, Equation of the required line passing through the point $\left(\stackrel{^}{i}+\stackrel{^}{j}-3\stackrel{^}{k}\right)$ and parallel to $\left(4\stackrel{^}{i}-5\stackrel{^}{j}+\stackrel{^}{k}\right)$ is $\stackrel{\to }{r}=\left(\stackrel{^}{i}+\stackrel{^}{j}-3\stackrel{^}{k}\right)+\lambda \left(4\stackrel{^}{i}-5\stackrel{^}{j}+\stackrel{^}{k}\right)$. #### Question 17: Find the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1). #### Answer: The direction ratios of the line joining the points (4, 3, 2), (1,$-$1, 0) and (1, 2, $-$1), (2, 1, 1) are $-$3, $-$4, $-$2 and 1, $-$1, 2, respectively. Let: Since the required line is perpendicular to the lines parallel to the vectors , it is parallel to the vector $\stackrel{\to }{b}=\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}$. Now, So, the direction ratios of the required line are proportional to $-$10, 4, 7. The equation of the required line passing through the point (1, $-$1, 1) and having direction ratios proportional to $-$10, 4, 7 is $\frac{x-1}{-10}=\frac{y+1}{4}=\frac{z-1}{7}$. #### Question 18: Determine the equations of the line passing through the point (1, 2, −4) and perpendicular to the two lines . #### Answer: We have $\frac{x-8}{8}=\frac{y+9}{-16}=\frac{z-10}{7}\phantom{\rule{0ex}{0ex}}\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$ Let: Since the required line is perpendicular to the lines parallel to the vectors , it is parallel to the vector $\stackrel{\to }{b}=\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}$. Now, The direction ratios of the required line are proportional to 24, 61, 112. The equation of the required line passing through the point (1, 2, $-$4) and having direction ratios proportional to 24, 61, 112 is $\frac{x-1}{24}=\frac{y-2}{61}=\frac{z+4}{112}$. #### Question 19: Show that the lines are perpendicular to each other. #### Answer: The direction ratios of the lines are proportional to 7, $-$5, 1 and 1, 2, 3, respectively. Let: Now, Hence, the given lines are perpendicular to each other. #### Question 20: Find the vector equation of the line passing through the point (2, −1, −1) which is parallel to the line 6x − 2 = 3y + 1 = 2z − 2. #### Answer: The equation of the line 6x − 2 = 3y + 1 = 2z − 2 can be re-written as Since the required line is parallel to the given line, the direction ratios of the required line are proportional to . The vector equation of the required line passing through the point (2, $-$1, $-$1) and having direction ratios proportional to is $\stackrel{\to }{r}=\left(2\stackrel{^}{i}-\stackrel{^}{j}-\stackrel{^}{k}\right)+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right)$. #### Question 21: If the lines are perpendicular, find the value of λ. #### Answer: The equations of the given lines are Since the given lines are perpendicular to each other, we have $-3\left(3\lambda \right)+2\lambda \left(1\right)+2\left(-5\right)=0\phantom{\rule{0ex}{0ex}}⇒-9\lambda +2\lambda -10=0\phantom{\rule{0ex}{0ex}}⇒\lambda =-\frac{10}{7}$ #### Question 22: If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD. #### Answer: The direction ratios of AB and CD are proportional to 3, 3, 4 and 6, 6, 8, respectively. Let $\theta$ be the angle between AB and CD. Then, #### Question 23: Find the value of λ so that the following lines are perpendicular to each other. #### Answer: The equations of the given lines  can be re-written as Since the given lines are perpendicular to each other, we have $\left(5\lambda +2\right)1-5\left(2\lambda \right)+1\left(3\right)=0\phantom{\rule{0ex}{0ex}}⇒5\lambda =5\phantom{\rule{0ex}{0ex}}⇒\lambda =1$ #### Question 24: Find the direction cosines of the line $\frac{x+2}{2}=\frac{2y-7}{6}=\frac{5-z}{6}$. Also, find the vector equation of the line through the point A(−1, 2, 3) and parallel to the given line.                                                              [CBSE 2014] #### Answer: The equation of the given line is $\frac{x+2}{2}=\frac{2y-7}{6}=\frac{5-z}{6}$. The given equation can be re-written as $\frac{x+2}{2}=\frac{y-\frac{7}{2}}{3}=\frac{z-5}{-6}$. This line passes through the point $\left(-2,\frac{7}{2},5\right)$ and has direction ratios proportional to 2, 3, −6. So, its direction cosines are $\frac{2}{\sqrt{{2}^{2}+{3}^{2}+{\left(-6\right)}^{2}}},\frac{3}{\sqrt{{2}^{2}+{3}^{2}+{\left(-6\right)}^{2}}},\frac{-6}{\sqrt{{2}^{2}+{3}^{2}+{\left(-6\right)}^{2}}}$ Or $\frac{2}{7},\frac{3}{7},\frac{-6}{7}$ The required line passes through the point having position vector $\stackrel{\to }{a}=-\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}$ and is parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}+3\stackrel{^}{j}-6\stackrel{^}{k}$. So, its vector equation is $\stackrel{\to }{r}=\left(-\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+3\stackrel{^}{j}-6\stackrel{^}{k}\right)$ #### Question 1: Show that the lines intersect and find their point of intersection. #### Answer: The coordinates of any point on the first line are given by The coordinates of a general point on the first line are . Also, the coordinates of any point on the second line are given by The coordinates of a general point on the second line are If the lines intersect, then they have a common point. So, for some values of , we must have #### Question 2: Show that the lines do not intersect. #### Answer: The coordinates of any point on the first line are given by The coordinates of a general point on the first line are . The coordinates of any point on the second line are given by The coordinates of a general point on the second line are . If the lines intersect, then they have a common point. So, for some values of , we must have #### Question 3: Show that the lines intersect. Find their point of intersection. #### Answer: The coordinates of any point on the first line are given by The coordinates of a general point on the first line are . The coordinates of any point on the second line are given by The coordinates of a general point on the second line are . If the lines intersect, then they have a common point. So, for some values of , we must have #### Question 4: Prove that the lines through A (0, −1, −1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (−4, 4, 4). Also, find their point of intersection. #### Answer: The coordinates of any point on the line AB are given by The coordinates of a general point on AB are . The coordinates of any point on the line CD are given by The coordinates of a general point on CD are . If the lines AB and CD intersect, then they have a common point. So, for some values of , we must have #### Question 5: Prove that the line intersect and find their point of intersection. #### Answer: The position vectors of two arbitrary points on the given lines are If the lines intersect, then they have a common point. So, for some values of , we must have Equating the coefficients of , we get Solving (2) and (3), we get Substituting the values in (1), we get Substituting $\mu =0$ in the second line, we get $\stackrel{\to }{r}=4\stackrel{^}{i}+0\stackrel{^}{j}-\stackrel{^}{k}$ as the position vector of the point of intersection. Thus, the coordinates of the point of intersection are (4, 0, $-$1). #### Question 6: Determine whether the following pair of lines intersect or not: (i) (ii) (iii) (iv) #### Answer: (i) The position vectors of two arbitrary points on the given lines are If the lines intersect, then they have a common point. So, for some values of , we must have Equating the coefficients of , we get Solving (2) and (3), we get . Substituting the values in (1), we get (ii) The coordinates of any point on the first line are given by The coordinates of a general point on the first line are . Also, the coordinates of any point on the second line are given by The coordinates of a general point on the second line are . If the lines intersect, then they have a common point. So, for some values of , we must have (iii) The coordinates of any point on the first line are given by The coordinates of a general point on the first line are . Also, the coordinates of any point on the second line are given by The coordinates of a general point on the second line are . If the lines intersect, then they have a common point. So, for some values of , we must have (iv) The coordinates of any point on the first line are given by The coordinates of a general point on the first line are . The coordinates of any point on the second line are given by The coordinates of a general point on the second line are . If the lines intersect, then they have a common point. So, for some values of , we must have Disclaimer: The question printed in the book is incorrect. Instead of z, 3 is printed. #### Question 7: Show that the lines are intersecting. Hence, find their point of intersection. #### Answer: The position vectors of two arbitrary points on the given lines are If the lines intersect, then they have a common point. So, for some values of , we must have Equating the coefficients of , we get Solving (1) and (2), we get $\lambda =-4\phantom{\rule{0ex}{0ex}}\mu =-2$. Substituting the values in (3), we get Substituting $\mu =-2$ in the second line, we get $\stackrel{\to }{r}=5\stackrel{^}{i}-2\stackrel{^}{j}-6\stackrel{^}{i}-4\stackrel{^}{j}-12\stackrel{^}{k}=-\stackrel{^}{i}-6\stackrel{^}{j}-12\stackrel{^}{k}$ as the position vector of the point of intersection. Thus, the coordinates of the point of intersection are ($-$1, $-$6, $-$12). #### Question 1: Find the perpendicular distance of the point (3, −1, 11) from the line $\frac{x}{2}=\frac{y-2}{-3}=\frac{z-3}{4}.$ #### Answer: Let the point (3, $-$1, 11) be P and the point through which the line passes be Q (0, 2, 3). The line is parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}-3\stackrel{^}{j}+4\stackrel{^}{k}$. Now, $\stackrel{\to }{PQ}=-3\stackrel{^}{i}+3\stackrel{^}{j}-8\stackrel{^}{k}$ Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book. #### Question 2: Find the perpendicular distance of the point (1, 0, 0) from the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}.$ Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular. #### Answer: Let the point (1, 0, 0) be P and the point through which the line passes be Q (1, $-$1, $-$10). The line is parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}-3\stackrel{^}{j}+8\stackrel{^}{k}$. Now, $\stackrel{\to }{PQ}=0\stackrel{^}{i}-\stackrel{^}{j}-10\stackrel{^}{k}$ Let L be the foot of the perpendicular drawn from the point P (1, 0, 0) to the given line. The coordinates of a general point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ are given by Let the coordinates of L be . The direction ratios of PL are proportional to . The direction ratios of the given line are proportional to 2 ,$-$3, 8, but PL is perpendicular to the given line. Substituting $\lambda =1$ in , we get the coordinates of L as (3, $-$4, $-$2). Equation of the line PL is given by $\frac{x-1}{3-1}=\frac{y-0}{-4-0}=\frac{z-0}{-2-0}\phantom{\rule{0ex}{0ex}}=\frac{x-1}{1}=\frac{y}{-2}=\frac{z}{-1}$ $⇒\stackrel{\to }{r}=\stackrel{^}{i}+\lambda \left(\stackrel{^}{i}-2\stackrel{^}{j}-\stackrel{^}{k}\right)$ #### Question 3: Find the foot of the perpendicular drawn from the point A (1, 0, 3) to the joint of the points B (4, 7, 1) and C (3, 5, 3). #### Answer: Let D be the foot of the perpendicular drawn from the point A (1, 0, 3) to the line BC. The coordinates of a general point on the line BC are given by Let the coordinates of D be . The direction ratios of AD are proportional to . The direction ratios of the line BC are proportional to 1, 2, $-$2, but AD is perpendicular to the line BC. $\therefore 1\left(\lambda +3\right)+2\left(2\lambda +7\right)-2\left(-2\lambda -2\right)=0\phantom{\rule{0ex}{0ex}}⇒\lambda =-\frac{7}{3}$ Substituting $\lambda =-\frac{7}{3}$ in , we get the coordinates of D as . #### Question 4: A (1, 0, 4), B (0, −11, 3), C (2, −3, 1) are three points and D is the foot of perpendicular from A on BC. Find the coordinates of D. #### Answer: Point D is the foot of the perpendicular drawn from the point A (1, 0, 4) to the line BC. The coordinates of a general point on the line BC are given by Let the coordinates of D be . The direction ratios of AD are proportional to . The direction ratios of the line BC are proportional to 2, 8,$-$2, but AD is perpendicular to the line BC. Substituting $\lambda =\frac{11}{9}$ in , we get the coordinates of D as . #### Question 5: Find the foot of perpendicular from the point (2, 3, 4) to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}.$ Also, find the perpendicular distance from the given point to the line. #### Answer: Let L be the foot of the perpendicular drawn from the point P (2, 3, 4) to the given line. The coordinates of a general point on the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$ are given by Let the coordinates of L be . The direction ratios of PL are proportional to . The direction ratios of the given line are proportional to $-$2, 6, $-$3, but PL is perpendicular to the given line. $\therefore -2\left(-2\lambda +2\right)+6\left(6\lambda -3\right)-3\left(-3\lambda -3\right)=0\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{13}{49}$ Substituting $\lambda =\frac{13}{49}$ in , we get the coordinates of L as  . . Hence, the length of the perpendicular from P on PL is . #### Question 6: Find the equation of the perpendicular drawn from the point P (2, 4, −1) to the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}.$ Also, write down the coordinates of the foot of the perpendicular from P. #### Answer: Let L be the foot of the perpendicular drawn from the point P (2, 4, $-$1) to the given line. The coordinates of a general point on the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$ are given by Let the coordinates of L be . The direction ratios of PL are proportional to . The direction ratios of the given line are proportional to 1, 4, $-$9, but PL is perpendicular to the given line. $\therefore 1\left(\lambda -7\right)+4\left(4\lambda -7\right)-9\left(-9\lambda +7\right)=0\phantom{\rule{0ex}{0ex}}⇒\lambda =1$ Substituting $\lambda =1$ in , we get the coordinates of L as . Equation of the line PL is $\frac{x-2}{-4-2}=\frac{y-4}{1-4}=\frac{z+1}{-3+1}\phantom{\rule{0ex}{0ex}}=\frac{x-2}{-6}=\frac{y-4}{-3}=\frac{z+1}{-2}$ #### Question 7: Find the length of the perpendicular drawn from the point (5, 4, −1) to the line $\stackrel{\to }{r}=\stackrel{^}{i}+\lambda \left(2\stackrel{^}{i}+9\stackrel{^}{j}+5\stackrel{^}{k}\right).$ #### Answer: Let the point (5, 4, $-$1) be P and the the point through which the line passes be Q (1, 0, 0). The line is parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}+9\stackrel{^}{j}+5\stackrel{^}{k}$. Now, $\stackrel{\to }{PQ}=-4\stackrel{^}{i}-4\stackrel{^}{j}+\stackrel{^}{k}$ #### Question 8: Find the foot of the perpendicular drawn from the point $\stackrel{^}{i}+6\stackrel{^}{j}+3\stackrel{^}{k}$ to the line $\stackrel{\to }{r}=\stackrel{^}{j}+2\stackrel{^}{k}+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right).$ Also, find the length of the perpendicular #### Answer: Let L be the foot of the perpendicular drawn from the point P ($\stackrel{^}{i}+6\stackrel{^}{j}+3\stackrel{^}{k}$) to the line $\stackrel{\to }{r}=\stackrel{^}{j}+2\stackrel{^}{k}+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right).$ Let the position vector L be $\stackrel{\to }{r}=\stackrel{^}{j}+2\stackrel{^}{k}+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right)=\lambda \stackrel{^}{i}+\left(1+2\lambda \right)\stackrel{^}{j}+\left(2+3\lambda \right)\stackrel{^}{k}$        ...(1) Now, Since $\stackrel{\to }{PL}$ is perpendicular to the given line, which is parallel to $\stackrel{\to }{b}=\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}$, we have Substituting $\lambda =1$ in (1), we get the position vector of L as $\stackrel{^}{i}+3\stackrel{^}{j}+5\stackrel{^}{k}$. Substituting $\lambda =1$ in (2), we get $\stackrel{\to }{PL}=-3\stackrel{^}{j}+2\stackrel{^}{k}$ Hence, the length of the perpendicular from point P on PL is . #### Question 9: Find the equation of the perpendicular drawn from the point P (−1, 3, 2) to the line $\stackrel{\to }{r}=\left(2\stackrel{^}{j}+3\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+\stackrel{^}{j}+3\stackrel{^}{k}\right).$ Also, find the coordinates of the foot of the perpendicular from P. #### Answer: Let L be the foot of the perpendicular drawn from the point P ($-$1, 3, 2) to the line $\stackrel{\to }{r}=\left(2\stackrel{^}{j}+3\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+\stackrel{^}{j}+3\stackrel{^}{k}\right).$ Let the position vector L be $\stackrel{\to }{r}=\left(2\stackrel{^}{j}+3\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+\stackrel{^}{j}+3\stackrel{^}{k}\right)=2\lambda \stackrel{^}{i}+\left(2+\lambda \right)\stackrel{^}{j}+\left(3+3\lambda \right)\stackrel{^}{k}$        ...(1) Now, Since $\stackrel{\to }{PL}$ is perpendicular to the given line, which is parallel to $\stackrel{\to }{b}=2\stackrel{^}{i}+\stackrel{^}{j}+3\stackrel{^}{k}$, we have Substituting $\lambda =-\frac{2}{7}$ in (1), we get the position vector of L as $-\frac{4}{7}\stackrel{^}{i}+\frac{12}{7}\stackrel{^}{j}+\frac{15}{7}\stackrel{^}{k}$. So, the coordinates of the foot of the perpendicular from P to the given line is L . Substituting $\lambda =-\frac{2}{7}$ in (2), we get $\stackrel{\to }{PL}=\frac{3}{7}\stackrel{^}{i}-\frac{9}{7}\stackrel{^}{j}+\frac{1}{7}\stackrel{^}{k}$ Equation of the perpendicular drawn from P to the given line is #### Question 10: Find the foot of the perpendicular from (0, 2, 7) on the line $\frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}.$ #### Answer: Let L be the foot of the perpendicular drawn from the point P (0, 2, 7) to the given line. The coordinates of a general point on the line $\frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}$ are given by Let the coordinates of L be . The direction ratios of PL are proportional to . The direction ratios of the given line are proportional to $-$1, 3, $-$2, but PL is perpendicular to the given line. $\therefore -1\left(-\lambda -2\right)+3\left(3\lambda -1\right)-2\left(-2\lambda -4\right)=0\phantom{\rule{0ex}{0ex}}⇒\lambda =-\frac{1}{2}$ Substituting $\lambda =-\frac{1}{2}$ in , we get the coordinates of L as . Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book. #### Question 11: Find the foot of the perpendicular from (1, 2, −3) to the line $\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}.$ #### Answer: Let L be the foot of the perpendicular drawn from the point P (1, 2, $-$3) to the given line. The coordinates of a general point on the line $\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}$ are given by Let the coordinates of L be . The direction ratios of PL are proportional to . The direction ratios of the given line are proportional to 2, $-$2, $-$1, but PL is perpendicular to the given line. Substituting $\lambda =1$ in , we get the coordinates of L as . #### Question 12: Find the equation of line passing through the points A (0, 6, −9) and B (−3, −6, 3). If D is the foot of perpendicular drawn from a point C (7, 4, −1) on the line AB, then find the coordinates of the point D and the equation of line CD. #### Answer: Equation of line AB passing through the points A(0, 6,$-$9) and B($-$3, $-$6, 3) is $\frac{x-0}{-3-0}=\frac{y-6}{-6-6}=\frac{z+9}{3+9}\phantom{\rule{0ex}{0ex}}=\frac{x}{1}=\frac{y-6}{4}=\frac{z+9}{-4}$ Here, D is the foot of the perpendicular drawn from C (7, 4, $-$1) on AB. The coordinates of a general point on AB are given by Let the coordinates of D be . The direction ratios of CD are proportional to . The direction ratios of AB are proportional to 1, 4, $-$4, but CD is perpendicular to AB. Substituting $\lambda =-1$ in , we get the coordinates of D as . Equation of CD is $\frac{x-7}{-1-7}=\frac{y-4}{2-4}=\frac{z+1}{-5+1}\phantom{\rule{0ex}{0ex}}=\frac{x-7}{4}=\frac{y-4}{1}=\frac{z+1}{2}$ #### Question 13: Find the distance of the point (2, 4, −1) from the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$.                  [NCERT EXEMPLAR] #### Answer: We know that the distance d from point P to the line l having equation $\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{b}$ is given by $d=\frac{\left|\stackrel{\to }{b}×\stackrel{\to }{\mathrm{PQ}}\right|}{\left|\stackrel{\to }{b}\right|}$, where Q is any point on the line l. The equation of the given line is $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$. Let P(2, 4, −1) be the given point. Now, in order to find the required distance we need to find a point Q on the given line. The given line passes through the point (−5, −3, 6). So, take this point as the required point Q(−5, −3, 6). Also, the line is parallel to the vector $\stackrel{\to }{b}=\stackrel{^}{i}+4\stackrel{^}{j}-9\stackrel{^}{k}$. Now, $\stackrel{\to }{\mathrm{PQ}}=\left(-5\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}\right)-\left(2\stackrel{^}{i}+4\stackrel{^}{j}-\stackrel{^}{k}\right)=-7\stackrel{^}{i}-7\stackrel{^}{j}+7\stackrel{^}{k}$ $\therefore \stackrel{\to }{b}×\stackrel{\to }{\mathrm{PQ}}=\left|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ 1& 4& -9\\ -7& -7& 7\end{array}\right|=-35\stackrel{^}{i}+56\stackrel{^}{j}+21\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}⇒\left|\stackrel{\to }{b}×\stackrel{\to }{\mathrm{PQ}}\right|=\sqrt{{\left(-35\right)}^{2}+{56}^{2}+{21}^{2}}=\sqrt{1225+3136+441}=\sqrt{4802}=49\sqrt{2}$ Let d be the required distance. $\therefore d=\frac{\left|\stackrel{\to }{b}×\stackrel{\to }{\mathrm{PQ}}\right|}{\left|\stackrel{\to }{b}\right|}=\frac{49\sqrt{2}}{\sqrt{1+16+81}}=\frac{49\sqrt{2}}{\sqrt{98}}=\frac{49\sqrt{2}}{7\sqrt{2}}=7$ Thus, the distance of the given point from the given line is 7 units. #### Question 14: Find the coordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, −1, 3) and C(2, −3, −1).                 [NCERT EXEMPLAR] #### Answer: The Cartesian equation of the line joining points B(0, −1, 3) and C(2, −3, −1) is Let L be the foot of the perpendicular drawn from the point A(1, 8, 4) to the line $\frac{x}{2}=\frac{y+1}{-2}=\frac{\mathrm{z}-3}{-4}$. The coordinates of general point on the line $\frac{x}{2}=\frac{y+1}{-2}=\frac{\mathrm{z}-3}{-4}$ are given by Let the coordinates of L be $\left(2\lambda ,-2\lambda -1,-4\lambda +3\right)$. Therefore, the direction ratios of AL are proportional to $2\lambda -1,-2\lambda -1-8,-4\lambda +3-4$ or $2\lambda -1,-2\lambda -9,-4\lambda -1$ Direction ratios of the given line are proportional to 2, −2, −4. But, AL is perpendicular to the given line. $\therefore 2×\left(2\lambda -1\right)+\left(-2\right)×\left(-2\lambda -9\right)+\left(-4\right)×\left(-4\lambda -1\right)=0\phantom{\rule{0ex}{0ex}}⇒4\lambda -2+4\lambda +18+16\lambda +4=0\phantom{\rule{0ex}{0ex}}⇒24\lambda +20=0\phantom{\rule{0ex}{0ex}}⇒\lambda =-\frac{5}{6}$ Putting $\lambda =-\frac{5}{6}$ in $\left(2\lambda ,-2\lambda -1,-4\lambda +3\right)$, we get $\left(2×\left(-\frac{5}{6}\right),-2×\left(-\frac{5}{6}\right)-1,-4×\left(-\frac{5}{6}\right)+3\right)=\left(-\frac{5}{3},\frac{2}{3},\frac{19}{3}\right)$ Thus, the required coordinates of the foot of the perpendicular are $\left(-\frac{5}{3},\frac{2}{3},\frac{19}{3}\right)$. #### Question 1: Find the shortest distance between the following pairs of lines whose vector equations are: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) $\stackrel{\to }{r}=\left(8+3\lambda \right)\stackrel{^}{i}-\left(9+16\lambda \right)\stackrel{^}{j}+\left(10+7\lambda \right)\stackrel{^}{k}$ and $\stackrel{\to }{r}=15\stackrel{^}{i}+29\stackrel{^}{j}+5\stackrel{^}{k}+\mu \left(3\stackrel{^}{i}+8\stackrel{^}{j}-5\stackrel{^}{k}\right)$              [NCERT EXEMPLAR] #### Answer: (i) Comparing the given equations with the equations , we get The shortest distance between the lines is given by (ii) Comparing the given equations with the equations , we get The shortest distance between the lines is given by (iii) Comparing the given equations with the equations, we get The shortest distance between the lines is given by (iv) The vector equations of the given lines can be re-written as Comparing the given equations with the equations , we get The shortest distance between the line is given by (v) The vector equations of the given lines can be re-written as Comparing the given equations with the equations , we get The shortest distance between the lines is given by (vi) Comparing the given equations with the equations , we get The shortest distance between the lines is given by (vii) Comparing the given equations with the equations, we get The shortest distance between the lines is given by (viii) The vector equations of the given lines can be re-written as $\stackrel{\to }{r}=8\stackrel{^}{i}-9\stackrel{^}{j}+10\stackrel{^}{k}+\lambda \left(3\stackrel{^}{i}-16\stackrel{^}{j}+7\stackrel{^}{k}\right)$ and $\stackrel{\to }{r}=15\stackrel{^}{i}+29\stackrel{^}{j}+5\stackrel{^}{k}+\mu \left(3\stackrel{^}{i}+8\stackrel{^}{j}-5\stackrel{^}{k}\right)$ Comparing the given equations with the equations , we get $\stackrel{\to }{{a}_{1}}=8\stackrel{^}{i}-9\stackrel{^}{j}+10\stackrel{^}{k}$ $\stackrel{\to }{{b}_{1}}=3\stackrel{^}{i}-16\stackrel{^}{j}+7\stackrel{^}{k}$ $\stackrel{\to }{{a}_{2}}=15\stackrel{^}{i}+29\stackrel{^}{j}+5\stackrel{^}{k}$ $\stackrel{\to }{{b}_{2}}=3\stackrel{^}{i}+8\stackrel{^}{j}-5\stackrel{^}{k}$ $\therefore \stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}=\left(15\stackrel{^}{i}+29\stackrel{^}{j}+5\stackrel{^}{k}\right)-\left(8\stackrel{^}{i}-9\stackrel{^}{j}+10\stackrel{^}{k}\right)=7\stackrel{^}{i}+38\stackrel{^}{j}-5\stackrel{^}{k}$ $\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}=\left|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ 3& -16& 7\\ 3& 8& -5\end{array}\right|=24\stackrel{^}{i}+36\stackrel{^}{j}+72\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}⇒\left|\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}\right|=\sqrt{{24}^{2}+{36}^{2}+{72}^{2}}=\sqrt{576+1296+5184}=\sqrt{7056}=84$ Also, $\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right).\left(\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}\right)\phantom{\rule{0ex}{0ex}}=\left(7\stackrel{^}{i}+38\stackrel{^}{j}-5\stackrel{^}{k}\right).\left(24\stackrel{^}{i}+36\stackrel{^}{j}+72\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}=7×24+38×36+\left(-5\right)×72\phantom{\rule{0ex}{0ex}}=168+1368-360\phantom{\rule{0ex}{0ex}}=1176$ We know that the shortest distance between the lines is given by . ∴ Required shortest distance between the given pairs of lines, #### Question 2: Find the shortest distance between the following pairs of lines whose cartesian equations are: (i) (ii) (iii) (iv) #### Answer: (i) The equations of the given lines are Since line (1) passes through the point (1, 2, 3) and has direction ratios proportional to 2, 3, 4, its vector equation is Also, line (2) passes through the point (2, 3, 5) and has direction ratios proportional to 3, 4, 5. Its vector equation is Now, The shortest distance between the lines is given by (ii) The equations of the given lines are Since line (1) passes through the point (1, $-$1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is Also, line (2) passes through the point ($-$1, 2, 2) and has direction ratios proportional to 3, 1, 0. Its vector equation is Now, The shortest distance between the lines is given by Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book. (iii) Since line (1) passes through the point (1, $-$2, 3) and has direction ratios proportional to $-$1, 1, $-$2, its vector equation is Also, line (2) passes through the point (1, $-$1, $-$1) and has direction ratios proportional to 1, 2, $-$2. Its vector equation is Now, The shortest distance between the lines is given by (iv) Since line (1) passes through the point (3, 5, 7) and has direction ratios proportional to 1, $-$2, 1, its vector equation is Also, line (2) passes through the point ($-$1, $-$1, $-$1) and has direction ratios proportional to 7, $-$6, 1. Its vector equation is Now, The shortest distance between the lines is given by #### Question 3: By computing the shortest distance determine whether the following pairs of lines intersect or not: (i) (ii) (iii) (iv) #### Answer: (i) Comparing the given equations with the equations , we get (ii) Comparing the given equations with the equations , we get (iii) Since the first line passes through the point (1, $-$1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is Also, the second line passes through the point ($-$1, 2, 2) and has direction ratios proportional to 5, 1, 0. Its vector equation is Now, (iv) Since the first line passes through the point (5, 7, $-$3) and has direction ratios proportional to 4, $-$5, $-$5, its vector equation is Also, the second line passes through the point (8, 7, 5) and has direction ratios proportional to 7, 1, 3. Its vector equation is Now, #### Question 4: Find the shortest distance between the following pairs of parallel lines whose equations are: (i) (ii) #### Answer: (i) The vector equations of the given lines are These two lines pass through the points having position vectors and are parallel to the vector $\stackrel{\to }{b}=\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}$. Now, $\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}=\stackrel{^}{i}-3\stackrel{^}{j}-4\stackrel{^}{k}$ and The shortest distance between the two lines is given by $\frac{\left|\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}\right|}{\left|\stackrel{\to }{b}\right|}=\frac{\sqrt{78}}{\sqrt{3}}=\sqrt{26}$ (ii) These two lines pass through the points having position vectors and are parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}$. Now, $\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}=\stackrel{^}{i}-\stackrel{^}{k}$ and The shortest distance between the two lines is given by $\frac{\left|\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}\right|}{\left|\stackrel{\to }{b}\right|}=\frac{\sqrt{11}}{\sqrt{6}}$ #### Question 5: Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines (i) (0, 0, 0) and (1, 0, 2) (ii) (1, 3, 0) and (0, 3, 0) #### Answer: (i) The equation of the line passing through the points (0, 0, 0) and (1, 0, 2) is $\frac{x-0}{1-0}=\frac{y-0}{0-0}=\frac{z-0}{2-0}\phantom{\rule{0ex}{0ex}}=\frac{x}{1}=\frac{y}{0}=\frac{z}{2}$ (ii) The equation of the line passing through the points (1, 3, 0) and (0, 3, 0) is $\frac{x-1}{0-1}=\frac{y-3}{3-3}=\frac{z-0}{0-0}\phantom{\rule{0ex}{0ex}}=\frac{x-1}{-1}=\frac{y-3}{0}=\frac{z}{0}$ Since the first line passes through the point (0, 0, 0) and has direction ratios proportional to 1, 0, 2, its vector equation is Also, the second line passes through the point (1, 3, 0) and has direction ratios proportional to $-$1, 0, 0. Its vector equation is Now, The shortest distance between the lines is given by #### Question 6: Write the vector equations of the following lines and hence determine the distance between them #### Answer: We have $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}\phantom{\rule{0ex}{0ex}}\frac{x-3}{4}=\frac{y-3}{6}=\frac{z+5}{12}$ Since the first line passes through the point (1, 2, $-$4) and has direction ratios proportional to 2, 3, 6, its vector equation is Also, the second line passes through the point (3, 3, $-$5) and has direction ratios proportional to 4, 6, 12. Its vector equation is These two lines pass through the points having position vectors and are parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}+3\stackrel{^}{j}+6\stackrel{^}{k}$. Now, $\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}=2\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}$ and The shortest distance between the two lines is given by $\frac{\left|\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}\right|}{\left|\stackrel{\to }{b}\right|}=\frac{\sqrt{293}}{7}\mathrm{units}$ #### Question 7: Find the shortest distance between the lines (i) (ii) (iii) (iv) #### Answer: (i) Comparing the given equations with the equations , we get The shortest distance between the lines is given by (ii) Since the first line passes through the point ($-$1, $-$1, $-$1) and has direction ratios proportional to 7, $-$6, 1, its vector equation is Also, the second line passing through the point (3, 5, 7) has direction ratios proportional to 1,$-$2, 1. Its vector equation is Now, The shortest distance between the lines is given by (iii) Comparing the given equations with the equations , we get The shortest distance between the lines is given by (iv) Comparing the given equations with the equations , we get The shortest distance between the lines is given by #### Question 8: Find the distance between the lines l1 and l2 given by #### Answer: Given: These two lines pass through the points having position vectors and are parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}+3\stackrel{^}{j}+6\stackrel{^}{k}$. Now, $\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}=2\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}$ and The shortest distance between the two lines is given by $\frac{\left|\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}\right|}{\left|\stackrel{\to }{b}\right|}=\frac{\sqrt{293}}{7}$ #### Question 1: Write the cartesian and vector equations of X-axis. #### Answer: Since x-axis passes through the the point (0, 0, 0) having position vector $\stackrel{\to }{a}=0\stackrel{^}{i}+0\stackrel{^}{j}+0\stackrel{^}{k}$ and is parallel to the vector $\stackrel{\to }{b}=1\stackrel{^}{i}+0\stackrel{^}{j}+0\stackrel{^}{k}$ having direction ratios proportional to 1, 0, 0, the cartesian equation of x-axis is $\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}\phantom{\rule{0ex}{0ex}}=\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$ Also, its vector equation is #### Question 2: Write the cartesian and vector equations of Y-axis. #### Answer: Since y-axis passes through the the point (0, 0, 0) having position vector $\stackrel{\to }{a}=0\stackrel{^}{i}+0\stackrel{^}{j}+0\stackrel{^}{k}$ and is parallel to the vector $\stackrel{\to }{b}=0\stackrel{^}{i}+1\stackrel{^}{j}+0\stackrel{^}{k}$ having direction ratios proportional to 0, 1, 0, the cartesian equation of y-axis is $\frac{x-0}{0}=\frac{y-0}{1}=\frac{z-0}{0}\phantom{\rule{0ex}{0ex}}=\frac{x}{0}=\frac{y}{1}=\frac{z}{0}$ Also, its vector equation is #### Question 3: Write the cartesian and vector equations of Z-axis. #### Answer: Since z-axis passes through the the point (0, 0, 0) having position vector $\stackrel{\to }{a}=0\stackrel{^}{i}+0\stackrel{^}{j}+0\stackrel{^}{k}$ and is parallel to the vector $\stackrel{\to }{b}=0\stackrel{^}{i}+0\stackrel{^}{j}+\stackrel{^}{k}$ having direction ratios proportional to 0, 0, 1, the cartesian equation of z-axis is $\frac{x-0}{0}=\frac{y-0}{0}=\frac{z-0}{1}\phantom{\rule{0ex}{0ex}}=\frac{x}{0}=\frac{y}{0}=\frac{z}{1}$ Also, its vector equation is #### Question 4: Write the vector equation of a line passing through a point having position vector and parallel to vector . #### Answer: The vector equation of the line passing through the point having position vector $\stackrel{\to }{\alpha }$ and parallel to vector $\stackrel{\to }{\beta }$ is $\stackrel{\to }{r}=\stackrel{\to }{\alpha }+\lambda \stackrel{\to }{\beta }$. #### Question 5: Cartesian equations of a line AB are $\frac{2x-1}{2}=\frac{4-y}{7}=\frac{z+1}{2}.$ Write the direction ratios of a line parallel to AB. #### Answer: We have $\frac{2x-1}{2}=\frac{4-y}{7}=\frac{z+1}{2}$ The equation of the line AB can be re-written as $\frac{x-\frac{1}{2}}{1}=\frac{y-4}{-7}=\frac{z+1}{2}$ The direction ratios of the line parallel to AB are proportional to 1, $-$7, 2. Also, the direction cosines of the line parallel to AB are proportional to #### Question 6: Write the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4. #### Answer: We have 6x − 2 = 3y + 1 = 2z − 4 The equation of given line can be re-written as $\frac{x-\frac{1}{3}}{\frac{1}{6}}=\frac{y+\frac{1}{3}}{\frac{1}{3}}=\frac{z-2}{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{x-\frac{1}{3}}{1}=\frac{y+\frac{1}{3}}{2}=\frac{z-2}{3}$ The direction ratios of the line parallel to AB are proportional to 1, 2, 3. Hence, the direction cosines of the line parallel to AB are proportional to . #### Question 7: Write the direction cosines of the line #### Answer: We have The equation of the given line can be re-written as The direction ratios of the given line are proportional to 4, $-$3, 0. Hence, the direction cosines of the given line are proportional to #### Question 8: Write the coordinate axis to which the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-1}{0}$is perpendicular. #### Answer: We have $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-1}{0}$ The given line is parallel to the vector $\stackrel{\to }{b}=3\stackrel{^}{i}+4\stackrel{^}{j}+0\stackrel{^}{k}$. Let $x\stackrel{^}{i}+y\stackrel{^}{j}+z\stackrel{^}{k}$ be perpendicular to the given line. Now, $3x+4y+0z=0$ It is satisfied by the coordinates of z-axis, i.e. (0, 0, 1). Hence, the given line is perpendicular to z-axis. #### Question 9: Write the angle between the lines #### Answer: We have The given lines are parallel to the vectors . Let $\theta$ be the angle between the given lines. Now, #### Question 10: Write the direction cosines of the line whose cartesian equations are 2x = 3y = −z. #### Answer: We have 2x = 3y = −z The equation of the given line can be re-written as $\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{-1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{x}{3}=\frac{y}{2}=\frac{z}{-6}$ The direction ratios of the line parallel to AB are proportional to 3, 2, $-$6. Hence, the direction cosines of the line parallel to AB are proportional to #### Question 11: Write the angle between the lines 2x = 3y = −z and 6x = −y = −4z. #### Answer: We have 2x = 3y = −z 6x = −y = −4z The given lines can be re-written as These lines are parallel to vectors . Let $\theta$ be the angle between these lines. Now, #### Question 12: Write the value of λ for which the lines are perpendicular to each other. #### Answer: We have The given lines are parallel to vectors . For $\stackrel{\to }{{b}_{1}}\perp \stackrel{\to }{{b}_{2}}$, we must have $\stackrel{\to }{{b}_{1}}.\stackrel{\to }{{b}_{2}}=0\phantom{\rule{0ex}{0ex}}⇒\left(-3\stackrel{^}{i}+2\lambda \stackrel{^}{j}+2\stackrel{^}{k}\right).\left(3\lambda \stackrel{^}{i}+\stackrel{^}{j}-5\stackrel{^}{k}\right)=0\phantom{\rule{0ex}{0ex}}⇒-7\lambda -10=0\phantom{\rule{0ex}{0ex}}⇒\lambda =-\frac{10}{7}$ #### Question 13: Write the formula for the shortest distance between the lines #### Answer: The shortest distance d between the parallel lines is given by $d=\frac{\left|\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}\right|}{\left|\stackrel{\to }{b}\right|}$ #### Question 14: Write the condition for the lines to be intersecting. #### Answer: The shortest distance between the lines is given by For the lines to be intersecting, $d=0$. #### Question 15: The cartesian equations of a line AB are $\frac{2x-1}{\sqrt{3}}=\frac{y+2}{2}=\frac{z-3}{3}.$ Find the direction cosines of a line parallel to AB. #### Answer: We have $\frac{2x-1}{\sqrt{3}}=\frac{y+2}{2}=\frac{z-3}{3}$ The equation of the line AB can be re-written as $\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{y+2}{2}=\frac{z-3}{3}\phantom{\rule{0ex}{0ex}}=\frac{x-\frac{1}{2}}{\sqrt{3}}=\frac{y+2}{4}=\frac{z-3}{6}$ Thus, the direction ratios of the line parallel to AB are proportional to $\sqrt{3}$, 4, 6. Hence, the direction cosines of the line parallel to AB are proportional to #### Question 16: If the equations of a line AB are $\frac{3-x}{1}=\frac{y+2}{-2}=\frac{z-5}{4},$ write the direction ratios of a line parallel to AB. #### Answer: We have $\frac{3-x}{1}=\frac{y+2}{-2}=\frac{z-5}{4}$ The equation of the line AB can be re-written as $\frac{x-3}{-1}=\frac{y+2}{-2}=\frac{z-5}{4}$ Thus, the direction ratios of the line parallel to AB are proportional to $-$1, $-$2, 4. #### Question 17: Write the vector equation of a line given by $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}.$ #### Answer: We have $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ The given line passes through the point (5, $-$4, 6) and has direction ratios proportional to 3, 7, 2. Vector equation of the given line passing through the point having position vector $\stackrel{\to }{a}=5\stackrel{^}{i}-4\stackrel{^}{j}+6\stackrel{^}{k}$ and parallel to a vector $\stackrel{\to }{b}=3\stackrel{^}{i}+7\stackrel{^}{j}+2\stackrel{^}{k}$ is $\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{b}\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{r}=5\stackrel{^}{i}-4\stackrel{^}{j}+6\stackrel{^}{k}+\lambda \left(3\stackrel{^}{i}+7\stackrel{^}{j}+2\stackrel{^}{k}\right)$ Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book. #### Question 18: The equations of a line are given by $\frac{4-x}{3}=\frac{y+3}{3}=\frac{z+2}{6}.$ Write the direction cosines of a line parallel to this line. #### Answer: We have $\frac{4-x}{3}=\frac{y+3}{3}=\frac{z+2}{6}$ The equation of the given line can be re-written as $\frac{x-4}{-3}=\frac{y+3}{3}=\frac{z+2}{6}$ The direction ratios of the line parallel to the given line are proportional to $-$3, 3, 6. Hence, the direction cosines of the line parallel to the given line are proportional to #### Question 19: Find the Cartesian equations of the line which passes through the point (−2, 4 , −5) and is parallel to the line $\frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}.$ #### Answer: The equation of the given line is $\frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}$ It can be re-written as $\frac{x+3}{3}=\frac{y-4}{-5}=\frac{z+8}{6}$ Since the required line is parallel to the given line, the direction ratios of the required line are proportional to 3, $-$5, 6. Hence, the cartesian equations of the line passing through the point ($-$2, 4, $-$5) and parallel to a vector having direction ratios proportional to 3, $-$5, 6 is $\frac{x+2}{3}=\frac{y-4}{-5}=\frac{z+5}{6}$. #### Question 20: Find the angle between the lines $\stackrel{\to }{r}=\left(2\stackrel{^}{i}-5\stackrel{^}{j}+\stackrel{^}{k}\right)+\lambda \left(3\stackrel{^}{i}+2\stackrel{^}{j}+6\stackrel{^}{k}\right)$ and $\stackrel{\to }{r}=7\stackrel{^}{i}-6\stackrel{^}{k}+\mu \left(\stackrel{^}{i}+2\stackrel{^}{j}+2\stackrel{^}{k}\right)$.                  [CBSE 2014] #### Answer: Let $\theta$ be the angle between the given lines. The given lines are parallel to the vectors $\stackrel{\to }{{b}_{1}}=3\stackrel{^}{i}+2\stackrel{^}{j}+6\stackrel{^}{k}$ and $\stackrel{\to }{{b}_{2}}=\stackrel{^}{i}+2\stackrel{^}{j}+2\stackrel{^}{k}$, respectively. So, the angle $\theta$ between the given lines is given by Thus, the angle between the given lines is ${\mathrm{cos}}^{-1}\left(\frac{19}{21}\right)$. #### Question 21: Find the angle between the lines $2x=3y=-z$ and $6x=-y=-4z$.           [CBSE 2015] #### Answer: The equations of the given lines can be re-written as $\frac{x}{3}=\frac{y}{2}=\frac{z}{-6}$ and $\frac{x}{2}=\frac{y}{-12}=\frac{z}{-3}$ We know that angle between the lines $\frac{x-{x}_{1}}{{a}_{1}}=\frac{y-{y}_{1}}{{b}_{1}}=\frac{z-{z}_{1}}{{c}_{1}}$ and $\frac{x-{x}_{2}}{{a}_{2}}=\frac{y-{y}_{2}}{{b}_{2}}=\frac{z-{z}_{2}}{{c}_{2}}$ is given by $\mathrm{cos}\theta =\frac{{a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}+{c}_{1}^{2}}\sqrt{{a}_{2}^{2}+{b}_{2}^{2}+{c}_{2}^{2}}}$. Let $\theta$ be the angle between the given lines. Thus, the angle between the given lines is $\frac{\mathrm{\pi }}{2}$. #### Question 1: The angle between the straight lines is (a) 45° (b) 30° (c) 60° (d) 90° #### Answer: (d) 90° We have The direction ratios of the given lines are proportional to 2, 5, 4 and 1, 2, $-$3. The given lines are parallel to the vectors . Let $\theta$ be the angle between the given lines. Now, #### Question 2: The lines $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and $\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6}$ are (a) coincident (b) skew (c) intersecting (d) parallel #### Answer: (a) coincident The equation of the given lines are Thus, the two lines are parallel to the vector $\stackrel{\to }{b}=\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}$ and pass through the points (0, 0, 0) and (1, 2, 3). Now, Since the distance between the two parallel line is 0, the given lines are coincident. #### Question 3: The direction ratios of the line perpendicular to the lines are proportional to (a) 4, 5, 7 (b) 4, −5, 7 (c) 4, −5, −7 (d) −4, 5, 7 #### Answer: (a) 4, 5, 7 We have The direction ratios of the given lines are proportional to 2, $-$3, 1 and 1, 2, $-$2. The vectors parallel to the given vectors are . Vector perpendicular to the given two lines is Hence, the direction ratios of the line perpendicular to the given two lines are proportional to 4, 5, 7. #### Question 4: The angle between the lines is (a) ${\mathrm{cos}}^{-1}\left(\frac{1}{65}\right)$ (b) $\frac{\mathrm{\pi }}{6}$ (c) $\frac{\mathrm{\pi }}{3}$ (d) $\frac{\mathrm{\pi }}{4}$ #### Answer: (c) $\frac{\mathrm{\pi }}{3}$ We have The direction ratios of the given lines are proportional to 1, 1, 2 and . The given lines are parallel to vectors . Let $\theta$ be the angle between the given lines. Now, #### Question 5: The direction ratios of the line xy + z − 5 = 0 = x − 3y − 6 are proportional to (a) 3, 1, −2 (b) 2, −4, 1 (c) (d) #### Answer: (a) 3, 1, −2 We have xy + z − 5 = 0 = x − 3y − 6 From (1) and (2), we get So, the given equation can be re-written as $\frac{x-6}{3}=\frac{y}{1}=\frac{z+1}{-2}$ Hence, the direction ratios of the given line are proportional to 3, 1,$-$2. #### Question 6: The perpendicular distance of the point P (1, 2, 3) from the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$ is (a) 7 (b) 5 (c) 0 (d) none of these #### Answer: (a) 7 We have $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$ Let point (1, 2, 3) be P and the point through which the line passes be Q (6, 7, 7). Also, the line is parallel to the vector $\stackrel{\to }{b}=3\stackrel{^}{i}+2\stackrel{^}{j}-2\stackrel{^}{k}$. Now, $\stackrel{\to }{PQ}=5\stackrel{^}{i}+5\stackrel{^}{j}+4\stackrel{^}{k}$ #### Question 7: The equation of the line passing through the points is (a) (b) (c) (d) none of these #### Answer: (c) Equation of the line passing through the points having position vectors is #### Question 8: If a line makes angles α, β and γ with the axes respectively, then cos 2 α + cos 2 β + cos 2 γ = (a) −2 (b) −1 (c) 1 (d) 2 #### Answer: (b) −1 If a line makes angles α, β and γ with the axes, then ${\mathrm{cos}}^{2}\alpha +{\mathrm{cos}}^{2}\beta +{\mathrm{cos}}^{2}\gamma =1$                           ...(1) We have #### Question 9: If the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines are (a) (b) (c) (d) #### Answer: (a) The direction ratios of the line are proportional to 1, $-$3, 2. $\therefore$ The direction cosines of the line are #### Question 10: If a line makes angle with x-axis and y-axis respectively, then the angle made by the line with z-axis is (a) π/2 (b) π/3 (c) π/4 (d) 5π/12 #### Answer: (b) π/3 If a line makes angles α, β and γ with the axes, then ${\mathrm{cos}}^{2}\alpha +{\mathrm{cos}}^{2}\beta +{\mathrm{cos}}^{2}\gamma =1$. Here, $\alpha =\frac{\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}\beta =\frac{\mathrm{\pi }}{4}$ Now, #### Question 11: The projections of a line segment on X, Y and Z axes are 12, 4 and 3 respectively. The length and direction cosines of the line segment are (a) (b) (c) (d) none of these #### Answer: (a) If a line makes angles α, β and γ with the axes, then Let r be the length of the line segment. Then, Substituting r = 13 in (2), we get Thus, the direction cosines of the line are . The lines are (a) parallel (b) intersecting (c) skew (d) coincident #### Answer: (d) coincident The equations of the given lines are Thus, the two lines are parallel to the vector $\stackrel{\to }{b}=\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}$ and pass through the points (0, 0, 0) and (1, 2, 3). Now, Since, the distance between the two parallel lines is 0, the given two lines are coincident lines. Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book. #### Question 13: The straight line $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$ is (a) parallel to x-axis (b) parallel to y-axis (c) parallel to z-axis (d) perpendicular to z-axis #### Answer: (d) perpendicular to z-axis We have $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$ Also, the given line is parallel to the vector $\stackrel{\to }{b}=3\stackrel{^}{i}+\stackrel{^}{j}+0\stackrel{^}{k}$. Let $x\stackrel{^}{i}+y\stackrel{^}{j}+z\stackrel{^}{k}$ be perpendicular to the given line. Now, $3x+4y+0z=0$ It is satisfied by the coordinates of z-axis, i.e. (0, 0, 1). Hence, the given line is perpendicular to z-axis. #### Question 14: The shortest distance between the lines is (a) $\sqrt{30}$ (b) $2\sqrt{30}$ (c) $5\sqrt{30}$ (d) $3\sqrt{30}$ #### Answer: (d) $3\sqrt{30}$ We have We know that line (1) passes through the point (3, 8, 3) and has direction ratios proportional to 3, $-$1, 1. Its vector equation is Also, line (2) passes through the point ($-$3, $-$7, 6) and has direction ratios proportional to $-$3, 2, 4. Its vector equation is Now, The shortest distance between the lines is given by #### Question 1: Find the vector and cartesian equations of the line through the point (5, 2, −4) and which is parallel to the vector $3\stackrel{^}{i}+2\stackrel{^}{j}-8\stackrel{^}{k}.$ #### Answer: We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to vector $\stackrel{\to }{b}$ is . Here, Vector equation of the required line is given by Reducing (1) to cartesian form, we get #### Question 2: Find the vector equation of the line passing through the points (−1, 0, 2) and (3, 4, 6). #### Answer: We know that the vector equation of a line passing through the points with position vectors $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ is , where $\lambda$ is a scalar. Here, Vector equation of the required line is #### Question 3: Find the vector equation of a line which is parallel to the vector $2\stackrel{^}{i}-\stackrel{^}{j}+3\stackrel{^}{k}$ and which passes through the point (5, −2, 4). Also, reduce it to cartesian form. #### Answer: We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is . Here, So, the vector equation of the required line is Reducing (1) to cartesian form, we get #### Question 4: A line passes through the point with position vector $2\stackrel{^}{i}-3\stackrel{^}{j}+4\stackrel{^}{k}$ and is in the direction of $3\stackrel{^}{i}+4\stackrel{^}{j}-5\stackrel{^}{k}.$ Find equations of the line in vector and cartesian form. #### Answer: We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is . Here, . So, the vector equation of the required line is Reducing (1) to cartesian form, we get #### Question 5: ABCD is a parallelogram. The position vectors of the points A, B and C are respectively, Find the vector equation of the line BD. Also, reduce it to cartesian form. #### Answer: We know that the position vector of the mid-point of $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ is $\frac{\stackrel{\to }{a}+\stackrel{\to }{b}}{2}$. Let the position vector of point D be $x\stackrel{^}{i}+y\stackrel{^}{j}+z\stackrel{^}{k}$. Position vector of mid-point of A and C = Position vector of mid-point of B and D The vector equation of line BD passing through the points with position vectors $\stackrel{\to }{a}$(B) and $\stackrel{\to }{b}$(D) is . Here, Vector equation of the required line is Reducing (1) to cartesian form, we get #### Question 6: Find in vector form as well as in cartesian form, the equation of the line passing through the points A (1, 2, −1) and B (2, 1, 1). #### Answer: We know that the vector equation of a line passing through the points with position vectors $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ is , where $\lambda$ is a scalar. Here, Vector equation of the required line is Reducing (1) to cartesian form, we get #### Question 7: Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector $\stackrel{^}{i}-2\stackrel{^}{j}+3\stackrel{^}{k}.$ Reduce the corresponding equation in cartesian from. #### Answer: We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is . Here, Vector equation of the required line is Reducing (1) to cartesian form, we get View NCERT Solutions for all chapters of Class 15

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# Thread: Intervals where the function is Increasing 1. ## Intervals where the function is Increasing The problem is F(x) = x + cosx Find intervals where the function is Increasing I took the first derivative and I'm lost as to where to go. Any help? F'(x)= 1 - Sinx 2. Set the first derivative equal to zero. the values you get for x will be your critical points. Then, test those critical points on a number line by plugging a number on either side of each critical number to test. If the answer is negative, it is decreasing. If the answer is positive, it is increasing. Use that to figure out max and min if you need. hope that helped. =] 3. So I get F'(x) = -Sinx/-1 = -1/-1 F'(x) = Sinx = 1 My trig skills are a bit rusty so I have no clue how to get the c.p. Any help ? 4. My brother just told me the C.P. for this would be Pi/2. So from there what do I do? He just told me the answer and walked off which is pi/2 + 2kpi. How did he arrive at that answer? 5. Originally Posted by Afterme The problem is F(x) = x + cosx Find intervals where the function is Increasing I took the first derivative and I'm lost as to where to go. Any help? F'(x)= 1 - Sinx Solve F'(x) > 0. In your case that means solving $\displaystyle 1 - \sin x > 0 \Rightarrow \sin x < 1$. So it should be very clear from a graph of $\displaystyle \sin x$ what the intervals will be ....

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# Beyond AlphaGo I have a tremendous respect for what they did. A machine learned a game and won against the human world champion in one of the most complicated games. Perhaps the most complicated game of them all, although I doubt that. It is not nearly enough, however. I want an algorithm, the simplest one of them all, which would tell me what the next optimal move is. It always exists, in every possible situation. Perhaps many possible moves exist, each optimal. But at least one is always guaranteed. We know this since John von Neumann. It doesn’t matter what your opponent might do after that, what strategy he might follow. Your move depends ONLY on the position you see, and nothing else. No game’s history matters either. Be it Go, chess or whichever finite game, there is always an optimal move you have to play to maximize your expectations. Therefore, an algorithm must calculate which one to play next. Nothing less. An explanation why this particular one – is just an interesting option. Your opponent may also see those reasons, but that will not help him much! This is very weird and contraintuitive, it goes against all our gut feelings. That’s why I wrote it here, anyway. This kind of algorithm which calculates the best move(s) solely from any given position is the holly grail. A human may or may not understand the procedure. A human may or  may not be able to follow its steps with pencil and paper. A machine as big as the Solar system may or may not be able to execute it for the game of Go or even chess. We don’t know that. I tend to think that a modern smartphone should suffice for an optimal game of Go, but it’s just a guess of mine. What is certain, at least every finite turn based game can be “hacked” this way. A simple to do list for every different set of possible positions, always exists. The upper bound for the winning algorithm size in Go is 3^361 IF clauses. It’s only about 500 steps when those IF clauses are sorted by the position number. An example line of this algorithm would be: IF (PositionNumber == 100000000400000000000007) MoveTo C9 The algorithm size in bytes is prohibitively large for the observable Universe to store it. So it’s useless, even if there are less than 1000 steps to execute  it. We can improve it, so that a typical line would be IF (F_function(PositionNumber) == 100006007) MoveTo C9 F_function is itself say a million lines long, which is not too bad, and then you can store all the IFs inside … well much smaller space. How far this optimization goes, nobody knows. My wild guess is – less than one million instructions of a smaller than one Giga byte large Phyton program. Maybe much less. Several kilobytes of code, and never more than a thousand steps algorithm to always win (or at least not to lose) a Go game – that would really impressed me. A very opaque Neural Network machine is just a very good step in the right direction. BTW .. IF (Game==Chess) & (Position==OpeningPosition) Move your right knight to f3! I can’t say I can prove this to you, but I came close. We currently have about 120 million such advices. Not enough on one hand, and not reduced enough on the other hand. But good enough to cheat successfully on http://www.chess.com. Standard ## 4 thoughts on “Beyond AlphaGo” • And what your phone says, about how Mr. White should always begin? With Nf3? And what about Mr. Black’s response to that move? 1. alpha007org says: Shall we try? My Nexus 5 with Stockfish vs your solution? Let’s test. When, date, time in UTC and where. Time constraints? I think 2 min per move is enough. • No, not yet. You will still win too easily. I believe my first few moves would be optimal, but soon I’ll be in the dark, for there will be no advice from my system for me what to do . At least in some positions. Which can already be deadly for a human against Stockfish. I could use Stockfish moves in the dark and then it would be a game between Stockfish and Stockfish+. I had an idea, to milk Stockfish for rules already. As I milked chess history database. But you see, people (and computer programs) don’t play the logically proven move. They don’t try to prove that a particular move is the best before playing it. I roughly estimate, that Stockfish average move is 1.05 the best. An optimal player would have 1.0 and a top human player has 1.1. When an amateur like me is lucky to have 2.0 or there about. For there is often the twentieth best move which I actually play. It looks good in my eyes, but it’s a death sentence in fact. I am not putting any manual effort in this. This chess engine might evolve to a full maturity one day, but it will be (as) a side effect. A chess toy model, since a tic-tac-toe toy model doesn’t sound too good.

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https://chloetoesboutique.com/little-baby/you-asked-can-i-measure-my-babys-feet-at-home.html

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# You asked: Can I measure my baby’s feet at home? Contents Firstly, find an appropriate room that has a hard flat surface and then stand your little one onto a piece of paper. Make sure that their feet are as flat as possible and not arched. Then, mark next to their heel and big toe. Take a note of this measurement in centimetres, to the nearest millimetre. ## How do I know my baby’s shoe size? From your baby’s big toe, measure to the tip of the shoe. The longest toe should be between the width of your pinkie and the width of your thumb from the tip of the shoe. ## How can I measure my child’s feet at home UK? Tip: Draw a straight line on a piece of paper and place the child’s foot on it. Mark both the heel end and toe end with a pen stroke and then measure the distance between these two markers. ## Is there an app to measure children’s feet? Using the RiteFit app, users take a photo of a child’s foot while standing on a reference object, such as an 8.5-inch by 11-inch piece of paper. The app then accurately measures the foot’s length and width, allowing parents to determine the correct shoe size. ## How fast do baby feet grow? The younger the foot, the faster it grows. On average, a child will grow up to 9 sizes in their first three years. Here’s a breakdown: From birth to 12 months, they will grow an average of 5 sizes, (from a 0 to a 5). Then from 12 months to 24 months, the average child will only grow two shoe sizes. ## What is a little kid size 1? Little Kid Size Conversions 4 – 7 years US Sizes Euro Sizes Inches 1 32 7.75″ 1.5 33 8″ 2 33 8.125″ 2.5 34 8.25″ ## Are Clarks measuring feet? Step 1: Buy a Clarks foot gauge Our foot measuring gauges come in two sizes depending on the age of your child and the size of their feet. To buy a foot measuring gauge, select the size you need. ## How do kids shoe sizes work? Kids’ Shoe Sizes by the Numbers Toddlers’ sizes run from 0–13: The smallest sizes in this range are tiny baby shoes, but the larger sizes often fit kids who are 4 or 5 years old or even older. Important: These shoes are often marked with “T” for toddler up to size 7 (e.g., 4T, 5T, 6T, 7T). ## Can you measure kids feet with tape measure UK? Use a tape measure or ruler to take the distance from back of heel to tip of their longest toe. Repeat for the other foot and record the longest of the two measurements – it’s normal for kids, and grown-ups for that matter, to have two slightly different-sized feet. 2. ## How do I measure my child’s foot? Measure length: IT IS INTERESTING:  Is it OK not to change baby at night? Use the ruler to measure from the outside of the heel to the tip of the big toe on both feet. Write the number down in centimeters, and label it as length. Remember – many kids’ feet are differently sizes as they grow. Measure each foot separately to find a size that’s sure to fit both! ## What happened to foot fairy from shark tank? It looks like the Foot Fairy deal didn’t happen, and the company failed after their Shark Tank appearance with only one comment in September of 2014. Their website is not functional. ## Are Startrite and Clarks sizes the same? What Is The Difference Between Clarks And Start-Rite Sizing? There are several opinions on this one – however, in general Start-Rite tend to be a little bit bigger than Clarks (some say a half size). However, just let us know and we can measure them for you! … Our son is the same size in both brands. ## Why are baby feet so fat? Baby’s feet normally appear flat because children are born with a pad of fat in the arch area. Also, their foot and leg muscles aren’t developed enough to support their arches when they first begin to stand. ## Can feet grow overnight? The gist: Your feet can grow up to half a size throughout the day, so it’s important to get fitted for shoes at the right time. Expert insight: … But since your feet will always be bigger later in the day, you’re best off getting fitted for sneakers in the afternoon, Positano says. IT IS INTERESTING:  What should my 1 week olds poop look like? ## How many cm do babies feet grow per month? A child’s foot grows approximately 2 cm a year between ages 0-2 – measure their foot length every 2 months. A child’s foot grows approximately 1.5 cm a year between ages 2-5 – measure their foot length every 3 months.

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https://www.educative.io/courses/visual-introduction-to-algorithms/insertion-sort

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# Insertion Sort There are many different ways to sort. As selection sort runs, the subarray at the beginning of the array is sorted, but the subarray at the end is not. Selection sort scans the unsorted subarray for the next element to include in the sorted subarray. Here's another way to think about sorting. Imagine that you are playing a card game. You're holding the cards in your hand, and these cards are sorted. The dealer hands you exactly one new card. You have to put it into the correct place so that the cards you're holding are still sorted. In selection sort, each element that you add to the sorted subarray is no smaller than the element already in the sorted subarray. But in our card example, the new card could be smaller than some of the cards you're already holding, and so you go down the line, comparing the new card against each card in your hand, until you find the place to put it. You insert the new card in the right place, and once again, your hand holds fully sorted cards. Then the dealer gives you another card, and you repeat the same procedure. Then another card, and another card, and so on, until the dealer stops giving you cards. This is the idea behind insertion sort. Loop over positions in the array, starting with index $1$. Each new position is like the new card handed to you by the dealer, and you need to insert it into the correct place in the sorted subarray to the left of that position. Here's a visualization that steps through that:

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# Waves - Multiple Choice Questions Q1. Two points on a progressive wave are one-eighth of a wavelength apart. The distance between them is 0.5 m, and the frequency of the oscillation is 10 Hz. What is the minimum speed of the wave? A 0.2 m s−1 B 10 m s−1 C 20 m s−1 D 40 m s−1 Q2. Which of the following waves cannot be polarised? A radio B ultrasonic C microwave D ultraviolet Q3. Which of the following is correct for a stationary wave? A Between two nodes the amplitude of the wave is constant. B The two waves producing the stationary wave must always be 180° out of phase. C The separation of the nodes for the second harmonic is double the separation of nodes for the first harmonic. D Between two nodes all parts of the wave vibrate in phase. Q4. Sound waves cross a boundary between two media X and Y. The frequency of the waves in X is 400Hz. The speed of the waves in X is 330 ms −1 and the speed of the waves in Y is 1320 ms –1 . What are the correct frequency and wavelength in Y? Frequency / Hz Wavelength / m A 100 0.82 B 400 0.82 C 400 3.3 D 1600 3.3

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http://gettingsharper.de/category/algorithms/

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It’s that time of the year again where F#ers come together and write a bit about what they enjoy. I really enjoyed this years Advent of code and while I only did a few of the exercises in F# I want to write about one algorithm that helped out in quite a few of the exercises. The Algorithm I’m talking about is the A* algorithm. # Ok what is this It’s a very general algorithm that helps you find an optimal path in some kind of graph and so it’s often corelated with path-finding algorithms. And indeed one of the AoC days (day 13) was all about finding the shortest Path in a maze. Here is what the program I’m gonna show you here found for my input to this day: #..###.#...#..#....##.#...#....#....#..#.#.##...## #O#..#.##.#######.#.#.####.##.###...##.#..#.###... #OO#.#.##.#....##...#..#.####.####...#..#..#..#.#. ##OO##.##.#..#...###.#.....#.....##.######....#..# ..#O##..########.#.#####...#..##.##.#.....##.###.# #..OOOOOOOOO.###.....#####.#####....#..##..#.###.. ###.##.####O..#.##....#..#.#..#.##.####.##....#### #.#..#.##.#O#.######..#.##..#.####..#.#######..### .###......#OO#..#..####.###..#..#.#.##..#.......## .####..###.#O##.#.#...#..#.#.##.##......#..####..# ..######.###O.###..##.#..###..##.#..##.####...#..# #..##..#.OOOO..###.#..###......#.#####..#.#...#... .......##O##.#.....#....#.##.#.###..#.#.####..##.# #####.#.#O##...###.##...##.#..#..##.##...#######.. ...##..##O#.###..#.###.#.##.#..#..##.#.....#..#### .#.###.#OO#...#.##...#..#.#..#.......##..#.##..##. ##..#..#O######.#####.#..###..##.###.####...#..... ....##.#O##..#...##.#.##.####..#.###....#...#..### ##...#..OOOO.#......#.....#.###...########.#####.# ###.#######O########.##.#.###.##...##..#.#.##..#.# .##.#...###O##OOO#.##.#..#.....##....#.##......#.# .##.#.....#OOOO#OO#.##.#.####.#.#.##..#.#####.##.# ...###.##.#.###.#O##.###...##..##.#.#......##.#..# .#..##.#..##..##.O.#.....#..##.#..#..##..#.##.#.## ###....#...###.#.O.##.#####.#..#..##.#####..###.#. .#.##.###.#..#.##O#.###..#..##.#####..........#.## .####..##..#.#.##OO#OOOOO#...#.#..#####.##.##.##.# ...#.#...#..##..##OOO###O##.##..#..##.#.##.#...#.# #..##.##..#.###.#.####.#O##.###.#.....#....#.#.### ##..#######..#..#.....##OOOO.#..######.##..#..#... .###..#......#..###.#.#.###OO#.....#.##.###.#..### #..##.#..########.###.###.##O##..#.#..###.#..#.##. .#...####....#.......#...#.#O#####.#.#...###.....# ...#....#..#.#.##..#.###..##O##...##.###.####.##.. ############.#.#####....#.#.OOOO#.##......#.#.#.## #..##..#....##..#....##.#.#..##O#....##.#.##..#.## #....#.##.#.#.#.#.###.#..######O#.###.#.##.#..#... ##.#..#.#.#.##..##..##.#.....#OO###.##...#.###.##. .##.#...#.##.#...#.#.######..#O#..##.#.#.#.#.##### #.#..###...#.##.##.##..##.#..#OO...#.#..##.....### .###.#.#.#.####.##..#.....##..##...#..#.####....## .###..##..#...#.....#..######..##.####...######..# ..#.#.#.#..##.#.##.#####..#.##.##.##.#....#...#... The path it’ll find are the O‘s – if you run code (available here) you will get it with some colors too. By the way the solution runs in about 200ms on my system which without even memoizing the wall-positions. Day 13 was not the only day where you could use this algorithm – there where many more (the famous Day11 comes to mind). # How does the algorithm work It’s a basic breadth-first search algorithm which means that it will put nodes it should look into in the future in a queue in such a way that it will first serach in notes on the same level, which has the big advantage that these way to search will find always find a shortest path first. The problem is that node-count you have to look at a level can become really huge fast. This is where A* shines: it uses a heuristic (just a functions you have to provide that will guess the distance/cost to the goal and then will first look at the nodes where this is minimal. As most of these algorithms it uses a visited set as well to mark nodes it already looked at (so you don’t revisit nodes if there are loops in the graph). ## some details The algorithm as I choose to implement it uses this structure: type private Runtime<'node when 'node : comparison> = { heuristic : 'node -> Score neighbours : 'node -> 'node seq isGoal : 'node -> bool distance : 'node -> 'node -> Score visitedNodes : Set<'node> openNodes : Priority<'node> gScores : Map<'node,Score> fScores : Map<'node,Score> cameFrom : Map<'node,'node> } to keep track of the search. • heuristic is the mentioned function provided by the user (based on the problem) that should guess the distance/cost to the goal-node. It’s important that you don’t overestimate the cost (better keep really conservative). Because if you overestimate the algorithm will degenerate into a depth-first- search and it’s likely that you don’t find a optimal solution (as the algorithm stops at the first found node that is recognized as a goal). • neighbours create the neighbours (children) of a node and is user-provided too – the algorithm uses this function to generate the next nodes. • isGoal is user-provided and is used to decide if a node is a goal (the algorithm will stop at the first such node it finds) • distance is the final user-provided function – it calculates the distance or cost between two nodes. The algorithm uses this only for a child and it’s parent and in the example bellow it’s just constant 1. • visitedNodes is the set of nodes the algorithm already visited – no node in here will be revisited. • openNodes is the set of nodes the algorithm can look at next. As stated above it’s important that the algorithm can get the minimal of these based on the heuristic and the cost of the path taken so far. This is why you should really use a good priority queue/heap here. For this example I just missused a F#-Map structure which is not optimal (it has no O(1) access to the min. element) but it’s sufficient 😉 • gScore: keeps track of the costs on the path taken to a node so far – the algorithm updates this when it finds better paths to a node as well. • fScore: is the acutal cost to a node plus the heuristic from that node to the goal. The algorithm uses these for the priority queue in openNodes The algorithm is initialized by providing the needed function in this structure: type Config<'node when 'node : comparison> = { heuristic : 'node -> Score neighbours : 'node -> 'node seq distance : 'node -> 'node -> Score isGoal : 'node -> bool } You can find the complete implementation on github and it’s a bit boring and really just the direct translation from the pseudocode on Wikipedia. # Example Day13 So the complete example for day 13 works like this: module Maze = let private bits n = let gen n = if n <= 0 then None else Some (n % 2, n / 2) Seq.unfold gen n let private oneBits n = bits n |> Seq.filter ( ( <> ) 0) |> Seq.length let isWall favNumber (x,y) = let nr = x*x + 3*x + 2*x*y + y + y*y + favNumber oneBits nr % 2 = 1 let private dist (x,y) (x',y') = abs (x'-x) + abs (y'-y) let private validCoord (x,y) = x >= 0 && y >= 0 let private neighbours favNumber (x,y) = [ (x-1,y); (x,y-1); (x+1,y); (x,y+1) ] |> Seq.filter (fun pos -> validCoord pos && not (isWall favNumber pos)) let findPath favNumber fromPos toPos = let config : Algorithm.Config<_> = { heuristic = fun pos -> dist pos toPos neighbours = neighbours favNumber distance = fun _ _ -> 1 isGoal = fun pos -> pos = toPos } in config |> Algorithm.aStar fromPos If you want to have some fun take this algorithm/file and see if you can tackle day 11 with it hint: you should override the comparision/equals for your node to take advantage of the strucutre of the problem – a simple heursitic is to count everyhting on floor 4 as 0, everything on floor 3 as 1, everything on floor 2 as 2 and floor 3 as 3. Happy Christmas everyone

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# §10.21 Zeros ## §10.21(i) Distribution The zeros of any cylinder function or its derivative are simple, with the possible exceptions of $z=0$ in the case of the functions, and $z=0,\pm\nu$ in the case of the derivatives. If $\nu$ is real, then $J_{\nu}\left(z\right)$, $J_{\nu}'\left(z\right)$, $Y_{\nu}\left(z\right)$, and $Y_{\nu}'\left(z\right)$, each have an infinite number of positive real zeros. All of these zeros are simple, provided that $\nu\geq-1$ in the case of $J_{\nu}'\left(z\right)$, and $\nu\geq-\tfrac{1}{2}$ in the case of $Y_{\nu}'\left(z\right)$. When all of their zeros are simple, the $m$th positive zeros of these functions are denoted by $j_{\nu,m}$, ${j^{\prime}_{\nu,m}}$, $y_{\nu,m}$, and ${y^{\prime}_{\nu,m}}$ respectively, except that $z=0$ is counted as the first zero of $J_{0}'\left(z\right)$. Since $J_{0}'\left(z\right)=-J_{1}\left(z\right)$ we have 10.21.1 $\displaystyle{j^{\prime}_{0,1}}$ $\displaystyle=0,$ $\displaystyle{j^{\prime}_{0,m}}$ $\displaystyle=j_{1,m-1},$ $m=2,3,\ldots$. When $\nu\geq 0$, the zeros interlace according to the inequalities 10.21.2 $\displaystyle j_{\nu,1}$ $\displaystyle $\displaystyle y_{\nu,1}$ $\displaystyle ⓘ Symbols: $j_{\NVar{\nu},\NVar{m}}$: zeros of the Bessel function $J_{\nu}\left(x\right)$, $y_{\NVar{\nu},\NVar{m}}$: zeros of the Bessel function $Y_{\nu}\left(x\right)$ and $\nu$: complex parameter A&S Ref: 9.5.2 Referenced by: §10.21(ii) Permalink: http://dlmf.nist.gov/10.21.E2 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 10.21(i), 10.21 and 10 For an extension see Pálmai and Apagyi (2011). The positive zeros of any two real distinct cylinder functions of the same order are interlaced, as are the positive zeros of any real cylinder function $\mathscr{C}_{\nu}\left(z\right)$ and the contiguous function $\mathscr{C}_{\nu+1}\left(z\right)$. See also Elbert and Laforgia (1994). When $\nu\geq-1$ the zeros of $J_{\nu}\left(z\right)$ are all real. If $\nu<-1$ and $\nu$ is not an integer, then the number of complex zeros of $J_{\nu}\left(z\right)$ is $2\left\lfloor-\nu\right\rfloor$. If $\left\lfloor-\nu\right\rfloor$ is odd, then two of these zeros lie on the imaginary axis. If $\nu\geq 0$, then the zeros of $J_{\nu}'\left(z\right)$ are all real. For information on the real double zeros of $J_{\nu}'\left(z\right)$ and $Y_{\nu}'\left(z\right)$ when $\nu<-1$ and $\nu<-\tfrac{1}{2}$, respectively, see Döring (1971) and Kerimov and Skorokhodov (1986). The latter reference also has information on double zeros of the second and third derivatives of $J_{\nu}\left(z\right)$ and $Y_{\nu}\left(z\right)$. No two of the functions $J_{0}\left(z\right)$, $J_{1}\left(z\right)$, $J_{2}\left(z\right),\ldots$, have any common zeros other than $z=0$; see Watson (1944, §15.28). ## §10.21(ii) Analytic Properties If $\rho_{\nu}$ is a zero of the cylinder function 10.21.4 $\mathscr{C}_{\nu}\left(z\right)=J_{\nu}\left(z\right)\cos\left(\pi t\right)+Y_% {\nu}\left(z\right)\sin\left(\pi t\right),$ where $t$ is a parameter, then 10.21.5 $\mathscr{C}_{\nu}'\left(\rho_{\nu}\right)=\mathscr{C}_{\nu-1}\left(\rho_{\nu}% \right)=-\mathscr{C}_{\nu+1}\left(\rho_{\nu}\right).$ ⓘ Symbols: $\mathscr{C}_{\NVar{\nu}}\left(\NVar{z}\right)$: cylinder function, $\nu$: complex parameter and $\rho_{\nu}(t)$: zero of cylinder function A&S Ref: 9.5.4 Referenced by: §10.21(ii) Permalink: http://dlmf.nist.gov/10.21.E5 Encodings: TeX, pMML, png See also: Annotations for 10.21(ii), 10.21 and 10 If $\sigma_{\nu}$ is a zero of $\mathscr{C}_{\nu}'\left(z\right)$, then 10.21.6 $\mathscr{C}_{\nu}\left(\sigma_{\nu}\right)=\frac{\sigma_{\nu}}{\nu}\mathscr{C}% _{\nu-1}\left(\sigma_{\nu}\right)=\frac{\sigma_{\nu}}{\nu}\mathscr{C}_{\nu+1}% \left(\sigma_{\nu}\right).$ ⓘ Symbols: $\mathscr{C}_{\NVar{\nu}}\left(\NVar{z}\right)$: cylinder function, $\nu$: complex parameter and $\sigma_{\nu}(t)$: zero of cylinder function A&S Ref: 9.5.5 Referenced by: §10.21(ii) Permalink: http://dlmf.nist.gov/10.21.E6 Encodings: TeX, pMML, png See also: Annotations for 10.21(ii), 10.21 and 10 The parameter $t$ may be regarded as a continuous variable and $\rho_{\nu}$, $\sigma_{\nu}$ as functions $\rho_{\nu}(t)$, $\sigma_{\nu}(t)$ of $t$. If $\nu\geq 0$ and these functions are fixed by 10.21.7 $\displaystyle\rho_{\nu}(0)$ $\displaystyle=0,$ $\displaystyle\sigma_{\nu}(0)$ $\displaystyle={j^{\prime}_{\nu,1}},$ then 10.21.8 $\displaystyle j_{\nu,m}$ $\displaystyle=\rho_{\nu}(m),$ $\displaystyle y_{\nu,m}$ $\displaystyle=\rho_{\nu}(m-\tfrac{1}{2})$, $m=1,2,\ldots$, 10.21.9 $\displaystyle{j^{\prime}_{\nu,m}}$ $\displaystyle=\sigma_{\nu}(m-1),$ $\displaystyle{y^{\prime}_{\nu,m}}$ $\displaystyle=\sigma_{\nu}(m-\tfrac{1}{2})$, $m=1,2,\ldots$. 10.21.10 $\displaystyle\mathscr{C}_{\nu}'\left(\rho_{\nu}\right)$ $\displaystyle=\left(\frac{\rho_{\nu}}{2}\frac{\mathrm{d}\rho_{\nu}}{\mathrm{d}% t}\right)^{-\frac{1}{2}},$ $\displaystyle\mathscr{C}_{\nu}\left(\sigma_{\nu}\right)$ $\displaystyle=\left(\frac{\sigma_{\nu}^{2}-\nu^{2}}{2\sigma_{\nu}}\frac{% \mathrm{d}\sigma_{\nu}}{\mathrm{d}t}\right)^{-\frac{1}{2}},$ 10.21.11 $2\rho_{\nu}^{2}\frac{\mathrm{d}\rho_{\nu}}{\mathrm{d}t}\frac{{\mathrm{d}}^{3}% \rho_{\nu}}{{\mathrm{d}t}^{3}}-3\rho_{\nu}^{2}\*\left(\frac{{\mathrm{d}}^{2}% \rho_{\nu}}{{\mathrm{d}t}^{2}}\right)^{2}-4\pi^{2}\rho_{\nu}^{2}\*\left(\frac{% \mathrm{d}\rho_{\nu}}{\mathrm{d}t}\right)^{2}+(4\rho_{\nu}^{2}+1-4\nu^{2})% \left(\frac{\mathrm{d}\rho_{\nu}}{\mathrm{d}t}\right)^{4}=0.$ The functions $\rho_{\nu}(t)$ and $\sigma_{\nu}(t)$ are related to the inverses of the phase functions $\theta_{\nu}\left(x\right)$ and $\phi_{\nu}\left(x\right)$ defined in §10.18(i): if $\nu\geq 0$, then 10.21.12 $\displaystyle\theta_{\nu}\left(j_{\nu,m}\right)$ $\displaystyle=(m-\tfrac{1}{2})\pi,$ $\displaystyle\theta_{\nu}\left(y_{\nu,m}\right)$ $\displaystyle=(m-1)\pi,$ $m=1,2,\ldots$, 10.21.13 $\displaystyle\phi_{\nu}\left({j^{\prime}_{\nu,m}}\right)$ $\displaystyle=(m-\tfrac{1}{2})\pi,$ $\displaystyle\phi_{\nu}\left({y^{\prime}_{\nu,m}}\right)$ $\displaystyle=m\pi,$ $m=1,2,\ldots$. For sign properties of the forward differences that are defined by 10.21.14 $\displaystyle\Delta\rho_{\nu}(t)$ $\displaystyle=\rho_{\nu}(t+1)-\rho_{\nu}(t),$ $\displaystyle\Delta^{2}\rho_{\nu}(t)$ $\displaystyle=\Delta\rho_{\nu}(t+1)-\Delta\rho_{\nu}(t),\ldots,$ ⓘ Symbols: $\Delta$: forward difference operator, $\nu$: complex parameter and $\rho_{\nu}(t)$: zero of cylinder function Permalink: http://dlmf.nist.gov/10.21.E14 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 10.21(ii), 10.21 and 10 when $t=1,2,3,\ldots$, and similarly for $\sigma_{\nu}(t)$, see Lorch and Szegő (1963, 1964), Lorch et al. (1970, 1972), and Muldoon (1977). Some information on the distribution of $\rho_{\nu}(t)$ and $\sigma_{\nu}(t)$ for real values of $\nu$ and $t$ is given in Muldoon and Spigler (1984). ## §10.21(iii) Infinite Products 10.21.15 $\displaystyle J_{\nu}\left(z\right)$ $\displaystyle=\frac{(\tfrac{1}{2}z)^{\nu}}{\Gamma\left(\nu+1\right)}\prod_{k=1% }^{\infty}\left(1-\frac{z^{2}}{{j_{\nu,k}^{2}}}\right),$ $\nu\geq 0$, 10.21.16 $\displaystyle J_{\nu}'\left(z\right)$ $\displaystyle=\frac{(\tfrac{1}{2}z)^{\nu-1}}{2\Gamma\left(\nu\right)}\prod_{k=% 1}^{\infty}\left(1-\frac{z^{2}}{{{j^{\prime}_{\nu,k}}^{\mspace{-16.0mu }2}}}% \right),$ $\nu>0$. ## §10.21(iv) Monotonicity Properties Any positive zero $c$ of the cylinder function $\mathscr{C}_{\nu}\left(x\right)$ and any positive zero $c^{\prime}$ of $\mathscr{C}_{\nu}'\left(x\right)$ such that $c^{\prime}>|\nu|$ are definable as continuous and increasing functions of $\nu$: 10.21.17 $\frac{\mathrm{d}c}{\mathrm{d}\nu}=2c\int_{0}^{\infty}K_{0}(2c\sinh t)e^{-2\nu t% }\mathrm{d}t,$ 10.21.18 $\frac{\mathrm{d}c^{\prime}}{\mathrm{d}\nu}=\frac{2c^{\prime}}{{c^{\prime}}^{2}% -\nu^{2}}\*\int_{0}^{\infty}({c^{\prime}}^{2}\cosh(2t)-\nu^{2})\*K_{0}(2c^{% \prime}\sinh t)e^{-2\nu t}\mathrm{d}t,$ where $K_{0}$ is defined in §10.25(ii). In particular, $j_{\nu,m}$, $y_{\nu,m}$, $j_{\nu,m}'$, and $y_{\nu,m}'$ are increasing functions of $\nu$ when $\nu\geq 0$. It is also true that the positive zeros $j^{\prime\prime}_{\nu}$ and $j^{\prime\prime\prime}_{\nu}$ of $J_{\nu}''\left(x\right)$ and $J_{\nu}'''\left(x\right)$, respectively, are increasing functions of $\nu$ when $\nu>0$, provided that in the latter case $j^{\prime\prime\prime}_{\nu}>\sqrt{3}$ when $0<\nu<1$. $j_{\nu,m}/\nu$ and $j_{\nu,m}'/\nu$ are decreasing functions of $\nu$ when $\nu>0$ for $m=1,2,3,\ldots$. For further monotonicity properties see Elbert (2001), Lorch (1990, 1993, 1995), Lorch and Muldoon (2008), Lorch and Szegő (1990, 1995), and Muldoon (1981). For inequalities for zeros arising from monotonicity properties see Laforgia and Muldoon (1983). ## §10.21(v) Inequalities For bounds for the smallest real or purely imaginary zeros of $J_{\nu}\left(x\right)$ when $\nu$ is real see Ismail and Muldoon (1995). ## §10.21(vi) McMahon’s Asymptotic Expansions for Large Zeros If $\nu$ $(\geq 0)$ is fixed, $\mu=4\nu^{2}$, and $m\to\infty$, then 10.21.19 $j_{\nu,m},y_{\nu,m}\sim a-\frac{\mu-1}{8a}-\frac{4(\mu-1)(7\mu-31)}{3(8a)^{3}}% -\frac{32(\mu-1)(83\mu^{2}-982\mu+3779)}{15(8a)^{5}}-\frac{64(\mu-1)(6949\mu^{% 3}-1\;53855\mu^{2}+15\;85743\mu-62\;77237)}{105(8a)^{7}}-\cdots,$ where $a=(m+\tfrac{1}{2}\nu-\tfrac{1}{4})\pi$ for $j_{\nu,m}$, $a=(m+\tfrac{1}{2}\nu-\tfrac{3}{4})\pi$ for $y_{\nu,m}$. With $a=(t+\tfrac{1}{2}\nu-\tfrac{1}{4})\pi$, the right-hand side is the asymptotic expansion of $\rho_{\nu}(t)$ for large $t$. 10.21.20 $j_{\nu,m}',y_{\nu,m}'\sim b-\frac{\mu+3}{8b}-\frac{4(7\mu^{2}+82\mu-9)}{3(8b)^% {3}}-\frac{32(83\mu^{3}+2075\mu^{2}-3039\mu+3537)}{15(8b)^{5}}-\frac{64(6949% \mu^{4}+2\;96492\mu^{3}-12\;48002\mu^{2}+74\;14380\mu-58\;53627)}{105(8b)^{7}}% -\cdots,$ where $b=(m+\tfrac{1}{2}\nu-\tfrac{3}{4})\pi$ for $j_{\nu,m}'$, $b=(m+\tfrac{1}{2}\nu-\tfrac{1}{4})\pi$ for $y_{\nu,m}'$, and $b=(t+\tfrac{1}{2}\nu+\tfrac{1}{4})\pi$ for $\sigma_{\nu}(t)$. For the next three terms in (10.21.19) and the next two terms in (10.21.20) see Bickley et al. (1952, p. xxxvii) or Olver (1960, pp. xvii–xviii). For error bounds see Wong and Lang (1990), Wong (1995), and Elbert and Laforgia (2000). See also Laforgia (1979). For the $m$th positive zero $j^{\prime\prime}_{\nu,m}$ of $J_{\nu}''\left(x\right)$ Wong and Lang (1990) gives the corresponding expansion 10.21.21 $j^{\prime\prime}_{\nu,m}\sim c-\frac{\mu+7}{8c}-\frac{28\mu^{2}+424\mu+1724}{3% (8c)^{3}}-\cdots,$ ⓘ Symbols: $\sim$: Poincaré asymptotic expansion, $m$: integer, $\nu$: complex parameter and $c$: zero of cylinder function Permalink: http://dlmf.nist.gov/10.21.E21 Encodings: TeX, pMML, png See also: Annotations for 10.21(vi), 10.21 and 10 where $c=(m+\tfrac{1}{2}\nu-\tfrac{1}{4})\pi$ if $0<\nu<1$, and $c=(m+\tfrac{1}{2}\nu-\tfrac{5}{4})\pi$ if $\nu>1$. An error bound is included for the case $\nu\geq\tfrac{3}{2}$. ## §10.21(vii) Asymptotic Expansions for Large Order Let $\mathscr{C}_{\nu}\left(x\right)$, $\rho_{\nu}(t)$, and $\sigma_{\nu}(t)$ be defined as in §10.21(ii) and $M\left(x\right)$, $\theta\left(x\right)$, $N\left(x\right)$, and $\phi\left(x\right)$ denote the modulus and phase functions for the Airy functions and their derivatives as in §9.8. As $\nu\to\infty$ with $t$ $(>0)$ fixed, 10.21.22 $\rho_{\nu}(t)\sim\nu\sum_{k=0}^{\infty}\frac{\alpha_{k}}{\nu^{2k/3}},$ 10.21.23 $\mathscr{C}_{\nu}'\left(\rho_{\nu}(t)\right)\sim\frac{(2/\nu)^{\frac{2}{3}}}{% \pi M\left(-2^{\frac{1}{3}}\alpha\right)}\sum_{k=0}^{\infty}\frac{\beta_{k}}{% \nu^{2k/3}},$ where $\alpha$ is given by 10.21.24 $\theta\left(-2^{\frac{1}{3}}\alpha\right)=\pi t,$ ⓘ Defines: $\alpha$: coefficients (locally) Symbols: $\pi$: the ratio of the circumference of a circle to its diameter and $\theta\left(\NVar{z}\right)$: Airy phase function Permalink: http://dlmf.nist.gov/10.21.E24 Encodings: TeX, pMML, png See also: Annotations for 10.21(vii), 10.21 and 10 and 10.21.25 $\displaystyle\alpha_{0}$ $\displaystyle=1,$ $\displaystyle\alpha_{1}$ $\displaystyle=\alpha,$ $\displaystyle\alpha_{2}$ $\displaystyle=\tfrac{3}{10}\alpha^{2},$ $\displaystyle\alpha_{3}$ $\displaystyle=-\tfrac{1}{350}\alpha^{3}+\tfrac{1}{70},$ $\displaystyle\alpha_{4}$ $\displaystyle=-\tfrac{479}{63000}\alpha^{4}-\tfrac{1}{3150}\alpha,$ $\displaystyle\alpha_{5}$ $\displaystyle=\tfrac{20231}{80\;85000}\alpha^{5}-\tfrac{551}{1\;61700}\alpha^{% 2},$ ⓘ Symbols: $\alpha$: coefficients Referenced by: §10.21(vii) Permalink: http://dlmf.nist.gov/10.21.E25 Encodings: TeX, TeX, TeX, TeX, TeX, TeX, pMML, pMML, pMML, pMML, pMML, pMML, png, png, png, png, png, png See also: Annotations for 10.21(vii), 10.21 and 10 10.21.26 $\displaystyle\beta_{0}$ $\displaystyle=1,$ $\displaystyle\beta_{1}$ $\displaystyle=-\tfrac{4}{5}\alpha,$ $\displaystyle\beta_{2}$ $\displaystyle=\tfrac{18}{35}\alpha^{2},$ $\displaystyle\beta_{3}$ $\displaystyle=-\tfrac{88}{315}\alpha^{3}-\tfrac{11}{1575},$ $\displaystyle\beta_{4}$ $\displaystyle=\tfrac{79586}{6\;06375}\alpha^{4}+\tfrac{9824}{6\;06375}\alpha.$ ⓘ Defines: $\beta$: coefficients (locally) Symbols: $\alpha$: coefficients Referenced by: §10.21(vii) Permalink: http://dlmf.nist.gov/10.21.E26 Encodings: TeX, TeX, TeX, TeX, TeX, pMML, pMML, pMML, pMML, pMML, png, png, png, png, png See also: Annotations for 10.21(vii), 10.21 and 10 As $\nu\to\infty$ with $t$ $(>-\tfrac{1}{6})$ fixed, 10.21.27 $\sigma_{\nu}(t)\sim\nu\sum_{k=0}^{\infty}\frac{\alpha^{\prime}_{k}}{\nu^{2k/3}},$ 10.21.28 $\mathscr{C}_{\nu}\left(\sigma_{\nu}(t)\right)\sim\frac{(2/\nu)^{\frac{1}{3}}}{% \pi N\left(-2^{\frac{1}{3}}\alpha^{\prime}\right)}\sum_{k=0}^{\infty}\frac{% \beta^{\prime}_{k}}{\nu^{2k/3}},$ where $\alpha^{\prime}$ is given by 10.21.29 $\phi\left(-2^{\frac{1}{3}}\alpha^{\prime}\right)=\pi t,$ ⓘ Defines: $\alpha^{\prime}$: coefficients (locally) Symbols: $\pi$: the ratio of the circumference of a circle to its diameter and $\phi\left(\NVar{z}\right)$: Airy phase function Permalink: http://dlmf.nist.gov/10.21.E29 Encodings: TeX, pMML, png See also: Annotations for 10.21(vii), 10.21 and 10 and 10.21.30 $\displaystyle\alpha^{\prime}_{0}$ $\displaystyle=1,$ $\displaystyle\alpha^{\prime}_{1}$ $\displaystyle=\alpha^{\prime},$ $\displaystyle\alpha^{\prime}_{2}$ $\displaystyle=\tfrac{3}{10}{\alpha^{\prime}}^{2}-\tfrac{1}{10}{\alpha^{\prime}% }^{-1},$ $\displaystyle\alpha^{\prime}_{3}$ $\displaystyle=-\tfrac{1}{350}{\alpha^{\prime}}^{3}-\tfrac{1}{25}-\tfrac{1}{200% }{\alpha^{\prime}}^{-3},$ $\displaystyle\alpha^{\prime}_{4}$ $\displaystyle=-\tfrac{479}{63000}{\alpha^{\prime}}^{4}+\tfrac{509}{31500}% \alpha^{\prime}+\tfrac{1}{1500}{\alpha^{\prime}}^{-2}-\tfrac{1}{2000}{\alpha^{% \prime}}^{-5},$ ⓘ Symbols: $\alpha^{\prime}$: coefficients Referenced by: §10.21(vii) Permalink: http://dlmf.nist.gov/10.21.E30 Encodings: TeX, TeX, TeX, TeX, TeX, pMML, pMML, pMML, pMML, pMML, png, png, png, png, png See also: Annotations for 10.21(vii), 10.21 and 10 10.21.31 $\displaystyle\beta^{\prime}_{0}$ $\displaystyle=1,$ $\displaystyle\beta^{\prime}_{1}$ $\displaystyle=-\tfrac{1}{5}\alpha^{\prime},$ $\displaystyle\beta^{\prime}_{2}$ $\displaystyle=\tfrac{9}{350}{\alpha^{\prime}}^{2}+\tfrac{1}{100}{\alpha^{% \prime}}^{-1},$ $\displaystyle\beta^{\prime}_{3}$ $\displaystyle=\tfrac{89}{15750}{\alpha^{\prime}}^{3}-\tfrac{47}{4500}+\tfrac{1% }{3000}{\alpha^{\prime}}^{-3}.$ ⓘ Symbols: $\beta$: coefficients and $\alpha^{\prime}$: coefficients Referenced by: §10.21(vii) Permalink: http://dlmf.nist.gov/10.21.E31 Encodings: TeX, TeX, TeX, TeX, pMML, pMML, pMML, pMML, png, png, png, png See also: Annotations for 10.21(vii), 10.21 and 10 In particular, with the notation as below, 10.21.32 $j_{\nu,m}\sim\nu\sum_{k=0}^{\infty}\frac{\alpha_{k}}{\nu^{2k/3}},$ 10.21.33 $y_{\nu,m}\sim\nu\sum_{k=0}^{\infty}\frac{\alpha_{k}}{\nu^{2k/3}},$ 10.21.34 $J_{\nu}'\left(j_{\nu,m}\right)\sim(-1)^{m}\frac{(2/\nu)^{\frac{2}{3}}}{\pi M% \left(a_{m}\right)}\sum_{k=0}^{\infty}\frac{\beta_{k}}{\nu^{2k/3}},$ 10.21.35 $Y_{\nu}'\left(y_{\nu,m}\right)\sim(-1)^{m-1}\frac{(2/\nu)^{\frac{2}{3}}}{\pi M% \left(b_{m}\right)}\sum_{k=0}^{\infty}\frac{\beta_{k}}{\nu^{2k/3}},$ and 10.21.36 $j_{\nu,m}'\sim\nu\sum_{k=0}^{\infty}\frac{\alpha^{\prime}_{k}}{\nu^{2k/3}},$ 10.21.37 $y_{\nu,m}'\sim\nu\sum_{k=0}^{\infty}\frac{\alpha^{\prime}_{k}}{\nu^{2k/3}},$ 10.21.38 $J_{\nu}\left(j_{\nu,m}'\right)\sim(-1)^{m-1}\frac{(2/\nu)^{\frac{1}{3}}}{\pi N% \left(a^{\prime}_{m}\right)}\sum_{k=0}^{\infty}\frac{\beta^{\prime}_{k}}{\nu^{% 2k/3}},$ 10.21.39 $Y_{\nu}\left(y_{\nu,m}'\right)\sim(-1)^{m-1}\frac{(2/\nu)^{\frac{1}{3}}}{\pi N% \left(b^{\prime}_{m}\right)}\sum_{k=0}^{\infty}\frac{\beta^{\prime}_{k}}{\nu^{% 2k/3}}.$ Here $a_{m}$, $b_{m}$, $a^{\prime}_{m}$, $b^{\prime}_{m}$ are the $m$th negative zeros of $\mathrm{Ai}\left(x\right)$, $\mathrm{Bi}\left(x\right)$, $\mathrm{Ai}'\left(x\right)$, $\mathrm{Bi}'\left(x\right)$, respectively (§9.9), $\alpha_{k}$, $\beta_{k}$, $\alpha^{\prime}_{k}$, $\beta^{\prime}_{k}$ are given by (10.21.25), (10.21.26), (10.21.30), and (10.21.31), with $\alpha=-2^{-\frac{1}{3}}a_{m}$ in the case of $j_{\nu,m}$ and $J_{\nu}'\left(j_{\nu,m}\right)$, $\alpha=-2^{-\frac{1}{3}}b_{m}$ in the case of $y_{\nu,m}$ and $Y_{\nu}'\left(y_{\nu,m}\right)$, $\alpha^{\prime}=-2^{-\frac{1}{3}}a^{\prime}_{m}$ in the case of $j_{\nu,m}'$ and $J_{\nu}\left(j_{\nu,m}'\right)$, $\alpha^{\prime}=-2^{-\frac{1}{3}}b^{\prime}_{m}$ in the case of $y_{\nu,m}'$ and $Y_{\nu}\left(y_{\nu,m}'\right)$. For error bounds for (10.21.32) see Qu and Wong (1999); for (10.21.36) and (10.21.37) see Elbert and Laforgia (1997). See also Spigler (1980). For the first zeros rounded numerical values of the coefficients are given by 10.21.40 $\displaystyle j_{\nu,1}$ $\displaystyle\sim\nu+1.85575\;71\nu^{\frac{1}{3}}+1.03315\;0\nu^{-\frac{1}{3}}% -0.00397\nu^{-1}-0.0908\nu^{-\frac{5}{3}}+0.043\nu^{-\frac{7}{3}}+\cdots,$ $\displaystyle y_{\nu,1}$ $\displaystyle\sim\nu+0.93157\;68\nu^{\frac{1}{3}}+0.26035\;1\nu^{-\frac{1}{3}}% +0.01198\nu^{-1}-0.0060\nu^{-\frac{5}{3}}-0.001\nu^{-\frac{7}{3}}+\cdots,$ $\displaystyle J_{\nu}'\left(j_{\nu,1}\right)$ $\displaystyle\sim-1.11310\;28\nu^{-\frac{2}{3}}\div(1+1.48460\;6\nu^{-\frac{2}% {3}}+0.43294\nu^{-\frac{4}{3}}-0.1943\nu^{-2}+0.019\nu^{-\frac{8}{3}}+\cdots),$ $\displaystyle Y_{\nu}'\left(y_{\nu,1}\right)$ $\displaystyle\sim 0.95554\;86\nu^{-\frac{2}{3}}\div(1+0.74526\;1\nu^{-\frac{2}% {3}}+0.10910\nu^{-\frac{4}{3}}-0.0185\nu^{-2}-0.003\nu^{-\frac{8}{3}}+\cdots),$ $\displaystyle{j^{\prime}_{\nu,1}}$ $\displaystyle\sim\nu+0.80861\;65\nu^{\frac{1}{3}}+0.07249\;0\nu^{-\frac{1}{3}}% -0.05097\nu^{-1}+0.0094\nu^{-\frac{5}{3}}+\cdots,$ $\displaystyle{y^{\prime}_{\nu,1}}$ $\displaystyle\sim\nu+1.82109\;80\nu^{\frac{1}{3}}+0.94000\;7\nu^{-\frac{1}{3}}% -0.05808\nu^{-1}-0.0540\nu^{-\frac{5}{3}}+\cdots.$ $\displaystyle J_{\nu}\left(j_{\nu,1}'\right)$ $\displaystyle\sim 0.67488\;51\nu^{-\frac{1}{3}}(1-0.16172\;3\nu^{-\frac{2}{3}}% +0.02918\nu^{-\frac{4}{3}}-0.0068\nu^{-2}+\cdots),$ $\displaystyle Y_{\nu}\left(y_{\nu,1}'\right)$ $\displaystyle\sim 0.57319\;40\nu^{-\frac{1}{3}}(1-0.36422\;0\nu^{-\frac{2}{3}}% +0.09077\nu^{-\frac{4}{3}}+0.0237\nu^{-2}+\cdots).$ For numerical coefficients for $m=2,3,4,5$ see Olver (1951, Tables 3–6). The expansions (10.21.32)–(10.21.39) become progressively weaker as $m$ increases. The approximations that follow in §10.21(viii) do not suffer from this drawback. ## §10.21(viii) Uniform Asymptotic Approximations for Large Order As $\nu\to\infty$ the following four approximations hold uniformly for $m=1,2,\ldots$: 10.21.41 $j_{\nu,m}=\nu z(\zeta)+\frac{z(\zeta)(h(\zeta))^{2}B_{0}(\zeta)}{2\nu}+O\left(% \frac{1}{\nu^{3}}\right),$ $\zeta=\nu^{-\frac{2}{3}}a_{m}$, 10.21.42 $J_{\nu}'\left(j_{\nu,m}\right)=-\frac{2}{\nu^{\frac{2}{3}}}\frac{\mathrm{Ai}'% \left(a_{m}\right)}{z(\zeta)h(\zeta)}\left(1+O\left(\frac{1}{\nu^{2}}\right)% \right),$ $\zeta=\nu^{-\frac{2}{3}}a_{m}$, 10.21.43 $j_{\nu,m}'=\nu z(\zeta)+\frac{z(\zeta)(h(\zeta))^{2}C_{0}(\zeta)}{2\zeta\nu}+O% \left(\frac{1}{\nu}\right),$ $\zeta=\nu^{-\frac{2}{3}}a^{\prime}_{m}$, 10.21.44 $J_{\nu}\left(j_{\nu,m}'\right)=\frac{h(\zeta)\mathrm{Ai}\left(a^{\prime}_{m}% \right)}{\nu^{\frac{1}{3}}}\left(1+O\left(\frac{1}{\nu^{\frac{4}{3}}}\right)% \right),$ $\zeta=\nu^{-\frac{2}{3}}a^{\prime}_{m}$. Here $a_{m}$ and $a^{\prime}_{m}$ denote respectively the zeros of the Airy function $\mathrm{Ai}\left(z\right)$ and its derivative $\mathrm{Ai}'\left(z\right)$; see §9.9. Next, $z(\zeta)$ is the inverse of the function $\zeta=\zeta(z)$ defined by (10.20.3). $B_{0}(\zeta)$ and $C_{0}(\zeta)$ are defined by (10.20.11) and (10.20.12) with $k=0$. Lastly, 10.21.45 $h(\zeta)=\left(4\zeta/(1-z^{2})\right)^{\frac{1}{4}}.$ ⓘ Defines: $h(\zeta)$: function (locally) Symbols: $z(\zeta)$: inverse of $\zeta(z)$ A&S Ref: 9.5.36 Permalink: http://dlmf.nist.gov/10.21.E45 Encodings: TeX, pMML, png See also: Annotations for 10.21(viii), 10.21 and 10 (Note: If the term $z(\zeta)(h(\zeta))^{2}C_{0}(\zeta)/(2\zeta\nu)$ in (10.21.43) is omitted, then the uniform character of the error term $O(\ifrac{1}{\nu})$ is destroyed.) Corresponding uniform approximations for $y_{\nu,m}$, $Y_{\nu}'\left(y_{\nu,m}\right)$, $y_{\nu,m}'$, and $Y_{\nu}\left({y^{\prime}_{\nu,m}}\right)$, are obtained from (10.21.41)–(10.21.44) by changing the symbols $j$, $J$, $\mathrm{Ai}$, $\mathrm{Ai}'$, $a_{m}$, and $a^{\prime}_{m}$ to $y$, $Y$, $-\mathrm{Bi}$, $-\mathrm{Bi}'$, $b_{m}$, and $b^{\prime}_{m}$, respectively. For derivations and further information, including extensions to uniform asymptotic expansions, see Olver (1954, 1960). The latter reference includes numerical tables of the first few coefficients in the uniform asymptotic expansions. ## §10.21(ix) Complex Zeros This subsection describes the distribution in $\mathbb{C}$ of the zeros of the principal branches of the Bessel functions of the second and third kinds, and their derivatives, in the case when the order is a positive integer $n$. For further information, including uniform asymptotic expansions, extensions to other branches of the functions and their derivatives, and extensions to half-integer values of the order, see Olver (1954). (There is an inaccuracy in Figures 11 and 14 in this reference. Each curve that represents an infinite string of nonreal zeros should be located on the opposite side of its straight line asymptote. This inaccuracy was repeated in Abramowitz and Stegun (1964, Figures 9.5 and 9.6). See Kerimov and Skorokhodov (1985a, b) and Figures 10.21.310.21.6.) See also Cruz and Sesma (1982), Cruz et al. (1991), Kerimov and Skorokhodov (1984c, 1987, 1988), Kokologiannaki et al. (1992), and references supplied in §10.75(iii). For describing the distribution of complex zeros by methods based on the Liouville-Green (WKB) approximation for linear homogeneous second-order differential equations, see Segura (2013). ### Zeros of $Y_{n}\left(nz\right)$ and $Y_{n}'\left(nz\right)$ In Figures 10.21.1, 10.21.3, and 10.21.5 the two continuous curves that join the points $\pm 1$ are the boundaries of $\mathbf{K}$, that is, the eye-shaped domain depicted in Figure 10.20.3. These curves therefore intersect the imaginary axis at the points $z=\pm\mathrm{i}c$, where $c=0.66274\ldots$. The first set of zeros of the principal value of $Y_{n}\left(nz\right)$ are the points $z=y_{n,m}/n$, $m=1,2,\ldots$, on the positive real axis (§10.21(i)). Secondly, there is a conjugate pair of infinite strings of zeros with asymptotes $\Im z=\pm\mathrm{i}a/n$, where 10.21.46 $a=\tfrac{1}{2}\ln 3=0.54931\ldots.$ ⓘ Symbols: $\ln\NVar{z}$: principal branch of logarithm function Permalink: http://dlmf.nist.gov/10.21.E46 Encodings: TeX, pMML, png See also: Annotations for 10.21(ix), 10.21(ix), 10.21 and 10 Lastly, there are two conjugate sets, with $n$ zeros in each set, that are asymptotically close to the boundary of $\mathbf{K}$ as $n\to\infty$. Figures 10.21.1, 10.21.3, and 10.21.5 plot the actual zeros for $n=1,5$, and $10$, respectively. The zeros of $Y_{n}'\left(nz\right)$ have a similar pattern to those of $Y_{n}\left(nz\right)$. ### Zeros of ${H^{(1)}_{n}}\left(nz\right)$, ${H^{(2)}_{n}}\left(nz\right)$, ${H^{(1)}_{n}}'\left(nz\right)$, ${H^{(2)}_{n}}'\left(nz\right)$ In Figures 10.21.2, 10.21.4, and 10.21.6 the continuous curve that joins the points $\pm 1$ is the lower boundary of $\mathbf{K}$. The first set of zeros of the principal value of ${H^{(1)}_{n}}\left(nz\right)$ is an infinite string with asymptote $\Im z=-\mathrm{i}d/n$, where 10.21.47 $d=\tfrac{1}{2}\ln 2=0.34657\ldots.$ ⓘ Defines: $d$ (locally) Symbols: $\ln\NVar{z}$: principal branch of logarithm function Permalink: http://dlmf.nist.gov/10.21.E47 Encodings: TeX, pMML, png See also: Annotations for 10.21(ix), 10.21(ix), 10.21 and 10 The only other set comprises $n$ zeros that are asymptotically close to the lower boundary of $\mathbf{K}$ as $n\to\infty$. Figures 10.21.2, 10.21.4, and 10.21.6 plot the actual zeros for $n=1,5$, and $10$, respectively. The zeros of ${H^{(1)}_{n}}'\left(nz\right)$ have a similar pattern to those of ${H^{(1)}_{n}}\left(nz\right)$. The zeros of ${H^{(2)}_{n}}\left(nz\right)$ and ${H^{(2)}_{n}}'\left(nz\right)$ are the complex conjugates of the zeros of ${H^{(1)}_{n}}\left(nz\right)$ and ${H^{(1)}_{n}}'\left(nz\right)$, respectively. ### Zeros of $J_{0}\left(z\right)-\mathrm{i}J_{1}\left(z\right)$ and $J_{n}\left(z\right)-\mathrm{i}J_{n+1}\left(z\right)$ For information see Synolakis (1988), MacDonald (1989, 1997), and Ikebe et al. (1993). ## §10.21(x) Cross-Products Throughout this subsection we assume $\nu\geq 0$, $x>0$, $\lambda>1$, and we denote $4\nu^{2}$ by $\mu$. The zeros of the functions 10.21.48 $J_{\nu}\left(x\right)Y_{\nu}\left(\lambda x\right)-Y_{\nu}\left(x\right)J_{\nu% }\left(\lambda x\right)$ ⓘ Symbols: $J_{\NVar{\nu}}\left(\NVar{z}\right)$: Bessel function of the first kind, $Y_{\NVar{\nu}}\left(\NVar{z}\right)$: Bessel function of the second kind, $x$: real variable and $\nu$: complex parameter A&S Ref: 9.5.27 Referenced by: §10.21(x) Permalink: http://dlmf.nist.gov/10.21.E48 Encodings: TeX, pMML, png See also: Annotations for 10.21(x), 10.21 and 10 and 10.21.49 $J_{\nu}'\left(x\right)Y_{\nu}'\left(\lambda x\right)-Y_{\nu}'\left(x\right)J_{% \nu}'\left(\lambda x\right)$ ⓘ Symbols: $J_{\NVar{\nu}}\left(\NVar{z}\right)$: Bessel function of the first kind, $Y_{\NVar{\nu}}\left(\NVar{z}\right)$: Bessel function of the second kind, $x$: real variable and $\nu$: complex parameter A&S Ref: 9.5.30 Referenced by: §10.21(x) Permalink: http://dlmf.nist.gov/10.21.E49 Encodings: TeX, pMML, png See also: Annotations for 10.21(x), 10.21 and 10 are simple and the asymptotic expansion of the $m$th positive zero as $m\to\infty$ is given by 10.21.50 $\alpha+\frac{p}{\alpha}+\frac{q-p^{2}}{\alpha^{3}}+\frac{r-4pq+2p^{3}}{\alpha^% {5}}+\cdots,$ ⓘ Symbols: $\alpha$, $p$, $q$ and $r$ A&S Ref: 9.5.28 Referenced by: §10.21(x) Permalink: http://dlmf.nist.gov/10.21.E50 Encodings: TeX, pMML, png See also: Annotations for 10.21(x), 10.21 and 10 where, in the case of (10.21.48), 10.21.51 $\displaystyle\alpha$ $\displaystyle=\frac{m\pi}{\lambda-1},$ $\displaystyle p$ $\displaystyle=\frac{\mu-1}{8\lambda},$ $\displaystyle q$ $\displaystyle=\frac{(\mu-1)(\mu-25)(\lambda^{3}-1)}{6(4\lambda)^{3}(\lambda-1)},$ $\displaystyle r$ $\displaystyle=\frac{(\mu-1)(\mu^{2}-114\mu+1073)(\lambda^{5}-1)}{5(4\lambda)^{% 5}(\lambda-1)},$ ⓘ Defines: $\alpha$ (locally), $p$ (locally), $q$ (locally) and $r$ (locally) Symbols: $\pi$: the ratio of the circumference of a circle to its diameter and $m$: integer A&S Ref: 9.5.29 Permalink: http://dlmf.nist.gov/10.21.E51 Encodings: TeX, TeX, TeX, TeX, pMML, pMML, pMML, pMML, png, png, png, png See also: Annotations for 10.21(x), 10.21 and 10 and, in the case of (10.21.49), 10.21.52 $\displaystyle\alpha$ $\displaystyle=\frac{(m-1)\pi}{\lambda-1},$ $\displaystyle p$ $\displaystyle=\frac{\mu+3}{8\lambda},$ $\displaystyle q$ $\displaystyle=\frac{(\mu^{2}+46\mu-63)(\lambda^{3}-1)}{6(4\lambda)^{3}(\lambda% -1)},$ $\displaystyle r$ $\displaystyle=\frac{(\mu^{3}+185\mu^{2}-2053\mu+1899)(\lambda^{5}-1)}{5(4% \lambda)^{5}(\lambda-1)}.$ ⓘ Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $m$: integer, $\alpha$, $p$, $q$ and $r$ A&S Ref: 9.5.31 Permalink: http://dlmf.nist.gov/10.21.E52 Encodings: TeX, TeX, TeX, TeX, pMML, pMML, pMML, pMML, png, png, png, png See also: Annotations for 10.21(x), 10.21 and 10 The asymptotic expansion of the large positive zeros (not necessarily the $m$th) of the function 10.21.53 $J_{\nu}'\left(x\right)Y_{\nu}\left(\lambda x\right)-Y_{\nu}'\left(x\right)J_{% \nu}\left(\lambda x\right)$ ⓘ Symbols: $J_{\NVar{\nu}}\left(\NVar{z}\right)$: Bessel function of the first kind, $Y_{\NVar{\nu}}\left(\NVar{z}\right)$: Bessel function of the second kind, $x$: real variable and $\nu$: complex parameter A&S Ref: 9.5.32 Permalink: http://dlmf.nist.gov/10.21.E53 Encodings: TeX, pMML, png See also: Annotations for 10.21(x), 10.21 and 10 is given by (10.21.50), where 10.21.54 $\displaystyle\alpha$ $\displaystyle=\frac{(m-\tfrac{1}{2})\pi}{\lambda-1},$ $\displaystyle p$ $\displaystyle=\frac{(\mu+3)\lambda-(\mu-1)}{8\lambda(\lambda-1)},$ $\displaystyle q$ $\displaystyle=\frac{(\mu^{2}+46\mu-63)\lambda^{3}-(\mu-1)(\mu-25)}{6(4\lambda)% ^{3}(\lambda-1)},$ $\displaystyle r$ $\displaystyle=\frac{(\mu^{3}+185\mu^{2}-2053\mu+1899)\lambda^{5}-(\mu-1)(\mu^{% 2}-114\mu+1073)}{5(4\lambda)^{5}(\lambda-1)}.$ ⓘ Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $m$: integer, $\alpha$, $p$, $q$ and $r$ A&S Ref: 9.5.33 Permalink: http://dlmf.nist.gov/10.21.E54 Encodings: TeX, TeX, TeX, TeX, pMML, pMML, pMML, pMML, png, png, png, png See also: Annotations for 10.21(x), 10.21 and 10 Higher coefficients in the asymptotic expansions in this subsection can be obtained by expressing the cross-products in terms of the modulus and phase functions (§10.18), and then reverting the asymptotic expansion for the difference of the phase functions. For further information see Cochran (1963, 1964, 1966a, 1966b), Kalähne (1907), Martinek et al. (1966), Muldoon (1979), and Salchev and Popov (1976). ## §10.21(xi) Riccati–Bessel Functions The Riccati–Bessel functions are $(\tfrac{1}{2}\pi x)^{\frac{1}{2}}J_{\nu}\left(x\right)$ and $(\tfrac{1}{2}\pi x)^{\frac{1}{2}}Y_{\nu}\left(x\right)$. Except possibly for $x=0$ their zeros are the same as those of $J_{\nu}\left(x\right)$ and $Y_{\nu}\left(x\right)$, respectively. For information on the zeros of the derivatives of Riccati–Bessel functions, and also on zeros of their cross-products, see Boyer (1969). This information includes asymptotic approximations analogous to those given in §§10.21(vi), 10.21(vii), and 10.21(x). ## §10.21(xii) Zeros of $\alpha J_{\nu}\left(x\right)+xJ_{\nu}'\left(x\right)$ For properties of the positive zeros of the function $\alpha J_{\nu}\left(x\right)+xJ_{\nu}'\left(x\right)$, with $\alpha$ and $\nu$ real, see Landau (1999). ## §10.21(xiii) Rayleigh Function The Rayleigh function $\sigma_{n}\left(\nu\right)$ is defined by 10.21.55 $\sigma_{n}\left(\nu\right)=\sum_{m=1}^{\infty}(j_{\nu,m})^{-2n},$ $n=1,2,3,\dots$. ⓘ Defines: $\sigma_{\NVar{n}}\left(\NVar{\nu}\right)$: Rayleigh function Symbols: $m$: integer, $n$: integer and $\nu$: complex parameter Permalink: http://dlmf.nist.gov/10.21.E55 Encodings: TeX, pMML, png See also: Annotations for 10.21(xiii), 10.21 and 10 For properties, computation, and generalizations see Kapitsa (1951b), Kerimov (1999, 2008), and Gupta and Muldoon (2000). See also Watson (1944, §§15.5, 15.51). ## §10.21(xiv) $\nu$-Zeros For information on zeros of Bessel and Hankel functions as functions of the order, see Cochran (1965), Cochran and Hoffspiegel (1970), Hethcote (1970), Conde and Kalla (1979), and Sandström and Ackrén (2007).

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On Wed, Mar 6, 2013 at 4:52 PM, Jaime Fernández del Río wrote: On Tue, Mar 5, 2013 at 5:23 AM, Charles R Harris wrote: On Tue, Mar 5, 2013 at 12:41 AM, Jaime Fernández del Río wrote: On Mon, Mar 4, 2013 at 8:37 PM, Charles R Harris wrote: There are actually seven versions of polynomial fit, two for the usual polynomial basis, and one each for Legendre, Chebyshev, Hermite, Hermite_e, and Laguerre ;) Correct me if I am wrong, but the fitted function is the same regardless of the polynomial basis used. I don't know if there can be numerical stability issues, but chebfit(x, y, n) returns the same as poly2cheb(polyfit(x, y, n)). In any case, with all the already existing support for these special polynomials, it wouldn't be too hard to set the problem up to calculate the right coefficients directly for each case. How do you propose to implement it? I think Lagrange multipliers is overkill, I'd rather see using the weights (approximate) or change of variable -- a permutation in this case -- followed by qr and lstsq. The weights method is already in place, but I find it rather inelegant and unsatisfactory as a solution to this problem. But if it is deemed sufficient, then there is of course no need to go any further. I hadn't thought of any other way than using Lagrange multipliers, but looking at it in more detail, I am not sure it will be possible to formulate it in a manner that can be fed to lstsq, as polyfit does today. And if it can't, it probably wouldn't make much sense to have two different methods which cannot produce the same full output running under the same hood. I can't figure out your "change of variable" method from the succinct description, could you elaborate a little more? I think the place to add this is to lstsq as linear constraints. That is, the coefficients must satisfy B * c = y_c for some set of equations B. In the polynomial case the rows of B would be the powers of x at the points you want to constrain. Then do an svd on B, B = u * d * v. Apply v to the design matrix of the unconstrained points A'  = A * v.T so that B' becomes u * d. The coefficients are now replaced by new variables c' with the contraints in the first two columns. If there are, say, 2 constraints, u * d will be 2x2. Solve that equation for the first two constraints then multiply the first two columns of the design matrix A' by the result and put them on the rhs, i.e., y = y - A'[:, :2] * c'[:2] then solve the usual l least squares thing with A[:, 2:] * c'[2:] = y to get the rest of the transformed coefficients c'. Put the coefficients altogether and multiply with v^T to get c = v^T * c' Very nice, and works beautifully! I have tried the method you describe, and there are a few relevant observations: 1. It gives the exact same result as the Lagrange multiplier approach, which is probably expected, but I wasn't all that sure it would be the case. It's equivalent, but I'm thinking in algorithmic terms, which is somewhat more specific than the mathematical formulation. 2. The result also seems to be to what the sequence of fits giving increasing weights to the fixed points converges to. This image http://i1092.photobucket.com/albums/i412/jfrio/image.png is an example. In there: * blue crosses are the data points to fit to * red points are the fixed points * blue line is the standard polyfit * red line is the constrained polyfit * cyan, magenta, yellow and black are polyfits with weights of 2, 4, 8, 16 for the fixed points, 1 for the rest Seeing this last point, probably the cleanest, least disruptive implementation of this, would be to allow np.inf values in the weights parameter, which would get filtered out, and dealt with in the above manner. Interesting idea, I like it. It is less general, but probably all that is needed for polynomial fits. I suppose that after you pull out the relevant rows you can set the weights to zero so that they will have no (order of roundoff) effect on the remaining fit and you don't need to rewrite the design matrix. So I have two questions: 1. Does this make sense? Or will it be better to make it more explicit, with a 'fixed_points' keyword argument defaulting to None? 2. Once I have this implemented, documented and tested... How do I go about submitting it for consideration? Would a patch be the way to go, or should I fork? A fork is definitely the way to go. That makes it easy for folks to review the code and tell you everything you did wrong ;) I think adding linear constraints to lstsq would be good, then the upper level routines can make use of them. Something like a new argument constraints=(B, y), with None the default. Chuck

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# Date Multiplying Challenge (Inspired by last week's Riddler on FiveThirtyEight.com. Sandbox post.) Given a year between 2001 and 2099, calculate and return the number of days during that calendar year where mm * dd = yy (where yy is the 2-digit year). 2018, for example, has 5: • January 18th (1 * 18 = 18) • February 9th (2 * 9 = 18) • March 6th (3 * 6 = 18) • June 3rd (6 * 3 = 18) • September 2nd (9 * 2 = 18) Input can be a 2 or 4-digit numeric year. Output should be an integer. Optional trailing space or return is fine. # Complete input/output list: Input = Output 2001 = 1 2021 = 3 2041 = 0 2061 = 0 2081 = 2 2002 = 2 2022 = 3 2042 = 4 2062 = 0 2082 = 0 2003 = 2 2023 = 1 2043 = 0 2063 = 3 2083 = 0 2004 = 3 2024 = 7 2044 = 3 2064 = 2 2084 = 5 2005 = 2 2025 = 2 2045 = 3 2065 = 1 2085 = 1 2006 = 4 2026 = 2 2046 = 1 2066 = 3 2086 = 0 2007 = 2 2027 = 3 2047 = 0 2067 = 0 2087 = 1 2008 = 4 2028 = 4 2048 = 6 2068 = 1 2088 = 3 2009 = 3 2029 = 1 2049 = 1 2069 = 1 2089 = 0 2010 = 4 2030 = 6 2050 = 3 2070 = 3 2090 = 5 2011 = 2 2031 = 1 2051 = 1 2071 = 0 2091 = 1 2012 = 6 2032 = 3 2052 = 2 2072 = 6 2092 = 1 2013 = 1 2033 = 2 2053 = 0 2073 = 0 2093 = 1 2014 = 3 2034 = 1 2054 = 4 2074 = 0 2094 = 0 2015 = 3 2035 = 2 2055 = 2 2075 = 2 2095 = 1 2016 = 4 2036 = 6 2056 = 4 2076 = 1 2096 = 4 2017 = 1 2037 = 0 2057 = 1 2077 = 2 2097 = 0 2018 = 5 2038 = 1 2058 = 0 2078 = 2 2098 = 1 2019 = 1 2039 = 1 2059 = 0 2079 = 0 2099 = 2 2020 = 5 2040 = 5 2060 = 6 2080 = 4 This is a challenge, lowest byte count in each language wins. Pre-calculating and simply looking up the answers is normally excluded per our loophole rules, but I'm explicitly allowing it for this challenge. It allows for some interesting alternate strategies, although its not likely a 98 99-item lookup list is going to be shortest. • If it makes it any easier in your language, the answer will be the same regardless of century; 1924 and 2124 have the same number of days as 2024. – BradC Apr 13 '18 at 14:41 • if the the result of mm*dd is bigger than 100 it is automatically filtered? – DanielIndie Apr 13 '18 at 16:03 • @DanielIndie Correct, no "wraparound" dates should be counted. In other words, Dec 12, 2044 doesn't count, even though 12 * 12 = 144. – BradC Apr 13 '18 at 16:13 • As we need only handle a limited number of inputs, I've edited them all in. Feel free to rollback or reformat. – Shaggy Apr 13 '18 at 17:20 • @gwaugh Just that you can decide which to accept as valid input (so you don't have to spend extra characters converting between the two). – BradC Feb 4 at 20:06 # Excel, 48 bytes Hooray! Finally something Excel is actually good at. =COUNT(VALUE(ROW(1:12)&"/"&A1/ROW(1:12)&"/"&A1)) Takes input from A1 in the form of an integer 1-99 representing the year, and outputs to wherever you enter this formula. It's an array formula, so use Ctrl-Shift-Enter instead of Enter to enter it. This takes advantage of the fact that COUNT ignores errors, so any errors that are caused by either the month not dividing the year (leading Excel to parse something like 2/12.5/25 or by the date not being valid, like 2/29/58, are just silently ignored. • Very nice. Probably worth mentioning it requires a 2-digit year in A1. Entering a 4-digit year simply returns 0. – BradC Apr 13 '18 at 19:14 • True! I'll edit that into the description. – Sophia Lechner Apr 13 '18 at 19:16 • Also I should mention that it's locale-specific; it depends on having a locale that uses a mm/dd/yy ordering. In a locale with dd/mm/yy ordering, the answer would be the same number of bytes, of course. – Sophia Lechner Apr 13 '18 at 19:20 • Super clever; I like how you're only testing 12 date candidates (one per month), instead of running through every day of the year. – BradC Apr 13 '18 at 19:26 • Fix the year to a non-leap one save byte? – l4m2 May 4 '18 at 2:59 # Python 2, 44 bytes [k/32%13*(k%32)for k in range(96,509)].count Try it online! An anonymous function given as a method object. Produces all products of (month, day) pairs (m, d) as encoded by k=32*m+d with 0≤m≤12, 0≤d≤31, wrapping around. Eliminates Feb 29-31 by excluding them from the range. # Java (JDK 10), 65 bytes y->{int m=13,c=0;for(;m-->1;)if(y%m<1&y/m<29+m%2*3)c++;return c;} Try it online! # Credits • No leap year fits 29*n, so no need the check – l4m2 Apr 13 '18 at 16:36 • Thanks for your input. I could remove 27 bytes in total with a bunch of other changes. – Olivier Grégoire Apr 13 '18 at 18:07 • Changing (m==2?29:32) to 29+m%2*3 still seems to give all OK results. Credit to @AsoneTuhid's Ruby answer. – Kevin Cruijssen Apr 16 '18 at 6:52 # PowerShell, 94 bytes param($a)for($x=Date 1/1/$a;$x-le(Date 12/9/$a);$x=$x.AddDays(1)){$z+=$x.Month*$x.Day-eq$a};$z Try it online! Takes input as a two-digit year, then constructs a for loop from 1/1/year to 12/9/year (because 12/10 and beyond will never count and this saves a byte). Each iteration, we increment $z iff the .Month times the .Day is equal to our input year. Outside the loop, $z is left on the pipeline and output is implicit. Edit - this is culture dependent. The above code works for en-us. The date format may need to change for other cultures. • "culture"? Did you mean "locale"? ... – user202729 Apr 13 '18 at 15:31 • @user202729 Colloquially, yes, but PowerShell documentation refers to it as "culture." – AdmBorkBork Apr 13 '18 at 15:33 • "Culture" is the word MS generally uses to talk about locale, eg. System.Globalization.CultureInfo in .NET. – sundar Jun 29 '18 at 18:31 # Ruby, 46 42 bytes ->y{(1..12).count{|j|y%j<1&&y/j<29+j%2*3}} Try it online! # JavaScript (Node.js), 4844 43 bytes F=(x,h)=>h>12?0:F(x,-~h)+(x/h<3*h%6+28>x%h) Try it online! # JavaScript (Node.js), 59 58 bytes x=>[a=0,...'030101001010'].map((k,m)=>x%m|x/m>31-k||a++)|a Try it online! • I knew a mathematical solution would be short, but I had fun with my version :) – Shaggy Apr 13 '18 at 16:48 # Jelly, 15 bytes 31x12_2¦3RĖP€Fċ Try it online! Take a number in range [0,100[ as input. # JavaScript (ES6), 91 bytes I was curious to know how hardcoding would compare to an iterative computation. It's definitely longer (see @Shaggy's answer), but not awfully longer. Edit: It is, however, much longer than a more direct formula (see @l4m2 answer). Takes input as an integer in [1..99]. n=>(k=parseInt('8ijkskercdtbnqcejh6954r1eb2kc06oa3936gh2k0d83d984h'[n>>1],36),n&1?k>>3:k&7) Try it online! ### How? Odd years have significantly less chance of having mm * dd = yy than even years. More concretely, odd years have 0 to 3 matches, while even years have 0 to 7 matches. This allows us to code each pair of years with just 5 bits, which can conveniently be represented as a single character in base 36. {+grep {$_%%$^m&&29+$m%2*3>$_/$m},1..12} Try it online! # Python 2 and 3, 55 52 bytes lambda y:sum(y%i<(y/i<29+i%2*3)for i in range(1,13)) Try it online! # Bash + GNU utilities, 57 • 1 byte saved thanks to @SophiaLechner seq -f1/1/$1+%gday 0 365|date -f- +%m*%d-%y|bc|grep -c ^0 Note that the seq command always produces a list of 366 dates - for non-leap years January 1st of the next year will be included. However in the date range 2001..2099, MM*DD will never be YY for January 1st of the next year for any of these years, so this extra day won't affect the result. Try it online! • Very nice - I hadn't even known date will do date math like that while parsing. seq doesn't need a space after the -f, so you can save a byte there. – Sophia Lechner Apr 16 '18 at 17:22 # T-SQL, 123 121 bytes Per our IO rules, input is taken via pre-existing table t with an integer field y, which contains a 2-digit year. WITH c AS(SELECT 1m UNION ALL SELECT m+1FROM c WHERE m<12) SELECT SUM(ISDATE(CONCAT(m,'/',y/m,'/',y)))FROM c,t WHERE y%m=0 Line break is for readability only. Inspired largely by Sophia's Excel solution. • The top line generates a 12-item number table c which is joined to the input table t. • If that passes I mash together a date candidate using CONCAT(), which does implicit varchar datatype conversions. Otherwise I'd have to do a bunch of CAST or CONVERT statements. • Found the perfect evaluation function ISDATE(), which returns 1 for valid dates and 0 for invalid dates. • Wrap it in a SUM and I'm done. • EDIT: Moved the integer division check (y%m=0) into the WHERE clause to save 2 bytes, thanks @RazvanSocol. Unfortunately, it's not much shorter than the lookup table version (using the string from osdavison's version): # T-SQL lookup, 129 bytes SELECT SUBSTRING('122324243426133415153317223416132126011504033106131204241006003213011306002122042005101305111014012',y,1)FROM t EDIT: Leaving my original above, but we can save a few bytes by using a couple of new functions: • STRING_SPLIT is available in MS SQL 2016 and above. • CONCAT_WS is available in MS SQL 2017 and above. • As above, replaced IIF with WHERE # MS-SQL 2017, 121 118 bytes SELECT SUM(ISDATE(CONCAT_WS('/',value,y/value,y))) FROM t,STRING_SPLIT('1-2-3-4-5-6-7-8-9-10-11-12','-') WHERE y%value=0 # MS-SQL 2017, extra cheaty edition: 109 bytes SELECT SUM(ISDATE(CONCAT_WS('/',number,y/number,y))) FROM t,spt_values WHERE number>0AND y%number=0AND'P'=TYPE Requires you to be in the master database, which contains a system table spt_values that (when filtered by TYPE='P'), gives you counting numbers from 0 to 2048. • Similar to other answers, the order I assemble the date (m/d/y) depends on the locality settings of the SQL instance. Other localities might require a different order, or a different separator, but I don't think would impact the code length. – BradC Apr 13 '18 at 21:47 • I guess you realized that converting '2/29/64' to a date gives 1964-02-29 (not 2064-02-29), but considering that the years 2000 and 2100 are excluded, it's a nice way to get a shorter code. – Razvan Socol Apr 15 '18 at 9:45 • Using SPLIT_STRING instead of a CTE brings it down to 120 bytes. Using CONCAT_WS instead of CONCAT saves another character, getting it to 119 bytes. – Razvan Socol Apr 15 '18 at 9:51 • @RazvanSocol Yeah, I'm not sure where the break is between 2-digit dates that map to 19xx vs 20xx, but both give the same answer. I'll give the other two suggestions a try, thanks! – BradC Apr 15 '18 at 18:08 • Replace IIF with WHERE. – Razvan Socol Apr 17 '18 at 19:24 # Julia 0.6, 4944 42 bytes y->sum(y/i∈1:28+3(i%2⊻i÷8)for i=1:12) Try it online! -5 bytes inspired by Asone Tuhid's Ruby answer. -2 bytes replacing count with sum ### Explanation: For each month i from 1 to 12, calculate y/i, and check if it's one of that month's days. Months with 31 days are 1, 3, 5, 7, 8, 10, 12 - so they're odd below 8 and even at and above 8. So either i%2 or i÷8 (which is 0 for i<8 and 1 for i>=8 here) should be 1, but not both - so we XOR them. If the xor result is true, we check dates 1:28+3 i.e. 1:31, otherwise we check just dates 1:28. 1:28 is sufficient for the rest of the months (this improvement inspired by Asone Tuhid's Ruby answer) because: • for February, the only possibility would have been 2*29 = 58, but 2058 is not a leap year, so we can assume Feb always has 28 days. • the other months with 30 days are month 4 and above - for which i*29 (and i*30) would be above 100, which can be ignored. Finally, we count out the number of times y/i belongs in this list of days (by using boolean sum here), and return that. # JavaScript, 91858281 77 bytes Takes input as a 2-digit string (or a 1 or 2 digit integer). Takes advantage of the fact that new Date will rollover to the next month, and continue doing so, if you pass it a day value that exceeds the number of days in the month you pass to it so, on the first iteration, it tries to construct the date yyyy-01-345 which becomes yyyy-12-11, or yyyy-12-10 on leap years. We don't need to check dates after that as 12*11+ results in a 3-digit number. y=>(g=d=>d&&([,M,D]=new Date(y,0,d).toJSON().split(/\D/),D*M==y)+g(--d))(345) 3 bytes saved thanks to Arnauld. ## Test It f= y=>(g=d=>d&&([,M,D]=new Date(y,0,d).toJSON().split(/\D/),D*M==y)+g(--d))(345) o.innerText=[...Array(99)].map((_,x)=>(2001+x++)+ = +f(x)).join\n pre{column-count:5;width:480px;} <pre id=o></pre> # Python 2, 898468 58 bytes f=lambda y,d=62:d-59and((d/31+1)*(d%31+1)==y)+f(y,-~d%372) Try it online! # Excel, 83 bytes {=SUM(IF(MONTH(DATE(A1,1,0)+ROW(1:366))*DAY(DATE(A1,1,0)+ROW(1:366))=A1-2000,1,0))} Input is in cell A1 in the format yyyy. This is an array formula and is entered with Ctrl+Shift+Enter to get the curly brackets {}. It is fairly straightforward and without any cleverness. When in an array formula, DATE(A1,1,0)+ROW(1:366) gives us an array of 366 date values. On non-leap years, this will include Jan 1 of the next year but that is not a problem because 1*1=1 and would only count as a false positive if the next year is 2001 but, since the required year range is 2001 - 2099, it will never arise as an issue. If you shorted that bit into simply ~, the formula because much easier to follow: {=SUM(IF(MONTH(~)*DAY(~)=A1-2000,1,0))} I tried using COUNTIF() instead of SUM(IF()) but Excel wouldn't even let me enter it as an array formula, much less give me a result. I did find a Google Sheets solution using CountIf() but the same method otherwise that turned out to be 91 bytes, mostly because it uses ArrayFormula() instead of simply { }. =CountIf(ArrayFormula(Month(Date(A1,1,0)+Row(1:366))*Day(Date(A1,1,0)+Row(1:366))),A1-2000) • I haven't seen a consensus, but I've generally not been including the outer curly brackets in my byte counts for Excel. They feel more like a way that Excel is formatting its display than part of the formula. Opinions? – Sophia Lechner Apr 13 '18 at 19:15 • @SophiaLechner I've included them instead of deciding how to include the extra keystrokes required to enter it as an array formula. There's a meta question about that and the only answer says that the CTRL+ALT+ENTER command would count as 1 keystroke. If you use the same as vim (per meta), then that keystroke would count as 1 byte. However, I don't usually count the ENTER at the end of entering a formula as 1 byte in other answers. – Engineer Toast Apr 13 '18 at 21:20 # Retina 0.8.2, 55 bytes .. $* (?<=^(1{1,12}))(?=(?(?<=^11$)\1{0,27}|\1{0,30})$) Try it online! Takes a two-digit year; add 1 byte to support 4-digit years. Explanation: The first stage simply converts to unary. The second stage starts by matching 1 to 12 characters before the match position, representing the month, and then attempts to look ahead for a whole number of repetitions of that month. However, the lookahead contains a conditional, which chooses between up to 27 or 30 further repetitions depending on the month. The count of match positions is then the desired result. # R, 22 122 bytes x=scan();substr("122324243426133415153317223416132126011504033106131204241006003213011306002122042005101305111014012",x,x) Try it online! Decided to go with a lookup table approach. Input year needs to be 2 digits. • Also on TIO you can hit the "link" button and it will auto-format the answer for you. Thanks to Dennis for setting it up! – Giuseppe Jul 19 '18 at 23:51 • You can remove the initial if, since input can be either 2-digit or 4-digit, your choice (so you can choose to accept only 2-digit input). But it looks like the code considers every month to contain 31 days, so for eg., 62 (for 2062) returns 1 where it should return 0. – sundar Jul 20 '18 at 13:16 • I misunderstood. That being said, I'm pretty sure the lookup table would have to be included in the byte count. – ngm Jul 20 '18 at 16:46 • @ngm I also was not sure if the lookup table needed to be included, but I will add it to the byte count just to be safe. – Robert S. Jul 20 '18 at 16:51 • I'm still not sure what all the rules are! – ngm Jul 20 '18 at 16:53 # C (gcc), 6560 59 bytes d(a,t,e){for(t=0,e=13;e-->1;)t+=a%e<1&a/e<(e-2?32:29);a=t;} Port of user202729's Java answer. Try it online here. Thanks to Jonathan Frech for golfing 1 byte. • a=0,m=13;for(; ~> for(a=0,m=13;. – Jonathan Frech Jun 30 '18 at 12:23 • @JonathanFrech Thanks for pointing that out, I have edited. – O.O.Balance Jun 30 '18 at 14:56 # J, 29 bytes 1#.(,x*i."+29+3*2~:x=.i.13)=] Try it online! ### How it works 1#.(,x*i."+29+3*2~:x=.i.13)=] Input: year (2-digit, 1..99) x=.i.13 array of 0..12; assign to x 2~: 1 1 0 1 .. 1 3* 3 3 0 3 .. 3 29+ 32 32 29 32 .. 32 i."+ for each item of above, generate a row 0 .. n ,x* each row times 0 .. 12 and flatten =] 1 for each item == input, 0 otherwise 1#. sum Tried hard to get under 2 times Jelly solution :) ### Side note If someone really wants to hardcode the 99-digit data, here's a bit of information: Divide the 99-digit into chunks of 2 digits. Then the first digit is <4 and the second <8, which means five bits can encode two numbers. Then the entire data can be encoded in 250 bits, or 32 bytes. # Python 3, 158 162215241 bytes Removed 4 Thanks to Stephen for golfing the conditionals. Removed 53 thanks Stephen for pointing out the white space Removed 26 thanks to the link provided by caird I'm pretty new at this. Couldn't think of how to do this without having the days in a month be described. r=range def n(Y): a,b,c=31,30,0 for x in r(12): for y in r([a,(28if Y%4else 29),a,b,a,b,a,a,b,a,b,a][x]): if(x+1)*(y+1)==int(str(Y)[2:]):c+=1 return c Try it online! • Welcome to the site, and nice first post! There are a fair few ways you can golf this, such as removing some whitespace, so be sure to check out these tips for golfing in Python – caird coinheringaahing Apr 13 '18 at 18:03 • @cairdcoinheringaahing Thanks for the advice the link was very useful! – akozi Apr 13 '18 at 21:16 • 158 bytes - golfed some away, but mostly you had a big long line of spaces on your third line, dunno how those got there – Stephen Feb 3 at 18:08 • @Stephen Thanks :) I added yours as two edits. One for the white space and the other as the golfing. – akozi Feb 3 at 18:17 • (28if Y%4else 29) can be shortened to [29,28][Y%4>0]. Also, the long list can be shortened to [a,...]+2*[a,b,a,b,a]. a,b,c can be added in the parameter list to save a line. int(str(Y)[2:]) can be shortened to Y%100. Finally, counter variables can mostly be shortened to lens of list comprehensions, this also allows n to be made a lambda. This makes 118. – Black Owl Kai Feb 4 at 0:19 # Forth (gforth), 60 59 bytes : f 0 13 1 do over i /mod swap 32 i 2 = 3 * + 0 d< - loop ; Try it online! This version takes advantage of the fact that there can't be more than 1 matching day per month, and that the year has to be divisible by the month for it to match. ### Explanation Iterates over the months, checks if year is divisible by month, and if the quotient is < 31 (28 for Feb) Months after March can't match for days greater than 25, so we can just assume all months (other than Feb) have 31 days for the purpose of the puzzle. ### Code Explanation : f \ start new word definition 0 \ set up counter 13 1 do \ start a counted loop from 1 to 12 over i /mod \ get the quotient and remainder of dividing year by month swap \ default order is remainder-quotient, but we want to swap them 32 i 2= 3 * + \ we want to compare to 32 days unless month is Feb 0 d<= \ treat the 4 numbers on the stack as 2 double-length numbers \ and compare [1] - \ subtract result from counter (-1 is true in Forth) loop \ end loop ; \ end word definition [1] - Forth has the concept of double length numbers, which are stored on the stack as two single-length numbers (of the form x y, where the value of the double = y * 2^(l) + xwhere l is the size in bits of a single in the forth implementation you're working with). In this case, I compared the quotient and remainder to 32 (or 29) 0. If the remainder was greater than 0 (year not divisble by month), the first double would be automatically bigger than 32 (or 29) 0 and the result would be false. If the remainder is 0, then it resolves to effectively a regular check of quotient <= 32 (or 29) # Forth (gforth), 61 bytes : f 0 13 1 do i 2 = 3 * 32 + 1 do over i j * = - loop loop ; Try it online! Saved some bytes by realizing that only February matters in terms of having the correct number of days in the month # Explanation Forth (at least gforth) comparisons return -1 for true and 0 for false 0 \ place 0 on top of the stack to use as a counter 13 1 do \ begin the outer loop from 1 to 12 i 2 = 3 * \ if i is 2 place -3 on top of the stack, otherwise 0 32 + 1 do \ begin the inner loop from 1 to 31 (or 28) over \ copy the year from stack position 2 and place it on top of the stack i j * = - \ multiply month and day compare to year, subtract result from counter loop \ end the inner loop loop \ end the outer loop # Java (JDK 10), 7972 70 bytes y->{int a=0,m=13;for(;m-->1;)a+=y%m<1&&y/m<(m==2?29:32)?1:0;return a;} Try it online! • @Shaggy Thanks! (leftover from previous version where I loop over all days) – user202729 Apr 13 '18 at 16:24 • If you change && to & it's the same answer as OlivierGrégoire's Java answer, although he has answered 19 minutes earlier. – Kevin Cruijssen Apr 16 '18 at 6:47 • @KevinCruijssen Random coincidence. Although my version is a bit worse (unnecessary use of ternary). – user202729 Apr 16 '18 at 11:30 # JavaScript (Node.js), 108 bytes a=>'0122324243426133415153317223416132126011504033106131204241006003213011306002122042005101305111014012'[a] f=a=>'0122324243426133415153317223416132126011504033106131204241006003213011306002122042005101305111014012'[a] o.innerText=[...Array(99)].map((_,x)=>(2001+x++)+ = +f(x)).join\n pre{column-count:5;width:480px;} <pre id=o></pre> • Brute-force lookup, nice! – BradC Apr 13 '18 at 20:07 • Welcome to PPCG! – Shaggy Apr 13 '18 at 21:27 # Perl 5, 68 bytes //;map$\+=$'%++$m==0&&$'/$m<$_,32,29,32,31,32,31,32,32,31,32,31,32}{ Try it online! # Python 3, 132 bytes This is really quite long of a program but I thought it might be of interest. All the values are between 0-7 so I encode each number with 3 bits in a long binary string. I tried pasting a raw binary string into my python program but I couldn't get it to work, so I settled on base64 in the file. I used the following string as a lookup table (ending 7 used for padding): 01223242434261334151533172234161321260115040331061312042410060032130113060021220420051013051110140127 The program takes this string and decodes it as a number, then uses bit shifting to extract the result. import base64 lambda n:int.from_bytes(base64.b64decode(b'pNRRxYtw01s9Jw4tFYE0QNkYsoRQgYBosEsYBFIRAUgsUkgwFQM='),'big')>>(6306-3*n)&7 # 66 bytes + 37 byte file = 103 bytes This reads a binary file called e and avoids using base64. lambda n:int.from_bytes(open('e','rb').read(),'big')>>(6306-3*n)&7 Here is a hexdump of the read file (without padding): 00000000 a4 d4 51 c5 8b 70 d3 5b 3d 27 0e 2d 15 81 34 40 |..Q..p.[='.-..4@| 00000010 d9 18 b2 84 50 81 80 68 b0 4b 18 04 52 11 01 48 |....P..h.K..R..H| 00000020 2c 52 48 30 0a |,RH0.| # Haskell, 61 51 bytes f y=sum[1|i<-[96..508],mod(div i 32)13*mod i 32==y] Try it online! Inspired by xnor's Python 2 answer and Laikoni. • 52 bytes: f y=sum[1|i<-[1..12],mod y i<1,div y i<29+mod i 2*3] Try it online! – Laikoni Apr 14 '18 at 11:04 # Oracle SQL, 115 bytes select count(decode(sign(decode(l,2,29,32)-y/l),1,1))from t, xmltable('1to 12'columns l int path'.')where mod(y,l)=0 We may note that it does not really matter how many days in April (and later months), since 100/4 < 28. Also it's not necessary to check whether year is leap or not. We just have to specify that there are 28 days in February (not 29 because that validation will be executed only for 2058 which is not leap) otherwise it may be just 31 for any month. Other approaches # Oracle SQL (12c Release 2 and later), 151 bytes select count(to_date(y/l||'-'||l||'-1' default '' on conversion error,'dd-mm-y'))from t, (select level l from dual connect by level<=12)where mod(y,l)=0 # Oracle SQL (12c Release 2 and later), 137 bytes select sum(validate_conversion(y/l||'-'||l||'-1'as date,'dd-mm-y'))from t, (select level l from dual connect by level<=12)where mod(y,l)=0 Both solution could have been 8 bytes shorter if we replace (select level l from dual connect by level<=12) with xmltable('1to 12'columns l int path'.') but Oracle throws an exception because of the bug (tested on versions 12.2.0.1.0, 18.3.0.0.0). SQL> select count(to_date(y/l||'-'||l||'-1' default '' on conversion error,'dd-mm-y'))from t, 2 xmltable('1to 12'columns l int path'.')where mod(y,l)=0 3 / select count(to_date(y/l||'-'||l||'-1' default '' on conversion error,'dd-mm-y'))from t, * ERROR at line 1: ORA-00932: inconsistent datatypes: expected DATE got DATE SQL> select sum(validate_conversion(y/l||'-'||l||'-1'as date,'dd-mm-y'))from t, 2 xmltable('1to 12'columns l int path'.')where mod(y,l)=0 3 / select sum(validate_conversion(y/l||'-'||l||'-1'as date,'dd-mm-y'))from t, * ERROR at line 1: ORA-43909: invalid input data type The only case in both solution when year does matter is 2058 which is non-leap so literal '-1' was used to specify non-leap year. # Oracle SQL, 128 bytes select count(d)from t,(select date'1-1-1'+level-1 d from dual connect by level<365) where to_char(d(+),'mm')*to_char(d(+),'dd')=y # Oracle SQL, 126 bytes select substr('122324243426133415153317223416132126011504033106131204241006003213011306002122042005101305111014012',y,1)from t Update # Oracle SQL, 110 bytes select-sum(least(sign(y/l-decode(l,2,29,32)),0))from t, xmltable('1to 12'columns l int path'.')where mod(y,l)=0 # Oracle SQL, 108 bytes select sum(decode(sign(decode(l,2,29,32)-y/l)*mod(y+1,l),1,1))from t, xmltable('1to 12'columns l int path'.') # Spark SQL, 137 bytes select count(cast(regexp_replace(concat_ws('-','2001',l,y/l),'.0$','')as date)) from(select explode(array(1,2,3,4,5,6,7,8,9,10,11,12))l),t # Spark 2.3+ SQL, 126 bytes (replace function becomes available) select count(cast(replace(concat_ws('-','2001',l,y/l),'.0')as date)) from(select explode(array(1,2,3,4,5,6,7,8,9,10,11,12))l),t # PHP, 73 bytes Using pipe input and php -nR: for(;++$m<13;$d=1)while($d<25+($m-2?$m>4?:7:4))$c+=$argn==$m*$d++;echo$c; Try it online! # PHP, 76 bytes Using command line arg input php dm.php 18: for(;++$m<13;$d=1)while($d<25+($m-2?$m>4?:7:4))$c+=$argv[1]==$m*$d++;echo$c; Try it online! Iterative approach. Since the only leap year to look at is 2 * 29 = 58, and 2058 is not a leap year there's no need to consider leap year in the Feb days. And since wraparound is not a concern -- from April on, any day greater than 25 will exceed 100 we just say the rest of the months only have 25 days. Input is 2 digit year via command line (-10 bytes as program, thx to suggestion from @Titus). OR: # PHP, 101 bytes $y=$argv[1];for($d=strtotime("$y-1");date(Y,$d)==$y;$d+=86400)eval(date("$\x+=j*n=='y';",$d));echo$x; Try it online! Still iterative but using PHP's timestamp functions. Accepts year as four digit number. Thx to @Titus for suggestion of using strtotime() instead of mktime(). • First version fails for 4-digit years, second fails for two-digit. But nice thinking. Try $m<5?$m-2?31:28:25 for the first and $d=strtotime("$y-1") for the second – Titus Feb 4 at 18:30 • Uhm ... Why did you put the y in the eval in quotes? – Titus Feb 4 at 18:37 • @Titus because PHP7 now treats digits preceeded with 0 as octal, so 08 and 09 are actually not valid. tio.run/##K8go@G9jXwAkU5Mz8hUMFWxtFQwsrP//BwA Using the date() function you can only get a 2 digit, zero-padded year. – gwaugh Feb 4 at 18:39 • Bummer! of course. – Titus Feb 4 at 18:41 • @Titus, updated second version using strtotime() instead of mktime() and re-implemented as program, -7 bytes. Also, I looked at the majority of the submissions, including the top voted ones, will only accept year as either 2 or 4 digits, so I'm going to take that to mean it is up to submitter. Thx again for the suggestions! – gwaugh Feb 4 at 19:56 # PHP, 74 70 bytes while(++$m<13)for($d=$m<5?$m-2?31:28:25;$d;)$n+=$d--*$m==$argn;echo$n; accepts two-digit years only. I adopted gwaugh´s considerations and golfed them; my 1st approach was longer than his (92 bytes): for($y=$argn%100;++$m<13;)for($d=$m-2?30+($m+$m/8)%2:29-($y%4>0);$d;)$n+=$d--*$m==$y;echo$n; %100 allows to use 4-digit years. Run as pipe with -nR or try them online.

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1. Jan 4, 2012 ### pc2-brazil Good evening, I have a doubt concerning the worst case scenario of the quicksort algorithm, based on the number of comparisons made by the algorithm. Every source about this subject seems to agree that the worst case scenario is when the list to be sorted is already sorted and the pivot is the smaller element on the list. In this case, the total number of comparisons is n(n - 1)/2, where n is the number of items in the list But it seems to me that the worst case happens when the list is sorted in decreasing order and the pivot is the first element in the list, which, in this case, would be the greatest element on the list. For example, consider the list {4, 3, 2, 1}. I will chose the pivot as the first element. Each step of the quicksort will divide the original list as follows: In the first step, {4, 3, 2, 1} is divided into two lists: {3, 2, 1, 4} (elements smaller than the pivot plus the pivot appended in the end) and {} (elements greater than the pivot: empty list). The first step requires 3 comparisons to find that 3, 2, 1 are smaller than 4. Similarly, in the second step, the list {3, 2, 1, 4} is divided into {2, 1, 3} and {4}, requiring 3 comparisons (to find that 2, 1 are smaller than 3 and 4 is greater than 3). The third step divides {2, 1, 3} into {1, 2} and {3}, requiring 2 comparisons. The last step divides {1, 2} into {1} and {2}, requiring 1 comparison. If I sum the number of comparisons for each step, I find 3 + 3 + 2 + 1 = 9. In fact, if I generalize the above situation to n elements sorted in decreasing order, with the first element as the pivot, I get n(n - 1)/2 + n - 1 comparisons. If I chose the pivot to be the smaller element on the list, and the list were sorted {1, 2, 3, 4}, the number of comparisons would be n(n - 1)/2 = 4(3)/2 = 6. 9 comparisons is greater than 6 comparisons. Why isn't it the worst case? 2. Jan 5, 2012 ### AlephZero There is a different way of thinking about at this. The worst choice of a pivot element splits a list of length n into lists of length 1 and n-1. This will happen if the pivot is either the smallest element or the largest element in the list. In practice, there are several ways to select the pivot element to avoid the the problem that sorting an already-sorted list is slow. For example you can select the element at a random position in the list, not the first or last element. Or you can select say 3 different elements at random positions in the list, and choose the middle-sized one as the pivot. 3. Jan 5, 2012 ### pc2-brazil The choice that I made to sort the list in decreasing order and choose the largest element as the pivot split the original list into lists of length 0 and n (because there is no element larger than the first element, so the second sublist at the first step is empty). It made necessary a larger number of comparisons than if the list was split in lists of length 1 and n-1. Why isn't this approach taken as the worst case? Is there any restriction that makes this approach incorrect?

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Courses Courses for Kids Free study material Offline Centres More Store # NCERT Solutions for Class 11 Maths Chapter 14: Mathematical Reasoning - Exercise 14.2 Last updated date: 09th Aug 2024 Total views: 606k Views today: 6.06k ## NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Free PDF download of NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.2 (Ex 14.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 14 Mathematical Reasoning Exercise 14.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails. Class: NCERT Solutions for Class 11 Subject: Class 11 Maths Chapter Name: Chapter 14 - Mathematical Reasoning Exercise: Exercise - 14.2 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes Competitive Exams after 12th Science ## NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning 1. Write the negation of the following statements. i. Chennai is the capital of Tamil Nadu. Ans: The negation of a statement means to negate the statement. In other words, it is writing the opposite of the given statement. Negation reverses the meaning of the statement. If a given statement is false then its negation will be true and vice versa. To write the negation of any statement, we usually add or remove the word ‘not’. The given statement is that, ‘Chennai is the capital of Tamil Nadu’. The given statement does not carry the word ‘not’. To form the required negation of the given statement, we will add the word ‘not’ to it. Therefore, the negation formed of the given statement is, ‘Chennai is not the capital of Tamil Nadu’. ii. $\sqrt 2$ is not a complex number. Ans: The negation of a statement means to negate the statement. In other words, it is writing the opposite of the given statement. Negation reverses the meaning of the statement. If a given statement is false then its negation will be true and vice versa. To write the negation of any statement, we usually add or remove the word ‘not’. The given statement is that, ‘$\sqrt 2$ is not a complex number’. The given statement carries the word ‘not’. To form the required negation of the given statement, we will remove the word ‘not’ from it. Therefore, the negation formed of the given statement is, ‘$\sqrt 2$ is a complex number’. iii. All triangles are not equilateral triangles. Ans: The negation of a statement means to negate the statement. In other words, it is writing the opposite of the given statement. Negation reverses the meaning of the statement. If a given statement is false then its negation will be true and vice versa. To write the negation of any statement, we usually add or remove the word ‘not’. The given statement is that, ‘All triangles are not equilateral triangles’. The given statement carries the word ‘not’. To form the required negation of the given statement, we will remove the word ‘not’ from it. Therefore, the negation formed of the given statement is, ‘All triangles are equilateral triangles’. iv. The number 2 is greater than 7. Ans: The negation of a statement means to negate the statement. In other words, it is writing the opposite of the given statement. Negation reverses the meaning of the statement. If a given statement is false then its negation will be true and vice versa. To write the negation of any statement, we usually add or remove the word ‘not’. The given statement is that, ‘The number 2 is greater than 7’. The given statement does not carry the word ‘not’. To form the required negation of the given statement, we will add the word ‘not’ to it. Therefore, the negation formed of the given statement is, ‘The number 2 is not greater than 7’. v. Every natural number is not an integer. Ans: The negation of a statement means to negate the statement. In other words, it is writing the opposite of the given statement. Negation reverses the meaning of the statement. If a given statement is false then its negation will be true and vice versa. To write the negation of any statement, we usually add or remove the word ‘not’. The given statement is that, ‘Every natural number is not an integer’. The given statement carries the word ‘not’. To form the required negation of the given statement, we will remove the word ‘not’ from it. Therefore, the negation formed of the given statement is, ‘Every natural number is an integer’. 2. Are the following pairs of statements negations of each other? i. The number $x$ is not a rational number. The number $x$ is not an irrational number. Ans: The negation of a statement means to negate the statement. In other words, it is writing the opposite of the given statement. Negation reverses the meaning of the statement. If a given statement is false then its negation will be true and vice versa. To write the negation of any statement, we usually add or remove the word ‘not’. Consider the first statement. The given statement is that, ‘The number $x$ is not a rational number’. The given statement carries the word ‘not’. To form the required negation of the given statement, we will remove the word ‘not’ from it. Therefore, the negation formed of the given statement is, ‘The number $x$ isa rational number’. Now this statement is similar to the second statement. This is because if a number is not an irrational number, then it will be rational. And this is what the negation of our first statement says. Therefore, the given pair of statements are the negations of each other. ii. The number $x$ is a rational number. The number $x$ is an irrational number. Ans: The negation of a statement means to negate the statement. In other words, it is writing the opposite of the given statement. Negation reverses the meaning of the statement. If a given statement is false then its negation will be true and vice versa. To write the negation of any statement, we usually add or remove the word ‘not’. Consider the first statement. The given statement is that, ‘The number $x$ is a rational number’. The given statement does not carry the word ‘not’. To form the required negation of the given statement, we will add the word ‘not’ to it. Therefore, the negation formed of the given statement is, ‘The number $x$ is not a rational number’. Now this statement is similar to the second statement. This is because if a number is not a rational number, then it will be irrational. And this is what the negation of our first statement says. Therefore, the given pair of statements are the negations of each other. 3. Find the component statements of the following compound statements and check whether they are true or false. i. Number 3 is prime or it is odd. Ans: Compound statements are those statements that are made up of two or more simpler or smaller statements. These smaller statements are complete in themselves and have their own independent meanings. Consider the given statement, ‘Number 3 is prime or it is odd’. We will determine the first component statement and check whether it is true or false. Let this first component statement be ${\text{p}}$. ${\text{p}}$: Number 3 is prime. This statement is true because 3 is a prime number as it has only two factors, 1 and itself. We will now determine the second component statement and check whether it is true or false. Let this second component statement be ${\text{q}}$. ${\text{q}}$: Number 3 is odd. This statement is true because 3 is an odd number as it is not completely divisible by 2. Therefore, the component statements for the given compound statement are, ${\text{p}}$: Number 3 is prime. ${\text{q}}$: Number 3 is odd. Both these above component statements are true. ii. All integers are positive or negative. Ans: Compound statements are those statements that are made up of two or more simpler or smaller statements. These smaller statements are complete in themselves and have their own independent meanings. Consider the given statement, ‘All integers are positive or negative’. We will determine the first component statement and check whether it is true or false. Let this first component statement be ${\text{p}}$. ${\text{p}}$: All integers are positive. This statement is false because integers are both positive and negative numbers. We will now determine the second component statement and check whether it is true or false. Let this second component statement be ${\text{q}}$. ${\text{q}}$:All integers are positive. This statement is false because integers are both positive and negative numbers. Therefore, the component statements for the given compound statement are, ${\text{p}}$: All integers are positive. ${\text{q}}$: All integers are negative. Both these above component statements are false. iii. 100 is divisible by 3, 11, and 5. Ans: Compound statements are those statements that are made up of two or more simpler or smaller statements. These smaller statements are complete in themselves and have their own independent meanings. Consider the given statement, ‘100 is divisible by 3, 11, and 5’. We will determine the first component statement and check whether it is true or false. Let this first component statement be ${\text{p}}$. ${\text{p}}$: 100 is divisible by 3. This statement is false because 100 is not divisible by 3. We will now determine the second component statement and check whether it is true or false. Let this second component statement be ${\text{q}}$. ${\text{q}}$: 100 is divisible by 11. This statement is false because 100 is not divisible by 11. We will now determine the third component statement and check whether it is true or false. Let this third component statement be ${\text{r}}$. ${\text{r}}$: 100 is divisible by 5. This statement is true because 100 is divisible by 5. Therefore, the component statements for the given compound statement are, ${\text{p}}$: 100 is divisible by 3. ${\text{q}}$: 100 is divisible by 11. ${\text{r}}$: 100 is divisible by 5. The statements ${\text{p}}$ and ${\text{q}}$ are false, whereas the statement ${\text{r}}$ is true. ## NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Opting for the NCERT solutions for Ex 14.2 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 14.2 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online. Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Subject Mathematical Reasoning textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 14 Exercise 14.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam. Besides these NCERT solutions for Class 11 Maths Chapter 14 Exercise 14.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 14 Exercise 14.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well. ## FAQs on NCERT Solutions for Class 11 Maths Chapter 14: Mathematical Reasoning - Exercise 14.2 1. Where can I find the NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning (Ex 14.2) Exercise 14.2? The NCERT solutions for the second exercise of chapter 14 Mathematical Reasoning have been carefully chosen by Vedantu teachers. The solutions were created with the concepts in mind and in accordance with the most recent CBSE patterns and guidelines. The Vedantu website or app makes it simple for students to access them. 2. What is the basis for NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning (Ex 14.2) Exercise 14.2? The second exercise in math chapter 14 is based on the ideas of Mathematical Reasoning when all the objects are distinct. Students can learn more about these subjects from the NCERT's explanations and examples that have been solved before moving on to the questions in exercise 14.2. 3. How do the NCERT solutions for Mathematical Reasoning Exercise 14.2 in Chapter 14 of Class 11 Math work? Students can improve their understanding of math concepts, increase their familiarity with the more important subjects, and get a sense of the kinds of questions that might be asked about these subjects in the examination by using the NCERT solutions for exercise 14.2. Students can download the full step-by-step solutions in pdf format from the Vedantu website for the entirety of Exercise 14.2. 4. How many questions are there in Exercise 14.2 of Chapter 14's Mathematical Reasoning? Exercise 14.2 from Class 11 Math's Chapter 14 Mathematical Reasoning consists of a total of 3 questions. The top online resource in India, Vedantu, is where you can find the NCERT solutions for Class 11 Math. At Vedantu, all of the chapter exercises are collected in one location and solved by a qualified teacher in accordance with the recommendations of the NCERT books. The solutions are complete, step-by-step, and 100% accurate. 5. Is it necessary for students to practice each of the exercises in NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning (Ex 14.2)? Students must undoubtedly practice each and every one of the NCERT textbook's questions if they want to get good grades. The Class 11 Maths NCERT solutions are the most helpful source because they contain a variety of questions that need the right concept and understanding to be answered. You can get ready for any challenging or uncommon exam questions by consistently practicing. For free, Vedantu offers comprehensive, step-by-step NCERT solutions for all math chapters.

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# What Fraction Is Equivalent To 0.375 Last Updated on September 3, 2022 by amin Contents ## How do multiply fractions? There are 3 simple steps to multiply fractions 1. Multiply the top numbers (the numerators). 2. Multiply the bottom numbers (the denominators). 3. Simplify the fraction if needed. ## How do u divide fractions? How to Divide Fractions 1. Rewrite the equation as in “Keep Change Flip” 2. Keep the first fraction. 3. Change the division sign to multiplication. 4. Flip the second fraction by switching the top and bottom numbers. 5. Multiply all numerators together. 6. Multiply all denominators together. 7. Reduce the result to lowest terms. ## What is .625 on a tape measure? FRACTIONS INCHES MILLIMETERS 5/8 .625 15.875 41/64 .640625 16.272 21/32 .65625 16.669 43/64 .671875 17.066 ## What is 30 20 as a decimal? 1.5Explanation: NOTE: 3020 is equivalent to 32 since the zeros on the end would cancel out. 32 expressed as a mixed number is 112 which as a decimal would be 1.5 . ## What is the reduced fraction 4 12? Therefore 4/12 simplified to lowest terms is 1/3. ## What is the result when we convert 0.625 into a fraction in its simplest form? 5/8In simplest form 0.625 is 5/8 as a fraction.See also what language is spoken in chad ## .375 as a fraction ||0.375 as a fraction ||What is 0.375 as a simplified fraction? 0.375 as a fraction||Decimal to Fraction ## What is 8125 in a fraction? FRACTIONS/DECIMALS/MILLIMETERS CONVERSION CHART Fraction Decimals Decimal 9/32 .2813 .7813 19/64 .2969 .7969 5/16 .3125 .8125 21/64 .3281 .8281 ## What is 3/8 in its simplest form? 38 is already in the simplest form. It can be written as 0.375 in decimal form (rounded to 6 decimal places). ## Is 0.375 a rational number? 0.375 is a rational no. … which can be expressed in form of p / q. ## What is .325 as a fraction? 13/40Answer: 0.325 as a fraction is expressed as 13/40. ## What is 3/8 as a decimal? 0.3753/8 as a decimal is 0.375. ## What is the simplest form of 21 56? Reduce 21/56 to lowest terms • Find the GCD (or HCF) of numerator and denominator. GCD of 21 and 56 is 7. • 21 ÷ 756 ÷ 7. • Reduced fraction: 38. Therefore 21/56 simplified to lowest terms is 3/8. ## What fraction is the same as 3 5? Decimal and Fraction Conversion Chart Fraction Equivalent Fractions 3/5 6/10 9/15 4/5 8/10 12/15 1/6 2/12 3/18 5/6 10/12 15/18 3 / 8 ## What should be used as the denominator to convert 0.625 to fraction form? Hence the simplest form of the fraction derived from the decimal 0.625 is 5/8. ## What is the reduced fraction 4 10? 410 is not in lowest terms. Since both the numerator and denominator are even numbers they can be divided by 2 . ## What is 0.0625 as a fraction in simplest form? 1/16Answer: 0.0625 as a fraction is 1/16. ## What percent is 11 out of 24? Now we can see that our fraction is 45.833333333333/100 which means that 11/24 as a percentage is 45.8333%. ## What is 0.625 as a fraction? 5/8Answer: 0.625 as a fraction is 625/100 which can be reduced to 5/8.See also how do sherpas get their first name ## What is the reduced fraction 6 10? Therefore 6/10 simplified to lowest terms is 3/5. ## Why is 0.375 a rational number? Any number which when written as a decimal form which terminates (that is ends with only 0 -value digits following the last digit) or any decimal form number that ends with a sequence which repeats indefinitely is rational. ## Can you simplify 9 24? The fraction 924 simplifies to 38 . ## Does 0.375 Repeat or terminate? Answer. A terminating decimal is a decimal that ends. … 3/8 has terminating decimal expansion because when we divide it we get 0.375. ## Is 0.375 a terminating decimal? The decimal is terminating.0.375 → 375. This requires moving decimal point to 3 places to the right to make it a whole number. Move the decimal point of the denominator to the right by the same number of places that the decimal point of the numerator moves which is 3. … Therefore 0.375 in fraction is 38. ## What is .375 inches as a fraction? INCHES DECIMAL OF AN INCH MILLIMETERS 3/8 .375 9.525000 25/64 .390625 9.921875 13/32 .40625 10.318750 27/64 .421875 10.715625 See also rock is broken down into soil by what process? ## What simplified 35 21? Therefore 35/21 simplified to lowest terms is 5/3. ## How do you find a fraction of a whole number? For finding a fraction of a whole number we multiply the numerator of the fraction by the given number and then divide the product by the denominator of the fraction. Solved examples for finding a fraction of a whole number: (i) Find 1/3 of 21. ## Can you simplify 11 24? As you can see 11/24 cannot be simplified any further so the result is the same as we started with. ## How do you work out 30 divided by 20? Using a calculator if you typed in 30 divided by 20 you’d get 1.5. You could also express 30/20 as a mixed fraction: 1 10/20. ## How do you turn 0.66 into a fraction? Answer: 0.66 as a fraction in simplest form is 33/50. ## Is 0.625 rational or irrational? 0.625 can be written as a fraction and the value does terminate therefore the number is rational. !

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GeeksforGeeks App Open App Browser Continue # Area of a triangle inscribed in a rectangle which is inscribed in an ellipse Given here is an ellipse with axes length 2a and 2b, which inscribes a rectangle of length l and breadth h, which in turn inscribes a triangle.The task is to find the area of this triangle. Examples: Input: a = 4, b = 3 Output: 12 Input: a = 5, b = 2 Output: 10 Approach We know the Area of the rectangle inscribed within the ellipse is, Ar = 2ab(Please refer here), also the area of the triangle inscribed within the rectangle s, A = Ar/2 = ab(Please refer here) Below is the implementation of the above approach: ## C++ // C++ Program to find the area of the triangle// inscribed within the rectangle which in turn// is inscribed in an ellipse#include using namespace std; // Function to find the area of the trianglefloat area(float a, float b){     // length of a and b cannot be negative    if (a < 0 || b < 0)        return -1;     // area of the triangle    float A = a * b;    return A;} // Driver codeint main(){    float a = 5, b = 2;    cout << area(a, b) << endl;    return 0;} ## Java // Java Program to find the area of the triangle// inscribed within the rectangle which in turn// is inscribed in an ellipse import java.io.*; class GFG { // Function to find the area of the trianglestatic float area(float a, float b){     // length of a and b cannot be negative    if (a < 0 || b < 0)        return -1;     // area of the triangle    float A = a * b;    return A;} // Driver code      public static void main (String[] args) {    float a = 5, b = 2;    System.out.println(area(a, b));    }}//This code is contributed by anuj_67.. ## Python3 # Python 3 Program to find the# area of the triangle inscribed# within the rectangle which in# turn is inscribed in an ellipse # Function to find the area# of the triangledef area(a, b):         # length of a and b cannot    # be negative    if (a < 0 or b < 0):        return -1     # area of the triangle    A = a * b    return A # Driver codeif __name__ == '__main__':    a = 5    b = 2    print(area(a, b))     # This code is contributed# by Surendra_Gangwar ## C# // C# Program to find the area of// the triangle inscribed within// the rectangle which in turn// is inscribed in an ellipseusing System; class GFG{     // Function to find the// area of the trianglestatic float area(float a, float b){     // length of a and b    // cannot be negative    if (a < 0 || b < 0)        return -1;     // area of the triangle    float A = a * b;    return A;} // Driver codestatic public void Main (){    float a = 5, b = 2;    Console.WriteLine(area(a, b));}} // This code is contributed by ajit ## Javascript Output: 10 Time complexity: O(1) Auxiliary Space: O(1) My Personal Notes arrow_drop_up

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Return to Handouts Some Notes on Toulmin Analysis Here are a few reminders that might make the Toulmin analysis a little clearer for you: Purposes of T.A.: Sometimes it's easiest for us to see if an argument is effective if we take it apart and examine its parts. Sometimes we can even tell where an argument is ineffective after we take it apart. When we examine other people's arguments using Toulmin, we use the "taking apart" to help us understand the argument more fully or to summarize it more accurately. But we can also use T.A. to help us revise our own arguments, and you'll use T.A. on your arguments as you prepare portfolio 2. Claim: Think of the claim in an argument as the most general statement in the argument. It may not be a particularly general statement all by itself, and some claims for arguments are very narrow indeed. But the claim in an argument is like the umbrella statement that all other parts of an argument have to fall under. If a reason (or evidence) doesn't fall under the umbrella of the claim, then it's all wet (i.e., irrelevant). Qualifier: We make lots of statements every day that we don't necessarily mean to apply to every human being everywhere. But in the language of logicians like Stephen Toulmin, all unqualified statements could have inserted in them the words all, every, never, none, always, and so on, without changing the statement. For example, I might say, Books by Anthony Trollope are fun to read. Stephen Toulmin would rewrite that sentence as All books by Anthony Trollope are always fun for everyone to read. I would consider defending some version of my statement, but I would never try to defend Toulmin's version of my statement. Most folks stating claims for arguments try not to get themselves into the position of defending a statement that always applies to everyone everywhere, so most claims will include qualifiers. These are often little words like some, most, many, in general, usually, typically, and so on. They may be little words, but their value to an argument is immeasurable. Exceptions: When a writer specifically excludes certain cases or situations from the argument, those are the exceptions. Reasons: Why does a writer believe the claim she makes? The reasons a writer gives are the first line of development of any argument. To use our umbrella image again, reasons are like the spokes that hold the umbrella open. If reasons are strong enough, the claim holds off the rain. If reasons aren't strong enough, the umbrella collapses. How can we tell if reasons are strong? According to Toulmin, we use two main questions: is the reason relevant? is the reason good? If a reason is relevant, it stays within the confines of the umbrella; if a reason isn't relevant, it pokes out beyond the edge of the umbrella or doesn't even fall under the cover of the umbrella. If a reason is good, it invokes a value we can believe in or agree with. Value judgment are often the most difficult to make in arguments, so be sure to restate the value as clearly as possible in your own terms. Then you'll be able to evaluate whether or not the value is good in itself or worth pursuing. Evidence: We'd all probably like to believe that the people we argue with will accept our claims and reasons as perfect and complete by themselves, but most readers just won't do that. They want evidence of some sort--facts, examples, statistics, expert testimony, among others--to back up our reasons. Take one more look at the umbrella. An argument without evidence is like an umbrella without a handle. It will keep the rain off your head but it sure is awkward to hold onto. To be believable and convincing, evidence should satisfy three conditions: • Sufficient: Is there enough evidence to convince a reasonable reader of the reason and the claim? • Credible: Does the evidence match with the readers' experience of the world? If not, does the evidence come from a source that readers would accept as more knowledgeable than they are? Evidence needs to avoid obvious bias. • Accurate: Does the evidence "tell the truth"? Are statistics gathered in verifiable ways from good sources? Are the quotations complete and fair (not out of context)? Are the facts verifiable from other sources? Objections: These are not your objections to an argument you read. Toulmin analysis doesn't call for you to get involved in the argument, except to analyze it. Rather, these are objections that the writer feels opponents on the other side of the argument might make. Usually, these are included in arguments so that writers can get to rebuttal or refutation. (Gentle reminder: writers don't help their overall credibility by misstating their opponents' objections or ignoring major objections in favor of minor objections that are easier to rebut.) Rebuttal: Once a writer identifies counter-arguments opponents might make, it would be silly to announce those counter-arguments and not argue against them. So after stating the objections of opponents, most writers will refute or rebut the objections. It's possible to construct an entire argument out of objections and rebuttal. Good rebuttal usually requires evidence, so don't forget to look for support for the rebuttal position in that part of an argument. Like all evidence, rebuttal evidence should be sufficient, accurate, and credible. Miscellaneous points: 1. There will be a range of responses on a Toulmin analysis because readers construct their own meanings as they read texts. When in doubt, always go back to the text. Also, don't assume that your first draft of a claim or reasons will be the best representation of the text's meaning. You will almost certainly see ways to capture the text's meaning more effectively (clearly, accurately) in a second or third draft of the claim, etc. 2. Different texts encourage more or less interpretation. At one end of a spectrum, we have highly metaphorical texts, like much poetry, that demand personal interpretation. At the other end, we have informative texts that direct you to, for instance, put together a bicycle. These should be so explicit that there's little variation in how readers read and respond to them. But there's lots of room in the middle for a wide variety of texts, especially arguments. Don't be surprised when another reader doesn't construct exactly the same meaning you have after reading an argument. So how can your teacher "grade" your Toulmin analysis? Your statement of the claim should cover all the reasons you state. Your explanation of the reasons should show how they support the claim. Your analysis of the evidence should show how sufficient, credible and accurate the evidence is for you. Teachers comment on the effectiveness of a Toulmin analysis, not on whether it's right or wrong.

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# ch24_p56 - 56(a Since the two conductors are connected V1... This preview shows page 1. Sign up to view the full content. 56. (a) Since the two conductors are connected V 1 and V 2 must be equal to each other. Let V 1 = q 1 /4 πε 0 R 1 = V 2 = q 2 /4 πε 0 R 2 and note that q 1 + q 2 = q and R 2 = 2 R 1 . We solve for q 1 and q 2 : q 1 = q /3, q 2 = 2 q /3, or (b) q 1 / q = 1/3 = 0.333, (c) and This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]} ### What students are saying • As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students. Kiran Temple University Fox School of Business ‘17, Course Hero Intern • I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero. Dana University of Pennsylvania ‘17, Course Hero Intern • The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time. Jill Tulane University ‘16, Course Hero Intern

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https://datascience.stackexchange.com/questions/65433/prediction-vs-causation-in-a-ml-project

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Prediction vs causation in a ML project I am performing a classification task and was able to identify significant predictors (important features using Random Forest) that can help separate the classes or influence the outcome. But I read online that prediction models are not causal models. Let's say if my prediction model says that Age is one of the significant factors that influence outcome (death), how can I prove that Age is the cause of death. I read that any intervention/change on strong predictors of your models, will not necessarily impact the outcome. How can I find out the list of factors that really cause change in the outcome? Currently what I do is run a RF model to identify the important features and communicate that these are the top 5 features that seem to influence the outcome. How can I prove that it is causation and not just correlation? ML questions are concerned mainly around predictions, but we can extrapolate to (in a certain sence) causality. First of all these are two different modelling approaches: Causal inference is focused on knowing what happens to $$Y$$ when you change $$X$$. Prediction is focused on knowing the next $$Y$$ given $$X$$ Some of the current causal approaches are randomised testing, do-calculus etc... So how can we extrapolate to causal inference of standard predictive ML models? Counterfactual Explanations we can simulate counterfactuals for predictions of machine learning models where we simply change the feature values of an instance before making the predictions and we analyze how the prediction changes. Read more about it here, and there is python library called alibi that implements it. • Thanks for the response. Upvoted. can PFI help us with causal inference as it is also about feature values (shuffling)? Commented Dec 26, 2019 at 10:30 • sure, yes it can Commented Dec 26, 2019 at 10:38 • Hi @Noah Weber, can you help me with this post? datascience.stackexchange.com/questions/69384/… Commented Mar 9, 2020 at 9:08 I would agree with your assessment of things. ML is much more concerned with making predictions and a discipline like Econometrics, or Statisitcs, for instance, strives to find causation between variables. ML excels at finding patterns in data and using these patterns for classification and prediction. Econometrics shares machine learners' interest in classification and prediction, as well as statisticians' concern with sample representativeness and sampling variance. As an aside, The discipline of statistics was born out of a desire to work with data efficiently, primarily by drawing relatively small samples from larger populations of interest instead of collecting data on everyone. As you know, people in the ML world will try to consume as much data as possible, whereas people in the statistics world, are taking samples of populations, with the understanding that a small subset is representative of an entire population, and that's good enough for the analysis that's being done. Now, back to your question about proving causation. Correlation is a statistical technique which tells us how strongly the pair of variables are linearly related and change together. Causation takes a step further than correlation; it says any change in the value of one variable will cause a change in the value of another variable, which means one variable makes another happen. This is referred as cause and effect. Essentially, we can infer causality from a well designed randomized controlled experiment. Randomized and controlled don't intuitively belong together, but it's a complex dynamic. Think of the predator-prey model. As the number of prey increase, more predators can exist, but too many predators will decimate the prey population, so the number of predators will diminish, and then the number of prey will increase. This cycle continues over and over. I did a quick Google search and came up with a couple of links, which seem to make a decent comparison between the two disciplines. https://towardsdatascience.com/why-correlation-does-not-imply-causation-5b99790df07e https://medium.com/causal-data-science/if-correlation-doesnt-imply-causation-then-what-does-c74f20d26438 Hope that helps!!! • Hi, Thanks for your response. Upvoted Commented Feb 29, 2020 at 3:06

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https://robotics.stackexchange.com/questions/16672/how-to-calculate-effort-torque-values-of-a-joint

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How to calculate effort/torque values of a joint I have a robotic arm with 6 DOF. For each joint, I get get the Joint Wrench, which is composed by the force and torque of the parent and child links of the joints. In other works, for joint 1 I can get ParentLinkForce, ParentLinkTorque, ChildLinkForce and ChildLinkTorque. Each of one is a 3D vector with x, y z. My question is, how can I transform this to the current effort of the joint in N.m? So I can directly get the effort value of the torque after applying a torque action in the joint. The wrench characterizes the forces and torques acting on the respective linkages. Part of this is motor torque, the other forces and torques are loading the mechanical structure (and motors down the chain). If you have followed the Denavit Hartenberg convention in defining the coordinate systems and you have a rotational joint the joint torque is the z component of the torque part either the parent of the child wrench (depending on the motor orientation). If you did not follow the DH convention, you should locate the axis which is co-located with the motor axis and use the torque component from the wrence corresponding to the co-located axis. If you do not have such an axis, you need to define a new coordinate system which has such an axis and make the coordinate system transformations form the parend/child coordinate system to the new one. Afterwards transform the wrench to the new coordinate system atteched to the motor and select the torque component corresponding to the co-located axis. Please make sure the regard any gearboxes which the motor might have when calculating any actual motor torques. • Hi,Please forgive my ignorance on this, my background is in AI and now I am working in robotics as well. Then I understand that if I follow the Denavit Hartenberg convention, the torque of a rotational joint is the Z component of the torque of the parent or the child? I am not sure what is the motor orientation since is a simulation :( Commented Nov 12, 2018 at 0:12 • Its just + or - the same value – 50k4 Commented Nov 12, 2018 at 5:06

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# Objects in mirror are closer than they appear For those without comedic tastes, the so-called experts at Wikipedia have an article about Objects in mirror are closer than they appear. Must. Go. Faster. Warning: Objects In Mirror Are Apache Helicopters. PIECE OF CRAP WARNING!This page is a piece of crap. The author acknowledges this fact. ## General Information/Bullcrap A standard warning blurb written on most right-hand side rear-view mirrors. Reasons for this are widely accepted as an S.E.P. (Somebody Else's Problem), but have you ever really thought about it? If you're sitting in the driver's seat on the left side of the car, and since the right-hand mirror is further away from you than your left-hand mirror, wouldn't you want objects in the right hand mirror to appear closer rather than further away (as in, Objects In Mirror Are Further Away Than They Appear)? I suppose that would be a little obvious, though. It's been determined mathematically that it would only take a total dumbfuck sitting in the driver's seat to actually believe the objects in the right-hand side rear view mirrors are the exact distance away from you that they claim to be, after all, you're veiwing them through a tiny mirror on the other side of the car, perilessly swerving to avoid the fallen branches on the muddy path, all with your gas pedal floored to stay well ahead of the rampaging t-rex hot on your heels. Must go faster. Furthermore, who in their right mind leans alllll the way over from the driver's side to actually read the tiny incoherent writing on the passenger side mirror? "Shit. What does that say? Objects in mirror are... what? Fuck I better pull over." Paradoxically, in cars in the UK and China that have their passenger sides and driver sides reversed, the blurb on the bottom of the left-hand side mirror states: Objects In This Mirror Are On The Other Side Of The Car. ## Historic Appearances • 450 A.D.: Julius Caesar invents the mirror, then writes: Res Intus Speculum Est Propinquus Quam Videor on it. He doesn't know why, as elliptical mirrors were only invented over a millenium later by Thomas Edison. • 1969: In this year, the drug known as LSD was still extremely popular among teenage hippies and philosophy professors. Accidents involving right-side mirrors at this time were caused by pie-eyed idiots seeing how close they could get to their own reflection, and either breaking the mirror with their faces and cutting themselves up pretty badly, or by crashing their car. • 1993: Jurassic Park Ranger Robert Muldoon spots the t-rex in his rear-view mirror, and because he is instantly reminded that Objects In Mirror Are Closer Than They Appear (the object being the t-rex), he is prompted to drive much faster and thus save everybody's life, especially Jeff Goldblum. Unfortunately, Muldoon is eaten by a raptor later in the movie. • 1998: The Narrator's face is thrown into a rear-view mirror by himself. He then checks himself out in the very same mirror to see how badly his mouth is cut. It is not known whether the mirror contained the Objects In Mirror statement or not. The first rule of Project Mayhem is You Do Not Ask Questions. • 2004: George W. Bush fails to heed the warning on his SUV's right-hand mirror and reverses over Condoleeza Rice. When questioned later by secret service, he explained that he's the president and he doesn't get a lot of time for reading.

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Game Development Reference In-Depth Information ( ) ( )   =+ 2 m rsE + m s 2 E kk 3 k 3 k k ( ) ( )   r s s r s s . (14.98) m m m kk kk k k k k Now, if the origin of the coordinate system coincides with the center of mass, then the summation k r is equal to the point 0, 0, 0 . This allows us to make a tremendous simplification because all of the terms in Equation (14.98) contain- ing this summation vanish. We therefore can use the formula kk ( )   ′ =+ ms 2 Ess −⊗ (14.99) 3 to transform an inertia tensor from a coordinate system in which the center of mass lies at the origin to another coordinate system in which the new origin lies at the point s in the original coordinate system. Note that the sign of s does not matter because it is squared in both terms where it appears in Equation (14.99). A translation by a certain distance in one direction produces the same inertia tensor as a translation by the same distance in the opposite direction. It's important to understand that Equation (14.99) can only be applied once to an inertia tensor in order to move it away from the center of mass. After the inertia tensor has been moved, it no longer uses a coordinate system in which the origin coincides with the center of mass, but that condition must be true for Equation (14.99) to be valid. However, it is possible to recover the inertia tensor from the offset inertia tensor if the vector s is known, once again allowing Equation (14.99) to be used to perform a new offset. Example 14.6. Determine the inertia tensor for an axis-aligned box of constant density ρ in a coordinate system where the box's center of mass lies at the origin. Solution. We already calculated the inertia tensor for a box in Example 14.4, but it was in a coordinate system where the origin coincided with one of the corners of the box. We can treat this as the transformed inertia tensor in Equation (14.99) and recover the inertia tensor about the center of mass by solving for with an offset = ,, . This gives us s a bc 222

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http://mathhelpforum.com/advanced-algebra/32387-rotation-matrix.html

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1. ## The Rotation Matrix The rotation matrix for an anti-clockwise rotation is |cos -sin| |sin cos| Find the matrix for a rotation in a clockwise direction using matrix methods. 2. Originally Posted by checky123 The rotation matrix for an anti-clockwise rotation is |cos -sin| |sin cos| Find the matrix for a rotation in a clockwise direction using matrix methods. $\displaystyle \begin{bmatrix} \cos(\theta) && -\sin(\theta) \\ \sin(\theta) && \cos(\theta) \\ \end{bmatrix}$ Is a counter clockwise rotation of angle theta note that $\displaystyle \cos(-\theta)=\cos(\theta) \mbox{ and } \sin(-\theta)=-\sin(\theta)$ so to rotate clockwise try negative theta $\displaystyle \begin{bmatrix} \cos(-\theta) && -\sin(-\theta) \\ \sin(-\theta) && \cos(-\theta) \\ \end{bmatrix} = \begin{bmatrix} \cos(\theta) && \sin(\theta) \\ -\sin(\theta) && \cos(\theta) \\ \end{bmatrix}$ I hope this helps Good luck

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https://sites.google.com/site/ransquared/cardiac/picco

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# PiCCO The most exciting thing about PiCCO for me is the ability to determine extravascular lung water and pulmonary vascular permeability. I don’t find GEDI terribly useful or even cardiac output as a number. I can already perform volume responsiveness estimation by manually measuring the sDown on an arterial waveform. Skip to the bottom to see extravascular lung water and pulmonary vascular permeability. But if you are interested in how PiCCO determines these magical numbers, read from the top for a detailed analysis. ## Concept and Definitions Extravascular Lung Water : EVLW Pulmonary blood volume + EVLW : Pulmonary Thermal Volume (PTV) RA + RV + LV + LA : Global End-Diastolic Volume (GEDV) PTV + GEDV : Intra-thoracic Thermal Volume (ITTV) ## Global End-Diastolic Volume Unlike a Swan, the cold saline bolus passes through the lungs before detection. This allows us to determine the total amount of fluid in the thorax also known as intrathoracic thermal volume (ITTV). Using the thermodilution curve, we can estimate the ITTV. ITTV = CO x MTt ITTI = ITTV / BSA (normal is 850-1000) More importantly, we can estimate the PTV because it is the largest mixing chamber and should produce the largest drop in the thermodilution curve (from 85% of max temperature response to 45% of max). The slope of this drop corresponds to the pulmonary thermal volume. PTV = CO x DSt ITTV is composed of the blood volumes in the heart (GEDV) as well as the total fluid volume in the lungs (PTV). This means that the GEDV can be calculated by calculated by simple arithmetic. GEDV = ITTV – PTV GEDV / BSA = GEDI (normal is 680-800) ## Cardiac Output Tb : Temperature of blood Ti : Temperature of infusate Vi : Volume of infusate Denominator is AUC ### Volume Responsiveness In mechanically ventilated patients with machine triggered and delivered breaths (8mL/kg IBW), one can use stroke volume variation or pulse pressure variation. PiCCO analyzes AUC of pulse pressure wave form to determine SVV. ## Extravascular Lung Water EVLW is my new favorite number. It is the most sensitive and specific measurement of pulmonary edema. It can even detect volumes left in the lung from a BAL within 30mL (Dres, CCM 2014). It is not affected by pleural effusions (Saugel, JCC 2013). There is a fantastic agreement between EVLW and postmortem (within 48h) lung weight (Tagami, CC 2010). There is great sensitivity and specificity with EVLW > 10 and there is 99% specificity with EVLW > 14. Measuring EVLW is based on the observation that the pulmonary vasculature hold 1.25x of the total blood volume in the heart (GEDV). Since EVLW + Pulmonary Blood Volume = PTV (see diagram) then simple substitution solves for EVLW: EVLW = PTV – 1.25x GEDV It’s also possible to use EVLW to calculate the pulmonary vascular permeability index (PVPI), which can help distinguish ARDS and hydrostatic edema. PVPI = EVLW / (1.25 x GEDV)

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This discussion is archived 1 2 Previous Next 27 Replies Latest reply: Jun 23, 2007 11:53 PM by 807600 Go to original post • ###### 15. Re: why   the  outprint  for  the  first number in loop only? Currently Being Moderated I'm sure it can be done more elegantly than this but thats up to you. Beautiful. Just f'n beautiful. This teaches the OP how to go sparrow hunting with an Uzi. Nuthin left but feathers. afaik its not homework and i purposly didn't made clean code so he can figureout how to do the for loop and whatever he wants to do with it. i don't see a problem with this. • ###### 16. Re: why   the  outprint  for  the  first number in loop only? Currently Being Moderated Biginteger? we don't want to mislead the OP either. • ###### 17. Re: why   the  outprint  for  the  first number in loop only? Currently Being Moderated ... //          Ark de Appel                www.engineeringserver.com // //---------------------------------------------------- ------------------*/ There is no need for such a disclaimer/header because AFAIK, what you just posted is no intellectual property of yours. You only demonstrated the use of some standard Java classes. Nothing more. import java.math.BigInteger; class Sun_Algho { public static void main(String [] args){ int isPrime = 137; int result = 0; int i = 1; System.out.println("Test Prime: "); for (int j = 0; j < 100; j++){ BigInteger bigInt = BigInteger.valueOf(j); if(bigInt.isProbablePrime(10)){ Why did you choose the number 10 in isProbablePrime(10)? ... System.out.println("Test# 2"); if (isPrime%i == 0){ result++; } if (result <=2){ System.out.println("Prime: " + isPrime + " is is Prime? Yes "); } else{ System.out.println("Prime: " + isPrime + " is is Prime? No "); ... What is this test #2? If you change isPrime to composite number, it'll be marked as a prime number! • ###### 18. Re: why   the  outprint  for  the  first number in loop only? Currently Being Moderated ... afaik its not homework and i purposly didn't made clean code so he can figureout how to do the for loop and whatever he wants to do with it. Sticking some lines of code in one (main) method is not helping the OP at all, IMHO. You could have at least used two different methods and pointed out that the OP needed to do some work with test #2. Just saying "it's not elegant code" doesn't help the OP much with what's wrong with it. i don't see a problem with this. I do. • ###### 19. Re: why   the  outprint  for  the  first number in loop only? Currently Being Moderated Its a template i use in eclipe for any new class that i make. Anyway i'll look into that. I probably forgot a check somewhere. • ###### 20. Re: why   the  outprint  for  the  first number in loop only? Currently Being Moderated Its a template i use in eclipe for any new class that i make. Anyway i'll look into that. I probably forgot a check somewhere. Yes, I figured that much. What I mean is you don't need to post it here if you're posting some lines of code. • ###### 21. Re: why   the  outprint  for  the  first number in loop only? Currently Being Moderated ... afaik its not homework and i purposly didn't made clean code so he can figureout how to do the for loop and whatever he wants to do with it. Sticking some lines of code in one (main) method is not helping the OP at all, IMHO. i don't see a problem with this. I do. Looking at my code again I suppose i could create a method for it and split the code so its easier to read and understand. • ###### 22. Re: why   the  outprint  for  the  first number in loop only? Currently Being Moderated Its a template i use in eclipe for any new class that i make. Anyway i'll look into that. I probably forgot a check somewhere. Yes, I figured that much. What I mean is you don't need to post it here if you're posting some lines of code. I totally agree on that. i'll work on my posting skill in my future posts :) Message was edited by: deAppel • ###### 23. Re: why   the  outprint  for  the  first number in loop only? Currently Being Moderated ... I totally agree on that. i'll work on my posting skill in my future posts :) • ###### 24. Re: why   the  outprint  for  the  first number in loop only? Currently Being Moderated • ###### 25. Re: why   the  outprint  for  the  first number in loop only? Currently Being Moderated BigInteger.valueOf(j); if(bigInt.isProbablePrime(10)){ Why did you choose the number 10 in isProbablePrime(10)? AFAIK, the bigger the argument in the isProbablyPrime method, the greater the accuracy, but the slower the calculation. But the OP doesn't need obscure approximation methods for calculating primes, the OP needs a precise simple algorithm. I liken the above to sparrow hunting w/ an uzi. Sure he can kill birds with it, but with what result? What useful information does he gain? • ###### 26. Re: why   the  outprint  for  the  first number in loop only? Currently Being Moderated ... Why did you choose the number 10 in isProbablePrime(10)? AFAIK, the bigger the argument in the isProbablyPrime method, the greater the accuracy, but the slower the calculation. ... Yes, I know. I was more interested in the motivation of the author of that code as to why s/he choose the number 10. ; ) But the OP doesn't need obscure approximation methods for calculating primes, the OP needs a precise simple algorithm. I liken the above to sparrow hunting w/ an uzi. Sure he can kill birds with it, but with what result? What useful information does he gain? Probably none. • ###### 27. Re: why   the  outprint  for  the  first number in loop only? Currently Being Moderated @vanpersie: to print prime elements between zero and 25 Then my suggestion is the following: Write a method that has the signature: boolean isPrime(int n) This method returns true if the number n is a prime and false otherwise. Use this method in the following code: ``````for (int i = 0; i < 26; i++) { if (isPrime(i)) { System.out.println("A prime: " + i); } }`````` Much clearer, don't you think? (It also seperates the showing of the result from your 'prime-determining-code' which allows you to focus better.) 1 2 Previous Next

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## Algebra 1 $x=6$ First square both sides: $x^2=5x+6$ Then move the $5x+6$ to the left-hand side to create a quadratic: $x^2-5x-6=0$ Use the quadratic formula to solve for possible values of $x$: The quadratic formula states that for a quadratic $ax^2+bx+c$, the two possible roots take the form $\frac{-b+\sqrt{b^2-4ac}}{2a}$ or $\frac{-b-\sqrt{b^2-4ac}}{2a}$. We see that the two possible roots of our equation are $\frac{5+\sqrt{49}}{2}=6$ and $\frac{5-\sqrt{49}}{2} = -1$. Looking back at our original equation, we see that $x$ is the square root of a number. Because the square root of a number cannot be negative, we see that $x=6$ is the only solution to this equation.

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# Godel's incompleteness; result of implication? You know, in all my discussions about Godel's theorem, I've never seen an example of a statement which is well formed and true, but not provable by that system. That would certainly help. I guess the trouble is as soon as you find one, it is immediately added as one of the axioms of a larger system. But that only argues for the definition of axioms to be such that each axiom is so fundamental it cannot be proven by the others. But's that's inherently obvious. Some other people have already shown theorems not provable in Peano arithmetic. For any "formal system" S; consider the assertion C(S) : "S is free of paradox". Proofs can be coded by finite sequences of symbols and the correctness of a proof can be checked by a Turing like machine. If S is strong enough to develop elementary arithmetic C(S) can be formalized in S. A paradox is a contradiction in the theory: i.e. the ability to proof a theorem and its contrary (and in this case we can show that the system can proof everything). Sure that if you consider a formal system S; you "believe" that S is free of paradox; so you believe that C(S) is true but it can be shown that C(S) is non provable in S (by Goedel). Of course since you have no proof you cannot be sure that C(S) is true. This has the consequence that if you want a formal system S in which you can develop all of mathematics; you cannot have a "convincing" proof P that S is consistent (because P would be reproducible in S; in contradiction with Goedel). The usual system that mathematician take to develop mathematics is ZFC (Zermelo Fraenkel set theory with the axiom of choice). Mathematicians "believe" it is consistent but it is exaggerate that they "know" it is consistent. The consistency of ZFC can be proved in stronger systems; say ZFC*; but it does not really help because if you consider ZFC* then you would be interested in the consistency of ZFC*; not ZFC alone and the problem appears again.

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• Page 2 out of 27 Search Results 'Linear Inequalities' • 1 Filter(s) ### All 27 results NURS3151 Lecture Notes Summary Foundations of Nursing Research.docx (0) \$15.49 NURS3151 Lecture Notes Summary Foundations of Nursing R NURS3151Foundations of Nursing ResearchLecture Notes (Week 1 – Week 9)Very Helpful For Midterm & Final ExamLecture Slides Week 1: Intro to nursing researchNursing ResearchSystematic inquiry designed to develop knowledge about phenomena important to nursing practice, profession, education.CASNCanadian Association of Schools of NursingBasic Nursing Research1. Extend base of knowledge
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3. nursing applicati... • Exam (elaborations) •  • 44 pages • • by WaldenUniversity • Quick View i x • MAT 540 Week 11 Final Exam Newly Taken 2016 (0) \$16.49 1. Which of the following could be a linear programming objective function? (Points : 5) Z = 1A 2BC 3D Z = 1A 2B 3C 4D Z = 1A 2B / C 3D Z = 1A 2B2 3D all of the above 2. Which of the following could not be a linear programming problem constraint? (Points : 5) 1A 2B 1A 2B = 3 1A 2B LTOREQ 3 1A 2B GTOREQ 3 3. Types of integer programming models are _____________. (Points : 5) total 0 – 1 mixed all of the above 4. The production manager for Beer etc. produces 2 kinds o... • Exam (elaborations) •  • 12 pages • • by mastersolution • Quick View i x • Saint GBA334 Final Exam 2015 (0) \$35.49 Question 1.1.Slack time in a network is the: (Points : 2) amount of time that an activity would take assuming very unfavorable conditions. shortest amount of time that could be required to complete the activity. amount of time that you would expect it would take to complete the activity. difference between the expected completion time of the project using pessimistic times and the expected completion time of the project using optimistic times. amount of time that an activity can be delayed witho... • Exam (elaborations) •  • 18 pages • • by mastersolution • Quick View i x • Specialized Maths D Syllabus Points Notes (0) \$21.47 This is very precious notes of mine which was created to study mathematics d syllabus easily it contains the following topics; 1. Decimals and standard form 2. Accuracy and Error 3. Powers and roots 4. Ratio & proportion visit the notes 5. Fractions, ratios 6. Percentages 7. Rational and irrational numbers 8. Algebra: simplifying and factorising 9. Equations: linear, quadratic, simultaneous 10. Rearranging formulae 11. Inequalities 12. Parallel lines, bearings, polygons 13. Areas and volumes, si... •  Book • Class notes •  • 67 pages • • by tomyjimy • Quick View i x • • College Algebra Section 1.1-2.4 (0) \$3.49 In class notes taken step by step along with professor with problems taken from "College Algebra by Gustofson and Hughes 12th edition". Subjects include: applications of quadratic equations, other types of equations, Inequalities, absolute value, Functions and graphs, difference quotient, rectangular coordinate system and graphing lines, linear functions and slopes. There is also an exam review at the very end of the PDF. This is from Alex Cuatok's math 105 class at MSJC. •  Book & Bundle • Class notes •  • 0 pages • • by lauranoelle • Quick View i x • • • #### Sell your knowledge on Stuvia Hundred thousands of people are searching for your content every day. You can easily upload your summaries to our platform and start earning money from your study notes. College Algebra Chapter 6.3--8.1 (0) \$3.49 Topics covered: Matrix algebra, determinants, Cramer's rule, partial fractions, graphs of inequalities, linear programming, conic section and quadratic system, ellipses, hyperbolas, solving nonlinear systems of equations, binomial theorem. Easy to read and there are plenty of notes I put beside the problem we were working on to remind me of rules-- all of these notes are included. 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The probability of completing the project in 7 or fewer days is the same as the probability of completing the project in 13 days or more. 1) True 2) False Question 2 The equation P(A|B) = P(AB)/P(B) is: 1) only relevant when events A and B are collectively exhaustive. 2) the marginal probability. 3) the formula for a cond... • Exam (elaborations) •  • 10 pages • • by mastersolution • Quick View i x • Chapter 1 - Lecture 1 (0) Free Lecture 1 of Chapter 1 College Algebra Course. Covers and provides examples basic equations, and inequalities, linear equations, rational equations, and lowest common denominator. • Class notes •  • 4 pages • • by haleyrohde • Quick View i x • Saint HCA498 Module 6 CBK-Post-Test latest 2016 (0) \$50.49 Question Question 1: Assume you have a normal distribution representing the likelihood of completion times. The mean of this distribution is 10, and the standard deviation is 3. The probability of completing the project in 7 or fewer days is the same as the probability of completing the project in 13 days or more. True False Question 2. Question : The difference in decision making under risk and decision making under uncertainty is that under risk, we think we know the probabilities of the stat...

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# How Do I Calculate a Mortgage PITI Payment? Calculating your total monthly PITI--principal, interest, taxes and insurance--payment is easy to do if you know all of the figures and percentages that affect your mortgage loan. Fixed-rate loans do not have any variance in principal and interest payments, but taxes and insurance can increase or decrease, which will make the monthly payment change. If you receive a statement from your lender and find discrepancies, call and ask for an explanation. Do some mortgage calculations of your own. #### 1 Find the correct loan amount, either by contacting your lender or your real estate agent. Decide whether you are going to use a 30-year mortgage term of 360 months or a 15-year mortgage term of 180 months. You also need to know the loan's actual interest rate. #### 2 Enter the loan amount, number of years and interest rate into a financial or mortgage calculator. Press the payment key to get the correct principal and interest payment. If you do not have a mortgage calculator, you can use an online version. You can also use mortgage factors to calculate your payment. Factors are math calculations that give you the amount of your payment per \$1,000, based on the number of years in the term and the interest rate. For example, a loan amount of \$200,000 divided by \$1,000 equals 200. The factor for 5 percent interest over 30 years is 5.36822; multiplied by 200, the result is \$1,073.64. This represents the monthly principal and interest payment for 30 years. #### 3 Find the insurance quote or the declarations page for your homeowners insurance policy, which is usually a yearly amount. Divide by 12 to get a monthly amount. For example, \$1,800 per year divided by 12 equals \$150 per month. Add this to the monthly mortgage payment. #### 4 Use your annual property tax bill, and divide by 12 to get a monthly amount. For example, a yearly tax amount of \$2,400 divided by 12 equals \$200. Add this to your monthly mortgage payment. #### 5 Use all of the above calculations to show a total monthly mortgage payment. Add the monthly payments of \$1,073.64 plus \$150 plus \$200 for a grand total of \$1,423.64. #### Tips • If you have an FHA loan, you will pay a monthly mortgage insurance premium. Using an FHA loan in the above example, multiply the loan amount of \$200,000 by 0.55 percent. This equals \$1,100 and represents the yearly premium. Divide by 12 to get the monthly mortgage insurance premium of \$91.67. Added to the above total payment, the entire monthly payment is now \$1,515.31. • Visit your local tax office to find if there are any tax credits or discounts that might reduce your property taxes. Also, speak with your homeowners insurance provider for possible discounts or reductions on your policy. #### Warnings • Making monthly mortgage payments on time or early is very important to your credit and scores. Should a late fee be assessed, it may be 5 percent of the principal and interest payment. In the above example, the late payment would add \$53.68 to your monthly payment. • Your lender may audit your escrow account of taxes and insurance each year. If there is not enough coming in to cover the yearly amount of taxes or insurance, your monthly payments will increase slightly to cover the deficit.

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Leaving a Cell Value Unchanged If a Condition Is False in Excel Introduction Excel is a powerful tool for organizing and analyzing data, but sometimes we encounter situations where we need to leave a cell value unchanged if a certain condition is false. Whether it's calculating formulas, applying conditional formatting, or creating complex spreadsheets, maintaining data integrity is crucial for accurate results and informed decision-making. Key Takeaways • Excel is a powerful tool for organizing and analyzing data, but maintaining data integrity is crucial for accurate results and informed decision-making. • The IF function in Excel allows us to conditionally change cell values based on certain conditions. • Cell references can be used within the IF statement to dynamically reference other cells in the spreadsheet. • Logical operators like AND and OR can be used in conjunction with the IF function to create more complex conditions. • Nested IF statements can be used to create multiple layers of conditions for more intricate data manipulation. • When using the IF function, it is important to organize data properly, avoid common mistakes, and consider alternative functions or methods for complex conditions. • Leaving cell values unchanged based on conditions helps maintain accuracy and efficiency in Excel data management. • Practicing and exploring different Excel functions can enhance data manipulation skills. Using the IF function The IF function is a powerful tool in Microsoft Excel that allows users to perform conditional calculations and execute specific actions based on the result of a logical test. This function helps in automating decision-making processes and creating more dynamic and adaptable spreadsheets. Overview of the IF function in Excel The IF function evaluates a specified condition and returns different values based on whether the condition is true or false. It follows a simple logic: if the condition is met, it executes one set of instructions; otherwise, it executes a different set of instructions. Syntax and usage of the IF function The syntax for the IF function is as follows: =IF(logical_test, value_if_true, value_if_false) • logical_test: This is the condition that we want to evaluate. It can be a comparison between two values, the result of a formula, or any logical expression that returns either true or false. • value_if_true: This is the value that Excel returns if the logical_test is true. It can be a number, text, formula, or a reference to another cell. • value_if_false: This is the value that Excel returns if the logical_test is false. It can also be a number, text, formula, or reference to another cell. The IF function can be nested to create more complex calculations by using multiple logical tests. It allows for the creation of intricate decision trees within a single formula. Example of using the IF function to conditionally change cell values Suppose we have a spreadsheet that contains a list of students and their corresponding test scores. We want to create a column that displays "Pass" if the score is above or equal to 70, and "Fail" otherwise. To achieve this, we can use the IF function in the following way: =IF(B2>=70, "Pass", "Fail") In this example, B2 is the cell containing the first test score. If the score in B2 is greater than or equal to 70, the formula will return "Pass". If the score is less than 70, it will return "Fail". When the formula is applied to the remaining cells in the column, it will dynamically update based on the value in each corresponding cell. The IF function can be further expanded to include additional logical tests and nested IF statements to incorporate multiple conditions and outcomes as required. Using the IF statement with cell references When working with Excel, you may often encounter situations where you need to conditionally change the value of a cell based on certain criteria. The IF statement in Excel is a powerful tool that allows you to do just that. By using cell references within the IF statement, you can create dynamic formulas that adapt to changes in your spreadsheet. Explanation of how to reference cells within the IF statement In order to reference cells within the IF statement, you need to use the cell address or range name as an argument of the function. The cell reference should be enclosed in square brackets and can include the sheet name, if necessary. For example, `=IF([Sheet1!A1]>5, "True", "False")` compares the value in cell A1 of Sheet1 with 5. Demonstrating the use of cell references in conditionally changing cell values Let's say you have a sales spreadsheet where you want to calculate a bonus for each salesperson based on their performance. You can use the IF statement with cell references to conditionally change the bonus values. • First, you would define the conditions for the bonus calculation. For example, if the sales amount is greater than \$10,000, the bonus would be 10% of the sales amount. If it is between \$5,000 and \$10,000, the bonus would be 5% of the sales amount. Otherwise, the bonus would be 2% of the sales amount. • Then, you would use the IF statement with cell references to implement these conditions. For example, the formula `=IF(B2>10000, B2*0.1, IF(B2>5000, B2*0.05, B2*0.02))` would calculate the bonus for the sales amount in cell B2. • The cell reference in the IF statement allows the formula to dynamically update the bonus calculation whenever the sales amount changes. Tips for correctly referencing cells in the IF statement • Always double-check the cell addresses or range names to ensure they are correctly referenced in the IF statement. A small mistake can lead to incorrect calculations. • Consider using named ranges instead of cell references to make your formulas more readable and easier to maintain. • If you are referencing cells in different sheets, make sure to include the sheet name in the cell reference to avoid errors. • Use relative cell references (e.g., A1) if you want the formula to adjust based on the position of the formula cells when copied or dragged. • Use absolute cell references (e.g., \$A\$1) if you want the formula to always refer to a specific cell, regardless of its position. Using the IF Function with Logical Operators Excel provides a powerful tool called the IF function that allows users to perform logical tests and return different values based on the results. By combining the IF function with logical operators such as AND and OR, you can create more complex conditions and make your formulas even more versatile. Introduction to Logical Operators in Excel Before diving into the examples, let's briefly discuss the logical operators available in Excel: • AND: The AND operator returns TRUE if all the conditions specified are true, and FALSE otherwise. • OR: The OR operator returns TRUE if at least one of the conditions specified is true, and FALSE otherwise. • NOT: The NOT operator reverses the logical value of its argument. For example, if a condition is true, NOT(condition) will return FALSE. Showcasing Examples of Using Logical Operators with the IF Function Now, let's explore some examples of how you can use logical operators in combination with the IF function: Example 1: Using the AND Operator Suppose you have a dataset with two columns: "Revenue" and "Expenses." You want to determine if both the revenue is greater than \$10,000 and the expenses are less than \$5,000. You can use the following formula: `=IF(AND(Revenue>10000, Expenses<5000), "Good", "Bad")` This formula will return "Good" if both conditions are met, and "Bad" otherwise. Example 2: Using the OR Operator Let's say you have a list of products, and you want to categorize them as either "Electronics" or "Non-electronics" based on whether the product is a smartphone OR a laptop. You can use the following formula: `=IF(OR(Product="Smartphone", Product="Laptop"), "Electronics", "Non-electronics")` This formula will return "Electronics" if the product is either a smartphone or a laptop, and "Non-electronics" otherwise. Utilizing Logical Operators to Leave a Cell Value Unchanged if a Condition is False One useful application of logical operators with the IF function is to leave a cell value unchanged if a condition is false. This can be achieved using a combination of the IF function and a simple logical operator. Example 3: Preserving the Original Value Suppose you have a dataset with a column called "Price" and another column called "Discount." You want to calculate the final price after applying the discount, but in cases where the discount is 0%, you want to leave the original price unchanged. You can use the following formula: `=IF(Discount=0%, Price, Price*(1-Discount))` This formula checks if the discount is 0%. If it is, the original price is returned. Otherwise, the discounted price is calculated. By understanding how to use logical operators with the IF function, you can leverage the power of Excel's built-in functions to create dynamic and intelligent spreadsheets. Whether you need to perform complex calculations or make decisions based on specific conditions, logical operators can help you achieve your goals efficiently and effectively. Using the IF function with nested IF statements The IF function is a powerful tool in Excel that allows you to perform logical tests and return different values based on the test results. One of the advanced features of the IF function is the ability to nest multiple IF statements within each other, creating a more complex condition and response structure. Definition and purpose of nested IF statements in Excel When a condition needs to be evaluated against multiple scenarios, the nested IF statement comes into play. It allows you to test multiple conditions in a hierarchical manner, executing different actions or returning different values based on the outcome of each condition. Step-by-step guide on constructing nested IF statements Constructing nested IF statements in Excel requires careful organization and attention to detail. Follow these steps to create effective nested IF statements: • Step 1: Determine the condition you want to evaluate. This can be based on specific cell values or logical expressions. • Step 2: Decide on the actions or values you want to return for each possible outcome of the condition. • Step 3: Begin the nested IF statement by writing the first IF function and specifying the condition to test. • Step 4: Define the action or value to return if the condition is true for the first IF statement. • Step 5: Use the nested IF function as the value_if_false argument for the previous IF statement, and repeat steps 3 and 4 for the next condition. • Step 6: Continue nesting IF statements as needed until all conditions have been evaluated. • Step 7: Specify the default action or value to return if none of the conditions are true. • Step 8: Close all parentheses and press Enter to complete the nested IF statement. Illustrating the use of nested IF statements to leave cell values unchanged The nested IF statements can be particularly useful when you want to leave a cell value unchanged if a certain condition is false. Here's an example: Let's say you have a column of numbers in cells A1 to A10, and you want to check if each number is greater than 5. If the number is greater than 5, you want to leave it unchanged. If the number is less than or equal to 5, you want to replace it with a blank cell. To achieve this, you can use a nested IF statement: ```=IF(A1>5, A1, "") ``` In this example, the first condition being tested is whether the value in cell A1 is greater than 5. If this condition is true, the value in cell A1 is returned. If the condition is false, an empty string is returned, effectively leaving the cell unchanged. You can then drag the formula down to apply it to the remaining cells in the column, automatically adjusting the cell references accordingly. This way, the nested IF statement will check and modify each cell value based on the defined condition. Best Practices for Leaving Cell Values Unchanged In Excel, it is often necessary to apply specific conditions to cell values and make changes accordingly. However, in some cases, you may want to leave a cell value unchanged if a certain condition is false. This can help maintain data integrity and prevent unintended modifications. Here are some best practices to follow when dealing with this situation: Organizing data in Excel to facilitate conditional formatting Before applying any conditions or formulas, it is important to organize your data in a way that makes it easy to identify and apply conditional formatting. This can be achieved by: • Using consistent naming conventions: Give meaningful names to your cells, ranges, and worksheets so that it becomes easier to reference them in formulas. • Adding headers and labels: Clearly label your columns and rows to make it easier to understand the purpose of each data field. • Sorting and filtering: Sort your data in a logical order and apply filters if necessary to focus on specific subsets of data. Avoiding common mistakes when using the IF function The IF function is a common tool for applying conditions in Excel. However, there are a few common mistakes to avoid when using it to leave cell values unchanged: • Not using the appropriate syntax: Make sure you are using the correct syntax for the IF function, including specifying the condition, value_if_true, and value_if_false arguments. • Forgetting to include the FALSE argument: If you want to leave the cell value unchanged when the condition is false, you need to include a value_if_false argument that references the original cell value. • Using incorrect cell references: Double-check that you are referencing the correct cells in your IF function to ensure that the condition is being applied to the desired cell and that the original value is correctly referenced. Considering alternative functions or methods for more complex conditions In some cases, the IF function may not be sufficient to address complex conditions that require multiple criteria or more advanced logic. In such scenarios, consider using alternative functions or methods, such as: • Nested IF functions: If your condition requires multiple criteria, you can nest multiple IF functions within each other to create more complex conditions. • AND or OR functions: The AND and OR functions can be used to combine multiple conditions within a single formula, allowing for greater flexibility in determining when to leave cell values unchanged. • Conditional formatting: Instead of using formulas, you can also utilize the conditional formatting feature in Excel to highlight or format cells based on specific conditions, leaving the actual cell values unchanged. By following these best practices, you can effectively leave cell values unchanged when certain conditions are false, ensuring the accuracy and integrity of your data in Excel. Conclusion Leaving a cell value unchanged if a condition is false is a crucial aspect of Excel data management. By understanding this concept and applying it effectively, users can maintain the accuracy and efficiency of their data analysis. It is important to recap the significance of leaving cell values unchanged based on conditions to ensure the integrity of the data. Additionally, we encourage users to continue practicing and exploring different Excel functions for data manipulation. Excel offers a wide range of powerful tools that can help streamline data management processes. By continuously learning and leveraging these functions, users can enhance their data analysis skills and achieve more precise results. In conclusion, maintaining accuracy and efficiency in Excel data management is a continuous journey that requires both knowledge and practice. SAVE \$698 ULTIMATE EXCEL TEMPLATES BUNDLE

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Cody # sea knowledge Rank Score 1 – 60 of 64 #### Commenter+10 Earned on 11 Dec 2012 for commenting on Solution 175748. #### Promoter+10 Earned on 13 Dec 2012 for liking Solution 175648. #### Creator+20 Earned on 11 Jan 2014 for creating Problem 2104. construct matrix with identical rows. #### Speed Demon+50 Earned on 11 Jan 2014 for first solving Problem 2104. construct matrix with identical rows. #### Quiz Master+20 Earned on 2 Dec 2017 for having 50 or more solvers for Problem 2104. construct matrix with identical rows. #### Community Group Solver+50 Solve a community group #### CUP Challenge Master+50 Solve all the problems in CUP Challenge problem group. #### Introduction to MATLAB Master+50 Solve all the problems in Introduction to MATLAB problem group. 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# Lower and Upper Bound Theory Lower and upper bound theory is a mathematical concept that involves finding the smallest and largest possible values for a quantity, given certain constraints or conditions. It is often used in optimization problems, where the goal is to find the maximum or minimum value of a function subject to certain constraints. 1. In mathematical terms, the lower bound of a set of numbers is the smallest number in the set, while the upper bound is the largest number. If a set has a lower and upper bound, it is said to be bounded. 2. Lower and upper bound theory can also be used to determine the range of possible values for a variable. For example, if we know that a certain variable must be between 0 and 1, we can say that its lower bound is 0 and its upper bound is 1. 3. The concept of lower and upper bounds is closely related to the concepts of infimum and supremum. The infimum of a set is the greatest lower bound of the set, while the supremum is the least upper bound of the set. 4. Overall, lower and upper bound theory is an important tool in mathematical analysis, optimization, and decision-making, as it allows us to determine the range of possible values for a quantity and identify the optimal value within that range. The Lower and Upper Bound Theory provides a way to find the lowest complexity algorithm to solve a problem. Before understanding the theory, first, let’s have a brief look at what Lower and Upper bounds are. • Lower Bound – Let L(n) be the running time of an algorithm A(say), then g(n) is the Lower Bound of A if there exist two constants C and N such that L(n) >= C*g(n) for n > N. Lower bound of an algorithm is shown by the asymptotic notation called Big Omega (or just Omega). • Upper Bound – Let U(n) be the running time of an algorithm A(say), then g(n) is the Upper Bound of A if there exist two constants C and N such that U(n) <= C*g(n) for n > N. Upper bound of an algorithm is shown by the asymptotic notation called Big Oh(O) (or just Oh). 1. Lower Bound Theory: According to the lower bound theory, for a lower bound L(n) of an algorithm, it is not possible to have any other algorithm (for a common problem) whose time complexity is less than L(n) for random input. Also, every algorithm must take at least L(n) time in the worst case. Note that L(n) here is the minimum of all the possible algorithms, of maximum complexity. The Lower Bound is very important for any algorithm. Once we calculated it, then we can compare it with the actual complexity of the algorithm and if their order is the same then we can declare our algorithm as optimal. So in this section, we will be discussing techniques for finding the lower bound of an algorithm. Note that our main motive is to get an optimal algorithm, which is the one having its Upper Bound the Same as its Lower Bound (U(n)=L(n)). Merge Sort is a common example of an optimal algorithm. Trivial Lower Bound – It is the easiest method to find the lower bound. The Lower bounds which can be easily observed based on the number of input taken and the number of output produced are called Trivial Lower Bound. Example: Multiplication of n x n matrix, where, ```Input: For 2 matrices we will have 2n2 inputs Output: 1 matrix of order n x n, i.e., n2 outputs ``` In the above example, it’s easily predictable that the lower bound is O(n2). Computational Model – The method is for all those algorithms that are comparison-based. For example, in sorting, we have to compare the elements of the list among themselves and then sort them accordingly. Similar is the case with searching and thus we can implement the same in this case. Now we will look at some examples to understand its usage. Ordered Searching – It is a type of searching in which the list is already sorted. Example-1: Linear search Explanation – In linear search, we compare the key with the first element if it does not match we compare it with the second element, and so on till we check against the nth element. Else we will end up with a failure. Example-2: Binary search Explanation – In binary search, we check the middle element against the key, if it is greater we search the first half else we check the second half and repeat the same process. The diagram below there is an illustration of binary search in an array consisting of 4 elements Calculating the lower bound: The max no of comparisons is n. Let there be k levels in the tree. 1. No. of nodes will be 2k-1 2. The upper bound of no of nodes in any comparison-based search of an element in the list of size n will be n as there are a maximum of n comparisons in worst case scenario 2k-1 3. Each level will take 1 comparison thus no. of comparisons k≥|log2n| Thus the lower bound of any comparison-based search from a list of n elements cannot be less than log(n). Therefore we can say that Binary Search is optimal as its complexity is Θ(log n). Sorting – The diagram below is an example of a tree formed in sorting combinations with 3 elements. Example – For n elements, finding lower bound using computation model. Explanation – For n elements, we have a total of n! combinations (leaf nodes). (Refer to the diagram the total combinations are 3! or 6) also, it is clear that the tree formed is a binary tree. Each level in the diagram indicates a comparison. Let there be k levels => 2k is the total number of leaf nodes in a full binary tree thus in this case we have n!≤2k. As the k in the above example is the no of comparisons thus by computational model lower bound = k. ```Now we can say that, n!≤2T(n) Thus, T(n)>|log n!| => n!<=nn Thus, log n!<=log nn Taking ceiling function on both sides, we get |-log nn-|>=|-log n!-| Thus complexity becomes Θ(lognn) or Θ(nlogn) ``` Using Lower bond theory to solve the algebraic problem: • Straight Line Program – The type of program built without any loops or control structures is called the Straight Line Program. For example, ## C `//summing to nos` `Sum(a, b)` `{` `    ``//no loops and no control structures` `    ``c:= a+b;` `    ``return` `c;` `}` • Algebraic Problem – Problems related to algebra like solving equations inequalities etc. come under algebraic problems. For example, solving equation ax2+bx+c with simple programming. ## C `Algo_Sol(a, b, c, x)` `{ ` `    ``//1 assignment` `    ``v:=a*x; `   `    ``//1 assignment` `    ``v:=v+b;`   `    ``//1 assignment` `    ``v:=v*x;`   `    ``//1 assignment` `    ``ans:=v+c;` `    ``return` `ans;` `}    ` The complexity for solving here is 4 (excluding the returning). The above example shows us a simple way to solve an equation for a 2-degree polynomial i.e., 4 thus for nth degree polynomial we will have a complexity of O(n2). Let us demonstrate via an algorithm. Example: x+a0 is a polynomial of degree n. ## C `pow``(x, n) ` `  ``{ ` `    ``p := 1; ` `  `  `    ``//loop from 1 to n ` `    ``for` `i:=1 to n ` `        ``p := p*x; `   `    ``return` `p; ` `  ``} `   `polynomial(A, x, n) ` `  ``{ ` `     ``int` `p, v:=0; ` `     ``for` `i := 0 to n ` `  `  `         ``//loop within a loop from 0 to n` `         ``v := v + A[i]*``pow``(x, i); ` `     ``return` `v; ` `  ``} ` Loop within a loop => complexity = O(n2); Now to find an optimal algorithm we need to find the lower bound here (as per lower bound theory). As per Lower Bound Theory, The optimal algorithm to solve the above problem is the one having complexity O(n). Let’s prove this theorem using lower bounds. Theorem: To prove that the optimal algo of solving a n degree polynomial is O(n) Proof: The best solution for reducing the algo is to make this problem less complex by dividing the polynomial into several straight-line problems. ```=> anxn+an-1xn-1+an-2xn-2+...+a1x+a0 can be written as ((..(anx+an-1)x+..+a2)x+a1)x+a0 Now, the algorithm will be as, v=0 v=v+an v=v*x v=v+an-1 v=v*x ... v=v+a1 v=v*x v=v+a0 ``` ## C `polynomial(A, x, n) ` `     ``{` `      ``int` `p, v=0; `   `      ``// loop executed n times` `      ``for` `i = n to 0 ` `             ``v = (v + A[i])*x;`   `         ``return` `v;` `      ``}` The complexity of this code is O(n). This way of solving such equations is called Horner’s method. Here is where lower bound theory works and gives the optimum algorithm’s complexity as O(n). 2. Upper Bound Theory: According to the upper bound theory, for an upper bound U(n) of an algorithm, we can always solve the problem at most U(n) time. Time taken by a known algorithm to solve a problem with worse case input gives us the upper bound. It’s difficult to provide a comprehensive list of advantages and disadvantages of lower and upper bound theory, as it depends on the specific context in which it is being used. However, here are some general advantages and disadvantages: 1. Provides a clear understanding of the range of possible values for a quantity, which can be useful in decision-making. 2. Helps to identify the optimal value within the range of possible values, which can lead to more efficient and effective solutions to problems. 3. Can be used to prove the existence of solutions to optimization problems. 4. Provides a theoretical framework for analyzing and solving a wide range of mathematical problems.

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Ask Dr. Math College Archive Dr. Math Home || Elementary || Middle School || High School || College || Dr. Math FAQ TOPICS This page:   higher-dimensional   geometry    Search   Dr. Math See also the Dr. Math FAQ:   geometric formulas Internet Library:   higher-dimensional   geometry COLLEGE Algorithms Analysis Algebra    linear algebra    modern algebra Calculus Definitions Discrete Math Exponents Geometry    Euclidean/plane      conic sections/        circles      constructions      coordinate plane      triangles/polygons    higher-dimensional      polyhedra    non-Euclidean Imaginary/Complex   Numbers Logic/Set Theory Number Theory Physics Probability Statistics Trigonometry Browse College Higher-Dimensional Geometry Stars indicate particularly interesting answers or good places to begin browsing. Changing Angle of a Tank [06/11/2003] Points A and B represent pressure sensors in fixed positions on the base of a round tank. 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How may the resulting ellipse be scribed? - i.e., how do I mark the hole to be cut in the deck? Definitions of Edge and Face in 2D and 3D [10/10/2008] What is the 'official' definition of 'edge'? Specifically, is an edge restricted to the intersection of two non-coplanar faces or do two- dimensional shapes have edges? I'm also curious about a definition of 'face'. How many faces does a two-dimensional shape have? Derivation of Formula for Surface Area of Torus [05/05/2005] I saw the formula for the surface area of a torus on your web site, and I'm wondering if you can show me how that formula was derived? Derivation of Geometric Formulas [5/29/1996] What are the formulas for the surface area, total surface area, and volume of a sphere, and volume of a pyramid and cone? Deriving the Area of a Sphere [10/21/2003] I know the area of a sphere is 4phi(r^2), but I'm wondering how to derive that formula. 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Equation of a Sphere [09/17/1997] We are trying to find an equation that would solve for the center point and radius of a sphere given three points in 3D space. Finding a Best-Fit Regression Plane [12/15/2005] I have a height map of a terrain where the x and y values are fixed. I can calculate best-fit slopes in the x and y directions, but I can't figure out how to combine them into a best-fit regression plane. Finding an Ellipse in 3-D Space [06/14/2005] How can I fit an ellipse to a set of 3-dimensional data points? Finding Dimensions of Elliptical Shadow Cast on Plane [01/19/2005] A sphere is positioned above a plane and light is shown on it from one side, resulting in an elliptical shadow on the plane. How can you determine the lengths of the axes of the resulting ellipse? Finding Points on the Earth [06/08/2001] Find the point that has latitude and longitude five miles north of a given point, and the other three points to the south, east, and west. 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# Search by Topic #### Resources tagged with Working systematically similar to Advent Calendar 2010: Filter by: Content type: Stage: Challenge level: ##### Other tags that relate to Advent Calendar 2010 Interactivities. Counting. Visualising. Dice. Cubes. Nets. Mathematical reasoning & proof. Working systematically. Networks/Graph Theory. Games. ### There are 320 results Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically ### Diagonal Sums Sudoku ##### Stage: 2, 3 and 4 Challenge Level: Solve this Sudoku puzzle whose clues are in the form of sums of the numbers which should appear in diagonal opposite cells. ### Intersection Sums Sudoku ##### Stage: 2, 3 and 4 Challenge Level: A Sudoku with clues given as sums of entries. ### Games Related to Nim ##### Stage: 1, 2, 3 and 4 This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning. ### Maths Trails ##### Stage: 2 and 3 The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind maths trails. ### Magnetic Personality ##### Stage: 2, 3 and 4 Challenge Level: 60 pieces and a challenge. What can you make and how many of the pieces can you use creating skeleton polyhedra? ### I've Submitted a Solution - What Next? ##### Stage: 1, 2, 3, 4 and 5 In this article, the NRICH team describe the process of selecting solutions for publication on the site. ### Colour in the Square ##### Stage: 2, 3 and 4 Challenge Level: Can you put the 25 coloured tiles into the 5 x 5 square so that no column, no row and no diagonal line have tiles of the same colour in them? ### The Naked Pair in Sudoku ##### Stage: 2, 3 and 4 A particular technique for solving Sudoku puzzles, known as "naked pair", is explained in this easy-to-read article. ### Sandwiches ##### Stage: 2, 3, 4 and 5 Challenge Level: Arrange the digits 1, 1, 2, 2, 3 and 3 so that between the two 1's there is one digit, between the two 2's there are two digits, and between the two 3's there are three digits. ### Creating Cubes ##### Stage: 2 and 3 Challenge Level: Arrange 9 red cubes, 9 blue cubes and 9 yellow cubes into a large 3 by 3 cube. No row or column of cubes must contain two cubes of the same colour. ### Two Dice ##### Stage: 1 Challenge Level: Find all the numbers that can be made by adding the dots on two dice. ### Find the Difference ##### Stage: 1 Challenge Level: Place the numbers 1 to 6 in the circles so that each number is the difference between the two numbers just below it. ### Are You Well Balanced? ##### Stage: 1 Challenge Level: Can you work out how to balance this equaliser? You can put more than one weight on a hook. ### Difference ##### Stage: 2 Challenge Level: Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it. ### Making Maths: Double-sided Magic Square ##### Stage: 2 and 3 Challenge Level: Make your own double-sided magic square. But can you complete both sides once you've made the pieces? ### Winning the Lottery ##### Stage: 2 Challenge Level: Try out the lottery that is played in a far-away land. What is the chance of winning? ### More Transformations on a Pegboard ##### Stage: 2 Challenge Level: Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle. ### Three Sets of Cubes, Two Surfaces ##### Stage: 2 Challenge Level: How many models can you find which obey these rules? ### Junior Frogs ##### Stage: 1 and 2 Challenge Level: Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible? ### Tetrafit ##### Stage: 2 Challenge Level: A tetromino is made up of four squares joined edge to edge. Can this tetromino, together with 15 copies of itself, be used to cover an eight by eight chessboard? ### Inky Cube ##### Stage: 2 and 3 Challenge Level: This cube has ink on each face which leaves marks on paper as it is rolled. Can you work out what is on each face and the route it has taken? ### Red Even ##### Stage: 2 Challenge Level: You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters? ### Page Numbers ##### Stage: 2 Short Challenge Level: Exactly 195 digits have been used to number the pages in a book. How many pages does the book have? ### A Bag of Marbles ##### Stage: 1 Challenge Level: Use the information to describe these marbles. What colours must be on marbles that sparkle when rolling but are dark inside? ### Fault-free Rectangles ##### Stage: 2 Challenge Level: Find out what a "fault-free" rectangle is and try to make some of your own. ### Twinkle Twinkle ##### Stage: 2 and 3 Challenge Level: A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour. ### A Square of Numbers ##### Stage: 2 Challenge Level: Can you put the numbers 1 to 8 into the circles so that the four calculations are correct? ### Mrs Beeswax ##### Stage: 1 Challenge Level: In how many ways could Mrs Beeswax put ten coins into her three puddings so that each pudding ended up with at least two coins? ### Counters ##### Stage: 2 Challenge Level: Hover your mouse over the counters to see which ones will be removed. Click to remover them. The winner is the last one to remove a counter. How you can make sure you win? ### Triangle Edges ##### Stage: 1 Challenge Level: How many triangles can you make using sticks that are 3cm, 4cm and 5cm long? ### 1 to 8 ##### Stage: 2 Challenge Level: Place the numbers 1 to 8 in the circles so that no consecutive numbers are joined by a line. ### Counting Cards ##### Stage: 2 Challenge Level: A magician took a suit of thirteen cards and held them in his hand face down. Every card he revealed had the same value as the one he had just finished spelling. How did this work? ### A Bit of a Dicey Problem ##### Stage: 2 Challenge Level: When you throw two regular, six-faced dice you have more chance of getting one particular result than any other. What result would that be? Why is this? ### Teddy Town ##### Stage: 1, 2 and 3 Challenge Level: There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules? ### Seven Flipped ##### Stage: 2 Challenge Level: Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time. ### Tessellate the Triominoes ##### Stage: 1 Challenge Level: What happens when you try and fit the triomino pieces into these two grids? ### Factor Lines ##### Stage: 2 and 3 Challenge Level: Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line. ### An Introduction to Magic Squares ##### Stage: 2, 3 and 4 Find out about Magic Squares in this article written for students. Why are they magic?! ### Arrangements ##### Stage: 2 Challenge Level: Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....? ### The Third Dimension ##### Stage: 1 and 2 Challenge Level: Here are four cubes joined together. How many other arrangements of four cubes can you find? Can you draw them on dotty paper? ### Making Trains ##### Stage: 1 Challenge Level: Can you make a train the same length as Laura's but using three differently coloured rods? Is there only one way of doing it? ### Combining Cuisenaire ##### Stage: 2 Challenge Level: Can you find all the different ways of lining up these Cuisenaire rods? ### Triangles All Around ##### Stage: 2 Challenge Level: Can you find all the different triangles on these peg boards, and find their angles? ### Code Breaker ##### Stage: 2 Challenge Level: This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code? ### One to Fifteen ##### Stage: 2 Challenge Level: Can you put the numbers from 1 to 15 on the circles so that no consecutive numbers lie anywhere along a continuous straight line? ### Cuisenaire Counting ##### Stage: 1 Challenge Level: Here are some rods that are different colours. How could I make a dark green rod using yellow and white rods? ### Nine-pin Triangles ##### Stage: 2 Challenge Level: How many different triangles can you make on a circular pegboard that has nine pegs? ### More on Mazes ##### Stage: 2 and 3 There is a long tradition of creating mazes throughout history and across the world. This article gives details of mazes you can visit and those that you can tackle on paper. ### Dice Stairs ##### Stage: 2 Challenge Level: Can you make dice stairs using the rules stated? How do you know you have all the possible stairs? ### Neighbours ##### Stage: 2 Challenge Level: In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?

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CC-MAIN-2017-47

latest

en

http://www.batterytender.com/blog/engineering-corner-batteries-are-constantly-going-dead

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Batteries are constantly going dead from vehicles with newer electronics or ones that are only being driven every few days... so naturally most people will assume that if they disconnect the battery from the vehicle that the drain on the battery will stop. While speaking with our Engineering department and technical writer Dan Williams he was able to clear up the common assumptions that lead to battery pitfalls. "There is a lot of misinformation out there. Just like the seemingly logical misconception that a 12 volt lead acid battery has a maximum voltage of 12 volts when it is fully charged. For a quick overview, disconnecting the battery from the motorcycle (or vehicle) electrical system does make perfect sense to avoid the continuous drain of electrical / electronic parasitic loads. The amount of some vehicle parasitic current draw is significant, relative to the size of the battery. Let's say that the load is 1 amp, which we would consider to be pretty high. A large motorcycle battery would have a capacity of say 25 to 30 amp-hours. With a 1 amp draw, that battery will be totally dead in 25 to 30 hours. If the battery is used to start a high compression engine, then with the capacity below 50%, in this case 12 to 15 amp hours, there may not be enough energy to support multiple engine start 3-5 second bursts to attempt to start the engine. I would guess that 30% State Of Charge (SOC) would normally be enough to start a normal vehicle engine. Let me just say that for a battery in reasonably good condition, there is a good correlation between the battery voltage measured at its terminals and the state of charge. So if we take 12.9V-11.4V = 1.5V Then that 1.5V represents the full 100% range of battery capacity, or stated as a ratio, then 0.015V on the battery correlates with 1% battery capacity. This is only true with no load and no charger connected. A typical 12V lead acid battery has 6 nearly identical cells that are internally connected in series. The nominal cell voltage ranges from 2.15V to 1.9V. That means a typical 12V battery has terminal voltages that ranges from 12.9V (fully charged = 6 x 2.15) to 11.4V (fully discharged = 6 x 1.9V). With that assumption, and my previous statement about 50% capacity, I would think that if the battery voltage is less than 12.1V, than you may not be able to start the vehicle. I would give it a 50/50 chance. A battery will support a higher voltage when it is connected to a charger. In fact, our Battery Tender battery chargers will take the charge voltage up to nearly 15 volts on a 12V lead acid battery, before dropping down to float levels of about 13.5 volts. When the charger is disconnected, it might take a few minutes for the battery voltage to settle at its fully charged voltage around 12.9 volts. You will see an immediate drop as soon as you disconnect the charger." This is why it is imperative to constantly have a vehicle or a battery that sits on a proper smart charger!

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CC-MAIN-2019-39

latest

en

http://golfsoftware.freshdesk.com/support/solutions/articles/4000038211-we-play-in-teams-and-award-point-s-for-lowest-team-aggregate-net-score-on-each-hole-and-lowest-team-

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crawl-data/CC-MAIN-2020-40/segments/1600400206329.28/warc/CC-MAIN-20200922161302-20200922191302-00678.warc.gz

To begin, make sure you are on event #1 in your League Manager Desktop Software. You will find the Point Calculation Parameters by clicking on Setup menu. STEP 1 - View Tab 2. Player vs Player • Delete the "1" from the points per hole box. This is a default. Type "0" (zero). • Click Next to advance to Tab 3. Step 2 - View Tab 3 Team vs Team • Note: in a team competition, we call it an aggregate competition because points are awarded based on the aggregate total of team member net scores on each hole. • In this example we have chosen to award 1 point per hole AND 2 points to the team with the best total net score for the round. • Notice you will need to enter your specific league parameters in the left and middle boxes that the arrows are pointing to. • Ties are divided evenly. • Notice the middle box gives you an additional option. You may select to have your team scores based on the lowest "x" out of "y" scores. Step 3 - Make sure there are no other points entered on tabs 4 through 6. Step 4 - View Tab 7 Other: • Click the Finish button to save your settings. Step 5 - Setting the correct option in Handicap Calculation Parameters. • Make sure you are viewing event #1. • From the Setup, click Handicap Calculation Parameters • Click the OK check mark at the lower right corner of the first page of the Setup Wizard on the first page. This will advance you to the screen that looks like the image below. • As shown in the image below, make sure your radio button is set to A. Player vs Player then click the Next button, • Click the Next button again and the Finish button. This will save your settings. • At Tab 2 - " Handicap Strokes for Teams" - you need to make a choice - Which way will your league distribute handicap strokes? • Use Team Strokes • Use Individual Strokes ADDENDUM: Scorecard: We play in teams and award point(s) for lowest team aggregate net score on each hole AND lowest team aggregate net score for round We will use two (2) scorecards to demonstrate the difference in results based on using team strokes or player strokes. Note that Hole 5 is where there is a difference between the two methods of awarding handicap strokes. Using TEAM STROKES - Example 1: • There are 4 strokes awarded to Team 11 based on handicaps and they are marked in red (+) to indicate team strokes and circled in red for demonstration purposes. • Notice under the team stroke concept, strokes are awarded on handicap holes 1, 3, 5, and 7. • Aggregate net hole by hole results are marked with red ovals: Team 1 and Team 11 tie and earn 4.5 points toward the match. • The winner of 2 points for lowest team aggregate net score for the round is Team 11. The net score for team 11 is 60 versus team 1 at 63. • The overall winner with 6.5 points is Team 11. Using Player Strokes - Example 2 • There are 4 strokes based on handicaps and they are marked in black (+) to indicate player strokes. These strokes are circled in red and the accompanying holes. • Notice under the player stroke concept, strokes are awarded on handicap holes 1, 3,and  5. • Aggregate net hole by hole results are marked with red ovals: Team 1 earns 4.0 points. Team 11 earns 5.0 points. • The winner of 2 points for lowest team aggregate net score for the round is Team 11. The net score for team 11 is 60 versus team 1 at 63. • The overall winner for the match with 7.0 points is Team 11.

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CC-MAIN-2020-40

latest

en

https://freegames-1.com/bets/what-numbers-have-the-best-odds-in-craps.html

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# What numbers have the best odds in craps? Contents You can bet on other point numbers but the probability of landing on these are lower and the house edge increases. On 4 0r 10, the probability is only 8.33% and the house edge is 6.7%. On a 5 or 9, the probability is 11.11% and the house edge is 4%, so the 6 and/or 8 is the best bet. ## What are the best odds bets in craps? The best craps bet in terms of odds is the don’t pass/don’t come bet, which gives the house an edge of just 1.36%. To make things even better, you can use the “free odds” option once the point is set and reduce the house edge to as low as 0.01%. ## What is the safest bet in craps? The simplest, most fundamental bet in the game of craps, the pass bet, is also one of the very safest, with a low house edge of 1.41%. Pass bets pay even money – in other words, if you bet \$10, you win \$10. With a pass bet, if the come out roll is 7 or 11, you win, while if the come out roll is 2, 3, or 12, you lose. IT IS INTERESTING:  How do you play Lucky Dip lottery? ## What is the best roll in craps? Among standard craps bets, the one-roll bet is the field. You get even money if the shooter rolls 3, 4, 9, 10 or 11. You also win on 2 and 12. If both of those pay 2-1, the house edge is 5.56 percent. ## How do you win at craps every time? Every time you have a winning pass line bet, raise it 30 percent and take double odds. Every time a place bet hits, take another number until you have them all covered. After that, every time a place bet hits, press it about 30 to 40 percent. ## What is the best strategy for winning at craps? The best strategy to win at craps relies on bets that minimize the house edge and give players the best odds of landing. To win the most amount of money while keeping the house edge low, we recommend players bet the minimum limit on Pass Line/Don’t Pass bets, and then lay the odds. ## What are the odds of rolling a 7 in craps? The odds of rolling a 7 are 6 out of 36 possible combinations, or 1 in 6. By contrast, there’s only one possible combination for a 2 or 12 when rolling the dice, meaning the odds for landing either are 1 out of 36. After pass bets, players should employ a betting strategy called laying the odds. ## How do you increase your odds in craps? Having a solid craps strategy is the best way to increase your odds of winning, and winning is the best way to increase your enjoyment of the game. But just like any skill, you’ll get better results with practice. ## Can you consistently win at craps? In the end, there’s nothing you can do to win consistently in craps, other than making pass line and don’t pass line bets backed with odds. ## What is the average number of rolls for craps? The average number of rolls per shooter is 8.525510. For the probability of exactly 2 to 200 rolls, please see my craps probability of survival page. ## What is the don’t pass line in craps? A Don’t Pass bet is a bet for the shooter to lose (“seven out, line away”) and is almost the opposite of the Pass line bet. Like the Pass bet, this bet must be at least the table minimum and at most the table maximum. If the come-out roll is 2 or 3, the bet wins. If the come-out roll is 7 or 11, the bet loses. ## What is the hardest number to roll on a dice? Probability of rolling more than a certain number (e.g. roll more than a 5). Roll more than a… Probability 1 5/6(83.33%) 2 4/6 (66.67%) 3 3/6 (50%) 4 4/6 (66.667%)

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CC-MAIN-2022-05

latest

en

https://www.enotes.com/homework-help/determine-total-distance-traveled-displacement-446010

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# determine the total distance traveled and the displacement made. A. 5.0 km 15 degrees east of north B. 5.0 km 20 degrees east of south C. 5.0 km east D. 5.0 km 30 degrees south of west E.... determine the total distance traveled and the displacement made. A. 5.0 km 15 degrees east of north B. 5.0 km 20 degrees east of south C. 5.0 km east D. 5.0 km 30 degrees south of west E. 5.0 km 20 degrees east of south F. 5.0 km 40 degrees north of west G. 5.0 km 40 degrees west of south H. 5.0 km 15 degrees east of north I. 5.0 km 60 degrees west of north J. 5.0 km east llltkl | Student Beginning the journey at the origin, and considering the directions and 5 kM distance travelled in each direction, the co-ordinates of points A through J can be worked out by adding/subtracting sine and cosine functions of the subtended angles, multiplied by 5 in each position. These coordinates are as follows: Point      x-coordinate       y-coordinate                                   --------- ---------------- ------------------- A      1.2941         4.8296 B      3.0042        0.1312 C      8.0042        0.1312 D      3.6741       -2.3688 E      5.3842        -7.0673 F       1.5539        -3.8534 G       -1.6600       -7.6836 H       -0.3659       -2.8540 I        -4.6960       -0.3540 J         0.3040       -0.3540 ----------------------------------------- Finally, the overall displacement (i.e. distance between origin, O and final point, J) is obtained from the two-point distance formula: OJ`= [sqrt((0.3040-0)^2+(-0.3540-0)^2)] =0.467` kM Distance travelled = 5*10=50 kM.

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CC-MAIN-2018-05

latest

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https://dsplog.com/2012/12/15/gate-2012-ece-q2-communication/

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crawl-data/CC-MAIN-2023-40/segments/1695233510427.16/warc/CC-MAIN-20230928162907-20230928192907-00187.warc.gz

Articles # GATE-2012 ECE Q2 (communication) Question 52 on communication from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper. ## Solution For a wide sense stationary function, the auto-correlation with delay $\tau$ is defined as, $R$$\tau$$=E$X(t+\tau)X^*(t)$$ From the Weiner-Kinchin theorem, the auto-correlation function $R$$\tau$$$ and power spectral density $S(f)$ are Fourier Transform pairs, i.e. $R$$\tau$$=\int_{-\infty}^{\infty}S$$f$$e^{j2\pi f\tau}df$ Expressing in terms of $\omega$, where $\omega=2\pi f$ $R$$\tau$$=\frac{1}{2\pi}\int_{-\infty}^{\infty}S$$w$$e^{jw\tau}dw$. When the delay $\tau=0$, the above equations simplifies to $\begin{array}R$$0$$&=&E$X(t)X^*(t)$&=&E$|X(t)|^2$&=&\frac{1}{2\pi}\int_{-\infty}^{\infty}S$$w$$dw\end{array}$. Applying this to the problem at hand, $\begin{array}{lll}E$|X(t)|^2$&=&\frac{1}{2\pi}\int_{-\infty}^{\infty}S$$w$$dw\\&=&\frac{2}{2\pi}\int_{0}^{\infty}S$$w$$dw\\&=&\frac{2}{2\pi}$$\frac{1}{2}6*2*10^3 + 400$$\\&=&\Large{\frac{6400}{\pi}}\end{array}$. Further, since the power spectral density $S(f)$ does not have any dc component, the mean of the signal $\|E$X(t)$\|=0$ Based on the above, the right choice is (B) $\Large{6400/\pi, \mbox{ } 0}$ ## References [1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

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CC-MAIN-2023-40

longest

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https://www.cfd-online.com/Forums/cfx/72392-cfx-emag.html

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# Cfx emag Register Blogs Members List Search Today's Posts Mark Forums Read February 4, 2010, 06:20 Cfx emag #1 New Member   Join Date: Dec 2009 Location: Berlin, Germany Posts: 5 Rep Power: 9 Sponsored Links Hi, I try to simulate the electric and magnetic field around a current-carrying copper wire in air (concentric cylindrical domains) as a preliminary test for magnetohydrodynamic simulations. I use version 12.0 with EMAG beta-feature. At the ends of the cylinders, boundary conditions (walls) are ground at the back side and a constant voltage at the front side, respectively, for both domains (domains differ in electric conductivity by a factor of 1000). The cylinder shell (wall) has constant temperature and zero flux for EMAG fields. The domain interface has conservative fluxes for heat and EMAG. The results for electric potential and field are ok but magnetic vector potential and magnetic induction look poor. Is there anybody with experience with electromagnetic modelling in CFX? Are there any special needs for EMAG simulation? I also did the same simulations for both domains solid with equal material properties except electric conductivity (again differ by factor 1000) and got no useful results. February 4, 2010, 09:52 #2 Senior Member     George Join Date: Mar 2009 Location: Birmingham, UK Posts: 257 Rep Power: 11 In studying the interaction between flow field and electromagnetic field, one needs to calculate the current density due to induction. There are two approaches are used to evaluate the current density. One is through the solution of a magnetic induction equation; the other is through solving an electric potential equation. CFX is using the first approach, and the magnetic induction equation is derived from Ohm's law and Maxwell's equation. note that current density in cfx is a vector (if you have access) the fluent magnetohydrodynamics module manual describes this a little better. bottom line cfx will calculate j for you. in cfx coulomb and lorentz forces are included in the momentum equation but (at least this was true in V11) cfx would not calculate joule heat in the energy equation, and you needed manually to define the joule heat as a source term. However i think in v12 Joule heat is calculated automatically. if you dont need to know the temperature within the solid you dont need to model the copper, you can just as easy define a flux on the wall boundaries. try to check the boundary conditions on your problem, especially if you only have voltage/current in your problem obviously you need to apply the the electric field however all other wall boundary the electric and magnetic field needs to be grounded and you need to make sure that the zero flux option is used. another possibility is that you have not defined how the electric and magnetic flux transfer though your interface boundaries. __________________ Top 4 tips 1. Knowledge is everything and Ignorance is dangerous. 2. Understand your limitations and try to eliminate them. 3. Get yerself a bike and hoon the chuffer. You will soon learn why dogs like to hang their heads out the car window. 4. Please before asking any questions on how to run simulations in CFX, go though all the tutorials February 4, 2010, 10:56 #3 New Member   Join Date: Dec 2009 Location: Berlin, Germany Posts: 5 Rep Power: 9 Thanks, I studied the fluent MHD manual. As far as I understand, it is only possible to calculate reasonable magnetic induction in cases of externally applied magnetic fields and/or the presence of fluid with non-zero velocities. Am I right? This would explain my problems with the current-carrying copper wire in surrounding quiescent air without externally applied magnetic fields. Altogether, it is always necessary to have external magnetic fields involved in the problem and is therefore called MHD- and not EHD-module?? Thanks again February 4, 2010, 11:50 #4 Senior Member     George Join Date: Mar 2009 Location: Birmingham, UK Posts: 257 Rep Power: 11 remember cfd is all about solving the navier stoke equations.... if an electromagnetic field is applied near a dielectric/permeable fluid this will cause a momentum change because of the coulomb and lorentz forces so essentially your fluid cannot have zero velocities. you cannot decouple the magnetic field with the electric field this is how the maxwell equations are solved, therefore whatever notation you use you will always solve the emag equations. __________________ Top 4 tips 1. Knowledge is everything and Ignorance is dangerous. 2. Understand your limitations and try to eliminate them. 3. Get yerself a bike and hoon the chuffer. You will soon learn why dogs like to hang their heads out the car window. 4. Please before asking any questions on how to run simulations in CFX, go though all the tutorials Last edited by ckleanth; February 4, 2010 at 17:39. February 5, 2010, 03:20 #5 New Member   Join Date: Dec 2009 Location: Berlin, Germany Posts: 5 Rep Power: 9 I'm familiar with the coupling of electric and magnetic field, but obviously CFX cannot properly calculate the magnetic field when there is an electric field as initial physical field with an initial zero-velocity field but without any magnetic quantities. That's why I thought it could work the other way round. February 5, 2010, 03:31 #6 Senior Member     George Join Date: Mar 2009 Location: Birmingham, UK Posts: 257 Rep Power: 11 I dont think the above is true and most likely you have a mistake with your BC and model setup. post a picture of your setup and post the model ccl; perhaps then the probelm will be clear __________________ Top 4 tips 1. Knowledge is everything and Ignorance is dangerous. 2. Understand your limitations and try to eliminate them. 3. Get yerself a bike and hoon the chuffer. You will soon learn why dogs like to hang their heads out the car window. 4. Please before asking any questions on how to run simulations in CFX, go though all the tutorials February 5, 2010, 05:48 #7 New Member Join Date: Dec 2009 Location: Berlin, Germany Posts: 5 Rep Power: 9 Hi, hopefully you are right Here there are the model setup and ccl Please tell me if there is something wrong Attached Images geometry.jpg (32.1 KB, 82 views) Attached Files model_ccl.txt (10.2 KB, 94 views) February 5, 2010, 06:07 #8 Senior Member     George Join Date: Mar 2009 Location: Birmingham, UK Posts: 257 Rep Power: 11 i think you did not ground the magnetic field. try to use Magnetic Potential = 0 [T m] on outer back boundary; if you want to get the magnetic field on the outer back boundary you need to add another domain and ground the magnetic field on that. __________________ Top 4 tips 1. Knowledge is everything and Ignorance is dangerous. 2. Understand your limitations and try to eliminate them. 3. Get yerself a bike and hoon the chuffer. You will soon learn why dogs like to hang their heads out the car window. 4. Please before asking any questions on how to run simulations in CFX, go though all the tutorials February 5, 2010, 08:10 #9 New Member   Join Date: Dec 2009 Location: Berlin, Germany Posts: 5 Rep Power: 9 You were right, I set Magnetic Vector Potential to 0 T m at the outer boundary and it now brings results comparable to analytical solutions for that configuration. Thanks again for your kind support and have a nice weekend, Marcel June 6, 2011, 10:19 2 phase flow #10 New Member   sarah Join Date: Oct 2009 Posts: 3 Rep Power: 9 Hi all, I am trying to investigate the effect of EHD on two phase flow in a dielectric fluid - however the MHD model seems to be problematic for two phase flow and when I use the VOF model I am having problems with my UDS function to calculate the potential field. Has anybody done any simulations along these lines that could point me in the right direction? My geometry is 2D concentric electrodes (ground on outer tube) Tags electric field, emag, magnetic field, magnetohydrodynamic, mhd Thread Tools Display Modes Linear Mode Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules Similar Threads Thread Thread Starter Forum Replies Last Post Val Main CFD Forum 3 June 10, 2011 02:20 Ema CFX 4 August 7, 2009 05:39 Alphonso CFX 1 August 1, 2008 14:01 Rui CFX 9 May 28, 2007 05:59 Pandu Sattvika CFX 1 December 1, 2001 05:07

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DiffEq Home ## Separable ODEs A separable ODE is of the form $\frac{dy}{dx}=\frac{M(x)}{N(y)}.$ Steps to solve: \begin{align*} N(y)\,dy&=M(x)\, dx\\ \int N(y)\,dy&=\int M(x)\, dx + c.\\ \end{align*} Example. Solve the following IVP (initial value problem) $$\label{7} \frac{dy}{dx}=\frac{2x-3}{y-5},\; y(0)=3$$ and find the valid interval of the solution. Find the maximum value of the solution. Solution. \begin{align*} \frac{dy}{dx}&=\frac{2x-3}{y-5} \\ (y-5)\,dy&=(2x-3)\,dx\\ \int (y-5)\,dy&=\int (2x-3)\,dx\\ \frac{y^2}{2}-5y&=2\frac{x^2}{2}-3x+c\\ y^2-10y&=2(x^2-3x+c)\\ y^2-10y-2x^2+6x-2c&=0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (8) \end{align*} Note that the initial condition is $$y(0)=3$$. So we have \begin{align*} 3^2-10\cdot 3-2c&=0\\ 2c&=-21 \end{align*} From (8) we get \begin{align*} y^2-10y-2x^2+6x+21&=0 \text{ (implicit solution)}\\ y&=\frac{10\pm \sqrt{100-4(-2x^2+6x+21)}}{2}\\ y&=\frac{10\pm 2\sqrt{25-(-2x^2+6x+21)}}{2}\\ y&=5\pm \sqrt{2x^2-6x+4} \end{align*} So we have two possible solutions: $y=5+ \sqrt{2x^2-6x+4} \text{ and } y=5- \sqrt{2x^2-6x+4}.$ Note that the first one does note satisfy the initial condition $$y(0)=3$$. Thus the solution is $y=5- \sqrt{2x^2-6x+4} \text{ (explicit solution)}.$ Obviously the maximum value of the solution is 5 because $$\sqrt{2x^2-6x+4}\geq 0$$. The solution is defined when \begin{align*} 2x^2-6x+4&\geq 0\\ 2(x^2-3x+2)&\geq 0\\ 2(x-1)(x-2)&\geq 0 \end{align*} The domain of $$y$$ is $$(-\infty,1]\cup [2,\infty)$$. Because of the initial condition $$y(0)=3$$, the valid interval of the solution is $$(-\infty,1]$$. Example. Solve the following ODE. $$$\frac{dy}{dx}=e^{-y}x\cos x$$$ Solution. \begin{align*} \frac{dy}{dx}&=\frac{x\cos x}{e^y}&&\\ e^y\, dy&=x\cos x\, dx&\\ \int e^y\, dy&=\int x\cos x\, dx&\\ e^y &=\int u\, dv && u=x,\, dv=\cos x\, dx\\ &=uv-\int v\, du && du=dx,\, v=\int dv=\int \cos x\, dx=\sin x\\ &=x\sin x-\int \sin x\, dx &&\\ &=x\sin x+\cos x +c && \end{align*} The general solution is $$y=\ln|x\sin x+\cos x +c|$$. Substitution for homogeneous ODEs: The ODE $$\displaystyle\frac{dy}{dx}=f(x,y)$$ is called homogeneous if $$f(x,y)$$ is a function of $$\displaystyle\frac{y}{x}$$. Steps to solve: Substitute $$v=y/x$$ or, $$y=vx$$. Then $$\displaystyle\frac{dy}{dx}=x\displaystyle\frac{dv}{dx}+v$$. Write $$f(x,y)$$ as a function of $$v$$. Then it becomes a separable ODE of $$v$$ and $$x$$. Example. Solve the following ODE. $\begin{equation*} \frac{dy}{dx}=\frac{x^2+3y^2}{2xy} \end{equation*}$ Solution. \begin{align} \frac{dy}{dx}&=\frac{(x^2+3y^2)/x^2}{2xy/x^2}\nonumber\\ \frac{dy}{dx}&=\frac{1+3(y/x)^2}{2(y/x)} \;\;\;\;\; (9) \end{align} Let $$v=y/x$$ or, $$y=vx$$. Then $$\displaystyle\frac{dy}{dx}=x\displaystyle\frac{dv}{dx}+v$$. So from (9) we get \begin{align} x\frac{dv}{dx}+v&=\frac{1+3v^2}{2v}\nonumber\\ x\frac{dv}{dx}&=\frac{1+3v^2}{2v}-v=\frac{1+v^2}{2v}\nonumber\\ \frac{2v}{1+v^2}\, dv&=\frac{dx}{x} \nonumber \end{align} \begin{align} \int\frac{2v}{1+v^2}\, dv&=\int \frac{dx}{x} \text{ (substitute } u=1+v^2) \nonumber\\ \ln|1+v^2|&=\ln|x|+\ln c \nonumber\\ \ln|1+v^2|&=\ln|cx| \nonumber\\ 1+v^2&=cx \nonumber\\ 1+\frac{y^2}{x^2}&=cx \nonumber\\ x^2+y^2&=cx^3 \nonumber \end{align} Last edited

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# Python Operators Operators are fundamental elements in any programming language, and Python provides a rich set of operators to perform various tasks, from simple arithmetic operations to complex logical evaluations. This comprehensive guide aims to unravel the intricacies of Python operators, covering everything from basic arithmetic to advanced concepts like bitwise operations and ternary operators. ## 1. Arithmetic Operators: ### Addition (+): sum_result = 5 + 3 # Output: 8 ### Subtraction (-): difference = 10 - 7 # Output: 3 ### Multiplication (*): product = 3 * 4.5 # Output: 13.5 ### Division (/): quotient = 15 / 3 # Output: 5.0 ### Floor Division (//): floor_result = 17 // 5 # Output: 3 ### Modulo (%): remainder = 17 % 5 # Output: 2 ### Exponentiation (**): power_result = 2 ** 3 # Output: 8 ## 2. Comparison Operators: ### Equal (==): x = 5 result = (x == 5) # Output: True ### Not Equal (!=): y = 10 result = (y != 5) # Output: True ### Greater Than (>), Less Than (<): a = 15 b = 20 result = (a > b) # Output: False ### Greater Than or Equal To (>=), Less Than or Equal To (<=): c = 10 result = (c >= 10) # Output: True ## 3. Logical Operators: ### and Operator: x = True y = False result = x and y # Output: False ### or Operator: a = True b = False result = a or b # Output: True ### not Operator: is_python_fun = True result = not is_python_fun # Output: False ## 4. Assignment Operators: ### = Operator: x = 5 ### +=, -=, *=, /=, //=, %= Operators: total = 10 total += 5 # Equivalent to total = total + 5 ## 5. Bitwise Operators: ### &, |, ^, ~, <<, >> Operators: Bitwise operators manipulate individual bits of binary representations of integers. num1 = 5 num2 = 3 bitwise_and = num1 & num2 # Output: 1 bitwise_or = num1 | num2 # Output: 7 ## 6. Membership Operators: ### in Operator: my_list = [1, 2, 3, 4, 5] result = 3 in my_list # Output: True ### not in Operator: result = 6 not in my_list # Output: True ## 7. Identity Operators: ### is Operator: a = [1, 2, 3] b = a result = a is b # Output: True ### is not Operator: c = [1, 2, 3] result = a is not c # Output: True ## 8. Ternary Operator: A concise way to express a conditional statement. age = 20 status = "Adult" if age >= 18 else "Minor" ## 9. Conclusion: Python operators are the building blocks of code, providing a wide array of functionalities to perform diverse operations. From basic arithmetic to advanced bitwise manipulations, understanding and mastering these operators is essential for effective Python programming. As you explore and apply these operators in your code, you’ll gain a deeper appreciation for their versatility and utility. Happy coding!

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# Math posted by on . A factory manufactures two products, each requiring the use of three machines. The first machine can be used at most 70 hours; the second machine at most 40 hours; and the third machine at most 90 hours. The first product requires 2 hours on machine 1, 1 hour on machine 2, and 1 hour on machine 3; the second product requires 1 hour each on machines 1 and 2, and 3 hours on machine 3. If the profit is \$40 per unit for the first product and \$60 per unit for the second product, how many units of each product should be manufactured to maximize profit? ### Related Questions More Related Questions Post a New Question

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Master of Information Technology @ The University of Melbourne 0% ## Data Structure Overview • Array • Stack • Queue • Priority Queue • Binary Tree • Hashing ## Array • Search Pseudocode (Recursive) “”” function find(A,x,lo,hi) lo->0 hi->n-1 ``````if lo > hi return -1 else if lo = A[lo] return lo else return find(A,x,lo+1,hi)`````` “”” • Node and Pointer(The pointer in last node is null) • Function • Value - ListName.val • Next Value - ListName.next • Search Pseudocode (Recursive) “”“ ``````if p = null then return p else if p.val = x return p else return find(p.next,x)`````` ”“” ## Stack • Property： Last In First Out • Function： • Initial - initilise st as an empty stack • Out - st.pop() • Null judgement - if st is empty ## Queue • Property： First In First Out • Function： • Initial - init(queue) • Out - eject(queue) • Null judgement - if queue is empty ## Binary Tree • Property： • Empty Tree Height = -1 • Traveral： • Pre-order • In-order • Post-order • Binary Search Tree [Based on Binary Tree] [Property] ``````Left Child is LESS THAN Parents; Right Child is LARGER (OR EQUAL) TO Parents;`````` [Count Quantity of Diff elements] ``B(n+1) = B(n) * B(0) + B(n-1) * (1) ... + B(0) * B(n)`` • (AVL)Balanced Binary Search Tree [Based on Binary Search Tree] [Property] ``````Left Child is LESS THAN Parents; Right Child is LARGER (OR EQUAL) TO Parents; [!]The |difference between two children| must be <= 1`````` [Balanced Rule] ``````L-Rotation R-Rotation LR-Rotation RL-Rotation`````` • (2-3Tree)Balanced Binary Search Tree [Based on Binary Search Tree] ## Priority Queue(max heap & min heap) • [Time Complexity] Delete the heighest priority element - O(logn) Add elements to empty heap - O(nlogn) Create a complete binary tree first and then construct from bottom up - O(n) ## Hashing (Use Space to Reduce Time) • [Handle Collision] • Separate Chain : Use link list Load Factor(alpha) = n/m ; n:the number of items stored, m: the table size Number of probes in successful search: (1+alpha)/2 ``Number of probes in unsuccessful search: alpha`` • Linear probing : Store in the next index Number of probes in successful search: 0.5+1/(2(1-alpha)) Number of probes in unsuccessful search: 0.5+1/(2(1-alpha)^2) #PS 在Linear Probe中查找一个已经被移过位置的key是unsuccessful probing • Double hashing : Use another hashing function s(k) to add to the original index[h(k)+s(k)]. If collision again, add 2s(k)…… ## Graphs • [Property] • Vertex[V], Edges[E], Weight —>>> G<V,E> • Edge(u,v) means the edge between node u and node v • two nodes is connected, then we call v and u are ADJACENT or NEIGHBOURS • Degree : NUMBER of EDGES connected on a node • In-degree : NUMBER of EDGES GOING TO v • Out-degree : NUMBER of EDGES LEAVING FROM v • [Graph Representation]

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Airplanes don't exist (a parody of the conspiracy theorists) Things that don't belong anywhere else. (Check first). Moderators: Moderators General, Prelates, Magistrates FlatAssembler Posts: 47 Joined: Fri Oct 27, 2017 7:42 pm UTC Airplanes don't exist (a parody of the conspiracy theorists) Our culture makes us hold many irrational beliefs. One of them is demonstrably the belief that airplanes exist. It's told us by our parents, told by our teachers, and most of us never really investigate it. And there is not much evidence of that. Most of the arguments we use to prove airplanes exist can be used to prove that dragons exist as well. We sometimes see white lines in the sky and we say they are evidence of jet airplanes. But saying they are the evidence of dragons is just as valid. There are people who say they have flown on an airplane, and use it as a proof that airplanes exist. But they could just as easily say it for dragons. And history tells us that before people claimed to have flown on a dragon just as often as people say today they have been on an airplane. In reality, what we usually mean when we say airplane is so called jet airplane, and they can be disproven with some basic physics. Jet airplanes are supposed to work by having water (or some other liquid) as a fuel and engines forcing that water to go out, so that that water accelerates and, by the Newton's third law, makes the airplane accelerate also. But remember the Torricelli's law? Most of the people have learned it school, they just have never really thought about it. If they have, they would realize that it makes the airplanes impossible. One of the well-known formulations of the Torricelli's law is that, when a liquid goes through a small hole (an outlet), its speed is determined by the formula: Code: Select all `v=sqrt(2*g*h)` But there is a pretty obvious implication here. That is: Code: Select all `a=0` The Newton's second law tells us: Code: Select all `F=m*a` Therefore: Code: Select all `F=m*a=m*0=0` So, by the Newton's third law: Code: Select all `F1=-F20=-F2F2=-0=0` So, the force acting on an airplane itself is zero, so by the Newton's first law: Code: Select all `F=0 | Va=0` So, how can jet airplanes work in reality if they don't even work on paper? You may give me some counter-example to the Torricelli's law. But do the counter-examples matter? They don't. The Torricelli's law is derived from the Bernoulli's equation, and it's derived right from the Newton's three axioms. Also, the burden of proof is definitely on you. You can't prove for anything that doesn't exist that it doesn't exist, but, in general, if something exists, you are able to prove it. And Occam's razor always favors more an explanation that involves someone lying or hallucinating than an explanation that involves something as complicated and as crazy sounding as airplanes. And you might ask me what if I am wrong. So what if I am wrong? At least I am thinking about whether airplanes exist, and other people aren't thinking about that at all, they just accept what most people believe as fact. And you are way more likely to be wrong if you aren't thinking than if you are thinking. (This is my parody of the Internet conspiracy theorists!)

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Electrostatic force between more than 2 charges hi i dont know if this has been posted or not but i am sort fo stuck on this one What is the force F_vec on the 1 nC charge at the bottom? View Figure Write your answer as two vector components, separated by a comma. Express each component numerically, in newtons, to three significant figures. the figure looks like this attempt to this problem: i found out the forces individually and added them together .. but it didnt work x-component: 6nC = 0 2nC = 5.09 x 10$$^{-5}$$ each y-component: 6nC = 2.16 x 10$$^{-5}$$ 2nc = 5.09 x 10$$^{-5}$$ each when i add them: 1.02 x 10 $$^{-5}$$ $$\hat{x}$$ , 7.76 x 10$$^{-5}$$ $$\hat{y}$$ but that was wrong.. i also tried the x-comp to be 0 .. but that was wrong too thanx in advance for the help Related Introductory Physics Homework Help News on Phys.org Doc Al Mentor How does the x-component of the force exerted by one 2 nC charge compare to that of the other? How does the x-component of the force exerted by one 2 nC charge compare to that of the other? they are opposite of each other .. which means there is no net x- component .. but thats not the right answer tho.. hi i dont know if this has been posted or not but i am sort fo stuck on this one What is the force F_vec on the 1 nC charge at the bottom? View Figure Write your answer as two vector components, separated by a comma. Express each component numerically, in newtons, to three significant figures. the figure looks like this View attachment 12308 attempt to this problem: i found out the forces individually and added them together .. but it didnt work x-component: 6nC = 0 2nC = 5.09 x 10$$^{-5}$$ each y-component: 6nC = 2.16 x 10$$^{-5}$$ 2nc = 5.09 x 10$$^{-5}$$ each when i add them: 1.02 x 10 $$^{-5}$$ $$\hat{x}$$ , 7.76 x 10$$^{-5}$$ $$\hat{y}$$ but that was wrong.. i also tried the x-comp to be 0 .. but that was wrong too thanx in advance for the help I see 2 things you might do differently to help. Make sure you're doing vectors (right is +, up is +). Keep the vector notation throughout. Check your use of scientific notation. If you're unsure, have your calculator switch it to regular decimal. I see 2 things you might do differently to help. Make sure you're doing vectors (right is +, up is +). Keep the vector notation throughout. Check your use of scientific notation. If you're unsure, have your calculator switch it to regular decimal. i tried a similar question, with 1 + and 1 - 2nC charge .. and that worked out but somehow i cant get this working .. i am not too sure if the y - component it right as well or not .. even tho it seems right Doc Al Mentor they are opposite of each other .. which means there is no net x- component .. but thats not the right answer tho.. Show how you calculated the vertical components of the forces from the 2 nC charges. Doc Al Mentor y-component: 6nC = 2.16 x 10$$^{-5}$$ 2nc = 5.09 x 10$$^{-5}$$ each What are the directions (signs) of these components? Show how you calculated the vertical components of the forces from the 2 nC charges. y-comp = Esin$$\vartheta$$ r= 5x10^-2 E = Kq$$_{1}$$q$$_{2}$$ / r$$^{2}$$ = (9.0x10^9)(2x10^-9)(1x10^-9)/ ( 2.5x10^-3) E= 7.2x10^-6 y-comp = E sin(45) = 5.09x10^-6 Show how you calculated the vertical components of the forces from the 2 nC charges. What are the directions (signs) of these components? 6nC = +y, no x 2nC = +y, +x (on the right) 2nC = +y, -x (on the left) and thats why my understand was that the x components would cancel out .. but i am not sure whats going on Doc Al Mentor 6nC = +y, no x 2nC = +y, +x (on the right) 2nC = +y, -x (on the left) Realize that the 2 nC charges are positive and the 6 nC charge is negative. Which way do their forces on the + 1 nC charge act? so the 2 +y .. should be -y ..? i did the calc again and i get x = 0 y = 1.14x10^-5 .. doe sthat seem right? Doc Al Mentor so the 2 +y .. should be -y ..? Correct. Since the charges repel, the y component of force is downward and thus negative. thank you ., thank you ... thank you.. Doc Al Mentor y-comp = Esin$$\vartheta$$ r= 5x10^-2 E = Kq$$_{1}$$q$$_{2}$$ / r$$^{2}$$ = (9.0x10^9)(2x10^-9)(1x10^-9)/ ( 2.5x10^-3) E= 7.2x10^-6 y-comp = E sin(45) = 5.09x10^-6 What's the distance between the charges? i did the calc again and i get x = 0 y = 1.14x10^-5 .. doe sthat seem right? what you said abt the charges repelling .. this answer is right .. thank you so much once again Doc Al Mentor what you said abt the charges repelling .. this answer is right .. thank you so much once again You are most welcome. What's the distance between the charges? Never mind. You had written the distance squared.

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# Distribution Analysis/Smart Grid Ηλεκτρονική - Συσκευές 21 Νοε 2013 (πριν από 4 χρόνια και 7 μήνες) 125 εμφανίσεις Distribution Analysis/Smart Grid ECE 445: Senior Design Spring 2010 By: Kenny Koester & Greg Gillespie Distribution Analysis/Smart Grid Objective of Project Distribution Analysis Overview Small - Scale Model Overview Challenges and Struggles Future work and Discussion Project Objective Perform a distribution and cost analysis for Coles Moultrie Electric Cooperative (CMEC). Design and construct a small - scale model of a portion of CMEC’s distribution system implementing “Smart Grid” technology. Distribution Analysis Overview CMEC will be adding a convention center to its distribution system. Giving an option of three substations, it was our job to determine the best substation to feed the center power. Segment of CMEC System Segment of CMEC Distribution Map PowerWorld Simulation of System Cost Analysis Underground Cost: Bore 100 ft.: \$3000.00 Trench 100ft.: \$2000.00 Terminator Pole: \$3552.78 Transformer Cost: \$50,000.00 Set Transformer: \$1450.00 Total Underground Cost: \$60,002.80 Cost Analysis Over - : Convert 2.5 mi. of 1/0 to 4/0: Poles with ground rod (10): \$56,845.50 Poles with out ground rod (28): \$332,629.22 Total Over - Overall Total Cost: \$449,479.00 Small - Scale Model Small - Scale Model Overview We downsized CMEC’s distribution system by a scale of 60:1. After downscaling we had the following measurements: Voltage Supply: 7200V 120V Line Impedance: Sarah Bush : Z = 1.894 + j1.859 R = .03157 South Mattoon : Z = 1.827 + j1.9745 R = .03045 Goals of Small - Scale Model Point - to - multipoint wireless communication to XBEEs through the use of LabView Implement smart grid technology through power factor correction. (P.F. = .9) - ends and fault switches throughout the model Demonstrate how an interruptible account works Small - Scale Model Overview PowerWorld Simulation of Small - Scale Model Components of Model 1/6 HP Single Phase AC Motors Resistor boxes (83.33 ) Voltage Sources (120V (AC) & DC supply) 12 gauge wire Sarah Bush: 16.88ft South Mattoon: 16.25ft PCBs Printed Circuit Boards Components XBEE 100 µ F Capacitor LED 15 Ω resistor 2N7000 MOSFET T75 series relay Banana Ports Picture of PCB (Single Relay) T75 RELAY 2N7000 MOSFET 15 Ω Resistor L.E.D XBEE Mount 100µF Capacitor Layout of Single Relay PCB in Eaglesoft PCBs Controlling Capacitor Banks 2N7000 MOSFETs 15 Ω Resistors L.E.D. XBEE Mount 100µF Capacitor T75 Relays Layout of Capacitor Bank PCBs in Eaglesoft Tests Ran on Motors Two motors using 120V Power = 192 W Current = 6.09A Power Factor = .263 One Motor using 120 V Power = 94 W Current = 3.11A Power Factor = .246 Four Motors using 120V P = 396 W Current = 12.65 Power Factor = .263 XBEE Communication XBEE: 802.15.4 XBEE Mounted on RS - 232 Interface Board On/Off Input Power RS - 232 Port Antenna XBEE Communication Star (point - to - multipoint) More Recent XBEEs use mesh communication XBEE Communication Our XBEEs are Programmed in a program called X - CTU. We programmed our XBEEs to communicate with HEX coding by enabling them in API mode. Our code sends out HI and LO signals that the XBEE can recognized in HEX coding XBEE Communication Using LabView The XBEEs on our circuit are controlled by a program created in LabView. In LabView we created a user interface. This was to make the communication between XBEEs easier for the operator. LabView User Interface Power Factor Correction This was done by adding capacitor banks in parallel with our motors We calculated the correct amount of capacitance by using the following equation: C = (VARs)/(ωV 2 ) - End/Fault Switches Goal : To keep the maximum amount of customers with power at all times. This helps to maximize a utility company’s income. Uses : To repair power lines when there is a fault. To work on substations. Interruptible Account Our model will have a load (motor) that will represent the convention center (new load on CMEC’s system). This motor will be set up on an interruptible account. This occurs during peak Benefits : The Convention Center will get a better rate on their electric bill CMEC will get billed less for not consuming as much Challenges and Struggles Deciding best way to model the loads. Obtaining a power factor of at least .9 . Creating a circuit that could switch our relays open or closed using the output from our XBEEs. Making our small - scale model work in PowerWorld . Getting our XBEEs to communicate. Getting LabView to communicate with our XBEEs Future Work By using XBEE Pro modules, we could set up a mesh network with all of the XBEEs on our model. There is also the possibility of reading in many different parameters on the line with the XBEE and sending them back to the main control Making the system much larger is also a possibility. Special Thanks to: Prof. Sauer and Prof. Garcia CMEC Kevin Colravy Tamer Rousan Ali Bazzi Jamie Weber (Power World) Prof. Carney Mark Smart Part Shop and Machine Shop

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https://forum.nutsvolts.com/viewtopic.php?f=45&t=2738

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## One-shot design needed This is the place for any magazine-related discussions that don't fit in any of the column discussion boards below. haklesup Posts: 3046 Joined: Thu Aug 01, 2002 1:01 am Location: San Jose CA Contact: ### One-shot design needed I need to design a simple circuit that will give me a 20us square pulse with variable amplitude (about 0V to 1V and at least 1mA). The event will be triggered by a button push.<p>I am familiar with variable pulse-width designs using the 555 but I am not sure how to vary the output amplitude without adding another stage like an op-amp to scale the signal.<p>At this point I am looking for suggestions for any similar design not just 555 based ones.<p>Chris desterline Posts: 86 Joined: Mon Jun 09, 2003 1:01 am Contact: ### Re: One-shot design needed The simplest idea I can think of is just a voltage divider. Two resistors in series between the output of the 555 and ground, take your output from the connection between the two resistors. The output voltage will be in proportion to the ratio of the two resistors, i.e. two 1k resistors will give you half voltage, ect.<p>Other ratios are redily creatable. If you use a pot, say 5k, with the wiper connected as your output, you can adjust it.<p>Good luck Denny bruinbear714 Posts: 26 Joined: Thu Apr 17, 2003 1:01 am Contact: ### Re: One-shot design needed I don't have my magazine in front of me, but in this month's issue of Nuts & Volts, there is a section on transistor circuits. In it, there are a few oscillator circuits that uses discrete devices, I believe with two transistors, a couple of caps and a couple of resistors.<p>Or you might want to look into ring oscillators. Chris Foley Posts: 146 Joined: Mon Jan 13, 2003 1:01 am Location: Chicago IL Contact: ### Re: One-shot design needed desterline has a good idea. You're already familiar with the 555; you might want to start with a standard 20 microsecond one-shot with a CMOS 555. Let's assume a regulated 5VDC supply. If you load the output pin with a 40K resistor in series with a 10K pot, you can adjust the wiper of the pot to give you pulses of 0 to 1V amplitude (the lightly loaded output of the CMOS 555 is practically equal to the power supply rails). If you then place that 0 to 1V signal as the input to a "rail-to-rail" op-amp configured as a voltage follower, you will have a low-impedance output. If you don't have a "rail-to-rail" op amp handy, you might want to try the LM6132 dual. It's 8 pins, and is available in the standard 8-pin DIP package. Don't use an LM358 -- it will accept your signal, but the rise time and fall time will probably be several microseconds. Happy hunting.<p>LM6132 Product Folder<p>[ July 03, 2003: Message edited by: Chris Foley ]</p> ### Who is online Users browsing this forum: No registered users and 29 guests

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# Fungrim entry: f1bd89 $z \left(1 - z\right) y''(z) + \left(c - \left(a + b + 1\right) z\right) y'(z) - a b y(z) = 0\; \text{ where } y(z) = \,{}_2F_1\!\left(a, b, c, z\right)$ Assumptions:$a \in \mathbb{C} \,\mathbin{\operatorname{and}}\, b \in \mathbb{C} \,\mathbin{\operatorname{and}}\, c \in \mathbb{C} \setminus \{0, -1, \ldots\} \,\mathbin{\operatorname{and}}\, z \in \mathbb{C} \setminus \left[1, \infty\right)$ TeX: z \left(1 - z\right) y''(z) + \left(c - \left(a + b + 1\right) z\right) y'(z) - a b y(z) = 0\; \text{ where } y(z) = \,{}_2F_1\!\left(a, b, c, z\right) a \in \mathbb{C} \,\mathbin{\operatorname{and}}\, b \in \mathbb{C} \,\mathbin{\operatorname{and}}\, c \in \mathbb{C} \setminus \{0, -1, \ldots\} \,\mathbin{\operatorname{and}}\, z \in \mathbb{C} \setminus \left[1, \infty\right) Definitions: Fungrim symbol Notation Short description ComplexDerivative$\frac{d}{d z}\, f\!\left(z\right)$ Complex derivative Hypergeometric2F1$\,{}_2F_1\!\left(a, b, c, z\right)$ Gauss hypergeometric function CC$\mathbb{C}$ Complex numbers ZZLessEqual$\mathbb{Z}_{\le n}$ Integers less than or equal to n ClosedOpenInterval$\left[a, b\right)$ Closed-open interval Infinity$\infty$ Positive infinity Source code for this entry: Entry(ID("f1bd89"), Formula(Where(Equal(Sub(Add(Mul(Mul(z, Sub(1, z)), ComplexDerivative(y(z), For(z, z, 2))), Mul(Sub(c, Mul(Add(Add(a, b), 1), z)), ComplexDerivative(y(z), For(z, z, 1)))), Mul(Mul(a, b), y(z))), 0), Equal(y(z), Hypergeometric2F1(a, b, c, z)))), Variables(a, b, c, z), Assumptions(And(Element(a, CC), Element(b, CC), Element(c, SetMinus(CC, ZZLessEqual(0))), Element(z, SetMinus(CC, ClosedOpenInterval(1, Infinity)))))) ## Topics using this entry Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub. 2019-12-30 15:00:46.909060 UTC

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https://dontjudgejustfeed.com/what-is-the-profit-maximizing-price-1200artists/

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# What is the profit-maximizing price? – dontjudgejustfeed.com A monopoly will charge what the market is willing to pay. The dotted line from the profit-maximizing quantity to the demand curve represents the profit-maximizing price.this price is Above average cost curvewhich indicates that the company is making a profit​​. ## How to calculate the profit maximizing price? The monopolist’s profit-maximizing choice will be to produce in quantities where marginal revenue equals marginal cost: that is, Mr = MC. If the monopoly produces fewer quantities, then at these output levels, MR > MC, the firm can make higher profits by expanding output. ## What is Profit Maximizing Pricing? Profit maximizing pricing Profit maximization is The short- or long-term process by which a firm determines the price and output level that returns the greatest profit. Any costs incurred by a company can be divided into two categories: fixed costs and variable costs. ## What is the profit-maximizing price for the firm? The profit-maximizing choice of a perfectly competitive firm will occur at the output level where marginal revenue equals marginal cost—that is, Mr = MC. This happens at Q = 80 in the figure. ## Are profit-maximizing prices and quantities effective? Profit is maximized when the marginal revenue (MR) of selling a product equals the marginal cost (MC) of producing the product.Economic benefits are maximized when prices are maximized (P) from Selling the product equals the marginal cost (MC) of producing the product. ## Profit Maximization | APⓇ Microeconomics | Khan Academy 32 related questions found ## How do you maximize profits? 12 Tips for Maximizing Business Profits 1. Evaluate and reduce operating costs. … 2. Adjusted Pricing/Cost of Goods Sold (COGS)… 3. View your product mix and pricing. … 4. Upsell, cross-sell, resell. … 5. Increase customer lifetime value. … 7. Optimize demand forecasting. … 8. Sell ​​old stock. ## What is perfect price discrimination? First-degree or outright price discrimination When a business charges the highest possible price for each unit consumed. Because prices differ between units, the firm captures all available consumer surplus for itself or for economic surplus. ## How to calculate profit? The formula for calculating profit is: Total Revenue – Total Expenses = Profit. Profit is determined by subtracting direct and indirect costs from all earned sales. ## What is the maximum profit? The maximum profit is MC is equal to the output level of MR. Producing an additional unit of production reduces profits when the production level reaches a point where the cost of producing the additional unit of production (MC) exceeds the revenue per unit of production (MR). ## How to calculate monopoly profit? A monopoly calculates its profit and loss by Use its average cost (AC) curve to determine its cost of production, then subtract that number from total revenue (TR). Recall from previous lectures that companies use average cost (AC) to determine profitability. ## What are the 5 pricing strategies? Consider these five common strategies that many new businesses use to attract customers. • Skimming price. Skimming involves setting a high price when a product is launched and then gradually lowering the price as more competitors enter the market. … • Market penetration pricing. … • Economical pricing. … • Bundled pricing. ## What is an example of competitive pricing? Competitive pricing involves setting prices at the same level as competitors. … E.g, A company needs to price a new coffee machine. The company’s competitors sell it for \$25, which the company believes is the best price for a new coffee maker. It decided to set this price for its own products. ## Does price equal marginal cost? In a perfectly competitive market, price equals marginal cost The economic profit of the company is zero. In a monopoly, the price is above marginal cost and the firm earns a positive economic profit. Perfect competition produces an equilibrium in which the price and quantity of a good are economically efficient. ## What is the formula for economic profit? Economy Profit = Total Revenue – (Explicit Costs + Implicit Costs). Accounting profit = total revenue – explicit cost. ## How is the total cost calculated? The formula for calculating the average total cost is: 1. (Total Fixed Cost + Total Variable Cost) / Number of Production Units = Average Total Cost. 2. (Total Fixed Cost + Total Variable Cost) 3. New cost – old cost = cost change. 4. New Quantity – Old Quantity = Quantity Change. ## What is normal profit? Normal profit is a profit metric that takes into account both explicit and implicit costs. It can be viewed in conjunction with economic profit.Normal profit occurs in The difference between the company’s total revenue and the combined explicit and implicit costs is zero. ## What is the optimal profit? The general rule is that firms produce Marginal revenue equals marginal cost…to maximize profits, firms should increase the use of inputs « until the marginal revenue product of the input equals its marginal cost ». ## How do you calculate maximum weekly profit? The weekly maximum profit appears When 4000 sandwiches are made and sold. The maximum weekly earnings can be found by plugging 4000 into the earnings function. We have to subtract the cost of making these sandwiches to make a profit. ## How to calculate earnings per share? Multiply the sale price per share by the number of shares sold to get your total sales proceeds. Subtract cost basis from total benefit Calculate your stock profit. Note that your answer will be negative if the cost basis is greater than the total gain from selling the stock. ## How to calculate profit percentage profit? The formula for calculating profit percentage is: Profit % = Profit / Cost Price × 100. ## How much do I need to sell to make a profit? Profit margin overview Subtract the cost from the sales price to obtain a profit margin, and Divide margin by sale price as profit margin percentage. For example, you sell a product for \$100 and your business costs \$60. The profit margin is \$40, or 40% of the selling price. ## Which companies use price discrimination? Industries that commonly use price discrimination include Tourism, Pharmaceuticals, Leisure and Telecommunications. Examples of forms of price discrimination include coupons, age discounts, career discounts, retail incentives, and gender-based pricing. ## What is the purpose of price discrimination? The goal of price discrimination is to To enable sellers to make the most profit possible and capture the consumer surplus of the market and generate as much revenue as possible for the goods sold. ## Is price discrimination profitable? Price discrimination is profitable only if there is an elasticity of demand in a market Unlike demand elasticity. Therefore, a monopolist will price discriminate between two markets only when it finds that the price elasticity of demand for its product in different submarkets is different. ## Why does Mr. 0 maximize returns? only when the marginal benefit Total income is zero maximized. Stopping below this amount means losing the opportunity to earn more revenue, while an increase in sales above this amount means MR goes negative and TR goes down.

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### Home > CCG > Chapter Ch10 > Lesson 10.2.1 > Problem10-69 10-69. When Erica and Ken explored a cave, they each found a gold nugget. Erica’s nugget is similar to Ken’s nugget. They measured the length of two matching parts of the nuggets and found that Erica’s nugget is five times as long as Ken’s. When they took their nuggets to the metallurgist to be analyzed, they learned that it would cost $$30$ to have the surface area and weight of the smaller nugget calculated, and$$150$ to have the same analysis done on the larger nugget. I won’t have that kind of money until I sell my nugget, and then I won’t need it analyzed!” Erica says. Wait, Erica. Don’t worry. I’m pretty sure we can get all the information we need for only $$30$.” 1. Explain how they can get all the information they need for$$30$. Since the two nuggets are similar, if they spend \$$30$ to get Ken's nugget analyzed, they can calculate the surface area and weight of Erica's nuggets using the ratios of similarity. 1. If Ken’s nugget has a surface area of $20\text{ cm}^2$, what is the surface area of Erica’s nugget? We know the linear scale factor is $5$, so the ratio of the areas is $5^2$, $25$. 2. If Ken’s nugget weighs $5.6$ g (about $0.2$ oz), what is the weight of Erica’s nugget? Since the nuggets are solid gold and have the same density throughout, calculate the weight using the ratio of the volumes. $7000\text{ g}\approx25\text{ ounces}$

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• anonymous Q=kor*(0.5)^2*(0.9)^1/2 kor=2.09 Q=2.09*(0.5)^2*(0.9)^1/2 Q=0.49568... Now say if I knew Q was 0.495.. and wanted to find kor, how would I rearrange the formula to get the answer 2.09? 0.4956=kor*(0.5)^2*(0.9)^1/2 Btw the above formula is related to Water Calcs for Orfices if it helps. Mathematics • Stacey Warren - Expert brainly.com Hey! We 've verified this expert answer for you, click below to unlock the details :) SOLVED At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat. Looking for something else? Not the answer you are looking for? Search for more explanations.

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# Plot Labels with arrows or lines pointing to different curves? Posted 6 months ago 627 Views | | 0 Total Likes | I'm trying to get plot labels with arrows or lines pointing to different curves. If I do A2 = 1; Plot[Evaluate@ Table[Labeled[Sqrt[x^2 - x^4]/Sqrt[ 1 + A2 \[Beta]], \[Beta]], {\[Beta], {.1, .5, 1, 5, 10, 20}}], {x, 0, 1.2}, AxesLabel -> {Subscript[k, x], Subscript[k, y]}, PlotRange -> All, LabelStyle -> Directive[14, Bold]] I get the following image.How do I make the lines point to a better part of the curves?I've also tried something along the lines of Plot[Evaluate@ Table[Sqrt[x^2 - x^4]/Sqrt[ 1 + \[Beta]], {\[Beta], {.1, .5, 1, 5, 10, 20}}], {x, 0, 1.2}, PlotLabels -> Placed[{"\[Beta] = .1", "\[Beta] = .5", "\[Beta] = 1", "\[Beta] = 5", "\[Beta] = 10", "\[Beta] = 20"}, Above], PlotRange -> All] I've tried using Scaled[] as well. Those options seem to move the labels around but remove the lines pointing to the curves. Sort By: Posted 6 months ago You may want to try experimenting with Callout[]. A2 = 1; Manipulate[ Plot[Evaluate@ Table[Callout[Sqrt[x^2 - x^4]/Sqrt[1 + A2 \[Beta]], \[Beta], pos, CalloutMarker -> "CirclePoint"], {\[Beta], {.1, .5, 1, 5, 10, 20}}], {x, 0, 1.2}, AxesLabel -> {Subscript[k, x], Subscript[k, y]}, PlotRange -> All, LabelStyle -> Directive[14, Bold], ImageSize -> Large], {{pos, 0.8, "Position"}, 0, 1, Appearance -> "Labeled"}] # Update The LeaderSize option will give you more control over the callout placement. You can experiment with the manipulate below. For me, the largest effects were adjusting the Neck parameters. A2 = 1; m = Manipulate[ Plot[Evaluate@ Table[Callout[Sqrt[x^2 - x^4]/Sqrt[1 + A2 \[Beta]], \[Beta], pos, Background -> White, neckangle Degree}}, CalloutMarker -> "CirclePoint"], {\[Beta], {.1, .5, 1, 5, 10, 20}}], {x, 0, 1.2}, AxesLabel -> {Subscript[k, x], Subscript[k, y]}, PlotRange -> All, LabelStyle -> Directive[14, Bold], ImageSize -> Large], {{pos, 1/Sqrt[2], "Position"}, 0, 1, Appearance -> "Labeled"}, {{horizangle, 225, "Horizontal Angle(\[Degree])"}, -360, 360, Appearance -> "Labeled"}, {{gap, 2, "Gap"}, 0, 10, Appearance -> "Labeled"}, {{neck, 20, "Neck Length"}, 0, 100, Appearance -> "Labeled"}, {{neckangle, 20, "NeckAngle(\[Degree])"}, -360, 360, Appearance -> "Labeled"}]

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## 5 gallon bucket dimensions lets to see 5 gallon bucket dimensions ## So 5 gallon bucket dimensions A gallon has 4 quarts. Each quart has 2 pints, so a gallon additionally contains 8 pints. Each pint is 2 cups, so a gallon has 16 cups, 5 gallon bucket dimensions. ## Lets to see 5 gallon bucket dimensions Each mug holds 8 liquid ounces. Therefore, there are 128 fluid ounces in 1 gallon. The basic recommendations are to consume 8 8-ounce glasses of water a day, which is equivalent to consuming half a gallon of water daily, due to the fact that a gallon amounts 64 ounces. So, 5 gallon bucket dimensions. ## How to know 5 gallon bucket dimensions So, 5 gallon bucket dimensions. When it comes to the U.K. standard is concerned, 1 gallon amounts 160 liquid oz, which is 4.546 liters of water. Significance, you would require 9.46 bottles of water to load a gallon. Consider this to be 9 containers or a little much less than 9 as well as a half containers. Individuals utilize gallons in order to carry a significant amount of liquid, 5 gallon bucket dimensions! ## United States gallon to mugs, 5 gallon bucket dimensions 1 Gallon (United States, fluid) = 16 Mugs (US). 1 Gallon (Dry, US) = 18.618355 Mugs (US). 5 gallon bucket dimensions

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0 # What is the name of a container that holds 5 to 15 gallons of corrosive liquid? Updated: 10/18/2022 Wiki User 10y ago Wiki User 10y ago Earn +20 pts Q: What is the name of a container that holds 5 to 15 gallons of corrosive liquid? Submit Still have questions? Related questions ### How many gallons in 72x72x24 container? A 72x72x24 foot container holds 930,696.3 gallons. 128 ### How many gallons doers a 48x12x12 container hold? A container that is 48x12x12 inches in size holds approximately 29 gallons of liquid. ### What is the amount a container holds? The amount a container will hold is the volume of the container. It is also the Capacity if the container has liquid in it. ### HOW MANY GALLONS DOES A 18X12X9 INCH CONTAINER HOLD? An 18x12x9 inch container holds approximately 3.56 gallons of liquid. ### What holds 108 gallons of water? A container with a volume of 14.44 cubic feet will hold 108 gallons of water. 20 ### What holds 180 gallons of water? A container with a volume of at least 24.06 cubic feet. ### A container that holds liquid that begins with the letter M? mason jarmouthmoat ### How much liquid will a school milk container hold? A typical school milk container holds around 8 ounces (240 milliliters) of liquid. ### How large of a container do you need to hold 1136 gallons? My swimming pool holds 1250 gallons. It is 12 feet diameter and 3 feet deep. ### How much liquid is in 32oz? There are 32 fluid ounces in a container that holds 32 oz.

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# McDermott and Shleifer double-team Taleb and Kahneman Not really. But I did enjoy reading the following passage from Andrei Shleifer's review of Daniel Kahneman's (superb) Thinking, Fast and Slow. To give context, he is talking about a key assumption of Prospect Theory (which Kahneman pioneered together with Amos Tversky); namely that individuals attach weights to objective probabilities in a way that causes them to overweight low probability events and underweight high probability ones. Shleifer writes: The evidence used to justify this assumption is the excessive weights people attach to highly unlikely but extreme events: they pay too much for lottery tickets, overpay for flight insurance at  the airport, or fret about accidents at nuclear power plants. Kahneman and Tversky use probability weighting heavily in their paper, adding several functional form assumptions (subcertainty, subadditivity) to explain various forms of the Allais paradox. In the book, Kahneman does not talk about these extra, assumptions, but without them Prospect Theory explains less. To me, the stable probability weighting function is problematic. Take low probability events. Some of the time, as in the cases of plane crashes or jackpot winnings, people put excessive weight on them, a phenomenon incorporated into Prospect Theory that Kahneman connects to the availability heuristic. Other times, as when investors buy AAA-rated mortgage-backed securities, they neglect low probability events, a phenomenon sometimes described as black swans (Taleb 2007). Whether we are in the probability weighting or the black swan world depends on the context: whether or not people recall and are focused on the low probability outcome. [Emphasis mine.] This is precisely the issue I was trying to point out here. Sometimes people greatly overweight the risks of low probability events (as suggested by Prospect Theory)... and yet at other times they completely underestimate them (as suggested by Taleb's black swan metaphor). As a result, we should be cautious in trying to make generalisable statements about human behaviour from either one of these theories alone. You may also recall that — for my temerity in pointing out this apparent tension between Kahneman and Taleb's theories — I was labelled an "idiot" by none other than Taleb himself. As I coyly suggested at the time, Taleb's knee-jerk tendency to accuse anyone who disagrees with him of idiocy meant that I was likely to be in good company at least. I am sure of that now having read Shleifer's article. Tags: Updated:

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# Jeopardy! All Star Games – Predictions Here are my predictions for the 2019 Jeopardy! All-Star Games! # Methodology & Player Rankings: For each of the 18 players, I analyzed every single game, regular play and tournament, and counted the number of low-valued and high-valued clues each player got correct (and incorrect) in each game. (In a regular, College, or Teachers, a low-valued clue is the top three rows on the board. In a Tournament of Champions/supertournament game, it is the top two rows. In a Teen tournament game, it was the top 4 rows. The high-valued clues were the rest). I then took an average for each player over their original run, and for each tournament they participated in. (I did not take stats for the Watson games). The average was then weighted to give 0.5 weight to a player’s original run, and then 1.0 weight to each tournament a player has since participated in. The “low” value is then multiplied by 4.5, and the “high” by 12 (to represent the average value of a low-valued and high-valued clue). That number was then multiplied by 100 and rounded to the nearest 200, so as to project a full game Coryat score for each player. Here are the calculations for Brad Rutter, for example. Original 5 games: 16.4 low, 8.8 high Tournament of Champions: 8.75 low, 12.75 high Million Dollar Masters: 11 low, 11 high Ultimate Tournament of Champions: 11.7 low, 11.4 high Battle of the Decades: 8.4 low, 13.6 high Weighted Average: 10.7 low, 11.8 high Projected Coryat: \$19,000 ## Player Rankings: Here is how the model ranked all 18 players: Captains: 2. Ken Jennings \$17,000 3. Colby Burnett \$14,600 4. Julia Collins \$13,200 5. Buzzy Cohen \$12,800 6. Austin Rogers \$12,200 Selections: 1. Alex Jacob \$20,000 2. Larissa Kelly \$18,800 3. Matt Jackson \$18,400 4. Roger Craig \$16,400 5. Ben Ingram \$15,400 6. Alan Lin \$14,600 7. Seth Wilson \$13,400 9. Pam Mueller \$12,400 10. Monica Thieu \$9,800 11. Jennifer Giles \$9,200 12. Leonard Cooper \$7,200 Daily Double and Final Jeopardy! performance were then also used for each team to create a “variance” score for each team for the model (as team good on Daily Doubles might not find them, or they could “crash and burn” on those clues, or in Final Jeopardy). Better Final Jeopardy! performance decreased variance, as does conservativity on Daily Doubles. For each team, I then prorated those scores to what rounds each player might play within the lineup composition rules of the tournament (in that each player may not play the same round in both halves of the match). Adjustments were also made at this point to team scores based on a player’s average gain on Daily Doubles per game and Final Jeopardy %age, as some teams were clearly picked based on Daily Double and Final Jeopardy performance. (Again, as an example, for Team Buzzy, I projected that Alex would play a Single and a Double Jeopardy, that Buzzy would play a Double and a Final, and that Jennifer would play a Single and a Final.) I then took the adjusted prorated scores and variance data and ran it through the tournament format one million times to come up with the tournament projections for each team. # Match #1 (Wednesday, February 20 – Friday, February 22): ## Team Colby (5th (Pam) & 8th (Alan) selections): Wins First Match: 9.860% of the time. Second in First Match: 30.554% of the time. Third in First Match: 59.586% of the time. 1st: 2.146% of the time. 2nd: 6.024% of the time. 3rd: 11.333% of the time. 4th: 20.085% of the time. 5th: 32.611% of the time. 6th: 27.800% of the time. Avg. Position: 4.584. Andy’s Thoughts: Having been paired against Team Buzzy and Team Brad, who were two of the favourites going into the event, Team Colby did not get a good draw, and that’s certainly had a negative effect on their tournament chances here. Also, the model only ranks Pam 9th of the 12 selectable players, which certainly is lower than many others have rated her. That being said, Colby did pick a team with a track record of out-playing their raw numbers in tournament play, so it is quite possible that this team outperforms its projection. With Teams Buzzy and Brad standing in their way, though, I’d be hard-pressed to see Team Colby finishing higher than 4th, though. ## Team Buzzy (1st (Alex) & 12th (Jennifer) selections): Wins First Match: 33.293% of the time. Second in First Match: 29.512% of the time. Third in First Match: 37.196% of the time. 1st: 17.509% of the time. 2nd: 16.759% of the time. 3rd: 19.192% of the time. 4th: 14.130% of the time. 5th: 16.055% of the time. 6th: 16.355% of the time. Avg. Position: 3.435. Andy’s Thoughts: I think that Alex Jacob is coming into this tournament with more to prove than anybody else. I’ve previously had the conversation on #JeopardyLivePanel as to who the #3 player of all-time is. If Alex has a good tournament, he may well enter the conversation of “Who is the #1 player of all time”, and I think Alex wants to be a part of that conversation. Looking back on Jennifer’s Tournament of Champions quarterfinal, she did very well at the bottom of the board against Matt Jackson, and I think that she doesn’t get as much respect as she deserves. I mean, I can see why Austin picked Leonard over Jennifer (I’ll get to that when I talk about Team Austin), but I think that she also feels like she has something to prove after being picked 12th. This is the only team with two Tournament of Champions winners on it, so there’s no mistaking that this team knows how to win. I’d personally be surprised if this team did not make the final, and at that point, anything can happen. ## Team Brad (6th (Larissa) & 7th (David) selections): Wins First Match: 50.241% of the time. Second in First Match: 36.291% of the time. Third in First Match: 13.467% of the time. 1st: 35.649% of the time. 2nd: 28.717% of the time. 3rd: 11.396% of the time. 4th: 15.714% of the time. 5th: 5.914% of the time. 6th: 2.610% of the time. Avg. Position: 2.354. Andy’s Thoughts: Well, this team is stacked. Larissa fell to 6th position for Brad, and if she plays anywhere near where the model is rating her, this tournament is Team Brad’s to lose. There are a couple of wild cards, here: David hasn’t played the game in 13 years, which is the longest layoff of anyone in the field. Also, Brad can’t both play Double Jeopardy and Final Jeopardy in Game 2 of a match; he’s going to need to pick one, and that may provide an opening for another team to take this team down. # Match #2 (Monday, February 25 – Wednesday, February 27): ## Team Austin (2nd (Roger) & 11th (Leonard) selections): Wins First Match: 36.677% of the time. Second in First Match: 15.635% of the time. Third in First Match: 47.689% of the time. 1st: 14.666% of the time. 2nd: 7.680% of the time. 3rd: 22.823% of the time. 4th: 3.977% of the time. 5th: 11.432% of the time. 6th: 39.422% of the time. Avg. Position: 4.081. Andy’s Thoughts: This is not a team built to finish second. It was built for aggressive betting and built in the same way Austin plays the game himself. Yes, the computer thinks he could have picked different people, but those others wouldn’t have fit into Team Austin as well. Will it work? Well, the computer thinks Team Austin is most likely to finish in 6th. But it also thinks they have a reasonable shot at winning – which is what will happen if the aggressive play works out. Regardless of the outcome, there will be excitement when Team Austin is on the stage. ## Team Julia (4th (Ben) & 9th (Seth) selections): Wins First Match: 18.703% of the time. Second in First Match: 46.870% of the time. Third in First Match: 34.428% of the time. 1st: 4.111% of the time. 2nd: 11.777% of the time. 3rd: 14.332% of the time. 4th: 27.167% of the time. 5th: 27.971% of the time. 6th: 14.642% of the time. Avg. Position: 4.07. Andy’s Thoughts: This team is probably stronger than the computer gives it credit for, as Julia did have the flu during her Tournament of Champions, and Ben is probably the strongest player in the field when it comes to Final Jeopardy. If things break the right way, there’s a very good chance of seeing Team Julia in the final (though they’re probably going to have to win the Wild Card game to do so). And as I said with Team Buzzy, at that point, anything can happen. ## Team Ken (3rd (Matt) & 10th (Monica) selections): Wins First Match: 44.621% of the time. Second in First Match: 37.495% of the time. Third in First Match: 17.884% of the time. 1st: 21.507% of the time. 2nd: 23.236% of the time. 3rd: 18.350% of the time. 4th: 15.511% of the time. 5th: 12.058% of the time. 6th: 9.339% of the time. Avg. Position: 3.014. Andy’s Thoughts: Much like the Battle of the Decades, it seems like the tournament is most likely going to be “Brad vs Ken vs someone” in the final. Matt Jackson is probably the most Ken-like player in the field and Monica pretty much won her tournament because she’s perfect in Final Jeopardy. Both Ken and Matt are likely stronger on the signaling device than anyone on the other teams in the quarterfinal, and that should see them through to the final easily. But once they get there, both players will likely face past nemeses (Ken in Brad and Matt in Alex). Will they be able to win, or will Ken and Matt see another 2nd place finish? So, there you have it. What will happen? Tune in February 20 to find out! There’s always something new coming into The Jeopardy! Fan Online Store, including something perfect for Boston baseball fans! Here are our top-selling items! Become a Supporter now! Make a monthly contribution to the site on Patreon! Contestant photo credit: jeopardy.com When commenting, please note that all comments on The Jeopardy! Fan must be in compliance with the Site Comment Policy. If you are going to quote any statistics from this page or this website, attribution is required. Have you had a chance to listen to our new podcast game show, Complete The List, yet? Check it out! It’s also available on Apple Podcasts. #### 1 Commenton "Jeopardy! All Star Games – Predictions" 1. One note: The “Correct” is “net correct answers” (correct answers count +1, incorrect counts -1).

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Try the Free Math Solver or Scroll down to Tutorials! Depdendent Variable Number of equations to solve: 23456789 Equ. #1: Equ. #2: Equ. #3: Equ. #4: Equ. #5: Equ. #6: Equ. #7: Equ. #8: Equ. #9: Solve for: Dependent Variable Number of inequalities to solve: 23456789 Ineq. #1: Ineq. #2: Ineq. #3: Ineq. #4: Ineq. #5: Ineq. #6: Ineq. #7: Ineq. #8: Ineq. #9: Solve for: Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg: ### Thousands of users are using our software to conquer their algebra homework. Here are some of their experiences: When kids who have struggled their entire lives with math, start showing off the step-by-step ability to solve complex algebraic equations, and smile while doing so, it reminds me why I became a teacher in the first place! Teron, PA The first time I used this tool I was surprised to see each and every step explained for each equation I entered. No other software I tried comes even close. This is great, finishing homework much faster! B.F., Vermont The Algebrator was very helpful, it helped me get back on track and bring back my skills for my next school season. The program shows step by step solutions which made learning easier. I think this would be very helpful to anyone just starting to learn algebra, or even if they already know it, it would sharpen their skills. Sharon Brightwell, WA Its a miracle! I looks like it's going to work. I get so close with solving almost all problems and this program has answered my prayers! James J Kidd, TX

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## Friday, April 18, 2014 ### Basic geometry - Sandown Meeting House revisited I have just revised the post I wrote on the Sandown, NH, Meeting House. http://www.jgrarchitect.com/2014/02/sandown-new-hampshire-meeting-house-1773.html I have learned a lot more about early New England geometry since I wrote about Sandown in February. I have been in the Sandown Meeting House, sat in the pews, climbed among the trusses, stood at the lectern in the pulpit. I have visited and explored the geometry of the Parson Capen House, 1683; and the Rockingham, VT, Meeting House, 1785-1800. I am reading Palladio's 4 Books. I revisited the Sandown measured drawings, found simpler, cleaner geometries which could have been employed. A new post would have let the old diagrams remain. They are better discarded. I really didn't want anyone to come across the old post through an internet search.  So I redrew and rewrote it. People had looked at that post 84 times before I revised. If you are one, please go back and read the new version. Some of what I learned: - How to divide a square into halves; more importantly, why. Follow the progression on the graph paper here: #1-#2-#3-#4 - counter-clockwise. I saw how that geometry informed the Parson Capen House. - What patterns happen when the square in question is divided both horizontally and vertically - see A, B, C, D, E. - And all the diagonals added. - And then, imagine 2 squares overlapping! Since this happens at all 3 of the meeting houses I've studied, I needed to think about it more carefully - see F, G. - And if overlapping is reasonable, how about turning the square 45*, on its point?  See H. The circle surrounding the square is really unnecessary, and, I think, probably not used. Imagine G or H with all its diagonals as in F! I presented some of that in my diagram for the doors of the Rockingham Meeting House. Finally I have been trying to understand the influence of Palladio on American vernacular architecture before the 1820's in terms of geometry. We seem to use squares as a base in New England, and daisy wheel circles in New York. We use 3-4-5 triangles, especially when building an addition. About 1800 the more urban builders seem to begin to explore the Golden Section and less traditional ways to use geometry. Currently, I do not see builders north of Boston using circles before 1790. And I will now be skeptical when I see what appears to be a Golden Section. The turned square - H above - uses diagonals just as does the Golden Section. But within the confines of the square, not as an extension. #### 1 comment: Unknown said... Jane, I enjoy your posts as I am a carpentry contractor and love geometry and proportions to be correct. I noticed reading this one that you wrote "the square 45*" If you want the real degree mark like this 45° you need to use the ascii character #248. You can enter that by holding down the alt key and type 248 on the numeric keypad. It doesn't work if you use the numbers above qwerty keyboard. You may also be interested in Alt+155 = ¢, 171 = ½, 172 = ¼, 227 = π, 232 = Φ, 237 = φ, 241 = ±, 242 = ≥, 243 = ≤, 246 = ÷, 251 = √. Best regards, Mike Rowan

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https://discuss.pytorch.org/t/inplace-operations-for-custom-layers-complicating-backpropagation/59987

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# Inplace operations for custom layers complicating backpropagation I’m trying to implement a custom version of ReLU that requires a bit more logic. It looks something like this ``````class ReLU(torch.nn.Module): def __init__(self, in_features): super(ReLU, self).__init__() # Parameterization of the special ReLU layer self.lambdas = Parameter(torch.rand(in_features)) def forward(self, x): # compute some upper and lower bounds on the input rows bounds = map(row_bound, x) _, epsilon_id = x.shape # loop over rows for i, (l, u), lmb in zip(range(x.shape[0]), bounds, self.lambdas): # check if the upper bound is <= 0 then the ReLU returns 0 if u <= 0: # Inplace op x[i] = x[i] * 0 # If the bounds cross 0 the ReLU implements some custom logic elif l < 0: x = torch.nn.ZeroPad2d((0, 1))(x) # Inplace op x[i] = x[i] * lmb if lmb >= u / (u - 1): # Inplace op x[i, epsilon_id] = -l * lmb / 2 else: # Inplace op x[i, epsilon_id] = u * (1 - lmb) # Inplace op x[i, 0] = x[i, 0] + x[i, epsilon_id] epsilon_id += 1 # If neither if branch ran the lower bound >= 0 and the ReLU returns identity return x `````` The input x comes in as an n by m tensor. The forward pass first computes upper and lower bounds for each row, then based on those bounds it either zeroes that row, leaves it, or applies a bit more logic based on the layer parameters lambda. The issue is I don’t know how to modify individual rows of x without performing in-place operations, and if I do those operations I get frequent errors doing backpropagation to learn the layer parameters. x[i] = x[i] * lmb RuntimeError: one of the variables needed for gradient computation has been modified by an inplace operation: [torch.FloatTensor [6]] is at version 3; expected version 0 instead. Hint: the backtrace further above shows the operation that failed to compute its gradient. The variable in question was changed in there or anywhere later. Good luck! My one attempt to fix this was to make whatever gets passed into the model, i.e `x`, require gradients, but this seems unnecessary since I don’t ever want to compute gradients of `x` and it still doesnt help because I get the following error likely because I’m directly manipulating trainable parameters RuntimeError: leaf variable has been moved into the graph interior My only other thought is that the logic is too complex for autograd and in need to implement this ReLU as a function with a proper backward pass method, but I don’t want to go there if I don’t have to. If anyone has some insight on how to solve this problem that would be much appreciated. Thanks Hi, One solution I would recommend to make this into out of place ops is to use list and torch.stack (or torch.cat if the dimension already exists): ``````res = [] for i in range(foo): new_x_i = bar(x[i]) res.append(new_x_i) `````` If you want to stick with inplace operations, and want to avoid the “RuntimeError: leaf variable has been moved into the graph interior” error, simply clone the input you get: `x = x.clone()` at the beginning of your function. 1 Like Thanks for the response. Ill give your first attempt a try but it seems like a bit of a pain. Cloning the variables doesn get rid of the leaf error, but unfortunately not the inplace operation error. For the inplace above, the problem is actuall with `x[i] = x[i] * lmb`. Since you need gradients for `lmb` as they are your parameters, you need the value `x[i]` to compute those. But you change `x[i]` inplace on the same line. So the value of `x[i]` needed to compute the gradients of `lmb` is overwritten hence the error you see. To remove the inplace error, you will need to at least make sure you don’t override this value. And potentially the same thing at other places. So do you suggest I use the method you gave above for all such operations? I don’t see why the operation being in place makes a huge difference. As a n example ``````x = torch.rand(1, in, requires_grad=False) x = nn.Linear(in, out)(x) loss = x[0,0] loss.backward() `````` Here we would get no errors even though x is needed to compute gradients of the layer parameters. The only difference is presumably that nn.Linear doesn’t do inplace modifications, but that seems like an almost irrelevant point. Wouldnt nn.ReLU run into the whole exact problem above since its needs to update each element of a tensor-based on its current value. `x = nn.Linear(in, out)(x)` is not an inplace operation ! What this does is • create a linear layer • forward x through it and get a new result • assign to this new result the variable name “x” You would do inplace operation if you use any of pytorch’s method that finish with a trailing underscore (like `mul_()`, `add_()`, `scatter_()` etc) or if you use python’s inplace ops (like `+=`, `/=` etc). For example if you do: ``````x = torch.rand(1, in, requires_grad=False) x_new = nn.Linear(in, out)(x) # This is out of place x += 1 # This is inplace loss = x_new[0,0] loss.backward() `````` So now we modify the input of the Linear inplace so this won’t run. The ReLU’s backward can be computed both with the input or the output of the function. So we can make inplace relu. 1 Like Hi. I had same issue and using your recommendation x = x.clone() readily solve it. Thanks.

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HSPICEtutorial # HSPICEtutorial - 1 INTRODUCTION HSPICE is a circuit... This preview shows pages 1–4. Sign up to view the full content. 1 INTRODUCTION HSPICE is a circuit simulation program. This tutorial is a guide to its use as a standalone tool for performing circuit simulation. In this tutorial HSPICE will be used to perform a transient analy- sis of several CMOS inverter models. These models vary in the complexity of the HSPICE mod- els used to describe the behaviour of the NMOS and CMOS transistors composing the inverter circuit. A simple inverter circuit will be simulated with the HSPICE program and the results will be ana- lyzed with the AvanWaves tool. This is a tool used to display, analyze, and print the results of HSPICE simulations. EXAMPLE 1: A simple CMOS inverter . Consider the basic CMOS inverter circuit shown below in Figure 1. Figure 1: Simple CMOS Inverter Circuit. + - + - VCC VIN IN M1 M2 OUT CLOAD 0.75 pF VCC This preview has intentionally blurred sections. Sign up to view the full version. View Full Document 2 The HSPICE netlist describing this inverter circuit is: Inverter Circuit .OPTIONS LIST NODE POST .TRAN 200P 20N .PRINT TRAN V(IN) V(OUT) M1 OUT IN VCC VCC PCH L=1U W=20U M2 OUT IN 0 0 NCH L=1U W=20U VCC VCC 0 5 VIN IN 0 0 PULSE .2 4.8 2N 1N 1N 5N 20N CLOAD OUT 0 .75P .MODEL PCH PMOS LEVEL=1 .MODEL NCH NMOS LEVEL=1 .END The PMOS and NMOS transistors are given the names M1 and M2 respectively. The two lines in the HSPICE netlist: M1 OUT IN VCC VCC PCH L=1U W=20U M2 OUT IN 0 0 NCH L=1U W=20U describe the connection of the drain , gate , source , and bulk(substrate) terminals, the name of the HSPICE model used to describe electrical characteristics of the transistor, and the length and width of the channel. For example, the transistor labelled M1 in the circuit (and the netlist) has its drain terminal connected to the node labelled OUT, the gate terminal connected to node IN, both the source and substrate terminals are connected to the node labelled VCC. The name of the HSPICE mosfet model used to describe this transistor’s electrical parameters is PCH. The two .MODEL lines tell HSPICE which transistor model to use. In this case, we are using two basic models called PCH and NCH. PERFORMING A TRANSIENT ANALYSIS 1. Type the above netlist into a file called inverter.sp . 2. Type the following to run HSPICE. hspice inverter.sp > inverter.lis This will run the HSPICE simulator, reading in the commands in the file inverter.sp and generat- ing an output file called inverter.lis. In addition, several intermediate files will be created which will be used by the AwanWaves tool to display the simulation results. 3. Once the simulation completes, the following message will be printed: >info: ***** hspice job concluded 3 real 0.5 user 0.2 sys 0.1 4. Examine the inverter.lis file. It will contain a listing of the voltages at nodes IN and OUT in the following tabular format: time voltage voltage in out 0. 200.0000m 4.9989 200.00000p 200.0000m 4.9989 400.00000p 200.0000m 4.9989 600.00000p 200.0000m 4.9989 800.00000p 200.0000m 4.9989 1.00000n 200.0000m 4.9989 1.20000n 200.0000m 4.9989 1.40000n 200.0000m 4.9989 1.60000n 200.0000m 4.9989 1.80000n 200.0000m 4.9989 2.00000n 200.0000m This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]} ### Page1 / 12 HSPICEtutorial - 1 INTRODUCTION HSPICE is a circuit... This preview shows document pages 1 - 4. Sign up to view the full document. View Full Document Ask a homework question - tutors are online

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Search All of the Math Forum: Views expressed in these public forums are not endorsed by NCTM or The Math Forum. Notice: We are no longer accepting new posts, but the forums will continue to be readable. Topic: Egyptian fraction identities Replies: 3   Last Post: May 26, 2013 2:05 PM Messages: [ Previous | Next ] James Waldby Posts: 545 Registered: 1/27/11 Re: Egyptian fraction identities Posted: May 26, 2013 11:26 AM On Sun, 26 May 2013 10:23:59 -0400, David Bernier wrote: > On 05/26/2013 02:10 AM, David Bernier wrote: >> John Baez on his Wordpress blog had an interesting fact about >> 42 today: [big snip] > If 1/2 = 1/p + 1/q + 1/r + 1/s + 1/t > > with p, q, r, s, t positive integers such that > p <= q <= r <= s <= t, > one possibility is given by: > 1/2 = 1/3 + 1/8 + 1/31 + 1/124 + 1/744 . > [ p=3 , q=8, r=31, s=124, t=744. ] > > Is it possible to have t > 744 ? Possibilities include sums like 1/3+1/8+1/26+1/313+1/97656 1/3+1/7+1/44+1/925+1/854700 1/3+1/7+1/43+1/1807+1/3263442 The last of those examples probably is extreme, since it is the result of a greedy approach; note that all of 1/3+1/6 1/3+1/7+1/42 1/3+1/7+1/43+1/1806 add up to 1/2. > This question is motivated by John Baez'Puzzle 1. > at his Azimuth Wordpress blog here: > < http://johncarlosbaez.wordpress.com/ > > > --> 42 > > Puzzle 1. > ``Consider solutions of 1/p + 1/q + 1/r = 1/2 > with positive integers p <= q <= r , and > show that the largest possible value of r is 42. " -- jiw Date Subject Author 5/26/13 David Bernier 5/26/13 David Bernier 5/26/13 James Waldby 5/26/13 David Bernier

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https://www.positiveachievementtutoring.com/Geometry-Trig

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# Learn Geometry and Intro to Trig Online and at Your Own Pace! ## How It Works #### INSTANT ACCESS ON DESKTOP AND MOBILE You do not need to spend hours on end working on this material. In fact it is more effective to spend frequent short periods of time. For this reason, your lessons are available in both Desktop and mobile versions. Do you have a few minutes before your practice or class begins? Check out a lesson. #### PRINTABLE NOTES AND EXAMPLES Each lesson starts with detailed notes covering concepts to be addressed and several examples of each process being covered. These notes are printable so that students can create their own book. Students are encouraged to use these pages to highlight topics for review as well as key concepts. #### VIDEOS OF PROBLEMS BEING SOLVED Sometimes the whole process can be overwhelming and you need to see things one step at a time. My videos provide step by step solutions to problems with  explanations for each step taken. Many students learn best when listening as well as seeing. This portion of the course provides both. #### QUIZZES AT THE END OF EACH LESSON Each lesson ends with a multiple choice quiz of the material presented. This provides students with immediate feedback regarding their mastery of the concepts. Students must score 80% or higher before being able to access the next lesson. This insures that a strong foundation is being built. Sometimes you just need to talk to somebody and ask a question. In that case, I offer webinars and Q&A sessions for an additional cost so that I can answer your questions “face to face". ## Course Syllabus Here's what's covered in this course (content is in both PDF and video format and each session ends with a quiz): Geometry Intro to Geometry • Points Lines and Planes, Measuring Segments, Measuring Angles, Angle Pairs Reasoning and Proofs • Patterns and Inductive Reasoning, Conditional Statements, Biconditionals and Definitions, Deductive Reasoning, Proving Angles Congruent Parallel Lines and Perpendicular Lines • Lines and Angles, Properties of Parallel Lines, Proving Lines Parallel, Parallel and Perpendicular Lines, Parallel Lines and Triangles, Constructions of Parallel and Perpendicular Lines, Linear Equations Expand for More Congruent Triangles • Congruent Figures, SSS, SAS, ASA, AAS, CPCTC, Isosceles and Equilateral Triangles, Congruence in Right Triangles, Congruence in Overlapping Triangles Relationships Within Triangles • Midsegments of Triangles, Perpendicular and Angle Bisectors, Bisectors in Triangles, Medians and Altitudes, Inequalities in One/Two Triangles • Angle Sum Theorem, Properties of Parallelograms, Proving a Parallelogram, Properties of Rectangles, Rhombus and Squares, Trapezoids and Kites, Coordinate Geometry Similarity • Ratio and Proportion, Similar Polygons, Proving Triangles Similar, Similarity in Right Triangles, Proportions in Triangles Right Triangles and Trig • Pythagorean Theorem, Special Right Triangles, Trigonometry, Angles of Elevations and Depression, Vectors Transformations • Translations, Reflections, Rotations, Symmetry, Dilations, Compositions of Reflections Area • Parallelograms, Triangles, Trapezoids, Rhombuses, Kites, Regular Polygons, Perimeter and Area of Similar Figures, Trig and Area, Circles and Arcs, Area of Circles and Sectors, Geometric Probability Surface Area and Volume • Space, Figures and Cross Sections, Surface Area of Prisms, and Cylinders, Surface Area of Pyramids and Cones, Volume of Prisms and Cylinders, Volume of Pyramids and Cones, Surface Area and Volume of Spheres, Areas and Volumes of Similar Solids Circles • Tangent Lines, Chords and Arcs, Inscribed Angles, Angle Measures and Segment Lengths, Circles in the Coordinate Plane Trigonometry Review • Solving equations, linear equations, graphs, • Functions, analyzing graphs of functions • Parent functions, transformations • Combinations of functions, inverse functions Trigonometry • Trig functions and the UNIT circle • Right triangle trig • Trig functions of any angle • Graphs of sine and cosine functions • Graphs of other trig functions • Inverse trig functions ## How We're Different I was a teacher for over 25 years. One of the hardest things for me as a classroom teacher was to realize that I had one or a few students that weren’t ready to move on to the next skill. Maybe they were missing a pre-requisite skill or maybe they just needed a bit more practice. But, as a classroom teacher, I had to move on. There was a whole list of topics to be presented that school year whether the students truly mastered them or not. We had to KEEP GOING! With this program you can spend as much time reviewing a skill as you need. You teacher may keep going but you can use this tool to look back and practice until you master a skill. Math is an upside down pyramid. You start with a few simple skills, adding, subtracting, multiplying and dividing. Each year the number of skills grows and grows. Why does that matter? if you have a gap in your knowledge from a prior year, it impacts your success more and more as you move on. For example, the first skill that give students trouble is often fractions. Building a strong foundation is invaluable in the mathematics. I believe it is those gaps in their foundations that causes so many students to think that they are “bad” at math or are “just not math people.” With the explosion of STEM careers, math literacy is a huge part of the future. By allowing students the ability to review topics as needed, this program provides a way to strengthen a student’s foundation and help them gain the confidence they need to become “math people.” Access to printable notes for each topic. Videos of problems being solved with step by step explanations. Quizzes to determine your mastery of each topic BEFORE you move on. If you need to review a previous lesson, you can always go back and access all of the material. You also have the option of attending a webinar (for an additional charge). These will be scheduled at least twice a week. Attendance is limited to 5 students and questions must be submitted 48 hours in advance.  For more information about our webinar options, click here. Math takes time and consistent effort. There is no lasting quick fix. If you have concerns as to whether this program is for you, start with the monthly option. There is no minimum number of months to purchase. That being said, please plan on working consistently 4 to 5 times a week for at least 15 to 20 minutes. That is only an hour to an hour and a half each week. The results will be worthwhile and lasting. I have developed comprehensive content for each of the courses. The number of lessons is different as the topics are different. Check the content listing for each course for a complete list of material covered. I over 25 years of teaching there have been less than 10 students who genuinely put in the effort and did not see results in a reasonable period of time. If you find that you are not making progress as you had hoped, you may want to join a webinar to ask your specific questions or make sure you have the prerequisite skills necessary for the topics you are studying. If you truly feel this product is not right for you, please see my cancellation policy. I am not able to meet in person. The joy of internet learning is that it allows teachers and students from anywhere in the world to work together. We may be 1000’s of miles apart but still be able to communicate. I do offer live webinars for members of my courses. Once you sign up for these (which will become available at checkout), you will receive an email of available times to choose from. If you'd like to see a sample schedule, please email us. ## Cancellation Policy Anything worthwhile takes work. You may wish that I could sprinkle magic pixy dust or use my Harry Potter wand and make your math problems disappear but that isn’t how this works. I am not suggesting that you will need to spend hours and hours each day to learn the material in this course. In fact, I often hear from my students that my explanations are clear, concise and help them understand the concepts more quickly and more fully. I wasn’t always a math person. I had to work at it. One day I was particularly frustrated and I went to see one of my college professors. He told me that math was his favorite subject because if he just worked at it hard enough for long enough he made saw results. Here he was teaching my class and admitting that he had to work at learning sometimes. I am a big believer in effort over outcome. If you put the effort in, you will achieve the outcome you desire. I ask you to work consistently (completing lessons 4 times a week) for a period of at least 2 months. If at that point you do not see improvement (your grade in school has not improved) I will refund 75% of your purchase price. Monthly memberships can be cancelled at any time. If you are concerned that this may not be a good fit for you, you might want to start with a monthly plan. I am confident that if you follow the program consistently, you WILL see improvement and your grades and confidence will increase more and more each day. ## Get Started Now! Ready to join? Just select your plan below to join and start improving your Geometry and Trig skills today! ## \$37 PER MONTH Printable Notes Videos with Step by Step Explanations Quizzes to Determine Mastery at the End of Each Lesson ## \$247 PER YEAR Printable Notes Videos with Step by Step Explanations Quizzes to Determine Mastery at the End of Each Lesson ## Webinars \$57 each (or get a package of 5 for \$249) Webinars are available for members upon checkout. Limited to 5 students per event to provide maximum support. Submit your specific questions in advance (48 hours prior to event). Close 50% Complete ### Two Step Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua.

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A CIESE Collaborative Project Fall 2006 Student Reports Hamden, Connecticut During the Boiling Point project, we came up with many different hypotheses about what factor most affected the boiling point of water, including the volume of water and the room temperature. Many of us did not even consider elevation to be a factor, much less a key factor. After looking at data from around the world, we noticed a consistent change of boiling points depending on the altitude. We learned that it is easier for the molecules of water to break apart from each other at higher elevations, because the gravitational pull on the air above them pushes down on them less when the air column above them is shorter. For instance, on Mt. Everest, because the altitude is higher, air pressure is lower than at sea level, so the liquid water molecules can escape as vapor at lower temperatures and with less energy than they would need to vaporize at sea level. We analyzed some sources of error in this lab, including problems with the calibration of the thermometers, having our thermometers too close to the hot plates or to the air, or having contaminants in the water. In the future, we would be interested in investigating the boiling point of other liquids such as soda, salt water, and vinegar. We would also be interested in looking at the correlation between boiling point and air pressure. Marlboro, New Jersey Our classes tested different volumes of water and found that all of the boiling points were about the same. The one thing that we did notice is that the boiling points of lab groups near the flow of the air conditioner were slightly lower than those groups on the other side of the room. In applied technology class we learned that flowing air causes low pressure so we guessed that that was why there was a difference. When we compared our elevation to other schools in the project we determined that the elevation affects the boiling point. Lower air pressure at higher elevations lets the water boil at a lower temperature, the same way the air conditioner flow in our lab affected the groups on that side of the room. We had alot of fun with this project. It helped us to practice the scientific method that we learned about in class. We felt like "real scientists" since we were part of a project with kids from all over the world! Camden High School (Grade 9 Honors Earth Science) Camden, New Jersey Our original hypothesis was that the heating device would have the strongest correlation to the boiling point of water. We were surprised at both our results and other classes results. When we analyzed the data we found the line of best fit on the graph for elevation vs. boiling point was closest to the data points. Therefore, elevation was the variable that showed the strongest correlation to the boiling point. We think there is a possibility that our boiling point temperatures were inaccurate, because when we reached our boiling point the temperature dropped slightly. We think this happened because we used a small volume of water (250 ml), and when some of the water evaporated and the water started to boil, some air may have touched the thermometer, making the temperature drop. If we did the project again we would change the volume of water that we use. Also, next time we would like to try boiling a different liquid. We learned a lot from this project. We hope you had fun because we did!! West Sussex, UK We were trying to find out what temperature water boils at by using a Bunsen burner. We found out that people closer to the sea had a higher boiling point temperature. We learnt that water boils at 100°C where we are. For next time on the website, it could show a picture of what type of heating device you can use so we can see it. We could do more experiments now, but use the same amount of water this time. Marlboro, New Jersey Just in from Marlboro Middle School Even with the different volumes of water, the boiling point for all of Mr. Conkling's classes were about the same. Period One had a boiling point of 99.73 degrees Celsius, Period Two had a boiling point of 99.33, Period Four had a boiling point of 99.00, Period Six had a boiling point of 99.86, and Period Seven had a boiling point of 99.75. Marlboro has an elevation of 61.876 meters above sea level, and after examining data from other schools around the world - WE STARTED TO NOTICE SOMETHING! Higher elevations had lower boiling points This also relates to what some of us are learning in applied technology (with planes). We, in Mr. Conkling's class, enjoyed performing the boiling point experiment. This experiment helped us see relation to places all over the world and it was cool to think that children all over the world were using our information - JUST LIKE WE WERE USING THEIRS. We also used the scientific method in a real-world setting (instead of boring notes ) Thank you to Stevens Institute for setting this experiment for the world and thank you to Marlboro Middle School - HOME OF THE HAWKS Loudonville, New York Upon analyzing the data from all over the world, it would seem that elevation had the most consistent effect on the boiling point. One thing that we found interesting in our own experiments using alcohol thermometers was that the boiling point of each of our individual groups (within the class) varied more than we thought it would. The average for the group (102.0 C) was reasonable, but we were surprised that all of our numbers weren't the same or closer together. It would seem that the measuring device can also influence how consistent the data is. We are planning on repeating the experiment using a temperature probe connected to a computer that is programmed to collect data continously.It will be interesting to compare that to our current data. We expect that we will have greater precision in our data. It was interesting to participate in this study because it allowed us to see how much more valuable data can be when there are numerous groups participating in the study. Short Hills, New Jersey Our class spent alot of time analyzing and graphing the data from The Boiling Point Project. We found that elevation has the greatest effect on boiling point. The higher the elevation, the lower the boiling point. Our graph of the type of heating device and boiling point suggested there might be some influence, but there was not enough data for all the different types of heating devices to make a conclusion. Our class hypothesis was that at our school elevation (41m) our boiling point would be close to 100C. In fact, the boiling point of water in our experiment was 100.5C, so our hypothesis was correct. We used the data from the project to discuss what differences there might be in cooking temperatures and times at different elevations. This project helps confirm the differences and similarities among different cultures around the globe. American School of Tampico (Grade 10) Tampico, Mexico After analyzing the data provided by the International Boiling Project, we found out that elevation had the greatest influence on the boiling point of water. The hypothesis was correct. The rest of the factors, heating device, volume of water, and room temperature didn’t have a great effect on the boiling point of water. Based on the results and in comparison with the other schools that were at different altitudes, the ones closes to sea level had a boiling point close to 100°C, whereas higher places had a boiling point as low as 92.1°C. We think that the accuracy in measurements is very important in all experiments. We all put a lot of effort into making our measurements as careful as possible, and since the same experiment was repeated by the five teams in each of our four science classes and then results were averaged, we believe that our final results are very accurate. We are 23 m above sea level and we obtained an average boiling point of 98.5°C. Another variable that can affect the boiling point of water is gravity. Although it would be very difficult to test this, we could research about this and find out if water has been boiled in outer space and how this happens. Thank you for the opportunity of participating with people from other places, ATS G10 Science Belle Chasse, Louisiana From looking at the data both classes concluded that our boiling point was low and that elevation seemed to have something to do with the boiling point. We discussed as a class what else might have made our boiling point lower and we also concluded that the humidity outside and inside that classroom might of played a role. Gurzler Miqdash Me'At (Grades 5 and 9) Drums, Pennsylvania Both children used the same volume of water (500 ML). Both children used the same heating device (a gas kitchen stove). The children did their experiments on the same days, at the same time (Monday, Tuesday, and Wednesday of this past week, after lunch). The temperature in the room was chillier than in most of your rooms, I see. (An average of 19.2 degrees C.) Our elevation is 323.088 Meters. The only differences were the types of pots used to heat the water: one child used an aluminum pot for his experiments, the other used a stainless steel one. The water in the aluminum pot boiled at an average temperature of 93.6 degrees C in an average of 25 minutes. The water in the stainless steel pot boiled at an average temperature of 98.3 degrees C in an average of 23 minutes. We were very surprised to see that the metal content of pot made a difference in how fast the water boiled. We hadn't realized that water boils at lower temperatures the higher up in elevation you go. Interesting. We just took for granted that water boils at 100 degrees C or 212 degrees F. We never thought about anything that might affect that number. The most important thing we learned was the importance of the accuracy of measurement, not just in the tools we used, but also in our handling of those tools. For instance, we learned that one of the thermometers in our home was off by 15 degrees. (No wonder the steaks we checked using this instrument were always far more underdone than we preferred!) Our younger son learned about the importance of keeping a consistent eye level when reading measurements. He had never considered that he must behave in a precise manner when doing experiments. Through the repetition of the experiment over the course of a few days, our older son gained a new respect for controlling variables that could affect scientific experiments. Both children enjoyed participating in the project and talked about what they were doing with their friends. A.J. Ferrell Middle Magnet School (Grade 8) Tampa, Florida Good Day from the Ferrell Roaring Tigers, We are proud to report our results, as surprising as we found them to be. We had nine groups complete this project and we found some amazing results. We found that after the first day we were quite able to predict just about to a tee the time that our water would boil and the temperature to. It was amazing, but we thought that if we started with a smaller amount of water in our air conditioned building, we would get to our boiling temperature quicker, and the temperature would be right on the acutal boiling point of water. Boy were we wrong. Our average boiling point was 97.8 degrees Celsius. We started off with our room temperature at 23.67 degrees Celsius. So we guessed maybe our room temperature might have played some type of role in the boiling point being lower. More than that we think that being at 53.55meters had something to do with our boiling point change. We all had a great time doing this experiment, and look forward to doing it again. We are sending in photos of our class in action. Can't wait for the next experiment. ROAR!!!!! Ms. Goellner's Roaring Tigers St. Louis King of France School (Grades 5, 7, and 8) Baton Rouge, Louisiana The average boiling point for our school was 93.65 degrees C. Our average room temperature was 22 degrees C. Most of us thought that the volume of water would have the greatest effect on the boiling point of the water. Some of us thought it would be the elevation that made the biggest difference. One thing we all learned is that the classroom temperature did not affect the boiling point at all. We also learned that the volume of water did not gratly affect the boiling point of the water either. Water boils at the same temperature no matter how much is boiling at once. We enjoyed doing this project because we learned a lot about controlling the variables in an experiment. Our classes have decided to continue this experiment by testing other variables like containers made of different materials and boiling other types of solutions. Roosevelt Junior High School (Grade 7) Roosevelt, Utah Period #1 There hypothesis was that the volume of water used affected the boiling point. By comparing one class to another we saw virtually no difference in boiling point. However we did notice that barometric pressure was different on two of the days and that could have made a difference in the in our class boiling pt data. Period #3 They stated that elevation is the biggest factor in boilingi point of water. Based on the data that seemed to be the most correct answer. We were at an elevation of 1554 meters and some schools were at an elevation of 0 meters and there was a difference in the boiling point of schools at a lower elevation then us. Period #7 We also stated that elevation is the biggest factor in boiling point. We concluded about the same things as 3 period. Period #8 We stated that the heating device used determined boiling point. After looking at the data we didn't see any relationship between boiling point at the type of heating device used. Altamont Jr./Sr. High School (Grade 7) Altamont, Utah Class #1 Heating device--a hot plate Elevation -- 6,390 ft (1947 m) Amount of water--250 mL Average room temperature--22 C Average boiling point -- 91.53 Class #2 Heating device--a hot plate Elevation--6,390 ft (1947 m) Amount of water--250 mL Average room temperature--22 C Average boiling point--92.1

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# Finding a Fraction of a Fraction Math Lesson Plan for Elementary Students Page content ## “Fraction of a Fraction” Math Problems Anticipatory Set: As a warm-up activity for students, put the problem “What is 1/2 of 1/4?” on the board. Ask students to solve the problem and then choose two students from each row or each table group to write their answer on the board. Have the students take their seats. Now engage in a classroom discussion by asking the class which answer is correct and circle the responses. Typical student answers might be: 1/2, 1/4. 1/6, 2/6, or 1/8 and your student responses may surprise you with most students choosing 1/6 or 1/8. So which one is correct: 1/8 of course. Lesson Objective: Problem solving “Fraction of a Fraction” math problems. There are fractions that start with basic problem solving such as 1/2 = 50% = 3/6 = 4/8 and so on. And then there are fraction of a fraction problems that require a different mathematical processing. So let’s get started by solving some of these problems. Directions: Solve for the following and put a brief explanation on how you got your final answer next to each problem. 1. 2/3 of 2/3 2. 1/4 of 1/3 3. 1/3 of 3/4 4. 1/5 of 1/5 5. 1/2 of 1/5 1. 2/3 of 2/3 = 4/9 because the “of” means you multiply “x” the two numerators and the denominators, so 2 x 2 = 4 and 3 x 3 =9 2. 1/4 of 1/3 = 1/12 because you multiply 1 x 1=1 and 4 x 3 = 12 3. 1/3 of 3/4 = 3/12 because you multiply 1 x 3 = 3 and 3 x 4 = 12, but you have to simplify, so you divide 3/3=1 and 3/12=4 - final answer = 1/4 4. 1/5 of 1/5 = 1/10 because you multiply 1 x 1 = 1 and 5 x 5 = 10 5. 1/2 of 1/5 = 1/10 because you multiply 1 x 1 = 1 and 2 x 5 = 10 Additional Strategies: You can also provide elementary students with manipulatives in different colors, so that visual and kinesthetic learners can see and touch math problem solving using groups of manipulatives to represent fractional parts. Another visual tool that is always fun and creative for students is allowing them to work collaboratively in groups of 2 and giving each group a piece of butcher paper and markers in five colors for each problem (groups can share the markers). Bonus Word Problem: To challenge students even further, provide a word problem, set a timer and the first two groups that problem solve correctly will get an extra 10 minutes on their class project on Friday or some other extrinsic reward that students are motivated to work hard to achieve in your classroom. 1. One-third of students in your class are males, and 1/2 of those male students have black hair. If there are 5 males in the class with black hair, how many students total are there in the classroom? Answer: 5 male students have black hair and 1/2 of these male students have black hair, so 5 x 2 = 10 males and since 1/3 are males, there are 10 X 3 = 30 students in the class which means there are 20 females in the class. The class is 1/3 males and 2/3 female. Checking the math processing might look like this: 10/30 males = 1/3 males and 20/30 females = 2/3 females in your class. A simpler method would be to count the number of males and females in the class, but then that’s too easy. Closure: Fraction of Fraction problems can provide lots of fun for younger students trying to understand how to figure out parts of wholes or 50% of the \$19.99 school backpack, which by the way is equal to a 1/2 of \$19.99 = \$9.995.

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0 # How many years ago was 8200 BC? Updated: 9/25/2023 Wiki User 8y ago 10,215 years ago. Wiki User 8y ago Earn +20 pts Q: How many years ago was 8200 BC? Submit Still have questions? Related questions ### How long ago was 8200 bce? Answering in 2018 CEAssuming that there is a 0 CE since 2000 CE was apparently the first year of the third millennium CE, it was 2018 - -8200 = 10,218 years ago (Unless there was also a 0 BCE, in which case it was 10,219 years ago.)Answering in 2018 ADThere was no year 0 AD (nor 0 BC) which means that 2000 AD was the LAST year of the second millennium AD, and 2001 AD was the FIRST year of the third millennium AD. Assuming you mean 8200 BC: &rarr; 8200 BC was 2018 - - 8200 - 1 = 10,217 years ago. ### How many years ago was 756 BC? 756 BC was 2767 years ago. 2012 years ago ### How many years ago was 1300 BC? 1300 BC was 3310 years from the present. ### How many years ago is 240 BC? 2035 years ago as of 2011 ### 460 bc means how many years ago? To determine how many years ago 460 BC was from the current year, you subtract 460 from the current year. Current year - 460 BC = 2024 - (-460) = 2484 years ago So, 460 BC was approximately 2484 years ago. ### How many years ago was 1800 BC? 2009+1800= 3809 years ago ### How many years ago was 8000 bc? 2010 + 8000 = 10,010 years ago ### How many years ago is 495 bc? 2508 years ago, as of 2014. 2000 years ? 1514 years 13010 years ago

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college math posted by . suppose that in a class of 50 students, 12 are married and 20 live at their parent's homes. two of the married students live at their parents' homes. a. draw a venn diagram of this information. b. how many students are single and don't live at their parents' homes? • college math - Box of all students with 2 intersecting circles in it. One circle is married students (total 12) One circle is parent homes(total 20) Intersection of circles contains 2 Outside both circles but inside box is single students living elsewhere. that is 50 -12 - 18 Similar Questions 1. math Approximately 5 out of every 10 students at Kendra's college live on campus. The college has 5,000 students. How many of them live on campus? 2. Math In a class of 550 students, students may take all, none, or a combination of courses as follows.? 3. English ll I dentify the complete adj. clause. 1. Many Californians who own homes live on hillsides. A: Who own homes live on hillsides. What word does the adj. clause modify? 4. Venn Diagrams In a class of 50 students, 36 students have a sandwich in their lunch box and 23 students have a piece of fruit and 7 have neither. Fill in a Venn diagram to represent this situation 5. Math 150 men live in a certain town: 85 are married 70 have a cell phone 75 own a car 55 are married and have a cell phone 35 have a cellphone and a car 40 are married and have car 30 are married, have a car and have a cellphone. How many … 6. Math 150 men live in a certain town: 85 are married 70 have a cell phone 75 own a car 55 are married and have a cell phone 35 have a cellphone and a car 40 are married and have car 30 are married, have a car and have a cellphone. How many … 7. Anonymous Life orientation task..My question is-What are any four kinds of fraud that students who receive and repay nsfas shoul guard against? 8. St Ann's College (DDL) Enugu In A Class Of 60 Students, 30 Offer Physcis And 40 Offer Chemistry, (a)Use Venn Diagram To Represent This Information (b) How Many Students Offer Both Subject 9. Math 3. After a study was completed at a local college, it was found that 386 1,112 student live with there parents. What percent of students live with their parents? 10. math In a freshman class of 3500 students at a particular college, 20000 are taking either a history of English class their first semester. Of these 2000 students, 1300 are taking both a history class and an English class, and 375 are taking … More Similar Questions

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# Related Videos ## Simplify $3+5+7$ Go! 1 2 3 4 5 6 7 8 9 0 a b c d f g m n u v w x y z . (◻) + - × ◻/◻ / ÷ 2 e π ln log log lim d/dx Dx |◻| = > < >= <= sin cos tan cot sec csc asin acos atan acot asec acsc sinh cosh tanh coth sech csch asinh acosh atanh acoth asech acsch ### Tutorial - Simplifying a complex rational expression, ((1/x) + (1/(2x+1))) / (4x/(2x+1)) $3+5+7$

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## If 18 is subtracted from twice the square of an integer, the result is equal to nine times the integer. Find the integer. [Only Question If 18 is subtracted from twice the square of an integer, the result is equal to nine times the integer. Find the integer. [Only an algebraic solution will be accepted.] in progress 0 3 years 2021-08-24T03:35:35+00:00 1 Answers 3 views 0 ## Answers ( ) -3/2 Step-by-step explanation: Let x = the integer: 2x^2 – 18 = 9x 2x^2 – 9x – 18 = 0 Now either factor that or use the quadratic formula if you’re lazy like me. x = -3/2 and 6, so x must be 6 because it has to be an integer.

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