url
stringlengths 7
1.6k
⌀ | fetch_time
int64 1,368,854,001B
1,726,893,573B
⌀ | content_mime_type
stringclasses 3
values | warc_filename
stringlengths 108
138
⌀ | warc_record_offset
int32 35.5k
1.73B
⌀ | warc_record_length
int32 630
783k
⌀ | text
stringlengths 77
589k
| token_count
int32 40
312k
⌀ | char_count
int32 77
526k
⌀ | metadata
stringlengths 439
444
⌀ | score
float64 3.5
5.13
⌀ | int_score
int64 4
5
⌀ | crawl
stringclasses 93
values | snapshot_type
stringclasses 2
values | language
stringclasses 1
value | language_score
float64 0.07
1
⌀ |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://goprep.co/q6-cos-4-a-sin-4-a-is-equal-toa-2-cos-2-a-1-b-2-cos-2-a-1-c-i-1nk9ms | 1,601,169,691,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400249545.55/warc/CC-MAIN-20200926231818-20200927021818-00450.warc.gz | 406,461,029 | 45,365 | Q. 6
# cos4 A − sin4 A is equal toA. 2 cos2 A + 1B. 2 cos2 A − 1C. 2 sin2 A − 1D. 2 sin2 A + 1
To find: cos4 A – sin4 A
Consider cos4 A – sin4 A = (cos2 A)2 – (sin2 A)2
a2 – b2 = (a – b) (a + b)
cos4 A – sin4 A = (cos2 A)2 – (sin2 A)2
= (cos2 A – sin2 A) (cos2 A + sin2 A)
= (cos2 A – sin2 A) [ cos2 A + sin2 A = 1]
= cos2 A – (1 – cos2 A) [ sin2A = 1 – cos2A]
= cos2 A – 1 + cos2 A = 2 cos2 A – 1
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better. | 241 | 534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-40 | latest | en | 0.621185 |
null | null | null | null | null | null | Chủ Nhật, ngày 16 tháng 12 năm 2012
Uvula is the fleshy mass, located in the back of the mouth, at the center. As you open your mouth and look in the mirror, you will notice it as a protruding structure than hangs from the soft palate. The main purpose of uvula is production of various sounds with the help of air coming in through the palate and throat. It also helps in minimizing throat infection and directing ingested foods to the digestive tract. Comprising muscle fibers and connective tissues, swollen uvula is a commonly reported problem.
What Causes a Swollen Uvula?
An inflamed uvula or uvulitis is not a disease, but swelling of this mass is caused due to other health problems. Any disease that is concentrated in the throat region can lead to a swollen uvula. Constant dry throat is a leading cause for abnormal swelling of uvula. Those who have the habit of breathing through mouth during sleep are more prone to this uvula condition than other people. The probable reasons for uvula swelling are many, some of which are listed below.
Less water intake and dehydration
Inhalation of pollutants and irritants
Cigarette smoking or inhalation of fumes
Allergy to ingested foods
Allergy to mouth cleansing substances
Uvula piercing
Physical trauma to the throat area
Persistent cough
Bacterial infection of the throat (e.g. strep throat)
Viral throat infection (e.g. common cold)
Having very hot foods
Ulceration of the throat that affects uvula
Canker sores in the throat region
The most notable symptom that accompanies a uvulitis is pain in the throat region. Pain intensity is more at the time of swallowing foods, talking, and while indulging in any activity that involves movement of the swollen area. Usually, patients experience increased redness and itching along with uvula swelling. Such symptoms are indications of bacterial or viral infections, and require medications for treatment.
Treating a Swollen Uvula
Effective tips for how to cure a swollen uvula are decided in accordance to the root causes. So, identifying the factors responsible for swelling of this fleshy mass is a prerequisite before proceeding for swollen uvula treatment. For example; if swelling is manifested due to dry throat, then avoid shouting, breathing through mouth, and refrain from processed food items that contain high salt. Very soon, uvula will restore its normal state. Some of the effectual home remedies for swollen uvula, which you can practice are mentioned below.
Drinking ample amounts of fluids and water is a simple, yet effectual way for addressing a swollen uvula. This is specially applicable for people, who experience uvula swelling due to dehydration. Overall, it reduces uvula swelling and keeps the body hydrated.
Gargling with lukewarm salt water is effectual in treating a swollen uvula, which is developed due to infections. The saline solution aids in reducing pain and swelling, and treats infection. However, make sure that water temperature is tolerable to the sensitive mouth parts.
Another remedy is gargling with dilute apple cider vinegar. Add a teaspoon of cider vinegar in a cup of lukewarm water, and gargle with this solution. The mild acid kills bacteria and virus, and soothes pain.
Taking honey, aloe vera juice, and sipping holy basil tea also works wonders in reducing uvula swelling. Nevertheless, these natural remedies for a swollen uvula may take time to show prompt results.
These were some effectual swollen uvula remedies. Fortunately, uvula swelling is treated effectually with home remedies, without causing no major complications. The swelling usually disappears within 1 - 2 days even without treatment. However, seek medical attention right away, if the discomfort signs disturb normal eating, or they are accompanied with severe pain. Also, call the doctor, if swelling increases in the following days.
According to the severity of swelling and pain, the doctor may recommend medications, like pain killers, antihistamines, and antibiotics. You can effectually prevent a swollen uvula condition by adopting effectual self-care tips. A simple tip is, avoid sleeping on your back, specially after having salty foods in dinner. Try to sleep with your head elevated and consider lying on your side. For those who complain of frequent uvulitis bouts, adrenaline injections are delivered to prevent recurrent incidences.
Read more at Buzzle: | null | null | null | null | null | null | null | null | null |
https://studdy.ai/shared-solution/f77fa7ed-bbd5-4543-b09d-44500f582f0c | 1,701,517,804,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100399.81/warc/CC-MAIN-20231202105028-20231202135028-00762.warc.gz | 634,924,149 | 8,398 | # Math Snap
## 2 Josh puts $\ 2.75$ into hiscoin bank each week. How many weeks will it take him to save enough money to buy a jet model kit that cost $\ 13.75$ ? A 3 weeks B 4 weeks C 5 weeks D 6 weeks
#### STEP 1
Assumptions1. Josh saves $.75 each week. . The cost of the jet model kit is$13.75. 3. Josh does not have any other source of income or savings.
#### STEP 4
Now, plug in the given values for the total amount and weekly savings to calculate the number of weeks. $Weeks = \frac{\13.75}{\2.75}$
##### SOLUTION
Calculate the number of weeks. $Weeks = \frac{\13.75}{\2.75} =5$It will take Josh5 weeks to save enough money to buy the jet model kit. So, the answer is C5 weeks. | 209 | 691 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2023-50 | latest | en | 0.880793 |
https://pleasefireme.com/blog/how-many-ml-is-4-mg/ | 1,709,423,783,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476137.72/warc/CC-MAIN-20240302215752-20240303005752-00875.warc.gz | 451,993,514 | 13,451 | # How many mL is .4 mg?
## How many mL is .4 mg?
How Many Milligrams are in a Milliliter?
Volume in Milliliters: Weight in Milligrams of:
Water All Purpose Flour
2 ml 2,000 mg 1,058 mg
3 ml 3,000 mg 1,587 mg
4 ml 4,000 mg 2,116 mg
## How do I convert mL to mg?
MilliLitre to Milligram formula: Since 1 milligram is equal to 0.001 milliliters, which can be written as 1 mg = 1/1000 mL. Deriving from this equation, 1/1000 mL = 1 mg, hence 1 mL = 1000 mg.
Is 0.1 ml the same as 1mL?
2. The total volume of the syringe is 1.0mL. Therefore, it is measured in ten 0.1mL increments.
### What is 0.25 mg converted to ml?
Milligram to Milliliter Conversion Table
Weight in Milligrams: Volume in Milliliters of:
Water All Purpose Flour
240 mg 0.24 ml 0.453686 ml
250 mg 0.25 ml 0.47259 ml
260 mg 0.26 ml 0.491493 ml
### Is 40 mg the same as 1 ml?
Notice there is an extra thousandth on the weight unit. Therefore, there must be 1,000 milligrams in a milliliter, making the formula for mg to ml conversion: mL = mg / 1000 .
Is 60 mg the same as ml?
Solution: Milligram to milliliter conversion, 1 milligram is equal to 0.001 milliliter., 60 milligram is equal to 0.06 milliliter.
## How do I measure 1 mg?
1) tare the balance with the plastic cap. 2) measure out 10 ul of solvent (methanol, acetone) and dissolve 1 mg of powder into it 3) you can then let the solvent evaporate and you then know you have 1 mg remaining.
## Is 1.5 ml the same as 1.5 mg?
Solution: Milliliter to Milligram conversion, 1 milliliter is equal to 1000 milligram, 1.5 milliliter is equal to 1500 milligram. | 520 | 1,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-10 | longest | en | 0.814455 |
https://www.studypage.in/general-science/resistors-resistance-and-ohm-s-law | 1,632,252,513,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057227.73/warc/CC-MAIN-20210921191451-20210921221451-00617.warc.gz | 1,032,356,851 | 4,935 | # Resistors, Resistance and Ohm's Law
The electrical resistance is the tendency to resist the flow of electric current. A wire having a desired resistance for use in an electric circuit is called a resistor.
Resistance can be both either desirable or undesirable in a conductor or circuit. In a conductor, to transmit electricity from one place to another place, the resistance is undesirable. Resistance in a conductor causes part of electrical energy to turn into heat, so some electrical energy is lost along the path.
On the other hand, it is the resistance which allows to use electricity for light and heat. For example, light that we receive from electric bulb and heat generated through electric heaters.
### Ohm's Law
The current flowing through a wire is directly proportional to the potential difference applied across its ends.
V ∝ i
V = Ri
Here, R is a constant of proportionality and is called the resistance of the given metallic wire. This observation was first made by Georg Simon Ohm and is known as Ohm's Law.
Ohm's Law states that the current flowing through a conductor is directly proportional to the potential difference applied across the ends of the conductor provided temperature of the conductor remains the same.
The law can be applied only to conducting wires and that too when its temperature and other physical conditions remain unchanged. If the temperature of the conductor increases its resistance also increases.
### Resistance of Wire
The R, resistance of wire, is a constant for a given wire. The resistance of a wire depends on:
• Its length - longer the wire, more the resistance
• Its thickness - thicker the wire, lesser the resistance
• Its width - more the width, lesser the resistance
Therefore, the resistance of the wire is directly proportional to the length and inversely proportional to the cross-sectional area.
The nature of material: Copper wire has lesser resistance than iron wire of same length and thickness. The resistance of a wire can never be negative.
Resistance is a scalar quantity and its SI unit is ohm denoted by the symbol Ω (omega). 1 ohm is the resistance of a wire across which when 1 V potential difference is applied, 1 A current flows through the wire.
1 ohm = 1 volt / 1 ampere
High resistances are measured in kilo ohm (kΩ) and mega ohm (MΩ)
1 kΩ = 103 Ω
1 MΩ = 106 Ω | 517 | 2,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2021-39 | latest | en | 0.91546 |
https://jmservera.com/evaluate-4-1-2-11-6-2/ | 1,670,350,657,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711111.35/warc/CC-MAIN-20221206161009-20221206191009-00478.warc.gz | 366,530,637 | 13,104 | # Evaluate (4-1/2)/(11/6)
4-12116
Multiply the numerator by the reciprocal of the denominator.
(4-12)611
To write 4 as a fraction with a common denominator, multiply by 22.
(4⋅22-12)611
Combine 4 and 22.
(4⋅22-12)611
Combine the numerators over the common denominator.
4⋅2-12⋅611
Simplify the numerator.
Multiply 4 by 2.
8-12⋅611
Subtract 1 from 8.
72⋅611
72⋅611
Cancel the common factor of 2.
Factor 2 out of 6.
72⋅2(3)11
Cancel the common factor.
72⋅2⋅311
Rewrite the expression.
7(311)
7(311)
Combine 7 and 311.
7⋅311
Multiply 7 by 3.
2111
The result can be shown in multiple forms.
Exact Form:
2111
Decimal Form:
1.90‾
Mixed Number Form:
11011
Evaluate (4-1/2)/(11/6)
## Our Professionals
### Lydia Fran
#### We are MathExperts
Solve all your Math Problems: https://elanyachtselection.com/
Scroll to top | 295 | 813 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-49 | latest | en | 0.701502 |
http://persson.berkeley.edu/Programming_for_Mathematical_Applications/content/Sparse_Matrices/Application_Google_Page_Rank.html | 1,709,374,399,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475806.52/warc/CC-MAIN-20240302084508-20240302114508-00314.warc.gz | 32,615,851 | 9,150 | Contents
PageRank (PR) is an algorithm used by Google Search to rank web pages in their search engine results. It works by counting the number and the quality of links to a page.
The PageRanks for an example network are shown below. Page C has a higher PageRank than Page E, even though there are fewer links to C; the one link to C comes from an important page and hence is of high value.
17.4.1. Algorithm#
The starting point is the webgraph, which is a graph where each web page is a vertex, and a link from one to another is an edge. The goal is to determine weights $$PR(p_j)$$ for each vertex (web page) $$p_j$$ (not for the edges, as in a weighted graph), which can be interpreted as a probability that someone clicking links at random will eventually reach the page. This can be used to derive the simple relationship between these probabilities shown below. Here it is also assumed that the person with probability $$1-d$$ stops clicking links and chooses a random page, where $$d$$ is a damping factor.
$PR(p_i) = \frac{1-d}{N} + d \sum_{p_j \in M(p_i)} \frac{PR(p_j)}{L(p_j)}$
$$N$$ is the total number of web pages, $$M(p_i)$$ is the set of pages that link to page $$p_i$$, and $$L(p_j)$$ is the number of outbound links from page $$p_j$$.
17.4.2. Iterative computation#
We can solve the equation for the PageRanks iteratively, by first writing it in matrix-vector form. Define the (column) vector of all PageRanks $$\boldsymbol{R} =\left[ PR(p_1), \ldots, PR(p_N)\right]$$. Then the equation can be written
$\boldsymbol{R} = d M \boldsymbol{R} + \frac{1-d}{N}\boldsymbol{1} = \mathrm{rhs}(\boldsymbol{R}\,)$
where $$\boldsymbol{1}$$ is a vector of all 1’s, and $$M=(K^{-1}A)^T$$ where $$A$$ is the regular adjancency matrix and $$K$$ is a diagonal matrix with the outdegrees on the diagonal. A page with no outbound links (such as page A in the example above) is a special case, which can be handled by adding links to all other pages.
To solve this equation, we use fixpoint iteration:
$\boldsymbol{R}^{\ n+1} = \mathrm{rhs}(\boldsymbol{R}^{\ n})$
until the difference between two iterations is smaller than a tolerance. For initial guess, we can use $$\boldsymbol{R}^{\ 0} = \boldsymbol{1} / N$$.
17.4.3. Example#
As an example, we reproduce the PageRanks for the graph shown above. First we create the adjancency matrix $$A$$, including the added links for pages without outbound links:
using PyPlot, SparseArrays
rows = [2,3,4,4,5,5,5,6,6,7,7,8,8,9,9,10,11]
cols = [3,2,1,2,2,4,6,2,5,2,5,2,5,2,5,5,5]
A = sparse(rows, cols, 1, 11, 11)
A[sum(A,dims=2)[:] .== 0,:] .= 1 # If a vertex has no outbound edges, all edges to all other vertices
spy(A, marker=".");
Next we implement a function for the algorithm. Note that $$K^{-1}A$$ is easiest evaluated by dividing the rows of $$A$$ by the vector of outward links $$L$$.
function pagerank(A, d=0.85, tol=1e-6)
N = size(A,2)
L = sum(A, dims=2)
M = (A ./ L)'
R = ones(N) / N
while true
newR = d * (M * R) .+ (1 - d) / N
if maximum(abs.(newR - R)) < tol
return newR
end
R = newR
end
end
pagerank (generic function with 3 methods)
Finally we can compute the PageRanks of the example network:
R = pagerank(A, 0.85)
round.(100R, digits=1) # Show percentages
11-element Vector{Float64}:
3.3
38.4
34.3
3.9
8.1
3.9
1.6
1.6
1.6
1.6
1.6 | 1,034 | 3,312 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-10 | latest | en | 0.891366 |
null | null | null | null | null | null | Take the 2-minute tour ×
Let $R$ be a Noetherian ring and $I$ an ideal of $R$. If $n$ is the cohomological dimension of $I$, then why is the following isomorphism true: $$H_{I}^{n}(M)\cong H_{I}^{n}(R)\otimes_R M.$$
The cohomological dimension of $I$ is defined to be the supremum of the set of integers $i$ such that $H_{I}^{i}(M)\neq 0$ for some $R$-module $M$.
share|improve this question
Also posted on MSE: math.stackexchange.com/questions/392162/… – user26857 Jul 7 '13 at 8:16
1 Answer 1
Assume $I$ is generated by $a_1,\dots,a_n$ (the cohomological dimension is $\ge$ the number of generators, so let's assume they are equal).
Let $C = C(a_1,\dots,a_n)$ be the Cech complex. Then, since $R$ is noetherian, for any complex $M$, we have that:
$R\Gamma_I M \cong C \otimes_R M$.
Thus, we want to calculate $H^n(C\otimes_R M)$.
As the cohomology of $C$ is supported in $[0,n]$, there is an isomorphism (See for example Lemma 15.3.6 of http://arxiv.org/pdf/1206.6632.pdf)
$H^n(C\otimes_R M) \cong H^n(C\otimes^L_R M) \cong H^n(C) \otimes_R H^0(M)$
And this is simply $H^n_I(R) \otimes_R M$.
share|improve this answer
Your Answer
| null | null | null | null | null | null | null | null | null |
https://www.yxgapp.com/regression-forecasting-and-predicting-practical-machine-learning-tutorial-with-python-p-5/ | 1,669,985,211,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710902.80/warc/CC-MAIN-20221202114800-20221202144800-00311.warc.gz | 1,167,114,431 | 49,446 | • 科普
SCIENCE
英语
ENGLISH
科技
TECHNOLOGY
MOVIE
FOOD
励志
INSPIRATIONS
社会
SOCIETY
TRAVEL
动物
ANIMALS
KIDS
卡通
CARTOON
计算机
COMPUTER
心理
PSYCHOLOGY
教育
EDUCATION
手工
HANDCRAFTS
趣闻
MYSTERIES
CAREER
GEEKS
时尚
FASHION
• 精品课
• 公开课
• 欢迎下载我们在各应用市场备受好评的APP
点击下载Android最新版本
点击下载iOS最新版本
扫码下载译学馆APP
《机器学习Python实践》#5 回归预测
Regression forecasting and predicting - Practical Machine Learning Tutorial with Python p.5
What’s going on everybody!
Welcome to the fifth Machine Learning and fourth regression tutorial.
In this tutorial we’ll be building on the last one
where we created this linear regression algorithm.
We found that it got great accuracy and all that.
And now we’re ready to actually predict
like out into the unknow. All right?
So it turns out that we actually already do have some unknown data.
Simply because we’re forecasting out the shift right which is about 30 days.
So we can actually work with that.
So we’re gonna do is where we define our…
our Xes…
X 的值
Let’s do the following.
Let’s actually cut this…come down here…
Paste.
And we’re gonna take this dropna cut and paste.
And now I am reminded why I was doing that negative forecast_out.
So…er…
So what we’re gonna do is
X equals X to the negative forecast_out.
X = X[-forecast_out]
Let’s see…negative forecast_out…to the point of negative forecast_out.
And then we’re gonna say…
We’re gonna do X_lately
equals X to the minus forecast_out colon.
And then we’re gonna drop
the missing data when we go to create the labels.
In this way we have both our Xes and our X_lately defined.
So the X_lately basically is the stuff we’re gonna actually predict again.
So we have the Xes.
And we just need to figure out
what the m and b is right? For y = mx + b
m 和 b 分别是什么就行了 对吧？就是这个函数 y = mx ＋ b
We get the answer for y.
We’ve done the linear regression.
So…so we’re gonna do against this X_lately
that’s we actually don’t have a y-value for.
Which is why we were not training or testing on that data.
So now we have X_lately.
So the next thing we’re gonna go ahead and do is basically
we’ll come down here and actually it’s going to run this really quick
just make sure we’re still getting
the accuracy we don’t have incorrect number of values.
OK no we don’t. So we’re good.
So we’ve got 96% accuracy.
Awsome~
So we come down. We’ll comment this out.
We don’t really need that anymore.
And now to predict stuff.
What you will do is also make sure we scale…
So
what we need to do is take this
Almost made a mistake there. And this…
Now let’s run that one more time.
I’m wanna make sure I don’t screw thing up.
Good. OK.
So now we’re gonna do is we’ll come down here.
And…
We need to predict based on the X data.
So the way that we can do this once you have a classifier
doing a prediction is super easy. So…
We’re gonna say forecast_set equals clf.predict.
And in here you’re gonna actually pass a single value.
Or you can pass like in
an array of values to predict, make a prediction, per value in that array.
And that’s what we’re gonna do right?
We’ve got this 30 days of database basically right here.
So…er…
X_lately rather so 30 days here. OK.
X_lately 应该就有30天的数据 好的
So last 30 days. So X_lately…
We want to create that with X_lately.
So then we have forecast_set. So now…
We can do…We can print…
er…forecast_set…
forecast_set
forecast_set
forecast_set
confidence and forecast_out
Just so we know how many days were forecasting out here.
Uh-oh…
confidence…do we…er…I’m sorry. So I change the set I usually use confidence…
So accuracy…try again…
Pull this up…
And yeah so there we go…
So we got our predict value.
So these are
the basically the next 30 days of unknown values for us.
That’s like these just straight up the stock prices
which is that pretty cool?
Because we…you know…that whole scaling part
is also playing a major role here and still outputting
You know…stock prices there
of decent value to us.
Anyway, I think it’s cool
So these are the next 30 days’ prices
So then let’s say you want to graph that.
So what we’re gonna do? We’re gonna come to the top
And again we’re gonna just blast through wrapping this
if you’re confused or whatever I have matplotLib tutorials
But otherwise we’re gonna import and in fact…
er…datetime?
datetime
And then we’re gonna
import matplotlib.pyplot as plt
import matplotlib.pyplot as plt
from matplotlib import style
from matplotlib import style
And we’re gonna say style.use(‘ggplot’)
This is just to plot stuff
This is how to make it look decent.
This is how to specify which decent looking thing you want.
So now what we’re gonna say is…we’ll come down…
er…Let’s come down here…And we’re gonna say
df[‘forecast’] equals np.nan
df[‘Forecast’] = np.nan
This is just specifys the entire column.
It’s just full of a lot number of data and you’ll see why at a moment.
But we actually puts some information there shortly.
Now we need to find out what the last day was.
This may not be the best-est way to do something like this
But this is what how we’re gonna actually plot this on the graph.
So we say the last_date equals df.iloc.
Oops [-1]. So this is the very the last day we’ll get the name of that.
And we’re gonna say the last_unix value is equal to last_date.timestamp.
And then one_day.
This is how many seconds in a day.
So you can just do the math there if you want.
But it’s 86400
And then the next_unix
would be like the next day, right?
And these are…we know these are daily prices
So we’re just gonna work kind of hard coding this part of it.
Just so we can create graph.
The last_unix + one_day.
last_unix + one_day
So when you do a prediction
the prediction has no idea like
what date that is…that’s like four…right?
So remember
when you doing machine learning X and y does not correspond to like
necessarily the Xes on the graph.
y 并不一定是 X 的函数值
In this case, it doesn’t. X are the features
y is the label. It just so happens the label is the price
y 是标签 那是因为 y 刚好就是价格这个标签
so y is correct.
But the X is correct? No, because the date is not a feature.
So that’s why we can’t have it work around here
Because we actually don’t have the date values.
I have lost my mouse…there we go
Anyway that’s unix. OK.
So now we get the dates.
And now we actually populate the data frame
with the new dates
and the forecast value. So…
the way we’re gonna do that is we’re gonna say
for i in forecast_set
for i in forecast_set
next_date equals datetme.datetime.fromtimestamp
next_date = datetime.datetime.fromtimestamp
next_unix
next_unix
And now we’re just gonna say
next_unix plus equals that value of one_day
next_unix += one_day 的值
So one_day
And then df.loc
and then next_date…oops…next_date
equals
and then we’re gonna do like one-liner for loop here
So we’re gonna say np.nan for something we don’t care about
in range()
in range()
len(df.columns)
len(df.columns)
er…let’s see…minus 1
And then plus i
So we’ll do is iterating through the forecast_set
taking each forecast and day
and then setting those as the values
in the data frame
basically making the features
the future features, not a number. OK.
And then the last line just takes all of the first columns
sets them to, not to numbers, and in the final columns
whatever i is. Which is the forecast in this case.
So now we’re gonna go ahead and do is
we’re gonna say df
And then we’re gonna say df[‘forecast’]
forecast.plot
df.[‘Forecast’].plot()
And then we’re just gonna do plt.legend
we’ll put that in the fourth location
That’s just like the bottom, right?
And then we’re gonna say plt.xlabel
And we’ll say that’s the date.
plt.ylabel 这就是价格
And finally, plt.show
OK. So we zoom to that. Hopefully that for loop is gonna work out
We’ll find out shortly.
See…and the graph here…OK
So this is our actual graph of the data here
Pull this up…
And as you can see. This is the known data here.
And then over here is our predicted data.
So let me zoom in to that spot
So this is like the future prediction here, the forecast. OK.
So it’s just like a really quick way…
to visualize the data.
And the really the complex part, the reason why we had all this nasty crap in here
We’re just simply so you can have dates on the Xes
Because that’s how I am. I want to have the dates there.
Anyway…Oh yeah…
So that’s how you can actually forecasted out
the data and actually do a prediction.
But the crux of doing prediction
with scikit-learn is right here
And just remember you can pass a single value
Or you can pass an array of values and it will
just output in the same order of the array of values
And then from there we just use logic
to know…because each investment is a day
right? Each price report was one day.
So then that just means
that each forecast was like one day later, right?
So we just kind of use our brians for that one.
So…anyway…
And I guess the other thing to think about df.loc…just in case…
I’m not sure we’re actually cover that in Pandas
But what happens there is basically .loc
is gonna referencing the index
for the data frame.
basically what that saying is that next_date
is a datestamp, right?
And that next_date is the…
index of the data frame. So…
Maybe it’ll help…just…
If you’re not confused in that at this point
Feel free to carry on to the next video. We’ll be talking about pickling.
But if you’re confused about that for loop. I just want to explain
that for loop just so everyone…no one is like:”what the hell?”
So anyway…So yeah…So here…
Right the date is the index. So when we say
df.loo(next_day) we’re saying the index.
df.loo(next_day)我们就是在用索引
And if that index doesn’t exist
It’s gonna created. And if it did exist we’re just gonna replace it.
OK. Then we’re saying np.nan for underscore in range(len(df.columns)-1)
What the heck is that?!
Well, that is just a list of values that are np.nan.
So basically we’re saying it’s np.nan for Adjust, High percent change.
All this stuff is just not a number, right?
Because this is in the future. We don’t have information on that data.
OK. Then…can we back down here?
Then we say + i
Remember i is the forecast, right?
i in forecast_set.
i in forecast_set
So when we’re just saying so basically it’s just the list
plus one value. So it’s just here like the huge list
Well not that huge…It’s just these many columns, right?
Well we just add the forecast at the very end.
So that’s just our super hacky way
of doing the following. I set to head there but probably more useful to set tail.
And so you can see the end of this data frame.
These are all those np.nans and then finally it’s just forecast, OK?
So that’s all that is.
Sorry I was just little confusing. Hopefully the explaination worked. If not
feel free to ask question wherever I’ll be happay to clarify.
Now…there…
There is actually one more thing I want to show you all before
we dive into the regression then actually write a regression algorithm all on our own
that’s pickling. The reason why you want to pickle is imagine
you have rather than training a classifier on this…you know
relatively small data set. We just had daily values for the last few years
But you know if you save this to a file. You know…it’s probably like
you know…500 killerbytes or something…who knows
But let’s say you have like intraday data. You’ve got like two gigabytes worth of data.
That’s gonna take a while to train the classifier on that data.
So won’t it be nice if eveytime you want to make a prediction
So just consider
making a prediction in using future data
Consider everytime you want to make a prediction you have to train the classifier
Is that not just crazy sounding?
So yes that’s crazy sounding.
So the next tutorial we’re gonna talking about pickling
which will let you save your classifier and then just quickly load it in
without any training time.
So definitely very useful with machine learning classifiers.
So anyway that’s what we’ll talk about the next video.
Questions comments leave them below. Otherwise as always thanks for watching
thanks for support and subscription and until next time.
[B]刀子
[B]刀子 | 3,092 | 12,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2022-49 | latest | en | 0.759937 |
wsgfl2.westsussex.gov.uk | 1,398,371,523,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223206672.15/warc/CC-MAIN-20140423032006-00596-ip-10-147-4-33.ec2.internal.warc.gz | 399,413,504 | 3,198 | # Mathematics
## Multicultural Mathematical Activities - Rangoli Patterns
This activity will help students
• Understand reflection in two dimensions.
• Construct the reflections of shapes placed at different angles.
Software Required: An Interactive Geometry package. (Instructions are given for Geometer’s Sketchpad – however they should apply to other packages with only slight modifications. If another program is used, a new template will need to be made. The instructions for this are given in the student help sheet.)
Organisation: Students can work in pairs on this activity.
Overview
Students construct Rangoli patterns. These traditional Indian designs make use of reflection in perpendicular and oblique (45° ) mirror lines. The completed designs can be printed and coloured in – they make attractive designs for greetings cards!
Differentiation
• All students draw lines, using the geometry program’s line tools. They use a series of reflections to produce a Rangoli design. The easiest examples are included in exercise 1.
• In the core activity (exercise 2), students attempt to reproduce given designs.
• The extension task involves ‘starting from scratch’, with students setting up their own patterns and experimenting with designs of different sizes.
Demonstration
First, demonstrate to the class how to produce a Rangoli pattern starting from the ready-made template. Explain how the pattern is formed.
• The basic pattern consists of 5 line segments. These segments are shown in the grey square in the corner of the screen.
• These line segments are reproduced (translated) to form the basis of the main pattern, where they are reflected in a diagonal mirror line to create the top left ‘quarter’ of the design.
• The top left quarter of the design is then reflected vertically and horizontally to complete the pattern.
To change the pattern, drag the line segments (or their end-points) in the grey square. Particularly pleasing patterns are formed when some of the end-points lie on the boundary of the grey square, and end-points within the square coincide; this will cause the completed pattern to ‘join up’.
Once students have had the opportunity to produce some Rangoli patterns of their own, they can start work on the worksheets.
Notes
• The diagram is set up so that points snap to the grid (Graph menu), with measurement units being centimetres (Preferences item from the Display menu.)
• The idea of displaying the ‘unreflected’ lines separately from the main pattern is that this makes it possible to see where these lines actually are; if the construction were set up so that the lines drawn formed the pattern directly, it would be almost impossible to distinguish between the original lines and their various reflections.
• Hiding unnecessary lines (like the mirror lines) and labels (just about everything except the labels on the ‘original’ line segments) will make the final construction much easier to understand.
Teaching Points
It is important to stress some key features of ‘mathematical mirrors’:
• They are ‘two-way’; objects on either side of the mirror line are reflected to the opposite side.
• Images of objects are produced the same distance from the mirror line as the original image, and the direction of reflection is perpendicular to the mirror line.
• Rangoli patterns do not have to be based on 5 line segments; however, 5 segments will produce some very aesthetically pleasing patterns, especially if care is taken to ensure that they ‘join up’ appropriately.
• Most students will enjoy making their own Rangoli patterns and colouring them in to form a symmetrical design. The core task introduces a further challenge, in the form of a game for two players. It is also possible to get students to construct their own version of the Rangoli patterns diagram, as indicated in the extension task. The necessary steps are set out in the student help sheet. Note, however, that this is a fairly demanding construction which is probably best suited to students who are quite confident with the software. | 800 | 4,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2014-15 | latest | en | 0.88984 |
https://windrawwin-prediction.com/betting/your-question-what-are-the-odds-of-rolling-three-of-a-kind-with-three-dice.html | 1,652,746,462,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662515466.5/warc/CC-MAIN-20220516235937-20220517025937-00044.warc.gz | 710,579,756 | 17,736 | # Your question: What are the odds of rolling three of a kind with three dice?
Contents
## What are the odds of rolling 3 sixes with 3 dice?
This seemed simple at first as the odds of rolling 1 six are 1/2, 2 sixes would be 1/14 and 3 sixes would be 1/216 so they would be our odds.
## What are the odds of rolling a pair with 3 dice?
Probability of all three dice the same = 1×1/6×1/6=1/36. Probability of no dice the same = 1×5/6×4/6=20/36. Probability of a pair = 1−1/36−20/36=15/36.
## What is the probability of 3 dice?
Two (6-sided) dice roll probability table
Roll a… Probability
3 3/36 (8.333%)
4 6/36 (16.667%)
5 10/36 (27.778%)
6 15/36 (41.667%)
## How many possibilities are there with 4 dice?
My initial reaction is to say that the answer is 64, since 4 dice can have 6 outcomes.
## What is the probability that all three dice show different numbers?
Summary : If three dice are tossed, then the probability that the numbers shown will all be different is 5/9.
## When three dice are rolled determine the number of different outcomes?
Total Number of possibilities=1×5×5×3C1=75. If there is only two 5. Total Number of possibilities=1×1×5×3C2=15. If there is only three 5.
## What is the probability of rolling a dice 3 times and getting a different number each time?
Thus, the actual probability of getting three different numbers is 56⋅23=59.
IT IS SURPRISING: How can I win America Visa Lottery? | 418 | 1,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2022-21 | latest | en | 0.907326 |
http://csegeek.com/csegeek/view/tutorials/algorithms/arrays_strings/arr_str_part3.php | 1,563,345,239,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525094.53/warc/CC-MAIN-20190717061451-20190717083451-00349.warc.gz | 35,887,188 | 6,866 | #### Arrays & Strings
Problem
Sort an array of 0s, 1s & 2s, i.e element at each index of the array is either 0, 1 or 2 and we need to place all 0s first, then 1s and finally all 2s.
Solution
We won’t consider the counting approach rather would go for a 1 pass approach in which the sorting would be done by traversing the array only once.
We need to divide the array into 3 parts.
Say ‘i’ points to index 0 & ‘j’ points to index n-1.
We will start traversing from the end of array & ‘k’ is the iterator starting at index n-1.
If arr(k) is 0, then swap arr(i) & arr(k) and increment ‘i’.
If arr(k) is 2, then swap arr(j) & arr(k) and decrement ‘j’.
If arr(k) is 1, then don’t do anything, simple decrement the iterator ‘k’.
In this way, the array will be divided into 3 portions.
```#include<iostream>
using namespace std;
// swap elements of two indexes in the array
void swap (int arr[], int pos1, int pos2) {
int temp = arr[pos1];
arr[pos1] = arr[pos2];
arr[pos2] = temp;
}
// sort the array of 0,1,2 in 1 pass with no extra space
void sort012(int arr[], int size) {
int start = 0;
int end = size-1;
int j = end;
// start traversing from end
// maintain two pointers
// start -> '0' boundary from left
// end -> '2' boundary from right
// j -> iterator
while (j >= start) {
if (arr[j] == 0) {
// move the 0 to the portion of array
// where all 0s present and increase the
// 0 boundary towards right
swap(arr,start,j);
start++;
}
else if (arr[j] == 2) {
// move the 2 to the portion of array
// where all 2s present and increase the
// '2' boundary towards left
swap(arr,j,end);
end--;
}
else {
// don't do anything if '1' is found
j--;
}
}
}
// display the array
void displayArray(int arr[],int size) {
int i;
for(i=0;i<size;i++)
cout<<arr[i]<<" ";
cout<<endl;
}
// main
int main() {
int arr[] = {1,2,0,1,0,1,2,0,2,0,0,2,1,0,1,0};
int size = sizeof(arr)/sizeof(arr[0]);
cout<<"\nOriginal Array :-";
displayArray(arr,size);
sort012(arr,size);
cout<<"\nSorted Array :-";
displayArray(arr,size);
cout<<endl;
return 0;
}``` | 631 | 2,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-30 | latest | en | 0.627298 |
http://zh.wikipedia.org/wiki/%E5%8F%8D%E5%8F%8C%E6%9B%B2%E5%87%BD%E6%95%B0 | 1,438,153,737,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042986148.56/warc/CC-MAIN-20150728002306-00148-ip-10-236-191-2.ec2.internal.warc.gz | 867,134,000 | 14,092 | # 反双曲函数
## 反双曲函数的导数
\begin{align} \frac{d}{dx} \operatorname{arsinh}\, x & {}= \frac{1}{\sqrt{1+x^2}}\\ \frac{d}{dx} \operatorname{arcosh}\, x & {}= \frac{1}{\sqrt{x^2-1}}, \qquad x>1\\ \frac{d}{dx} \operatorname{artanh}\, x & {}= \frac{1}{1-x^2}, \qquad |x| <1\\ \frac{d}{dx} \operatorname{arcoth}\, x & {}= \frac{1}{1-x^2}, \qquad |x| >1\\ \frac{d}{dx} \operatorname{arsech}\, x & {}= \frac{-1}{x\sqrt{1-x^2}}, \qquad x \in (0,1)\\ \frac{d}{dx} \operatorname{arcsch}\, x & {}= \frac{-1}{|x|\sqrt{1+x^2}}, \qquad x \text{ ≠ }0\\ \end{align}
$\frac{d\,\operatorname{arsinh}\, x}{dx} = \frac{d \theta}{d \sinh \theta} = \frac{1} {\cosh \theta} = \frac{1} {\sqrt{1+\sinh^2 \theta}} = \frac{1}{\sqrt{1+x^2}}$
## 幂级数展开式
$\operatorname{arsinh}\, x$
$= x - \left( \frac {1} {2} \right) \frac {x^3} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {x^5} {5} - \left( \frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} \right) \frac {x^7} {7} +\cdots$
$= \sum_{n=0}^\infty \left( \frac {(-1)^n(2n)!} {2^{2n}(n!)^2} \right) \frac {x^{2n+1}} {(2n+1)} , \qquad \left| x \right| < 1$
$\operatorname{arcosh}\, x$
$= \ln 2x - \left( \left( \frac {1} {2} \right) \frac {x^{-2}} {2} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {x^{-4}} {4} + \left( \frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} \right) \frac {x^{-6}} {6} +\cdots \right)$
$= \ln 2x - \sum_{n=1}^\infty \left( \frac {(-1)^n(2n)!} {2^{2n}(n!)^2} \right) \frac {x^{-2n}} {(2n)} , \qquad x > 1$
$\operatorname{artanh}\, x = x + \frac {x^3} {3} + \frac {x^5} {5} + \frac {x^7} {7} +\cdots = \sum_{n=0}^\infty \frac {x^{2n+1}} {(2n+1)} , \qquad \left| x \right| < 1$
$\operatorname{arcsch}\, x = \operatorname{arsinh}\, x^{-1}$
$= x^{-1} - \left( \frac {1} {2} \right) \frac {x^{-3}} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {x^{-5}} {5} - \left( \frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} \right) \frac {x^{-7}} {7} +\cdots$
$= \sum_{n=0}^\infty \left( \frac {(-1)^n(2n)!} {2^{2n}(n!)^2} \right) \frac {x^{-(2n+1)}} {(2n+1)} , \qquad \left| x \right| < 1$
$\operatorname{arsech}\, x = \operatorname{arcosh}\, x^{-1}$
$= \ln \frac{2}{x} - \left( \left( \frac {1} {2} \right) \frac {x^{2}} {2} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {x^{4}} {4} + \left( \frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} \right) \frac {x^{6}} {6} +\cdots \right)$
$= \ln \frac{2}{x} - \sum_{n=1}^\infty \left( \frac {(-1)^n(2n)!} {2^{2n}(n!)^2} \right) \frac {x^{2n}} {2n} , \qquad 0 < x \le 1$
$\operatorname{arcoth}\, x = \operatorname{artanh}\, x^{-1}$
$= x^{-1} + \frac {x^{-3}} {3} + \frac {x^{-5}} {5} + \frac {x^{-7}} {7} +\cdots$
$= \sum_{n=0}^\infty \frac {x^{-(2n+1)}} {(2n+1)} , \qquad \left| x \right| > 1$
$\operatorname{arcosh}(2x^2-1) = 2\operatorname{arcosh} x$
$\operatorname{arcosh}(2x^2+1) = 2\operatorname{arsinh} x$
## 反双曲函数的不定积分
\begin{align} \int \operatorname{arsinh}\,x\,dx &{}= x\,\operatorname{arsinh}\,x - \sqrt{x^2+1} + C\\ \int \operatorname{arcosh}\,x\,dx &{}= x\,\operatorname{arcosh}\,x - \sqrt{x^2-1} + C,\qquad x >1\\ \int \operatorname{artanh}\,x\,dx &{}= x\,\operatorname{artanh}\,x + \frac{1}{2}\ln\left|1-x^2\right| + C,\qquad |x| <1\\ \int \operatorname{arcoth}\,x\,dx &{}= x\,\operatorname{arcoth}\,x + \frac{1}{2}\ln\left|1-x^2\right| + C,\qquad |x| >1\\ \int \operatorname{arsech}\,x\,dx &{}= x\,\operatorname{arsech}\,x - \arcsin\,x + C,x \in (0,1)\\ \int \operatorname{arcsch}\,x\,dx &{}= x\,\operatorname{arcsch}\,x + \left(\sgn\,x\right) \operatorname{arsinh}\,x + C \end{align} | 1,745 | 3,500 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 49, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2015-32 | latest | en | 0.155107 |
http://mathhelpforum.com/trigonometry/183859-calculating-value-cotangent.html | 1,526,847,392,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863684.0/warc/CC-MAIN-20180520190018-20180520210018-00280.warc.gz | 197,223,340 | 10,557 | # Thread: Calculating the value for cotangent.
1. ## Calculating the value for cotangent.
My question says to "calculate the value for cotangent (0) if (0)=270 degrees, given in standard position"
Just calculating it and getting 1.8370 I got it counted wrong, and when I asked on another forum about it they had gone through and what seems like to me, they used quotient identities and put the cos(270)/(sin(270) to equal 0.
Is that the right way to go about this? It didn't say to use quotient identities so I wasn't sure.
There's a second question with the same degree but its asking for the csc (cosecant) of 270, given in standard position.
2. ## Re: Calculating the value for cotangent.
Originally Posted by sterces
My question says to "calculate the value for cotangent (0) if (0)=270 degrees, given in standard position"
Just calculating it and getting 1.8370 I got it counted wrong, and when I asked on another forum about it they had gone through and what seems like to me, they used quotient identities and put the cos(270)/(sin(270) to equal 0.
Is that the right way to go about this? It didn't say to use quotient identities so I wasn't sure.
There's a second question with the same degree but its asking for the csc (cosecant) of 270, given in standard position.
...some smart people you found out there...
It's true.
$\displaystyle \theta=270^{\circ}$
$\displaystyle \cot(\theta)=\frac{\cos(\theta)}{\sin(\theta)}=\fr ac{\cos(270^{\circ})}{\sin(270^{\circ})}=\frac{0}{-1}=0$
EDIT:
$\displaystyle \sec(\theta)=\frac{1}{\cos(\theta)}$
So, what is $\displaystyle \sec(270^{\circ})$ ?
3. ## Re: Calculating the value for cotangent.
Ah, alright, thanks for verifying
sec(270) would be -1 then.
4. ## Re: Calculating the value for cotangent.
Originally Posted by sterces
Ah, alright, thanks for verifying
sec(270) would be -1 then.
Yes!(But only by my definition of sec function)
5. ## Re: Calculating the value for cotangent.
Originally Posted by Also sprach Zarathustra
$\displaystyle \sec(\theta)=\frac{1}{\sin(\theta)}$
$\displaystyle \sec(x) = \dfrac{1}{\cos(x)}$ rather than 1/sin(x).
6. ## Re: Calculating the value for cotangent.
Originally Posted by sterces
Ah, alright, thanks for verifying
sec(270) would be -1 then.
No!
sec(270)=1/cos(270)=1/0
7. ## Re: Calculating the value for cotangent.
Originally Posted by Also sprach Zarathustra
No!
sec(270)=1/cos(270)=1/0
But the question doesn't ask for $\displaystyle \sec(270^\circ)$, it asks for $\displaystyle \csc(270^\circ) = \frac1{\sin(270^\circ)} = \frac1{-1} = -1$. | 724 | 2,565 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-22 | latest | en | 0.90062 |
null | null | null | null | null | null | A Hamburger Today
Snapshots from Asia: Jellied Bean Curd
Photograph by Shimin Wong
Smooth as a baby's bottom. That's how silken we want our tau foo fa (bean curd blossoms, also known as "jellied bean curd"). So incredibly delicate that a quivering spoonful will slide deliciously down your throat with the barest tilt of your head—no teeth required.
Made of soy milk that's been coagulated with a tiny amount of gypsum powder (calcium sulphate) and cornstarch to form curds, it is, at its most basic, served in a pool of simple syrup. Dressed up for company, it's been known to sport a crown of gingko nuts and lotus seeds, occasionally bobbing in a sweet, gingery broth. And yes, it's essentially sweetened tofu for dessert—and breakfast, and supper, and any time in between really (midday merienda, anyone?).
I remember a friend's bemused expression when roaming an Asian grocery store a while back. He gaped at the candied lotus roots, winter melon strips, and water chestnuts, before asking, "This is candy?"
And then, when he spotted the glazed crabs and cuttlefish strips, he asked, "Do you just add tons of sugar to everything?"
It was hard to keep from laughing, and even harder to keep a straight face in (truthfully) responding: "Well, except for the crabs and cuttlefish, which, yes, we do pop like candy, the rest are usually served in soup—for dessert."
Which brings us back to the Chinese belief that food is medicine—including dessert. Remember the heating and cooling nature of foods? Sweets are said to have a cold and damp nature, which may weaken the spleen if taken in excess. In fact, a craving for sweets is supposedly a tell-tale sign that you aren't regularly "strengthening your spleen" with the natural sweetness of grains as the basis of your diet.
I confess to being fairly addicted to the creamy, pleasantly nutty lightness of tau foo fa, and the way it soothes your throat going down. So it may be that I ought to have my spleen looked at. It probably won't stop me stalking the old Chinese man who hawks it from a giant, ambulatory, metal vat-on-wheels in New York City's Chinatown, though. And I heartily urge you to do the same if you're in the neighborhood.
About the author: Wan Yan Ling, Serious Eats's overseas summer intern, is an impoverished grad student and sourdough finger-crosser living in Singapore. She can usually be found in the kitchen procrastinating on "real work," or online tracking down obscure recipes. Ling thinks eating alone is no fun, and she still believes in hand-mixing.
Printed from http://www.seriouseats.com/2007/08/snapshots-from-asia-jellied-be.html
© Serious Eats | null | null | null | null | null | null | null | null | null |
https://us.metamath.org/mpeuni/elrestr.html | 1,680,286,794,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949678.39/warc/CC-MAIN-20230331175950-20230331205950-00439.warc.gz | 675,979,270 | 5,720 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > elrestr Structured version Visualization version GIF version
Theorem elrestr 16449
Description: Sufficient condition for being an open set in a subspace. (Contributed by Jeff Hankins, 11-Jul-2009.) (Revised by Mario Carneiro, 15-Dec-2013.)
Assertion
Ref Expression
elrestr ((𝐽𝑉𝑆𝑊𝐴𝐽) → (𝐴𝑆) ∈ (𝐽t 𝑆))
Proof of Theorem elrestr
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 eqid 2825 . . . 4 (𝐴𝑆) = (𝐴𝑆)
2 ineq1 4036 . . . . 5 (𝑥 = 𝐴 → (𝑥𝑆) = (𝐴𝑆))
32rspceeqv 3544 . . . 4 ((𝐴𝐽 ∧ (𝐴𝑆) = (𝐴𝑆)) → ∃𝑥𝐽 (𝐴𝑆) = (𝑥𝑆))
41, 3mpan2 682 . . 3 (𝐴𝐽 → ∃𝑥𝐽 (𝐴𝑆) = (𝑥𝑆))
5 elrest 16448 . . 3 ((𝐽𝑉𝑆𝑊) → ((𝐴𝑆) ∈ (𝐽t 𝑆) ↔ ∃𝑥𝐽 (𝐴𝑆) = (𝑥𝑆)))
64, 5syl5ibr 238 . 2 ((𝐽𝑉𝑆𝑊) → (𝐴𝐽 → (𝐴𝑆) ∈ (𝐽t 𝑆)))
763impia 1149 1 ((𝐽𝑉𝑆𝑊𝐴𝐽) → (𝐴𝑆) ∈ (𝐽t 𝑆))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 386 ∧ w3a 1111 = wceq 1656 ∈ wcel 2164 ∃wrex 3118 ∩ cin 3797 (class class class)co 6910 ↾t crest 16441 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1894 ax-4 1908 ax-5 2009 ax-6 2075 ax-7 2112 ax-8 2166 ax-9 2173 ax-10 2192 ax-11 2207 ax-12 2220 ax-13 2389 ax-ext 2803 ax-rep 4996 ax-sep 5007 ax-nul 5015 ax-pr 5129 ax-un 7214 This theorem depends on definitions: df-bi 199 df-an 387 df-or 879 df-3an 1113 df-tru 1660 df-ex 1879 df-nf 1883 df-sb 2068 df-mo 2605 df-eu 2640 df-clab 2812 df-cleq 2818 df-clel 2821 df-nfc 2958 df-ne 3000 df-ral 3122 df-rex 3123 df-reu 3124 df-rab 3126 df-v 3416 df-sbc 3663 df-csb 3758 df-dif 3801 df-un 3803 df-in 3805 df-ss 3812 df-nul 4147 df-if 4309 df-sn 4400 df-pr 4402 df-op 4406 df-uni 4661 df-iun 4744 df-br 4876 df-opab 4938 df-mpt 4955 df-id 5252 df-xp 5352 df-rel 5353 df-cnv 5354 df-co 5355 df-dm 5356 df-rn 5357 df-res 5358 df-ima 5359 df-iota 6090 df-fun 6129 df-fn 6130 df-f 6131 df-f1 6132 df-fo 6133 df-f1o 6134 df-fv 6135 df-ov 6913 df-oprab 6914 df-mpt2 6915 df-rest 16443 This theorem is referenced by: firest 16453 restbas 21340 tgrest 21341 resttopon 21343 restcld 21354 restfpw 21361 neitr 21362 restntr 21364 ordtrest 21384 cnrest 21467 lmss 21480 connsubclo 21605 restnlly 21663 islly2 21665 cldllycmp 21676 lly1stc 21677 kgenss 21724 xkococnlem 21840 xkoinjcn 21868 qtoprest 21898 trfbas2 22024 trfil1 22067 trfil2 22068 fgtr 22071 trfg 22072 uzrest 22078 trufil 22091 flimrest 22164 cnextcn 22248 trust 22410 restutop 22418 trcfilu 22475 cfiluweak 22476 xrsmopn 22992 zdis 22996 xrge0tsms 23014 cnheibor 23131 cfilres 23471 lhop2 24184 psercn 24586 xrlimcnp 25115 xrge0tsmsd 30326 ordtrestNEW 30508 pnfneige0 30538 lmxrge0 30539 rrhre 30606 cvmscld 31797 cvmopnlem 31802 cvmliftmolem1 31805 poimirlem30 33978 subspopn 34085 iocopn 40536 icoopn 40541 limcresiooub 40663 limcresioolb 40664 fourierdlem32 41144 fourierdlem33 41145 fourierdlem48 41159 fourierdlem49 41160
Copyright terms: Public domain W3C validator | 1,799 | 3,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-14 | longest | en | 0.121765 |
http://www.enotes.com/homework-help/2y-3-3xy-3-3x-2-y-4-2x-3-z-2-3-x-3-y-4-z-2-x-4-z-220029 | 1,477,210,438,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719192.24/warc/CC-MAIN-20161020183839-00305-ip-10-171-6-4.ec2.internal.warc.gz | 437,443,591 | 9,993 | # Rewrite using only positive exponent: (2y^3*3xy^3)/ (3x^2*y^4) (2x^3 *z^2)^3 / (x^3*y^4 *z^2 * x^-4*z^3)
### 2 Answers |Add Yours
hala718 | High School Teacher | (Level 1) Educator Emeritus
Posted on
Let E = (2y^3*3xy^3)/ (3x^2*y^4) (2x^3 *z^2)^3 / (x^3*y^4 *z^2 * x^-4*z^3)
==> E = (x^3*z^2*x^-4 *z^3)(2y^3*3xy^3)/(3x^2*y^4(2x^3*z^2)^3
==> E = (x^3*z^2*x^-4*z^3)(2y^3*3xy^3)/(3x^2*y^4)(8x^9*z^6)
We will use the exponent properties to simplify:
we know that: a^b/ a^c = a^(b-c)
also, we know that: a^b * a^c = a^(b+c)
==> E = 6x^0 * y^6*z^5)/(24x^11*y^4*z^5)
= y^(6-2) *z^5-5)/ 4x^11
= y^4/ 4x^11
kjcdb8er | Teacher | (Level 1) Associate Educator
Posted on
Rewrite:
(2y^3*3xy^3)/ (3x^2*y^4) (2x^3 *z^2)^3 / (x^3*y^4 *z^2 * x^-4*z^3)
= (2y^3*3xy^3)/ (3x^2*y^4) (2x^3 *z^2)^3 * 1/(x^3*y^4 *z^2 * x^-4*z^3)
= 16 x^8 y^2 z^6 * 1/(x^3*y^4 *z^2 * x^-4*z^3)
=16 x^8 y^2 z^6 *x/(y^4 z^5)
= (16 x^9 z)/y^2
We’ve answered 317,506 questions. We can answer yours, too. | 548 | 980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2016-44 | latest | en | 0.506768 |
https://math.stackexchange.com/questions/890242/how-do-you-find-the-modular-inverse-of-5-pmod-11?noredirect=1 | 1,576,330,346,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541157498.50/warc/CC-MAIN-20191214122253-20191214150253-00145.warc.gz | 446,383,883 | 33,400 | # How do you find the modular inverse of $5\pmod{\!11}$
I need to find out the modular inverse of 5(mod 11), I know the answer is 9 and got the following so far and don't understand how to than get the answer. I know how to get the answer for a larger one such as 27(mod 392) but am stuck because they are both low numbers.
11=5 (2)+1
5=1 (5)
In finding a modular inverse, you are trying to solve the modular equation $$ax\equiv 1\pmod n.$$ Ordinarily, you use the Extended Euclidean Algorithm for this to solve the equation $ax+ny=1$. If the numbers $a$, and $n$ are small, then simple trial and error is probably just as fast or faster.
For your example, we have $a=5$ and $n=11$, which means would could just use trial and error. \begin{align} 1\cdot 5 &\equiv 5\\ 2\cdot 5 &\equiv 10\\ 3\cdot 5 &\equiv 15\equiv 4 \\ 4\cdot 5 &\equiv 20\equiv -2\\ 5\cdot 5 &\equiv 25\equiv 3 \\ 6\cdot 5 &\equiv 30\equiv -3 \\ 7\cdot 5 &\equiv 35\equiv 2 \\ 8\cdot 5 &\equiv 40\equiv -4 \\ 9\cdot 5 &\equiv 45\equiv 1 \\ 10\cdot 5 &\equiv 50\equiv -5 \\ \end{align}
Since $9\cdot 5 \equiv 1$ then we have found the modular inverse to be 9.
When looking at those numbers on the far right side, keep in mind that any multiple of 11 made be added or subtracted to the modulus and it is still equivalent. That is, $-3\equiv 30\equiv 8$ since $-3+3(11)=30$ and $8+2(11)=30$.
• ok I haven't seen this way of doing it before so could you help clarify something for me: on the left you have (x) x a than the sum of it in the middle, and on the right you have how much its over a multiple of 11 ( a positive number) or how much it is under a multiple of 11 ( a negative number) – user2956865 Aug 7 '14 at 17:43
• Read the line from left-to-right. So we could have (for the line $4\cdot 5\equiv 20\equiv -2$), 4 times 5 equals 20. But we want the remainder after this is divided by 11. Since $20 = 11+9$, this means $20\equiv 9$ (the part that isn't a multiple of 11). But I like smaller numbers, so I'm going to subtract 11 from this to get $9-11=-2$ which means $20\equiv -2$. Since it's not 1, then this means that 4 is not the modular inverse. – Lee Aug 7 '14 at 17:51
• ok thanks lee that helped to clear it up – user2956865 Aug 7 '14 at 17:57
Hint $$\ \ \ \ \ \,\bmod n = ab\!+\!1\!:\,\, ab \equiv -1 \,\Rightarrow\, a(-b) \equiv 1.\,\$$ OP is $$\,a,b = 5,2.$$
So the inverse is computable by a $$\rm\color{#c00}{single\ division}$$ $$\,a^{-1}\equiv -b\equiv (n\!-\!1)/a\,$$ if $$\,n\equiv 1\pmod{\!a},\,$$ and similarly $$\bmod n = ab\!-\!1\!:\,\ ab\,\equiv\, 1\ \Rightarrow\$$ $$\,a^{-1}\equiv b\equiv (n\!+\!1)/a\,\pmod{\!n}$$
Generally we can use the Extended Euclidean Algorithm or Gauss's algorithm to compute modular inverses and fractions, e.g. see here and here. The above easy inverse formula is an optimization of the case when the algorithms terminate in a $$\rm\color{#c00}{single\ step}$$). The analogous double step optimization is inverse reciprocity. Both prove convenient in manual calculations.
You can use your work to get $1=11-5(2)=11+5(-2)$,
so an inverse of 5 (mod 11) is given by $-2\equiv9\pmod {11}$.
• Ok I've been trying to work it out differently trying to use the this way I saw in youtube video, so how do you go from 1=11-5(2)=11+5(-2) to -2= 9 (mod 11) – user2956865 Aug 7 '14 at 17:31
• The equation $1=11+5(-2)$ shows that -2 is an inverse of 5 (mod 11), and then you can add 11 (or any multiple of 11) to get other inverses of 5 (mod 11). – user84413 Aug 7 '14 at 17:48
Here is a piece of C code that you might find useful:
int Inverse(int n,int a)
{
int x1 = 1;
int x2 = 0;
int y1 = 0;
int y2 = 1;
int r1 = n;
int r2 = a;
while (r2 != 0)
{
int r3 = r1%r2;
int q3 = r1/r2;
int x3 = x1-q3*x2;
int y3 = y1-q3*y2;
x1 = x2;
x2 = x3;
y1 = y2;
y2 = y3;
r1 = r2;
r2 = r3;
}
return y1>0? y1:y1+n;
}
Calling Inverse(11,5) returns 9.
Calling Inverse(392,27) returns 363.
• thanks that could come in handy in future, I just needed to know how to do it by hand so I'm able to do it in a exam – user2956865 Aug 7 '14 at 17:55
• @user2956865: You're welcome. I think that the algorithm above is the same thing that you'll be doing manually. – barak manos Aug 7 '14 at 17:56 | 1,469 | 4,202 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2019-51 | latest | en | 0.883705 |
http://leetcode.com/2010/05/problem-snapper-chain-gcj-qualification.html | 1,368,920,304,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382989/warc/CC-MAIN-20130516092622-00063-ip-10-60-113-184.ec2.internal.warc.gz | 154,637,856 | 10,921 | ## Problem A: Snapper Chain Solution (Google Code Jam Qualification Round 2010)
May 8, 2010
Read the question here from GCJ Qualification Round 2010:
» Problem A: Snapper Chain
Once you understand how the snapper works, it is easy. This problem can be solved in various ways, but the main observations are:
Let k be the number of times you snap your finger, n be the light’s position, and assume 0 = OFF and 1 = ON.
The configuration of the first five snappers (with the first snapper on the far left side) as k increases are:
00000 k = 0
10000 k = 1
01000 k = 2
11000 k = 3
00100 k = 4
10100 k = 5
01100 k = 6
11100 k = 7
00010 k = 8
See the pattern? The configuration of the snapper for any k is the binary representation of k itself!
For example, when k=3 and n=3, we know that the light is OFF because the snapper is in the OFF position (because the 3rd bit of k=3 is 0). When k=3 and n=2, the light is ON. However, when k=5 and n=3, the light is OFF even though the 3rd bit is 1. As the 2nd bit is 0, the electric couldn’t “flow” to the light bulb.
Therefore, in order for a bulb to light, it requires all of the bits from 1 to n to be all 1s.
The problem can still be solved using arrays representing the bits and iterate through them to check, but it is more efficient using bit manipulation. If you are familiar with bit manipulation, you can check if the bits from 1 to n are all 1s using the XOR operation and some bit shifting. My solution is shown below. This is just one sample solution. If you have a more elegant solution, you are welcome to add to that in the comments section.
VN:F [1.9.22_1171]
Rating: 0.0/5 (0 votes cast)
### 12 responses to Problem A: Snapper Chain Solution (Google Code Jam Qualification Round 2010)
1. This site is fantastic. Thank you very much.
Here's an approach that may be a bit simpler, but I might have overlooked something because I'm not using XOR at all.
if( ( ((1<<n)-1) & k ) == ((1<<n)-1) ){
}
VA:F [1.9.22_1171]
0
2. You can solve this in O(1):
boolean isPowered(int n, int k) {
return k % (1<<n) == (1<<n) – 1;
}
Slightly simpler.
VA:F [1.9.22_1171]
0
3. @1337:
why this can’t work?
k & ((1<<(n+1))-1) == ((1<<(n+1))-1)
as all it needs is to find if k==pow(2,n)-1
VA:F [1.9.22_1171]
0
4. ((k&((1<<n)-1)) == ((1<<n)-1))
VA:F [1.9.22_1171]
0
5. k & ( (1 << (n-1) ) – 1 ) == ( (1 << (n-1) ) – 1)
VA:F [1.9.22_1171]
0
6. Will the following also be a solution ?
the above function will return the no of snaps required for a particular value of n. If it is the same as k then Snapper Chain is ON else OFF.
VA:F [1.9.22_1171]
0
7. via GCJ handle :neal.wu
return (k+1)%(1<<n) == 0 ? ON : OFF
VN:F [1.9.22_1171]
0
8. int steps=(int)Math.pow(2, K)-1;
VA:F [1.9.22_1171]
0
9. can pleae anyone explain the logic of (((1 << (n-1)) ^ (k & ((1 << n)-1))) == ((1 << (n-1))-1)) in the above solution
VA:F [1.9.22_1171]
0
• The logic seems to be wrong. the right side should be just 1<<(n-1)
VA:F [1.9.22_1171]
0
10. int S(int n, int k)
{
return (k >> (n – 1)) & 1;
}
VN:F [1.9.22_1171]
0 | 1,034 | 3,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2013-20 | longest | en | 0.914192 |
https://plainmath.org/algebra-i/103000-a-boat-goes-downstream-and-cov | 1,701,899,853,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100603.33/warc/CC-MAIN-20231206194439-20231206224439-00130.warc.gz | 540,482,484 | 20,909 | Bobby Gross
2023-02-26
A boat goes downstream and covers the distance between two ports in 96 minutes, while it covers the same distance in upstream in 120 minutes. If the speed of stream is 1km/hr, Find the speed of boat in still water?
Malia Klein
Let speed of boat be X kmph.
down stream- in 96 minute boat will travel $96/60\left(X+1\right)$ km
As per question $\frac{96}{60}\left(X+1\right)=\frac{120}{60}\left(X-1\right)$
$⇒96x+96=120x-120\phantom{\rule{0ex}{0ex}}⇒24x=216\phantom{\rule{0ex}{0ex}}x=9$
Do you have a similar question? | 183 | 544 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 20, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-50 | latest | en | 0.845114 |
https://www.doubtnut.com/question-answer/if-is-proposed-to-build-a-single-circular-park-equal-in-area-to-the-sum-of-areas-of-two-circular-par-642508310 | 1,642,921,414,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304134.13/warc/CC-MAIN-20220123045449-20220123075449-00037.warc.gz | 789,155,297 | 81,334 | If is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 25-2-2021
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Get Answer to any question, just click a photo and upload the photo
and get the answer completely free,
Watch 1000+ concepts & tricky questions explained!
Text Solution
10 m 15 m20 m24 m
Area of first circular park, whose diameter is 16m <br> = pir^(2) = pi((16)/(2))^(2)= 64 pim^(2) [because r=(d)/(2)=(16)/(2)=8m] <br> Area of second circular park, whose diameter is 12m <br> pi((16)/(2))^(2)= 64 pim^(2) [because r=(d)/(2)=(16)/(2)=8m] <br> According to the given condition, <br> Area of single circular park = Area of first circular park + Area of second circular park <br> piR^(2)=64 pi +36pi [because R be the radius of single circular park] <br> rArr piR^(2)= 100pi rArr R^(2)= 100 <br> :. R = 10m | 322 | 1,073 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2022-05 | latest | en | 0.854684 |
https://usmam.org/10-less-than-a-number-n/ | 1,660,144,103,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571190.0/warc/CC-MAIN-20220810131127-20220810161127-00481.warc.gz | 546,633,753 | 4,408 | Hope you have actually a great experience with this site and also recommend to her friends too.
You are watching: 10 less than a number n
bring away 8,083 time great 1 (738) Ten much more or Ten much less than a Number Repeat
bring away 5,556 times grade 1 (834) add and Subtract tens following Quiz
To find 10 an ext than a number, add 1 to the digit at the 10s place.
To find 10 much less than a number, subtract 1 native the number at the 10s place.
For example: Let's find 10 less than the number 47.
This number deserve to be created in the ar value table as:
Tens Ones 4 7
Notice the there are four tens in the number 47.
Tens Ones
To uncover 10 less than 47, subtract 1 indigenous the digit at its tens place.
4 – 1 = 3
Tens Ones 3 7
So, 10 less than the number 47 is 37.
Shapes
Equation | 228 | 811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2022-33 | latest | en | 0.903418 |
http://www.jiskha.com/display.cgi?id=1184970838 | 1,495,967,437,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609613.73/warc/CC-MAIN-20170528101305-20170528121305-00223.warc.gz | 678,673,778 | 3,992 | # Algebra
posted by on .
(x)/(x-3)-4-(2x-5)/(x+2)
I posted something before this, so now this is how far I got.
The answer I know is -(5x^2-17x-9)/(x-3)(x+2)
first I get
(x)(x+2)/(x-3)(x+2)- (4)(x-3)(x+2)/(x-3)(x+2)-(2x-5)(x-3)/(x-3)(x+2)
Then I get
x^2+2x-4x^2-4x-24-2x^2-11x+15 (for the top)
When I put everything together I get
(-5x^2 -13x -9)/ (x-3)(x+2)
as you can see however I am having a problem with + and - because I get -13x instead of -17x so could you look at what I am doing wrong and explain it to me?
The top should be this:
x^2 + 2x - 4x^2 + 4x + 24 - 2x^2 + 11x - 15
Combining like terms:
-5x^2 + 17x + 9
Factoring out the negative gives this:
-(5x^2 - 17x - 9)
-(5x^2-17x-9)/(x-3)(x+2)
I hope this helps. Watch those negative signs because they can be tricky on these types of problems! | 340 | 822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-22 | latest | en | 0.858774 |
https://brighterly.com/math/9500-in-words/ | 1,716,866,266,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059067.62/warc/CC-MAIN-20240528030822-20240528060822-00052.warc.gz | 114,517,258 | 12,141 | # 9500 in Words
The number 9500 is spelled as “nine thousand five hundred”. It is five hundred more than nine thousand. For example, if you have nine thousand five hundred coins, you have nine thousand coins and then five hundred extra coins, making nine thousand five hundred.
Thousands Hundreds Tens Ones 9 5 0 0
## How to Write 9500 in Words?
Writing 9500 in words is straightforward. It consists of a ‘9’ in the thousands place, a ‘5’ in the hundreds place, and ‘0’ in both the tens and ones places, making it ‘Nine Thousand Five Hundred’. If you have nine thousand five hundred books, you say, “I have nine thousand five hundred books.” Therefore, 9500 is written as ‘Nine Thousand Five Hundred’.
Example:
1. Place Value Chart:
Thousands: 9, Hundreds: 5, Tens: 0, Ones: 0
2. Writing it down: 9500 = Nine Thousand Five Hundred
Understanding this helps in grasping the basics of representing larger numbers in words.
## FAQ on 9500 in Words
### How do you spell the number 9500 in words?
The number 9500 is spelled as ‘Nine thousand five hundred’.
### Write 9500 in word form.
In word form, 9500 is written as ‘Nine thousand five hundred’.
### If you have 9500 apples, how would you write the number?
Nine thousand five hundred apples are written as ‘Nine thousand five hundred’. | 321 | 1,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-22 | latest | en | 0.894864 |
koderunners.ml | 1,563,220,408,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524111.50/warc/CC-MAIN-20190715195204-20190715221204-00199.warc.gz | 447,817,459 | 8,974 | / #MachineLearning #OptimizationTechniques
# Genetic Algorithm: Part 4 - CartPole-v0
So far, we have learned the basics of Genetic Algorithm(GA) and solved a classical problem using GA. GA can be applied to a variety of real world problems.
So, today we will use Open AI Gym environment to simulate a simple game known as CartPole-v0 and then we will use GA to automate the playing of the game. Sounds fun…..
So, lets jump right into it.
# Problem Statement
A pole is standing upright on the cart. The goal is to balance the pole in an upright position by moving the cart left or right. You lose the game if the angle of the pole with the cart is more than 15 degrees. You win the game if you manage to keep the pole balanced for given number of frames. For every frame you mange to keep the pole in upright position you get a ‘+1’ score.
# Solution
To solve this problem we will first create some random game data and then we will feed it to our GA model which in turn will predict the movement of cart(left or right) for every frame. For those of you new to GA, do refer to my previous tutorials.
As usual, we begin with importing libraries and making necessary initializations.
Code:
``````import gym
import random
import numpy as npimport matplotlib.pyplot as plt
from random import randint
from statistics import mean, median
from collections import Counter
``````
``````env = gym.make("CartPole-v0")
env.reset()
#Number of frames
goal_steps = 500
score_requirement = 50
initial_games = 10000
``````
If you are new to Open AI then you can check this out. It will give you a better intuition of its terminologies.
Now we will collect the data by running the game environment 10000 times for radom moves i.e. for every frame it will be randomly decided whether our cart goes left or right. If our score is greater than or equal to 50 then we are going to store every move we made thereafter.
Note that we are storing current action with previous observation. The reason behind it is that at first as the cart was still, there was no observation but when there was an action the observation was returned from the action. Hence, pairing the previous observation to the action we’ll take.
We are considering ‘0’ for ‘left’ and ‘1’ for ‘right’.
Code:
``````def create_data():
training_data, scores, accepted_scores = [], [], []
for _ in range(initial_games):
score = 0
#Moves from current environment and previous observations
game_memory, prev_observation = [], []
for _ in range(goal_steps):
action = random.randrange(0,2)
observation, reward, done, info = env.step(action)
if len(prev_observation) > 0:
game_memory.append([prev_observation, action])
prev_observation = observation
score += reward
if done:
break
if score >= score_requirement:
accepted_scores.append(score)
for data in game_memory:
training_data.append(data)
env.reset()
scores.append(score)
print('Average accepted score:', mean(accepted_scores))
print('Median accepted score:', median(accepted_scores))
return training_data
``````
We will follow the same flow chart for GA as we did earlier.
Initial population will be 8 randomly created set of genes/weights.
Code:
``````def create_initial_pop(pop_size):
initial_pop = np.random.uniform(low = -2.0, high = 2.0, size = pop_size)
print('Initial Population:\n{}'.format(initial_pop))
return initial_pop
``````
``````def sigmoid(z):
return 1/(1+np.exp(-z))
``````
``````def predict(X):
pred = np.empty((X.shape[0], 1))
for i in range(X.shape[0]):
if X[i] >= 0.5:
pred[i] = 0
else:
pred[i] = 1
return pred
``````
The fitness function that we will be using is :
where,
w = weights
x = observations
Code:
``````def cal_fitness(population, X, y, pop_size):
fitness = np.empty((pop_size[0], 1))
for i in range(pop_size[0]):
hx = X@(population[i]).T
fitness[i][0] = np.sum(hx)
return fitness
``````
We will select the fittest individuals as paretns i.e. solutions with the highest fitness value.
Code:
``````def selection(population, fitness, num_parents):
fitness = list(fitness)
parents = np.empty((num_parents, population.shape[1]))
for i in range(num_parents):
max_fitness_idx = np.where(fitness == np.max(fitness))
parents[i,:] = population[max_fitness_idx[0][0], :]
fitness[max_fitness_idx[0][0]] = -999999
return parents
``````
Selected parents will be mated to produce offsprings.
Code:
``````def crossover(parents, num_offsprings):
offsprings = np.empty((num_offsprings, parents.shape[1]))
crossover_point = int(parents.shape[1]/2)
crossover_rate = 0.8
i=0
while (parents.shape[0] < num_offsprings):
parent1_index = i%parents.shape[0]
parent2_index = (i+1)%parents.shape[0]
x = random.random()
if x > crossover_rate:
continue
parent1_index = i%parents.shape[0]
parent2_index = (i+1)%parents.shape[0]
offsprings[i,0:crossover_point] = parents[parent1_index,0:crossover_point]
offsprings[i,crossover_point:] = parents[parent2_index,crossover_point:]
i=+1
return offsprings
``````
Now we will mutate few offsprings to create diversity in solutions and for that we will add some noise to randomly selected gene of individuals selected for mutation.
Code:
``````def mutation(offsprings):
mutants = np.empty((offsprings.shape))
mutation_rate = 0.4
for i in range(mutants.shape[0]):
random_value = random.random()
mutants[i,:] = offsprings[i,:]
if random_value > mutation_rate:
continue
int_random_value = randint(0,offsprings.shape[1]-1)
mutants[i,int_random_value] += np.random.uniform(-1.0, 1.0, 1)
return mutants
``````
Now we will call the functions defined above in the order of flow chart.
Code:
``````def GA_model(training_data):
X = np.array([i[0] for i in training_data])
y = np.array([i[1] for i in training_data]).reshape(-1, 1)
weights = []
num_solutions = 8
pop_size = (num_solutions, X.shape[1])
num_parents = int(pop_size[0]/2)
num_offsprings = pop_size[0] - num_parents
num_generations = 50
population = create_initial_pop(pop_size)
for i in range(num_generations):
fitness = cal_fitness(population, X, y, pop_size)
parents = selection(population, fitness, num_parents)
offsprings = crossover(parents, num_offsprings)
mutants = mutation(offsprings)
population[0:parents.shape[0], :] = parents
population[parents.shape[0]:, :] = mutants
fitness_last_gen = cal_fitness(population, X, y, pop_size)
max_fitness = np.where(fitness_last_gen == np.max(fitness_last_gen))
weights.append(population[max_fitness[0][0],:])
return weights
``````
``````def GA_model_predict(test_data, weights):
hx = sigmoid(test_data@(weights).T)
pred = predict(hx)
pred = pred.astype(int)
return pred[0][0]
``````
``````training_data = create_data()
weights = GA_model(training_data)
print('Weights: {}'.format(weights))
weights = np.asarray(weights)
``````
Output:
``````Initial Population:
[[ 0.67999273 1.20045524 -0.31810563 -1.14804361]
[-1.51475165 -1.42250336 0.03428274 0.63371852]
[-0.8970108 1.03936397 1.84329259 1.72682724]
[ 1.94204407 -0.77717282 0.14019162 -1.1903907 ]
[ 0.41835458 -1.22852332 0.9296547 -1.12009693]
[-0.65292285 -1.40827788 1.55964313 -0.23029554]
[-1.44485637 0.02821767 0.48371509 -0.67509993]
[ 1.51550571 -1.66566025 0.17737747 -1.76249427]]
Weights: [array([ 7.53967288, 14.12987549, -2.29013767, -8.39335431])]
``````
Lets visualize the fitness growth with respect to generations.
Code:
``````fitness_history_mean = [np.mean(fitness) for fitness in fitness_history]
fitness_history_max = [np.max(fitness) for fitness in fitness_history]
plt.plot(list(range(num_generations)), fitness_history_mean, label = 'Mean Fitness')
plt.plot(list(range(num_generations)), fitness_history_max, label = 'Max Fitness')
plt.legend()
plt.title('Fitness through the generations')
plt.xlabel('Generations')
plt.ylabel('Fitness')
plt.show()
``````
Output:
So everything is set and good to go. We will feed our model with training data and then run the game environment 10 times. Initially, we will decide a random move and based on that our model will play the game.
Code:
``````scores, choices = [], []
for each_game in range(10):
score = 0
game_memory, prev_obs = [], []
env.reset()
for _ in range(goal_steps):
env.render()
if len(prev_obs) == 0:
action = random.randrange(0,2)
else:
action = GA_model_predict(prev_obs, weights)
choices.append(action)
new_observation, reward, done, info = env.step(action)
prev_obs = new_observation
game_memory.append([new_observation, action])
score += reward
if done:
break
scores.append(score)
print('Required Score:',str(score_requirement))
print('Average Score Achieved:',sum(scores)/len(scores))env.close()
``````
Output:
``````Required Score: 50
Average Score Achieved: 140.2
``````
Note that everytime you run the code, its not necessary that you get a good score. But if you combine this model with some other algorithm you can make the model more robust.
Thank you for reading this article. If you like it then mention the use cases you think Genetic Algorithm can have in the comment section below. This article was contributed by Satvik Tiwari. Stay tuned for more Machine Learning stuff…. :)
#### KoderunnersML
Official Machine Learning Developers' Pool of the Koderunners Community | 2,439 | 9,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-30 | longest | en | 0.916881 |
https://voer.edu.vn/profile/140?types=1 | 1,708,937,292,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474653.81/warc/CC-MAIN-20240226062606-20240226092606-00818.warc.gz | 616,588,758 | 10,504 | # Sy Hien Dinh
Sy Hien Dinh
Quốc gia:
##### Electronic Circuit Analysis
This textbook introduces DC and AC circuits and analog and digital circuits
##### Introduction
This module introduces approach to order of presentation of topics of electronic circuit analysis.
##### Applications and design of integrated circuits
As an example of the application of op-amps in area of active filters, we will discuss the Butterwort filter. The discussion is only an introduction to the subject of the filter theory design. We will also discuss various types of oscillators, Schmitt trigger circuits, and nonsinusoidal oscillators.
##### Basic Concepts
We introduce some basic concepts such as charge, current, voltage, power, energy, circuit elements, electric circuit.
##### First order circuits
We shall examine two types of simple circuits: a circuit comprising a resistor and capacitor and a circuit comprising a resistor and an inductor. These are called RC and RL circuits, respectively. As simple as these circuits are, they find continual applications in electronics, communications, and control system. We carry out the analysis of RC and RL circuits by applying Kirchhoff’s laws. The only difference is that applying Kirchhoff’s laws to purely resistive circuits, results in algebraic differential equations, which are more difficult to solve than algebraic equations. The differential equations resulting from analyzing RC and RL circuits are of the first order. Hence, the circuits are collectively known as first-order circuits.
##### Capacitors and Inductors
We shall introduce two important passive linear circuit elements: the capacitor and the inductor. With the introduction of capacitors and inductors, we will be able analyze more important and practical circuits. We begin by introducing capacitors and describing how to combine them in series or in parallel. Later, we do the same for inductors. As typical applications, we explore how capacitors are combined with op amp to form integrators, differentiators.
##### Second order circuits
We will consider circuits containing two storage elements. These are known as second order circuits because their responses are described by differential equations that contain second derivatives.
##### Method of analysis
We will develop two powerful techniques for circuit analysis: nodal analysis, which is based on a systematic application of Kirchhoff’s current law (KCL) and mesh analysis which based on a systematic application of Kirchhoff’s voltage law (KVL). With the two techniques to be developed we can analyze any linear circuit by obtaining a set of simultaneous equation that are then solved to obtain the required values of current or voltage. One method of solving simultaneous equations involves Cramer’s rule, which allow us to calculate circuit variables as a quotient of determinants. Finally, we apply the technique learned to analyze transistor circuits.
##### Basic Laws
We introduce some fundamental laws govern electric circuits. These laws known as Ohm’s law and Kirchhoff’s laws, from the foundation upon which electric circuit analysis is build. In addition to these laws we shall discuss some techniques commonly applied in circuit design and analysis. These techniques include combining resistors in series or parallel, voltage division, current division and delta-to-wye and wye-to-delta transformations
##### Operational amplifiers
We begin by discussing the ideal op amp and later consider the nonideal op amp. Using nodal analysis as a tool, we consider ideal op amp circuits such as the inverter, voltage follower, summer, and difference amplifier. Finally, we learn an op amp is used in digital-to-analog converters and instrumentation amplifiers.
##### Frequency response
We begin by considering the frequency response of simple circuits using their transfer functions. We then consider Bode plots which are the industry-standard way of presenting frequency response. We also consider series and parallel resonant circuits and encounter important concepts such as resonance, quality factor, cutoff frequency and bandwidth. We discuss different kinds of filters and network scaling. In the last section, we consider one practical application of resonant circuits and two applications of filters.
##### Digital circuits
This module presents the basic concepts of MOSFET digital logic circuits. We will examine NMOS logic circuits, which contain only n-channel transistors, and complementary MOS, or CMOS, logic circuits, which contain both n-channel and p-channel transistors. The NMOS inverter is the basic of NMOS digital logic circuits. We will analyze the dc voltage transfer characteristics of several inverter designs. We will also define and develop the noise margin of NMOS digital circuits in terms of the inverter voltage transfer curve. We will then determine the impact of the body effect on the dc voltage transfer curve and the logic levels. A transient analysis of the NMOS inverter will determine the propagation delay time in NMOS logic circuits. We will then develop and analyze basic NMOS NOR and NAND logic gates, as well as circuits that perform more complex logic functions. The CMOS inverter is the basic of CMOS logic gates. We will analyze the inverter dc voltage transfer characteristics, and will determine the power dissipation in the CMOS inverter, demonstrating the principle advantage of CMOS inverter over NMOS circuits. Next, we will develop the noise margin of the CMOS inverter, and then will develop and analyze basic CMOS NOR and NAND logic gates. Finally, we will look at more advanced clocked CMOS logic circuits which eliminate almost half of transistors in a conventional CMOS logic design while maintaining the low power advantage of the CMOS technology. In a digital system, a transistor can act as a switch between a driving circuit and a load circuit. An NMOS transistor that performs this function is called an NMOS transmission gate; the corresponding CMOS configuration is a CMOS transmission gate. These transmission gates, or pass transistors, can also be configured to perform logic functions, and the circuits are called NMOS-pass or CMOS-pass networks. We will discuss the basics of these networks. Finally in this module, we will consider a few examples of sequential logic circuits. Two dynamic shift registers are defined and analyzed, and a static R-S flip-flop are defined and analyzed.
LỌC THEO
Tất cả
Tài liệu
Viết giáo trình
Tất cả ngôn ngữ
Tiếng Việt
Tiếng Anh
Tất cả chủ đề | 1,275 | 6,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-10 | latest | en | 0.881518 |
http://mathhelpforum.com/calculus/219323-integral-understanding-question-print.html | 1,526,951,510,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864572.13/warc/CC-MAIN-20180521235548-20180522015548-00396.warc.gz | 176,855,309 | 4,535 | # integral understanding question
• May 25th 2013, 02:15 PM
ryu1
integral understanding question
Hello,
$\displaystyle f(x) = \frac{1}{a^2x^2+b^2}$
I know the correct answer but I don't understand EXACTLY why this is incorrect:
$\displaystyle \int\frac{1}{a^2x^2+b^2} dx = \int\frac{1}{(ax)^2+b^2} dx$
$\displaystyle = \frac{1}{b} \arctan(\frac{ax}{b}) +C$
I understand it has something to do with the integral "in regards to x" (because we write the dx) there, but could you explain that to me a little better please?
Thanks a lot!
Thanks to Wolfram I avoided sending a wrong answer in this on my homework.(really powerful tool, too bad we can't use the internet on tests :( (but we can use it in real life though...(Headbang)))
• May 25th 2013, 03:31 PM
Shakarri
Re: integral understanding question
The variable has to be on its own in the equation, otherwise you are integrating a variable ax with respect to dx and they are different.
You can do a change of variables, let y=ax
$\displaystyle dx=\frac{dy}{a}$
Then the integral
$\displaystyle \int\frac{1}{(ax)^2+b^2} dx$
becomes
$\displaystyle \int\frac{1}{y^2+b^2}\cdot \frac{dy}{a}$
$\displaystyle =\frac{1}{a}\cdot \frac{1}{b} \arctan(\frac{y}{b}) +C$
Presumably in the derivation of the solution to the integral, x2 is not multiplied by a constant and the proof cannot be done the same way when x2 is multiplied by a constant
• May 25th 2013, 03:53 PM
Plato
Re: integral understanding question
Quote:
Originally Posted by ryu1
$\displaystyle f(x) = \frac{1}{a^2x^2+b^2}$
I know the correct answer but I don't understand EXACTLY why this is incorrect:
$\displaystyle \int\frac{1}{a^2x^2+b^2} dx = \int\frac{1}{(ax)^2+b^2} dx$
$\displaystyle = \frac{1}{b} \arctan(\frac{ax}{b}) +C$
I understand it has something to do with the integral "in regards to x" (because we write the dx) there, but could you explain that to me a little better please?
To understand, do the derivative of that answer. Carefully do ALL the algebra to simply that derivative.
That process will show you how to reverse engineer the problem.
Quote:
Originally Posted by ryu1
Thanks to Wolfram I avoided sending a wrong answer in this on my homework.(really powerful tool, too bad we can't use the internet
Frankly, I see absolutely no reason that you should not be allowed to use Wolframalpha on a test.
If you were asked to find $\displaystyle \sqrt{13}$ on a test would you not be allowed to use a calculator?
Tell me, what is the difference? This webpage sums up exactly what I think of the matter.
BTW. There is a cell phone app and tablet app for Wolfframalpha.
• May 25th 2013, 03:54 PM
Soroban
Re: integral understanding question
Hello, ryu1!
Quote:
I had this problem: .$\displaystyle \int \frac{dx}{a^2x^2+b^2}$
I know the correct answer, but I don't understand EXACTLY why this is incorrect:
$\displaystyle \int\frac{dx}{a^2x^2+b^2} \:=\: \int\frac{dx}{(ax)^2+b^2} \:=\:\frac{1}{b} \arctan(\frac{ax}{b}) +C$
If you know the correct answer, you can see what you are missing, can't you?
Okay, back to basics.
Formula: .$\displaystyle \int\frac{du}{u^2+b^2} \;=\;\frac{1}{b}\arctan\left(\frac{u}{b}\right) + C$
We have: .$\displaystyle \int\frac{dx}{(ax)^2 + b^2}$
That is: .$\displaystyle u \,=\,ax \quad\Rightarrow\quad du \,=\,a\,dx \quad\Rightarrow\quad dx \,=\,\tfrac{1}{a}du$
Substitute: .$\displaystyle \int\frac{\frac{1}{a}du}{u^2+b^2} \;=\;\frac{1}{a}\int\frac{du}{u^2+b^2}$
. . . . . . . . $\displaystyle =\;\frac{1}{a}\cdot\frac{1}{b}\arctan\left(\frac{u }{b}\right)+C \;=\;\frac{1}{ab}\arctan\left(\frac{u}{b}\right)+C$
Back-substitute: .$\displaystyle \frac{1}{ab}\arctan\left(\frac{ax}{b}\right)+C$
• May 25th 2013, 04:26 PM
ryu1
Re: integral understanding question
Quote:
Originally Posted by Soroban
Hello, ryu1!
If you know the correct answer, you can see what you are missing, can't you?
Okay, back to basics.
Formula: .$\displaystyle \int\frac{du}{u^2+b^2} \;=\;\frac{1}{b}\arctan\left(\frac{u}{b}\right) + C$
We have: .$\displaystyle \int\frac{dx}{(ax)^2 + b^2}$
That is: .$\displaystyle u \,=\,ax \quad\Rightarrow\quad du \,=\,a\,dx \quad\Rightarrow\quad dx \,=\,\tfrac{1}{a}du$
Substitute: .$\displaystyle \int\frac{\frac{1}{a}du}{u^2+b^2} \;=\;\frac{1}{a}\int\frac{du}{u^2+b^2}$
. . . . . . . . $\displaystyle =\;\frac{1}{a}\cdot\frac{1}{b}\arctan\left(\frac{u }{b}\right)+C \;=\;\frac{1}{ab}\arctan\left(\frac{u}{b}\right)+C$
Back-substitute: .$\displaystyle \frac{1}{ab}\arctan\left(\frac{ax}{b}\right)+C$
There is some saying that I can't remember but it's something like an animal seeing the reflection of the moon in the lake, trying to touch it but can't really reach it.
(or maybe something else, point is, even if I can see it, I can't always understand it, unfortunately, but then I can ask someone and then he tell me, if he is nice, then I think about what he said a little, and accept or not accept his explanation, if Plato explains it then I will most likely accept it :)
I see what you did there, I missed the part with the du = a dx , because I was just blindly using the formula wrong...
Quote:
Originally Posted by Plato
Frankly, I see absolutely no reason that you should not be allowed to use Wolframalpha on a test.
If you were asked to find $\displaystyle \sqrt{13}$ on a test would you not be allowed to use a calculator?
Tell me, what is the difference? This webpage sums up exactly what I think of the matter.
BTW. There is a cell phone app and tablet app for Wolfframalpha.
I guess they think it has something to do with like-cheating, like here, I had to do this problem, and I would make a mistake, but thanks to that tool I "didnt"... but on the contrary, graphing calculator is allowed (ti-83 also has some apps for it to help with many calculations)
But then again, most of the problems they give us don't require complex computations, it requires mostly understanding and making good use of mathematical theorems etc.
Also wolfram has a step by step tool that makes it an "almost" no brainer, which I think is bad, but it's really good and useful :)
As for the app, my cellphone doesn't have always on internet connection (data plan too expensive).
P.S.
I learned to use Latex there :) | 1,925 | 6,220 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-22 | latest | en | 0.873207 |
https://gmatclub.com/forum/two-points-are-on-the-opposite-sides-of-a-square-who-sides-175119.html | 1,542,209,849,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742117.38/warc/CC-MAIN-20181114150002-20181114172002-00557.warc.gz | 642,778,548 | 55,819 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 14 Nov 2018, 07:37
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
## Events & Promotions
###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • ### Free GMAT Strategy Webinar November 17, 2018 November 17, 2018 07:00 AM PST 09:00 AM PST Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. # Two points are on the opposite sides of a square who sides new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Intern Joined: 24 May 2014 Posts: 13 Location: Brazil Two points are on the opposite sides of a square who sides [#permalink] ### Show Tags 26 Jul 2014, 10:24 3 00:00 Difficulty: 25% (medium) Question Stats: 73% (01:20) correct 27% (01:20) wrong based on 113 sessions ### HideShow timer Statistics Two points are on the opposite sides of a square who side measures 4. Which of the following could be the distance between these two points? (A) 3.8 (B) 5 (C) 6 (D) 7 (E) 9 Math Expert Joined: 02 Sep 2009 Posts: 50583 Re: Two points are on the opposite sides of a square who sides [#permalink] ### Show Tags 26 Jul 2014, 10:46 2 1 Reni wrote: Two points are on the opposite sides of a square who side measures 4. Which of the following could be the distance between these two points? (A)3.8 (B)5 (C)6 (D)7 (E)9 The shortest distance between any two points which are on opposite sides of a square is the length of the side of the square. The longest distance between any two points which are on a square is the length of the diagonal of the square. Attachment: Untitled.png [ 3.63 KiB | Viewed 1713 times ] So, the distance between these two points must be more than or equal to 4 and less than or equal to $$\sqrt{4^2+4^2}<6$$. The only option which fall into this range is B. Answer: B. _________________ EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12853 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Two points are on the opposite sides of a square who sides [#permalink] ### Show Tags 03 Feb 2016, 23:00 1 Hi All, With a bit of logic, and using the answer choices as reference, you can answer this question without having to do ANY calculations. We're told that two points are on the opposite sides of a square whose sides are 4. We're asked which of the following COULD be the distance between these two points. Since we don't know exactly where the two points are, we should start with what would be the SHORTEST distance between them, then we can work 'up' from there. If the points were at the exact some 'place' on opposite sides, then they would be the same distance as the length of a side: 4. That length is the SHORTEST distance between possible points, so Answer A is impossible. As we slide either (or both) points along the sides of the square, the distance will become GREATER than 4. So we could move one (or both) of the points so that they were 4.1 apart (or 4.2 or 4.325, etc.). Logically, the answer CANNOT be 6, because if 6 were a possibility, then 5 would ALSO be a possibility. Since there's only one correct answer to this question, it MUST be the smallest number that is greater than 4... Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****
Non-Human User
Joined: 09 Sep 2013
Posts: 8772
Re: Two points are on the opposite sides of a square who sides [#permalink]
### Show Tags
27 Jul 2018, 15:39
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Two points are on the opposite sides of a square who sides &nbs [#permalink] 27 Jul 2018, 15:39
Display posts from previous: Sort by
# Two points are on the opposite sides of a square who sides
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,487 | 5,711 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-47 | latest | en | 0.851509 |
http://65.2.130.20/introduction-to-shapes/ | 1,686,313,884,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656675.90/warc/CC-MAIN-20230609100535-20230609130535-00071.warc.gz | 478,022 | 11,586 | Introduction To Shapes
• A shape can be simply explained as the boundary or outline of an object.
• Everything around us can easily be defined in a shape.
• Learning about basic 2D shape will help kids to increase their observation skills, communication skills as well it helps kids to learn more about their surroundings.
Circle
• It is a closed round shape made up of curved lines.
• It has no sides and corners.
• Example of circle around us are clock, tyre, sun, moon, wheel, ball, etc.
Triangle
• It is a closed shape made up of straight lines.
• It has 3 sides and 3 corners (vertices).
• All sides of a triangle can be equal or of different size.
• Examples of triangle around us are mountains, hills, piece of Pizza, nachos, etc.
Square
• It is a closed shape made up of straight lines.
• It has 4 sides and 4 corners (vertices).
• All sides of a square are equal.
• Opposite sides of a square are parallel to each other.
• Examples of square around us are cheese slice, chess board, photo frame, tile, bread slice, etc.
Rectangle
• It is a closed shape made up of straight lines.
• It has 4 sides and 4 corners (vertices).
• Opposite sides of a rectangle are equal.
• Opposite sides of a rectangle are parallel to each other.
• Examples of rectangle around us are books, curtain, door, cupboard, chocolate, table top, bed, etc.
Oval
• It is a closed shape made up of curved lines.
• Oval Shape is an elongated circle like shape.
• It has no straight sides.
• It has no corners like squares or rectangles.
• Few Examples around us are egg, watermelon, beetle, mirror, racket, balloon, etc.
Semi-Circle
• It is a closed shape made up of curved and straight lines.
• When a circle is cut into 2 parts it forms a semi-circle.
• Half of the circle is called a semi-circle.
• Examples of semi-circle around us are umbrella, half pizza, parachute, rainbow, turtle, tunnel, japanese fan, etc.
Star
• It is a closed shape made up of straight lines.
• It has 10 sides.
• It has 10 corners.
• Examples of star shapes around us are stars in the night sky, star on the christmas tree, star shaped cookies, star on the wand, star fruit, star in the medal, starfish, etc.
Heart
• It is a closed shape made up of curved and straight lines.
• This shape is broader from the top and narrows at the bottom.
• It has no straight side.
• Examples of heart shapes around us are strawberry, peepal leaf, heart shaped balloon, heart shaped pendant, etc.
Let’s begin by discovering some ways to teach shapes to our young learners in a fun and interesting ways :
Stories
Teaching Shapes through stories from Uncle Math School by Fun2Do Labs :
Text of Stories
Song
Teaching Shapes through song from Uncle Math School by Fun2Do Labs :
Activities
Teaching shapes by play – way method :
Sort and Learn :
Gather items from around the surrounding area and sort them by shape. A fun way for kids to discover circles, squares, triangles, and more in the world around them.
Hop into shapes :
Create a shape maze on the playground with colored chalk. Hop from one shape to another, or call out a different shape each time!
Teaching shapes by hands – on activities :
Stamping by blocks :
Kids love stamping so this activity proves to be real fun for them. Kids can be instructed to make a design or picture using shape blocks and washable paint.
Worksheets
Help your kids to practise shapes by using the following interesting and engaging fun worksheets and solutions from Uncle Math School by Fun2Do Labs.
Worksheet 012 : Trace The Circles
Solution 012 : Trace The Circles
Worksheet 013 : Identify All The Circles On The Path
Solution 013 : Identify All The Circles On The Path
Worksheet 014 : Trace The Squares
Solution 014 : Trace The Squares
Worksheet 015 : Identify All The Squares On The Path
Solution 015 : Identify All The Squares On The Path
Worksheet 016 : Trace The Triangle
Solution 016 : Trace The Triangle
Worksheet 017 : Colour All Triangles
Solution 017 : Colour All Triangles
Worksheet 018 : Identify All The Triangles On The Path
Solution 018 : Identify All The Triangles On The Path
Worksheet 019 : Trace The Rectangles
Solution 019 : Trace The Rectangles
Worksheet 020 : Trace All The Rectangles On The Path
Solution 020 : Trace All The Rectangles On The Path
Worksheet 021 : Trace The Ovals
Solution 021 : Trace The Ovals
Worksheet 022 : Identify Ovals On The Path
Solution 022 : Identify Ovals On The Path
Worksheet 023 : Identify Stars On The Path
Solution 023 : Identify Stars On The Path
Worksheet 024 : Trace The Semicircle
Solution 024 : Trace The Semicircle
Worksheet 025 : Identify The Hearts
Solution 025 : Identify The Hearts
Worksheet 026 : Identify The Diamonds
Solution 026 : Identify The Diamonds
Worksheet 027 : Trace And Colour The Diamond Shapes
Solution 027 : Trace And Colour The Diamond Shapes
Worksheet 032 : Match Objects With Shapes
Solution 032 : Match Objects With Shapes
Worksheet 033 : Match Objects With Shapes
Solution 033 : Match Objects With Shapes
Worksheet 034 : Match Objects With Shapes
Solution 034 : Match Objects With Shapes
Worksheet 035 : Identify The Shapes
Solution 035 : Identify The Shapes
Worksheet 036 : Identify And Count The Shapes
Solution 036 : Identify And Count The Shapes | 1,216 | 5,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-23 | latest | en | 0.943122 |
null | null | null | null | null | null | Entries tagged with “Social Networking”.
ProPay has published a whitepaper titled Zumogo- The Rise of Social M-Payments. The whitepaper discusses the aspects of secure, social mobile payments and provides an overview of the Social M-Payments concept and the Zumogo Platform. To read the whitepaper please use this link.
Lehi, Utah – April xx, 2011 – ProPay (www.propay.com), an industry leader in Merchant Services, End-to-End Payment Security, credit card processing, and electronic payment services is pleased to announce ProPay’s Zumogo has been named a finalist at the Stoel Rives Utah Innovation Awards™. The annual Utah Innovation Awards are presented by Stoel Rives LLP and the Utah Technology Council. The program recognizes significant innovations and the Utah companies that created them. Winners of the Innovation Awards will be announced at a luncheon on May 3, 2011.
Zumogo (pronounced “Zoo-MOE-Go”) is a social mobile payment platform that enables a new connection between merchants and consumers. Using bi-directional communication and geo-location, Zumogo enables commerce through social technology. “Zumogo marries the best of secure mobile payment platforms and social media, enabling communication and payment between the consumer and the merchant,” stated Chris Mark, Executive Vice President of Emerging Markets at ProPay. “The ability to communicate with a merchant before you are in the store is priceless. Zumogo allows this social functionality to occur and ProPay is excited to lead the market.”
Zumogo is a new mobile payment technology allowing Smartphones to be used, not only as a payment device, but also as a social technology allowing merchants and consumers to communicate with each other in real-time. Additionally, the social m-payment platform allows the consumer to find merchants, contact them for information, and make payment all from the same device. Zumogo eliminates sensitive information from the transaction process, meaning no payment data is passed through the merchant’s system or stored on the consumer’s smartphone.
“ProPay is proud to have been named a finalist in the Utah Innovation Awards,” said Gary Goodrich, ProPay’s Chief Executive Officer. “Our company takes pride in offering products and services that help companies keep pace with technological trends in the market place. Our selection as a finalist for the Awards is a valuable recognition of our efforts and we are excited to bring Zumogo to the market.”
name="Media Suite Flash Media Player"
There has recently been a rise in the rate of social networking or recruiting scams in which a fraudster attempts to dupe an unsuspecting individual into opening merchant accounts under the individual’s own name that can then be used to process stolen credit cards.
Due to the availability of information on many social networks it is relatively easy for a fraudster to find those people who are unemployed or need extra income to make ends meet. Once the fraudster identifies a target, the fraudster presents the target with an opportunity to make money quickly and easily. Essentially the fraudster will convince the victim to open one or more merchant accounts under the victim’s personal information and provide the victim with stolen credit card information claiming the charges are for software licenses or some other non-tangible product. The victim is told to process transactions through the victim’s own merchant account, keep a certain percentage of all sales, and then is asked to send the remaining funds to the fraudster via a money transmitter that is difficult to trace.
Clearly this creates a unique type of fraud to detect and prevent. Since the individual signing up for the account is truly who they say they are, few red flags go up during a routine sign up and identity validation process. However, these scams are certainly not undetectable. The key, as with most fraud scams, is to understand how it works. It is crucial that situations like these are detected and resolved quickly before the chargebacks come in and the victims find themselves owing funds they don’t have.
This, like many fraudulent situations, can generally be resolved through due diligence and a little common sense. Research any potential opportunity that comes your way, and make sure you verify its legitimacy. Remember, the old adage is often correct, if something seems too good to be true it probably is. | null | null | null | null | null | null | null | null | null |
http://www.expertsmind.com/questions/binary-tree-and-binarytree-parts-3016392.aspx | 1,656,522,160,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00245.warc.gz | 75,717,208 | 13,960 | Binary tree and binarytree parts, Data Structure & Algorithms
Assignment Help:
Q. What do you understand by the term Binary Tree? What is the maximum number of nodes which are possible in a Binary Tree of depth d. Explain the terms given below with respect to the Binary trees
1)Strictly Binary Tree
2) Complete Binary Tree
3) Almost Complete Binary Tree
Ans:
A binary tree is a tree in which no nodes can posses' more than two children.
Figure drawn below shows us that the binary tree consists of the root and two subtrees, Tl and Tr, both of which could possibly be empty.
The highest numbers of nodes a binary tree of depth d can have is 2 d+1-1.
(i) Strictly Binary Tree:- If every non leaf node in binary tree has neither the left tree nor the right tree empty subtrees , then the tree is known as a strictly binary tree.
(ii) Complete Binary Tree:- The complete binary tree of depth d is that strictly binary tree whose all the leaves are at level D.
(iii) Almost Complete Binary Tree:- The binary tree of depth d is an almost complete binary tree if and only if:
1.Any node end at level less than d-1 has two children.
2. for any node nd in the tree with
(iv) a right descendant at the level d, nd should have a left child and every descendant of the nd is either a leaf at level d or has two children.
Maximum degree of any vertex in a simple graph, The maximum degree of any v...
The maximum degree of any vertex in a simple graph with n vertices is (n-1) is the maximum degree of the vertex in a simple graph.
Dynamic data structure, advanatges of dynamic data structure in programming...
advanatges of dynamic data structure in programming
Representation of max-heap sequentially, Q. How do we represent a max-heap ...
Q. How do we represent a max-heap sequentially? Explain by taking a valid example. Ans: A max heap is also called as a descending heap, of size n is an almos
Visual Basic Assignment, When writing a code for a program that basically a...
When writing a code for a program that basically answers Relative Velocity questions how do you go at it? How many conditions should you go through?
Searhing and sorting algorithms, how I can easily implement the bubble,sele...
how I can easily implement the bubble,selection,linear,binary searth algorithms?
Applications of b-trees, A database is a collection of data organized in a ...
A database is a collection of data organized in a manner that facilitates updation, retrieval and management of the data. Searching an unindexed database having n keys will have a
space, What is Space complexity of an algorithm? Explain
What is Space complexity of an algorithm? Explain.
Estimate cost of an optimal diapath, Normally a potential y satisfies y r ...
Normally a potential y satisfies y r = 0 and 0 ³ y w - c vw -y v . Given an integer K³0, define a K-potential to be an array y that satisfies yr = 0 and K ³ y w - c vw -y v
Implementation of dequeue, Dequeue (a double ended queue) is an abstract da...
Dequeue (a double ended queue) is an abstract data type alike to queue, where insertion and deletion of elements are allowed at both of the ends. Like a linear queue & a circular q
How do you find the complexity of an algorithm, How do you find the complex...
How do you find the complexity of an algorithm? Complexity of an algorithm is the measure of analysis of algorithm. Analyzing an algorithm means predicting the resources that | 800 | 3,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2022-27 | latest | en | 0.891429 |
https://www.helpteaching.com/lessons/1465/electric-charge-and-electric-force | 1,717,050,019,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059506.69/warc/CC-MAIN-20240530052602-20240530082602-00020.warc.gz | 679,468,721 | 8,652 | • ### Browse All Lessons
##### Assign Lesson
Help Teaching subscribers can assign lessons to their students to review online!
Tweet
# Electric Charge and Electric Force
This lesson aligns with Next Generation Science Standards (NGSS) PS3.C
Introduction
Has this ever happened to you? You walk through a garage, come in to touch a metal doorknob, and get an unpleasant electric shock. The reason for the shock is because of moving electric charges. Moving electric charges are also responsible for lightning streaks and the electric current that flows through cables and wires. Understanding the basic concepts of electric charge and electric force is crucial in comprehending the workings of electricity. In this article, we will learn about the fundamentals of electric charge and electric force, exploring their properties, interactions, and practical applications.
Electric Charge
Electric charge is a basic property of matter. It is a property of subatomic particles like protons and electrons, which are the building blocks of atoms. Electric charge is based on two types: positive and negative. Protons carry a positive charge, while electrons possess a negative charge. Objects can become charged when they gain or lose electrons, altering the balance of positive and negative charges within them.
Coulomb
The electric charge is measured in coulomb (C), the S.I. unit of electric charge. One coulomb is taken as the charge of approximately $6.24 x 10^18$ electrons. It is important to note that like charges repel each other, while opposite charges attract.
Electric Force
The electric force is the force exerted between charged objects. It is responsible for various phenomena, such as the attraction or repulsion between charged particles and the flow of electric current. The strength of the electric force is based on the following factors:
• the magnitude of electric charges involved
• the distance between the charges
The higher and closer the charges are, the greater the electric force will be.
Coulomb's Law
Coulomb's Law is a fundamental principle that describes the mathematical relationship between electric charge and electric force. It states that the force between two charged particles, known as electric force (F), is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Mathematically, Coulomb's Law can be expressed as:
$F = k * (q_1 * q_2) / r^2$
Where F represents the electric force, $q_1$ and $q_2$ are the magnitudes of the charges, r is the distance between the charges, and k is the proportionality constant. The value of k depends on the medium in which the charged objects are located.
Electric Fields
Electric fields are regions of space around charged objects where electric forces are exerted on other charged objects. When a charged object is placed in an electric field, a force is exerted due to the interaction between its charge and the electric field. The direction of the force depends on the nature of the charges involved.
The unit used to measure the strength of the electric field is newtons per coulomb (N/C). The direction and strength of the electric field is represented by the electric field lines. These lines are directed away from the positive charges and towards the negative charges. The more field lines are close to each other, the stronger the electric field in that region.
Applications of Electric Charge and Electric Force
Understanding electric charge and electric force is essential in comprehending various real-life applications. Here are a few examples:
Electric Circuits
Electric charge is the driving force behind the flow of electric current in circuits. The principles of electric charge and electric force enable us to understand how circuits work and how components like resistors, capacitors, and batteries interact.
Static Electricity
The accumulation and discharge of static electricity are familiar phenomena. Understanding electric charge and electric force helps explain how objects become charged through friction, induction, or conduction, and how this charge can be transferred or neutralized.
Lightning
Lightning is a powerful natural phenomenon resulting from the discharge of built-up electric charge in the atmosphere. It occurs due to the interaction between the positive and negative charges in clouds and on the ground.
Electrostatic Precipitators
These devices use the principles of electric charge and electric force to remove particulate matter from industrial emissions. Charged plates attract and collect the particles, allowing cleaner air to be released.
Summary
• Electric charge is a physical property of particles that causes them to attract or repel each other without touching.
• The force of attraction or repulsion exerted between charged particles is called electric force.
• An electric field is a space around a charged particle where the particle exerts an electric force on other particles.
• The strength of an electric field is measured in newtons per coulomb (N/C).
Related Worksheets: | 984 | 5,095 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-22 | latest | en | 0.934688 |
https://ccssanswers.com/angles/ | 1,723,021,155,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640690787.34/warc/CC-MAIN-20240807080717-20240807110717-00804.warc.gz | 125,010,265 | 45,668 | # Angles – Definition, Types, Properties, Examples
The angle is defined as a figure formed by two rays meeting at a common endpoint. Learn Angle Definition, Parts of an angle, Measure of an angle, etc., Learn the complete concept of an angle along with sample examples. We have explained every part of an angle clearly in this article. Angle denotes with the symbol ‘∠’.
The name of the angle is ∠ABC or ∠CAB
Lines and Angles
## Parts of an Angle
Arms: The two rays that are interset each other to form an angle are called Arms of an angle. From the above-given figure, BA and BC are the arms of the Angle ABC.
Vertex: The endpoint where two arms of an angle interest each other is called the vertex. From the given figure, point B is the vertex of Angle ABC.
Size of an Angle: The amount of turn from one arm of the angle to the other arm of the angle is said to be the size of an angle. The size of the angles represents in degrees. The symbol is º.
### Types of Angles
There are different types of angles available based on the measure of an angle. They are,
1. Acute Angle
2. Obtuse Angle
3. Right Angle
4. Straight Angle
5. Reflex Angle
6. Full Rotation
1. Acute Angle: An angle that forms less than 90 degrees is called an acute angle. The measure of an acute angle is between 0º and 90º.
2. Obtuse Angle: An angle that forms greater than 90 degrees is called an Obtuse angle. The measure of an Obtuse angle is between 90º and 180º.
Obtuse Angle Measure = (180 – acute angle measure)
3. Right Angle: An angle that forms exactly 90 degrees is called a right angle. The measure of a right angle is exactly 90º. The two lines are perpendicular to each other at the right angle.
4. Straight Angle: The angle that forms 108º is called the Straight Angle.
5. Reflex Angle: An angle that forms greater than 180 degrees and less than 360 degrees is called a Reflex angle.
6. Full Rotation: The angle that forms 360º called Full Rotation.
### Interior and Exterior Angles
In a polygon such as quadrilateral, triangle, hexagon, pentagon, etc., the interior and exterior angles are formed. Interior angles formed inside the polygon or a closed shape. Exterior angles formed outside the given shape.
### Positive & Negative Angles
An Angle that calculated in an Anti-Clockwise direction is called a Positive Angle and an angle Clockwise direction is called a Negative Angle.
### Angle Measurement
The angle is measured in degrees. The degree symbol is º.
#### Degree of an Angle
The full angle is 360. Firstly, the full angles are divided using the angle of an equilateral triangle. If the angle is 1, then it is 1/360 of the full rotation. The degree is divided into minutes and seconds.
1°= 60′ = 3600”
The SI unit of angle is the radian of an angle. The Radian of an Angle used in Calculus. All the derivatives and integrals are calculated in the terms of a radian. The radian of an angle is denoted by ‘rad’.
Example: In a complete circle, there are 2π radians available.
Gradian of an Angle is also called a grade or a gon. The angle is equal to 1 gradian if the rotation starts and moved up to 1/400 of the full rotation. It is denoted by ‘grad’.
### Frequently Asked Questions on Angles
1. What is an angle?
The angle is formed by intersecting two rays at a single point. The two rays are called the sides of an angle and the common point is called the vertex.
2. What are the six types of angles?
There are six types of angles. They are
• Acute angle
• Obtuse angle
• Right angle
• Straight angle
• Reflex angle
• Full rotation
3. How angles are calculated?
The angles are calculated with the degree.
4. What is a zero angle?
An angle that consists of zero degree is called zero angle. | 893 | 3,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-33 | latest | en | 0.901707 |
https://www.vedantu.com/question-answer/how-do-you-solve-7n-5-5n-+-7-class-10-maths-cbse-601161bd925e103366c11f81 | 1,709,610,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948217723.97/warc/CC-MAIN-20240305024700-20240305054700-00793.warc.gz | 1,030,933,490 | 28,443 | Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# How do you solve $7n - 5 = 5n + 7$?
Last updated date: 04th Mar 2024
Total views: 340.8k
Views today: 3.40k
Verified
340.8k+ views
Hint: In the given problem we need to solve this for ‘n’. We can solve this using the transposition method. The common transposition method is to do the same thing (mathematically) to both sides of the equation, with the aim of bringing like terms together and isolating the variable (or the unknown quantity). That is we group the ‘n’ terms one side and constants on the other side of the equation.
Complete step-by-step solution:
Given, $7n - 5 = 5n + 7$.
We transpose ‘5n’ which is present in the right hand side of the equation to the left hand side of the equation by subtracting ‘5n’ on the left hand side of the equation.
$7n - 5n - 5 = 7$
$2n - 5 = 7$
We transpose negative 5 to the right hand side of the equation by adding 5 on the right hand side of the equation.
$2n = 7 + 5$.
$2n = 12$
Divide the whole equation by 2,
$n = \dfrac{{12}}{2}$
$\Rightarrow n = 6$
$7(6) - 5 = 5(6) + 7$
$42 - 5 = 30 + 7$
$\Rightarrow 37 = 37$ | 364 | 1,142 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-10 | longest | en | 0.841428 |
https://metanumbers.com/4388 | 1,606,563,304,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195417.37/warc/CC-MAIN-20201128095617-20201128125617-00582.warc.gz | 382,323,382 | 7,391 | ## 4388
4,388 (four thousand three hundred eighty-eight) is an even four-digits composite number following 4387 and preceding 4389. In scientific notation, it is written as 4.388 × 103. The sum of its digits is 23. It has a total of 3 prime factors and 6 positive divisors. There are 2,192 positive integers (up to 4388) that are relatively prime to 4388.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 23
• Digital Root 5
## Name
Short name 4 thousand 388 four thousand three hundred eighty-eight
## Notation
Scientific notation 4.388 × 103 4.388 × 103
## Prime Factorization of 4388
Prime Factorization 22 × 1097
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 2194 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 4,388 is 22 × 1097. Since it has a total of 3 prime factors, 4,388 is a composite number.
## Divisors of 4388
1, 2, 4, 1097, 2194, 4388
6 divisors
Even divisors 4 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 6 Total number of the positive divisors of n σ(n) 7686 Sum of all the positive divisors of n s(n) 3298 Sum of the proper positive divisors of n A(n) 1281 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 66.242 Returns the nth root of the product of n divisors H(n) 3.42545 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 4,388 can be divided by 6 positive divisors (out of which 4 are even, and 2 are odd). The sum of these divisors (counting 4,388) is 7,686, the average is 1,281.
## Other Arithmetic Functions (n = 4388)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 2192 Total number of positive integers not greater than n that are coprime to n λ(n) 1096 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 601 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 2,192 positive integers (less than 4,388) that are coprime with 4,388. And there are approximately 601 prime numbers less than or equal to 4,388.
## Divisibility of 4388
m n mod m 2 3 4 5 6 7 8 9 0 2 0 3 2 6 4 5
The number 4,388 is divisible by 2 and 4.
• Arithmetic
• Deficient
• Polite
## Base conversion (4388)
Base System Value
2 Binary 1000100100100
3 Ternary 20000112
4 Quaternary 1010210
5 Quinary 120023
6 Senary 32152
8 Octal 10444
10 Decimal 4388
12 Duodecimal 2658
20 Vigesimal aj8
36 Base36 3dw
## Basic calculations (n = 4388)
### Multiplication
n×i
n×2 8776 13164 17552 21940
### Division
ni
n⁄2 2194 1462.67 1097 877.6
### Exponentiation
ni
n2 19254544 84488939072 370737464647936 1626795994875143168
### Nth Root
i√n
2√n 66.242 16.3715 8.13892 5.35123
## 4388 as geometric shapes
### Circle
Diameter 8776 27570.6 6.04899e+07
### Sphere
Volume 3.53906e+11 2.4196e+08 27570.6
### Square
Length = n
Perimeter 17552 1.92545e+07 6205.57
### Cube
Length = n
Surface area 1.15527e+08 8.44889e+10 7600.24
### Equilateral Triangle
Length = n
Perimeter 13164 8.33746e+06 3800.12
### Triangular Pyramid
Length = n
Surface area 3.33498e+07 9.95712e+09 3582.79
## Cryptographic Hash Functions
md5 473803f0f2ebd77d83ee60daaa61f381 6407473af8e07df6e215922010d1c4db8760f7db e753559bf52404ed6877db546a80e189e22c6ff1ca965d7115a882d4335c9be7 3b724e2e12781e389e84b85fe82b69a6ab5567e98bc973a7b2e036005fe699015e06b887bf51034db43b7f88431d189033ce1ce077f405117e51f727b899ca34 a34547228de09816245bfc518ec9a7c936223990 | 1,407 | 3,976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2020-50 | longest | en | 0.820721 |
http://gmatclub.com/forum/in-the-attached-figure-point-d-divides-side-bc-of-triangle-2998.html?sort_by_oldest=true | 1,467,263,832,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783398075.47/warc/CC-MAIN-20160624154958-00031-ip-10-164-35-72.ec2.internal.warc.gz | 131,328,547 | 43,467 | Find all School-related info fast with the new School-Specific MBA Forum
It is currently 29 Jun 2016, 22:17
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# In the attached figure, point D divides side BC of triangle
Author Message
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 17
Kudos [?]: 143 [0], given: 0
In the attached figure, point D divides side BC of triangle [#permalink]
### Show Tags
15 Oct 2003, 01:57
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
In the attached figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given angle ADC = 60 deg and angle ABD = 45, what is angle x? Please explain how you got your answer. (Figure is not drawn to scale.)
(A) 55
(B) 60
(C) 70
(D) 75
(E) 90
Attachments
triangle1.jpg [ 15.46 KiB | Viewed 2948 times ]
_________________
Best,
AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
Last edited by AkamaiBrah on 15 Oct 2003, 23:38, edited 2 times in total.
CEO
Joined: 15 Aug 2003
Posts: 3461
Followers: 67
Kudos [?]: 810 [0], given: 781
### Show Tags
15 Oct 2003, 03:14
AkamaiBrah wrote:
In the attached figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given angle ADC = 60 deg and angle ABD = 45, what is angle x? Please explain how you got your answer. (Figure is not drawn to scale.)
(A) 55
(B) 60
(C) 70
(D) 75
(E) 90
I think its D.
i wish i had time to explain this.
thanks
praetorian
SVP
Joined: 03 Feb 2003
Posts: 1603
Followers: 8
Kudos [?]: 181 [0], given: 0
### Show Tags
15 Oct 2003, 05:23
brain torture again... Akamaibrah, is it possible to solve this problem without employing theorem of sines?
My long way: angle DAB=15 deg. and angle ADB=120. DB=1. So, in triangle ADB we know all the angles and one side. Therefore, we can calculate via the theorem of sines all the other elements including side AD. Since CD=2, in triangle ACD we know two sides and angle CDA=60 deg. and so may calculate X using the above theorem.
Honestly, up to now, I see no other methods.
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 17
Kudos [?]: 143 [0], given: 0
### Show Tags
15 Oct 2003, 13:51
stolyar wrote:
brain torture again... Akamaibrah, is it possible to solve this problem without employing theorem of sines?
My long way: angle DAB=15 deg. and angle ADB=120. DB=1. So, in triangle ADB we know all the angles and one side. Therefore, we can calculate via the theorem of sines all the other elements including side AD. Since CD=2, in triangle ACD we know two sides and angle CDA=60 deg. and so may calculate X using the above theorem.
Honestly, up to now, I see no other methods.
It is possible to solve this without using any trigonometry or other fancy math whatsoever. After all, this is a GMAT forum. (BTW, drawing the figure carefully, then using a protractor to measure x is not an acceptable GMAT technique).
Use the force, Luke....
_________________
Best,
AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
Intern
Joined: 03 May 2003
Posts: 28
Followers: 0
Kudos [?]: 0 [0], given: 0
### Show Tags
15 Oct 2003, 19:06
logic:
CDA is 60 so ADB is 120
This makes DAB = 15 - Side with length 1 is opposite this angle (1)
Side CD is 2 so the angle opposite is must be 30 - from (1)
Making CDA a right 30-60-90 triangle
Correct?
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 17
Kudos [?]: 143 [0], given: 0
### Show Tags
15 Oct 2003, 20:37
exy18 wrote:
logic:
CDA is 60 so ADB is 120
This makes DAB = 15 - Side with length 1 is opposite this angle (1)
Side CD is 2 so the angle opposite is must be 30 - from (1)
Making CDA a right 30-60-90 triangle
Correct?
Doubling the angle does not double the length of the opposite side.
Hint: there is a 60 degree angle here. What shapes have 60 degree angles and what special properties of such shapes can we use to add more information to the problem?
_________________
Best,
AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
Manager
Joined: 27 Jul 2003
Posts: 122
Location: Singapore
Followers: 1
Kudos [?]: 37 [0], given: 0
### Show Tags
15 Oct 2003, 21:20
ans is b. 60
Exterior angle ADB is equal to the sum of interior opposite angles(
DAC + ACD).
since, angle CAD = 15+45 ==>60 (sum of interior opposite angles for triangle
X = 180 - angle ADB+Angle CAD ===> 180- (60+60) = 60
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 17
Kudos [?]: 143 [0], given: 0
### Show Tags
15 Oct 2003, 21:55
araspai wrote:
ans is b. 60
Exterior angle ADB is equal to the sum of interior opposite angles(
DAC + ACD).
since, angle CAD = 15+45 ==>60 (sum of interior opposite angles for triangle
X = 180 - angle ADB+Angle CAD ===> 180- (60+60) = 60
Sorry, you cannot conclude that CAD = 15 + 45 = 60. Those two angles are not interior opposite angles. Moreover, an "exterior" angle to a triangle must have one of its sides common with the triangle.
_________________
Best,
AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 17
Kudos [?]: 143 [0], given: 0
### Show Tags
15 Oct 2003, 23:32
Solution:
As first, it appears that there is not enough information to compute the rest of the angles in the diagram. especially since you are given only one angle and one side in triangle ACD and no obvious way of getting any more information short of actually measuring something or using trigonometry tricks.
In a geometry problem, when faced with situations such as this, you should look for ways to add construction lines to exploit any special properties of the given diagram and expose "concealed information."
For example, note than the figure contains a 60 deg angle, and two lines with lengths of ratio of 2 to 1. How can we use this information? Remember that a 30-60-90 right-triangle also has a ratio of 2 to 1 for the ratio of its hypotenuse to its short leg. This suggests that by dropping a construction line from C to line AD and forming a right triangle we identify another segment of length 1 and perhaps have a foothold into deducing more information about the diagram.
LetтАЩs drop a line from C to point E to form a right triangle, then connect points E and B as follows:
From the properties of a 30-60-90 triangle, we can set segment ED equal to 1. You will see that the addition of this simple bit of knowledge opens the floodgates to complete information about the diagram!!!
We can now note that triangle EDB is an isosceles triangle. angle EDB = 180 тАУ 60 = 120;, hence angles DEB and EBD are both 30;. Since angle ECD is also 30, triangle CEB is isosceles and segment CE = EB. Angle AEB = 180 тАУ 30 = 150, angle ABE = 45 тАУ 30 = 15, and angle BAE = 180 тАУ (15 + 150) = 15. This means than triangle AEB is isosceles and AE = EB. Since CE is also equal to EB, AE = CE and triangle ACE is isosceles with angle CAE = angle ACE = 45; (recall that angle DEC = angle AEC = 90). Hence, x = angle ACE + angle ECD = 45 + 30 = 75 and the answer is (D).
Hope you liked this problem. Big Kudos for everyone who made the attempt -- i hope you learned something from it!
Attachments
triangle2.jpg [ 17.63 KiB | Viewed 2893 times ]
_________________
Best,
AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
Display posts from previous: Sort by | 2,659 | 9,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2016-26 | longest | en | 0.868084 |
http://physicscatalyst.com/maths/inequalities.php | 1,521,351,591,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645538.8/warc/CC-MAIN-20180318052202-20180318072202-00312.warc.gz | 233,362,481 | 6,969 | # Introduction to inequalities,symbol and rules of inequality for Class 11 ,CBSE Board, IITJEE maths and other exams
What is inequalities
In mathematics, an inequality is a relation that holds between two values when they are different
Solving linear inequalities is very similar to solving linear equations, except for one small but important detail: you flip the inequality sign whenever you multiply or divide the inequality by a negative
## Symbols used in inequalities
The symbol < means less than. The symbol > means greater than.
The symbol < with a bar underneath means less than or equal to. Usually this is written as $\leq$
The symbol > with a bar underneath means greater than or equal to. Usually this is written as $\geq$
The symbol $\neq$ means the quatities on left and right side are not equal
Examples
1)a < b means a is less then b or b is greater a
2) $a \leq b$ means a is less then or equal to b
3) a > b means a is greater than b
4) $a \geq b$ means a is greater or equal to b
Things are which are safe to do in inequality which does not change in direction
1) addition of same number on both sides
a > b
=>a+c > b +c
2) Substraction of same number on both sides
a > b
=>a-c > b-c
3) Multipication/Division by same positive number on both sides
a > b
if c is positive number then
ac > bc
or
a/c > b/c
## Things which changes the direction of the inequality
1) swapping the left and right sides
2) Multiplication/Division by negative number on both sides
3) Dont multiple by variable whose values you dont know as you dont know the nature of the variable
## Concept Of Number line
A number line is a horizontal line that has points which correspond to numbers. The points are spaced according to the value of the number they correspond to; in a number line containing only whole numbers or integers, the points are equally spaced.
It is very useful in solving problem related to inequalities and also representing it
Suppose x >2(1/ 3), this can represent this on number line like that
## Linear Inequation in One Variable:
A equation of the form
ax+b> 0
or
$ax+b \geq 0$
or
ax+b< 0
or
$ax+b \leq 0$
are called the linear equation in One Variable
Example:
x-2 < 0
or 3x +10 > 0
or $10x-17 \geq 0$
## Linear Inequation in Two Variable:
A equation of the form
ax+by> c
or
$ax+by \geq c$
or
ax+by< c
or
$ax+by \leq c$
are called the linear equation in two Variable
Example:
x-2y < 0
or 3x +10y > 0
or $10x-17y \geq 0$ | 663 | 2,449 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2018-13 | latest | en | 0.929933 |
http://clay6.com/qa/9306/find-the-coordinates-of-the-foot-of-the-perpendicular-drawn-from-the-origin | 1,481,423,272,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543782.28/warc/CC-MAIN-20161202170903-00116-ip-10-31-129-80.ec2.internal.warc.gz | 54,336,255 | 27,215 | Browse Questions
# Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $(c)\; x + y + z = 1$
This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com
Toolbox:
• The equation of the form $lx+my+nz=d$ where $l,m,n$ are the direction cosines of the normal to the plane and $d$ is the distance of the normal from the origin.
• The coordinate of the foot of the perpendicular are $(ld,md,nd)$.
Step 1:
The given equation of the plane is $x+y+z=1$------(1)
Hence the direction cosines of the normal are $(1,1,1)$
Therefore $\sqrt{1^2+1^2+1^2}=\sqrt 3$
Divide both sides of the equation by $\sqrt 3$
$\large\frac{1}{\sqrt 3}$$x+\large\frac{1}{\sqrt 3}$$y+\large\frac{1}{\sqrt 3}$$z=\large\frac{1}{\sqrt 3}$
Step 2:
Therefore the coordinates of the foot of the perpendicular are $(ld,md,nd)$
(i.e)$\big(\large\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3}\big)$
$\Rightarrow (\large\frac{1}{ 3},\large\frac{1}{ 3},\large\frac{1}{ 3})$
Hence the coordinates are $(\large\frac{1}{ 3},\large\frac{1}{ 3},\large\frac{1}{ 3})$ | 412 | 1,165 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2016-50 | longest | en | 0.702085 |
http://clay6.com/qa/222/differentiate-the-following-w-r-t- | 1,516,194,515,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886939.10/warc/CC-MAIN-20180117122304-20180117142304-00625.warc.gz | 71,201,411 | 27,171 | # Differentiate the following w.r.t. $x : \log\;( cos \: e^x )$
$\begin{array}{1 1} - e^x\; \tan{e^x} \\ e^x\; \tan{e^x} \\ e^x\; \tan{e^{-x}} \\ -e^x\; \tan{e^{-x}} \end{array}$
Toolbox:
• According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
• $\; \large \frac{d(cosx)}{dx} $$=-sinx • \; \large \frac{d(e^{x})}{dx}$$= e^{x}$
• $\; \large \frac{d(\log x)}{dx} $$=\large\frac{1}{x} Given y = \log (\cos e^x) This is of the form y = f(g(x), where g(x) = \cos e^{x}, so we can apply the chain rule of differentiation. Now g(x) = \cos e^x itself is of the form g(x) = f_1(f_2(x)), where f_2 = e^{x}, so let us apply the chain rule and calculate g'(x) According to the Chain Rule for differentiation, given two functions f(x) and g(x), and y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x). \; \large \frac{d(e^{x})}{dx}$$= e^{x}$ and $\; \large \frac{d(cosx)}{dx} $$=-sinx \Rightarrow g'(x) = -sin(e^x)\; e^x \; \large \frac{d(\log x)}{dx}$$=\large\frac{1}{x}$ | 420 | 1,034 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2018-05 | longest | en | 0.507335 |
https://proofwiki.org/wiki/Motion_of_Cart_attached_to_Wall_by_Spring | 1,590,858,140,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347410284.51/warc/CC-MAIN-20200530165307-20200530195307-00007.warc.gz | 493,530,459 | 10,587 | Motion of Cart attached to Wall by Spring
Theorem
Problem Definition
Consider a cart $C$ of mass $m$ attached to a wall by means of a spring $S$.
Let $C$ be free to move along a straight line with zero friction.
Let the force constant of $S$ be $k$.
Let the displacement of $C$ at time $t$ from the equilibrium position be $\mathbf x$.
Then the motion of $C$ is described by the second order ODE:
$\dfrac {\mathrm d^2 \mathbf x} {\mathrm d t^2} + \dfrac k m \mathbf x = 0$
Proof
By Newton's Second Law of Motion, the force on $C$ equals its mass times its acceleration:
$\mathbf F = m \mathbf a$
$\mathbf a = \dfrac {\mathrm d^2 \mathbf x}{\mathrm d t^2}$
By Hooke's Law:
$\mathbf F = -k \mathbf x$
So:
$\displaystyle m \mathbf a$ $=$ $\displaystyle -k \mathbf x$ $\displaystyle \implies \ \$ $\displaystyle m \dfrac {\mathrm d^2 \mathbf x}{\mathrm d t^2}$ $=$ $\displaystyle -k \mathbf x$ $\displaystyle \implies \ \$ $\displaystyle \dfrac {\mathrm d^2 \mathbf x} {\mathrm d t^2} + \dfrac k m \mathbf x$ $=$ $\displaystyle 0$
$\blacksquare$ | 342 | 1,058 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2020-24 | latest | en | 0.660459 |
null | null | null | null | null | null | HOME > Chowhound > China & Southeast Asia >
Thailand? Does anyone eat there? [moved from International Bd.]
• 6
I've an upcoming trip to Thailand and I can't find a single reference to Thailand anywhere on chowhound. Does anyone have any recommendations/experiences they can share?
1. Click to Upload a photo (10 MB limit)
1. The original comment has been removed
1. The original comment has been removed
1. In Bangkok, stay or eat at the Emerald Hotel. Comfortable, nicely decorated large rooms at very low prices include an amazing buffet breakfast, and the hotel restaurant is a good food find all by itself.
1. Living in Bangkok half the year, I eat often in Thailand.
I have posted ad naseum about dining around Bangkok.
I leave other countries and cuisines to the experts!
I'm into local carts, stalls and small shops. | null | null | null | null | null | null | null | null | null |
https://steemkr.com/steemstem/@jakipatryk/algorithmic-approach-to-real-world-problems-rook-polynomials-1574524642811 | 1,597,288,032,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738950.61/warc/CC-MAIN-20200813014639-20200813044639-00396.warc.gz | 511,614,214 | 26,603 | # Algorithmic approach to real world problems - rook polynomials
9개월 전
Certain problems appear to be quite challenging, especially when the "input" is large enough so that the problem cannot be solved by a human in a reasonable time. But thankfully we have computers and algorithms. In this article I'll introduce rook polynomials, very simple algorithm that calculates them, and we'll solve a few real words problems.
## Defining the problem
Imagine a chessboard. Although there are a few pieces, we'll focus on rooks. Rooks can move through any number of squares in a straight line. We say that two rooks are attacking if they stand in one line (vertical or horizontal). We will be interested in counting the number of ways k rooks can be placed on n by m chessboard, so that none are attacking. Moreover, not every square will be allowed to use. If a square is not allowed to be used, we say it's forbidden. For example:
SOURCE
is a correct placement of 8 non-attacking rooks on 8x8 board, where the middle 4 squares are forbidden. Example that ilustrates incorrect placement:
SOURCE
Now we are ready to define rook polynomial Rb(x) of a n by m board b with some of its squares forbidden. It is just a polynomial of degree at most min(n,m), where k-th coefficient is a number of ways to place k non-attacking rooks on the board. Notice that there is one way to place zero rooks on any board, so 0-th coefficent of any Rb(x) is 1.
## Towards an algorithmic solution
Take any allowed square, call it s. Notice that every placement which is of our interest either has a rook ats or it doesn't. When it doesn't use s, then the board is equivalent to one with s forbidden. When it does, in none of squares in the same row and column as s other rook can be placed. We have just proved following theorem:
Theorem: Let b be a n by m chessboard. Let s be any allowed square. Let c be a board obtained from b by making s forbidden, and d be a board obtained from b by removing column and row of s. Then Rb(x)=Rc(x)+x·Rd(x).
This simple theorem gives us a way to algorithmically calculate any rook polynomial. Let's move to implementation.
## Implementing the algorithm
We'll use Haskell. Why Haskell? You might not like this language, but it makes implementing such a recursive problem trivial.
### What data structure to use?
A first thought would be a matrix, with 'X' when a square is forbidden, and 'O' when a square is allowed. However, as we are only interested in allowed squares, a more wise data structure would just be a list of their indices. So for a board:
``````X O X O
X X O X
X X X X
``````
our representation will be:
``````[(0,1), (0,3), (1,2)]
``````
To represent polynomial we'll just use a list.
### Removing row and column
In the theorem, we consider board d obtained from b by removing row and column containing s. So we need a function that does that:
``````removeRowAndColumn :: (Int, Int) -> [(Int, Int)] -> [(Int, Int)]
removeRowAndColumn (x, y) = filter (\(i, j) -> i /= x && j /= y)
``````
Nothing interesting happens here, we just pass x and y (position) of s and a list of allowed squares to the function, and it returns a list of allowed squares without column and row of s
### Calculate rook polynomial
Let's jump straight into code and then analyse what's going on:
``````calculateRookPolynomial :: [(Int, Int)] -> [Int]
calculateRookPolynomial [] = [1]
calculateRookPolynomial (s:c) = addPolynomials (calculateRookPolynomial c) (0 : (calculateRookPolynomial d))
where d = removeRowAndColumn s c
addPolynomials :: [Int] -> [Int] -> [Int]
addPolynomials a b = go a b []
where
go [] [] current = current
go xs [] current = current ++ xs
go [] ys current = current ++ ys
go (x:xs) (y:ys) current = go xs ys (current ++ [x+y])
``````
The base case for this recursion is empty list, so a situation, when there are no allowed squares. As we observed at the beginning, a rook polynomial for such a board is 1. For a recursive step, we use the theorem: (s:c) decomposes list into the square s and the rest of list c. Then we recursively calculate the rook polynomial for c and for d, where d is the board without column and row of s. Next, we multiply the rook polynomial of d by x, which in our representation of polynomials is equivalent to appending 0 at the beginning of the list. At the end, we add these polynomials.
This simple and elegant code computes a rook polynomial of any chessboard. Nice, isn't it? I guess it's time to solve some problems!
## Assigning employees to jobs
Suppose we have 5 employees (E0, ..., E4), and 7 jobs (J0, ..., J6). How many ways are there to assign a job to each employee such that no two jobs are assigned to one employee and no two employees are assigned to one job?
Suppose employees don't have any preferences (every employee can do every job). This situation clearly corresponds to a 5 by 7 board, where each row is one employee and each column is a job:
``````O O O O O O O
O O O O O O O
O O O O O O O
O O O O O O O
O O O O O O O
``````
what in our representation is:
``````[(0,0),(0,1),(0,2),(0,3),(0,4),(0,5),(0,6),(1,0),(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,0),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,0),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,0),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)]
``````
Let's use it in our algorithm! The output is the following list (polynomial):
``````[1,35,420,2100,4200,2520]
``````
What does it say? Well, we are interested in the 5th coefficient, which is 2520. There is 2520 ways to place 5 non-attacking rooks on this board, so there is 2520 ways to assign jobs to employees as we wanted.
Now let's do something more interesting. Suppose that employee E0 doesn't want to to job J3; E1 doesn't want to do jobs J0, J1, J6; E3 doesn't want to do jobs J1, J3; and E4 doesn't want to do job J0. This situation is more complicated, but not for our algorithm. We just change the board into:
``````O O O X O O O
X X O O O O X
O O O O O O O
O X O X O O O
X O O O O O O
``````
in our representation:
``````[(0,0),(0,1),(0,2),(0,4),(0,5),(0,6),(1,2),(1,3),(1,4),(1,5),(2,0),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,0),(3,2),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)]
``````
Now the output is:
``````[1,28,266,1042,1617,747]
``````
so there are 747 ways to assign jobs this time.
Let's eliminate even more squares, consider this board:
``````X X O X X O X
X X O X X O X
X O X X X X X
O X O X X X X
X X O X X X X
``````
in our representation:
``````[(0,2),(0,5),(1,2),(1,5),(2,1),(3,0),(3,2),(4,2)]
``````
This time the output is:
``````[1,8,18,15,4]
``````
Note that this list (polynomial) contains only 5 values, so the 5th coefficeint of this rook polynomial is 0! Thus there is no way to assign jobs to employees as we want when E0 only wants to do J2 and J5, E1 wants J2 and J5, E2 wants J1, E3 wants J0 and J2, E4 wants J2.
#### Exercise
This is for those who want to check their understanding of the topic and Haskell.
Imagine you are on SteemFest. You with 5 other Steemians go to a 40's-like restaurant, where you have to leave your phone in the entrance. It turns out that you'll get your phones back, but each one of you will get a random one. What is a probability that none of you will get his own phone? Use rook polynomial to solve that, and share your solution in comments :)
## Conclusion
Turns out that problems that seems challenging at first can be easly solved, making a simple observation, as we did in the theorem. We managed to implement the algorithm, and test it on "real world" problems. | 2,203 | 7,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2020-34 | latest | en | 0.939131 |
https://mpboardsolutions.guru/mp-board-class-8th-maths-solutions-chapter-14-ex-14-2/ | 1,721,502,031,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00262.warc.gz | 354,400,057 | 11,107 | # MP Board Class 8th Maths Solutions Chapter 14 गुणनखंडन Ex 14.2
## MP Board Class 8th Maths Solutions Chapter 14 गुणनखंडन Ex 14.2
प्रश्न 1.
निम्नलिखित व्यंजकों के गुणनखण्ड कीजिए –
1. a2 + 8a + 16
2. p2 – 10p + 25
3. 25m2 + 30m + 9
4. 49y2 + 84yz + 36z2
5. 4x2 – 8x + 4
6. 121b2 – 88bc + 16c2
7. (l + m)2 – 4lm
8. a4 + 2a2b2 + b4
हल:
1. a2 + 8a + 16 = (a)2 + 2 x a x 4 + (4)2
[∴ a2 + 2ab + b2 = (a + b)2]
= (a + 4)2
2. p2 – 10p + 25 = (p)2 – 2 x p x 5 + (5)2
[∴ a2 – 2ab + b2 = (a – b)]
= (p – 5)2
3. 25m2 + 30m + 9 = (5m)2 + 2 x 5m x 3 + (3)2
= (5m + 3)2
4. 49y2 + 84yz + 36z2
= (7y)2 + 2 x 7y x 6z + (6z)2
= (7y + 6z)2
5. 4x2 – 8x + 4 = (2x)2 – 2 x 4x × 2 + (2)2
= (2x – 2)2
6. 121b2 – 88bc + 16c2
= (11b)2 – 2 x 11b x 4c + (4c)2
= (11b – 4c)
7. (l + m)2 – 4lm = l2 + 2lm + m2 – 4lm
= l2 – 2lm + m2
= (l)2 – 2 x 1 x m + (m)2 = (1 – m)
8. a4 + 2a2b2 + b4 = (a2)2 + 2 x a2 x b2 + (a)2
= (a + b)2
प्रश्न 2.
गुणनखण्ड कीजिए –
1. 4p2 – 9q2
2. 63a2 – 112b2
3. 49x2 – 36
4. 16x5 – 144x3
5. (l + m)2 – (l – m)
6. 9x2y2 – 16
7. (x2 – 2xy +y2) – z2
8. 25a2 – 4b2 + 28bc – 49c2
हल:
1. 4p2 – 9q2
a2 – b2 = (a – b) (a + b)
4p2 – 9q2 = (2p)2 – (3q)2
= (2p – 3q) (2p + 3q)
2. 63a2 – 112b2 = 7 (9a2 – 16b2)
= 7 {(3a)2 – (4b)2}
= 7 (3a – 4b) (3a + 4b)
3. 49x2 – 36 = (7x)2 – (6)2
= (7x – 6) (7x + 6)
4. 16x5 – 144x3 = 16x3 (x2 – 9)
= 16x3 (x2 – 32)
= 16x3 (x – 3) (x + 3)
5. (l + m)2 – (l – m) = [(l + m) – (l – m)][(l + m) – (l – m)]
= (l + m – 1 + m) (l + m + l – m)
= 2m x 2l = 4lm
6. 9x2y2 – 16 = (3xy)2 – (4)2
= (3xy – 4) (3xy + 4)
7. x2 – 2xy + y2 – z2 = (x – y)2 – z2
= [(x – y) – z] [(x – y) + z]
= (x – y – z) (x – y + z)
8. 25a2 – 4b2 + 28bc – 49c2
= 25a2 – (4b2 – 28bc + 49c2)
= 25a2 – [(2b)2 – 2 x 26 x 7c + (7c)2]
= (5a)2 – (2b – 7c)2
= [5a – (2b – 7c)] [5a + (2b – 7c)]
= (5a – 25 + 7c) (5a + 2b – 7c)
प्रश्न 3.
निम्नलिखित व्यंजकों के गुणनखण्ड कीजिए –
1. ax2 + bx
2. 7p2 + 21q2
3. 2x2 + 2xy2 + 2xz2
4. am2 + bm2 + bn2 + an2
5. (lm + 1) + m + 1
6. y (y + z) + 9 (y + z)
7. 5y2 – 20y – 8z + 2yz
8. 10ab + 4a + 5b + 2
9. 6xy – 4y + 6 – 9x.
हल:
1. ax2 + bx = x (ax + b)
2. 7p2 + 21q2 = 7 (p2 + 3q2)
3. 2x2 + 2xy2 + 2xz2 = 2x (x2 + y2 + z2)
4. amv + bm2 + bn2 + an2
= (am2 + bm2) + (bn2 + an2)
= m2 (a + b) + n2 (b+ a)
= (a + b) (m2 + n2)
5. (lm + 1) + m + 1 = 1(m + 1) + 1 (m + 1)
= (m + 1) (1 + 1)
6. y (y + z) + 9 (y + z) = (y + z) (y + 9)
7. 5y2 – 20y – 8z + 2yz
= (5y2 – 20y) + (2yz – 8z)
= 5y (y – 4) + 2z (y – 4).
= (y – 4) (5y + 2z)
8. 10ab + 4a + 56+2
= (10ab + 5b) + (4a + 2)
= 5b (2a + 1) + 2 (2a + 1)
= (2a + 1) (5b + 2)
9. 6xy – 4y + 6 – 9x
= (6xy – 4y) – (9x – 6)
= 2y (3x – 2) – 3 (3x – 2)
= (3x – 2) (2y – 3)
प्रश्न 4.
गुणनखण्ड कीजिए –
1. a4 – b4
2. p4 – 81
3. x4 (y + z)4
4. x4 – (x – z)4
5. a2 – 2a2b2 + b4
हल:
1. a4 – b4 = (a2)2 – (b2)2
= (a2 + b2) (a2 – b2)
= (a2 + b2) (a + b) (a – b)
2. p4 – 81 = (p2)2 – (9)2
= (p2 + 9) (p2 – 9)
= (p2 + 9) (p + 3) (p – 3)
3. x4 – (y + z)4 = (x2)2 – [(y + 2)2]2
= [x2 – (y + z)2] [x2 + (y + z)2]
= [x – (y + z)] [x + (y + z)] [x2 + (y + z)2]
= (x – y – z) (x + y + z) [x2 + (y + z)2]
4. x4 – (x – z)4 = (x2)2 – [(x – z)2]2
= [x2 – (x – z)2] [x2 + (x – z)]
= [x – (x – z)] [x + (x – z)] [x2 + (x2 – 2x2 + z2)] = (x – x + z) (x + x – z)(2x2 – 2xz + z2)
= z(2x – z) (2x2 – 2xz + z2)
5. a4 – 2a2b2 + b2 = (a2)2 – 2 x a2 x b2 + (b2)2
= [a2 – b2]2
= [(a – b) (a + b)]2
= (a – b)2 (a + b)2
प्रश्न 5.
निम्नलिखित व्यंजकों के गुणनखण्ड कीजिए –
1. p2 + 6p + 8
2. q2 – 10q + 21
3. p2 + 6p – 16
हल:
1. P2 + 6p + 8 = p2 + (4 + 2) p + 8
(∴8 = 4 x 2)
= p2 + 4p + 2p +8
= p (p + 4) + 2 (p + 4)
= (p + 4) (p + 2)
2. q2 – 10q + 21 = q2 – (7 + 3) q + 21
(∴ 21 = 3 x 7)
=q2 – 7q – 3q + 21
= q(q – 7) – 3 (q – 7)
= q (q – 7) (q – 3)
= (q – 3) (q – 7)
3. p2 + 6p – 16 = p2 + (8 – 2)p – 16
(∴ 16 = 8 x 2)
= p2 + 8p – 2p – 16
= p (p + 8) – 2 (p + 8)
= (p + 8) (p – 2)
पाठ्य-पुस्तक पृष्ठ संख्या # 234
प्रयास कीजिए (क्रमांक 14.2)
प्रश्न 1.
भाग दीजिए –
1. 24xy23z3 को 6yz2 से
2. 63a2b4c6 को 7a2b2c3 से।
हल:
1. 24xy2z3 + 6yz2
= 4xyz
2. 63a2b4c6 ÷ 7a2b2c3
= 9b2c3 | 2,756 | 4,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-30 | latest | en | 0.346073 |
https://www.jiskha.com/display.cgi?id=1359662274 | 1,511,462,951,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806856.86/warc/CC-MAIN-20171123180631-20171123200631-00698.warc.gz | 821,862,924 | 3,955 | # Math PLZ HELP
posted by .
1. Which of the following expressions is written in scientific notation?
1. 73.4 x 105
2. 0.09 × 107
3. 80 x 103
4. 4.22 x 10–3
2. Which of the following is 0.0000000708 written in scientific notation?
1. 7.08 x 10–8
2. 7.8 x 10–8
3. 708 x 10–10
4. 70.08 x 10–9
3. Which expression represents the largest number?
1. 40.1 x 10–6
2. 4.1 x 10–7
3. 0.411 x 10–7
4. 0.04001 x 10–5
4. Which expression is equal to 1/8000 written in scientific notation?
1. 8.0 x 103
2. 1.25 x 10–4
3. 125 x 10–5
4. 1.25 x 104
• Math PLZ HELP -
if you know anything about scientific notation, #1,2 should be easy.
For #3, convert all to same power of 10, then it is easy to compare.
#4: 1/8000 = .000125
• Math PLZ HELP -
1.)A
2.)C
3.)B
4.)B
## Similar Questions
1. ### Math
Use scientific notation to divide the following two numbers. Express the answer using scientific notation; retain at least three decimal places. 6.2 • 107 / 4.15 • 103
11. 17, 200 written in scientific notation is 1.7 x 10^5 21. 0.00105 written in scientific notation is 1.05 x 10^-3 are these correct?
3. ### Math
Which of the following expressions is written in scientific notation?
4. ### Math
Which of the following is 0.0000000708 written in scientific notation?
5. ### math
1. Which of the following expressions is written in scientific notation?
6. ### Math! Please Check My Awnser!
Hello! I need some help with math! I have tried to do this but I am unsure about my answers. My I will identify my answers with a "<--". Also, if I have missed one, please tell me the right answer! Also, Thanks for your help in …
7. ### Homework help plz( Steve or Reiny)
Determine if the number is written in scientific notation. If not, explain 32 * 10^4. (1 point) No; it is not written as a power of 10. No; the first factor is not a number between 1 and 10. Yes; the number is written in scientific …
8. ### Math
Hi everyone! I am extremely struggling with this questions and was hoping someone would go over them with me. Thank you! 1. Which of the following expressions is written in scientific notation?
9. ### Algebra
Which of the following expressions is written in scientific notation?
10. ### Algebra
Which of the following is 0.0000000708 written in scientific notation?
More Similar Questions | 705 | 2,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-47 | latest | en | 0.887319 |
http://mymathclub.blogspot.com/2016/12/winter-break-cyclic-3-4-5.html | 1,539,696,763,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510749.55/warc/CC-MAIN-20181016113811-20181016135311-00140.warc.gz | 246,348,733 | 14,570 | ## Monday, December 19, 2016
### Winter break cyclic 3-4-5
This is another exercise in documenting geometry problem solving. I chose this problem because again it has a 3-4-5 triangle within it and the overall setup is very simple. It also shows how one can circle around the correct solution, as it were, before coming to it.
Given cyclic quadrilateral ABCD where ABC is a right triangle and AB = BC = 5 and
the diagonal BD = 7 find the area of ADC.
#### Thought process
1. Angle chasing: $\triangle{ABC}$ is right isosceleses. Since ABCD is a cyclic quadrilateral, that means we can angle chase a bunch i.e. $\angle{BDC} = \angle{BAC} =45^\circ$.
2. Cyclic also means that the product of diagonal portions is equal i.e. $AE \cdot EC = DE \cdot BE$
3. Also there are a bunch of similar triangles: $\triangle{AED} \sim \triangle{BCE}$ and $\triangle{CDE} \sim \triangle{ABE}$
4. Realize 2 and 3 are the same basically saying the same thing.
5. At this point I had a general idea that I'd need to use the length of both diagonals as well as the similarity of the triangles to find the area.
5. I started the algebra with the Pythagorean theorem and similar triangles to find the lengths of AE and ED. This looked fairly messy even going in although generally solvable.
6. So I stopped and checked with the angle bisector theorem around $\angle{ADC}$ and realized that produced nothing new.
7. I already strongly suspected just on visual inspection that the missing part was a 3-4-5 triangle. So once again I paused and checked in geogebra that my hunch was correct. (it was).
8. Then I started grinding through my first approach: let a = BE and b = CE then:
$\frac{BE}{EC} = \frac{AE}{ED}$ or $\frac{a}{b}=\frac{5 \sqrt{2} - b}{7-a}$
Which simplifies to $7a - a^2 = 5\sqrt{2}b - b^2$
In addition using the Pythagorean theorem: $\overline{AD}^2 + \overline{CD}^2 = 50$
Then applying triangle proportionality $\overline{AD} = \frac{5}{a} * (5\sqrt(2) - b))$ and $\overline{CD} = \frac{5}{a} * b$
When combined you finally end up with:
$$50 = \frac{25}{a^2}((5\sqrt(2) - b)^2 + b^2)$$
$$2a^2 = 50 - 10\sqrt{2}b + 2b^2$$
$$a^2 = 25 - (5\sqrt{2}b - b^2)$$
$$a^2 = 25 - 7a + a^2$$
$$a = \frac{25}{7}$$
9. I started to solve for b after which point I could find AD and CD. But this was even messier looking than above and was heading towards an ugly quadratic equation.
$$\frac{25 \cdot 24}{49} = (5\sqrt{2} -b) \cdot b$$
10. That looked super messy but I realized what I really wanted was $1/2 * AD * DC$ which
comes out to $1/2 \cdot \frac{25}{a^2} (5\sqrt{2} - b) \cdot b$ and you can substitute with the two derived results above to find the answer.
11. I was dissatisfied with the opaqueness and algebra of this method plus it never showed the 3-4-5 clearly so I started from scratch.
12. This time I played with the $\angle{ADC}$. First I stated thinking about breaking up the 3-4-5 into a square and the various 1:2 and 1:3 triangles. But then I realized that I had two 45 degree angles that could be extended and we know the lengths of the sides of those triangle and from there the drawing below immediately fell out. Framed this way the problem was much simpler.
* | 976 | 3,192 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2018-43 | longest | en | 0.895446 |
https://bizfluent.com/how-10016801-calculate-point-margin.html | 1,603,592,039,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107885126.36/warc/CC-MAIN-20201025012538-20201025042538-00471.warc.gz | 228,528,409 | 32,657 | How to Calculate Point Margin
Share It
In the sales world, there are two ways of looking at profit. You can look at profit compared to your costs, or you can look at profit based on your sales. When you compare profit to costs, you are looking at markup, but when you compare profit to sales, you are looking at margin.
Margin is just a way of calculating profit, while point margin is margin expressed as a percentage point. So, if you hear someone comparing margin vs. profit or margin points vs. percent of margin, these terms mean the same thing. Point generally represents 1%.
To calculate point margin, first subtract your cost from the sales price to determine the margin and then divide that margin by the sales price.
Calculating Margin From the Sales Price
Margin is essentially the same as profit. Margin is the amount you have in your pocket after selling an item and subtracting your cost:
Margin = Sales Price - Cost
For example, if you sell a sweater for \$50 and your cost for that sweater was \$30, then your margin is \$20. That's your profit in the transaction. If you sold the same sweater for \$40, then your margin would be \$10.
Calculating Point Margin
Point margin is simply your margin expressed as a percentage point instead of in dollars. You can calculate the point margin by dividing your margin by the sales price:
Point Margin = Margin / Sales Price
So, if you sold the sweater for \$50 with a \$20 margin, then your point margin was 40 points (\$20 / \$50 = 0.40). If you had sold that same sweater for \$40, then your point margin would have been only 25 points (\$10 / \$40 = 0.25).
Calculating a Sales Price Based on Point Margin
Many businesses make it a matter of policy to sell items at a specific point margin. Suppose, for example, before you put your sweater up for sale, you decided you wanted to sell it at a 45-point margin. In this case, you can use the following formula:
Sales Price = Cost / (1 - Margin)
If the sweater cost you \$30, then to sell it at a 45-point margin, the sales price would have to be \$54.54:
Sales Price = \$30 / (1-0.45)
Sales Price = \$30/0.55
Sales Price = \$54.54
Calculating Prices With Markup
While margin looks at profit based on the selling price, markup looks at profit based on the cost. Companies that use markup to calculate price simply add their markup to the cost of the item. For example, if you bought a shirt from a wholesaler for \$10 and want to add a flat markup of \$9, then the sales price would be \$19:
Sales Price = Cost + Markup
Like margin, markup can also be expressed as a percentage, but it is a percentage based on the cost rather than the selling price. For example, if your company's policy is to mark up shirts by 40%, then you would multiply the \$10 cost by 140% to get a \$14 sales price.
Sales Price = Cost x (1+Markup Percentage)
Note that markup must always be more than 100% when you are doing the calculation, or you will lose money.
Margin vs. Markup
You should have probably noticed that when expressed in dollars, markup and margin work out to the same thing. That's because they both represent profit. If the cost is \$8 and you sell an item for \$20, then your markup and margin are both the same \$12.
It's only when you look at the percentages or points that markup and margin vary considerably. As a final example, suppose you are selling a hat for \$54 that cost you \$30. Both markup and margin are \$20. This is a markup of 180%, but it's a 44-point margin. | 809 | 3,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-45 | longest | en | 0.926024 |
http://slideplayer.com/slide/719256/ | 1,542,612,552,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745486.91/warc/CC-MAIN-20181119064334-20181119085708-00002.warc.gz | 309,944,015 | 18,414 | # Constant Velocity The position increases by 2 meters every second. The position decreases by 2 meters every second.
## Presentation on theme: "Constant Velocity The position increases by 2 meters every second. The position decreases by 2 meters every second."— Presentation transcript:
Constant Velocity The position increases by 2 meters every second. The position decreases by 2 meters every second.
Example An object starts at a position of -5 meters and travels with a constant velocity of 3m/s for 5 seconds
Diagram – Motion Map Like the washers in the hallway. v v v v v v t = 2s x = 1m t = 3s x = 4m t = 4s x = 7m t = 5s x = 10m t = 0s x = -5m t = 1s x = -2m
Numerical t (s) x (m) 012345012345 -5 -2 7 4 1 10
t (s) x (m) 5432 0 1 -10 -5 5 10 t (s) v (m/s) 5432 0 1 -10 -5 5 10 Algebraic Graphical 3m/s Δx = 6m Δt = 2s Area = Δx
no direction Velocity – The rate at which the position changes. +/- sign gives direction Displacement – The change in position.
v = velocity (m/s) t = time (s) x = position (m) Δx = x 2 – x 1 (m)
Similar presentations | 342 | 1,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-47 | latest | en | 0.8631 |
https://www.physicsforums.com/threads/clever-coordinate-substitution-for-linear-pde.644736/ | 1,508,349,468,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823016.53/warc/CC-MAIN-20171018161655-20171018181655-00195.warc.gz | 994,584,794 | 15,948 | # Clever coordinate substitution for linear PDE
1. Oct 17, 2012
### sith
Hi! I am currently working with a linear PDE on the form
$\frac{\partial f}{\partial t} + A(v^2 - v_r^2)\frac{\partial f}{\partial \phi} + B\cos(\phi)\frac{\partial f}{\partial v} = 0$.
$A$ and $B$ are constants. I wish to find a clever coordinate substitution that simplifies, or maybe even solves the problem. So far I have tried with an action-angle approach that reduces the dimensionality of the problem. I first assume a general substitution $\lbrace\phi, v\rbrace \rightarrow \lbrace r(\phi, v), s(\phi, v)\rbrace$, and choose
$r(\phi, v) = A\left(\frac{1}{3}v^3 - v_r^2 v\right) - B\sin(\phi)$.
This choice then cancels all the $\partial/\partial r$ terms, and one is left with
$\frac{\partial f}{\partial t} + \left[A(v^2 - v_r^2)\frac{\partial s}{\partial\phi} + B\cos(\phi)\frac{\partial s}{\partial v}\right]\frac{\partial f}{\partial s} = 0$.
Now I am left with the choice of $s(\phi, v)$. Is there maybe a way to choose $s$ such that the expression in front of $\partial f/\partial s$ becomes constant? Am I even on the right track, or are there much more clever ways to solve the problem?
Simon
2. Oct 17, 2012
### sith
Sorry, I forgort to mention also that $v_r$ is constant.
3. Oct 17, 2012
### Mute
You basically need to solve the PDE
$$A(v^2 - v_r^2)\frac{\partial s}{\partial\phi} + B\cos(\phi)\frac{\partial s}{\partial v} = \mbox{const}.$$
Note that, however, solving this equation is the same as solving the time-independent version of your original equation. e.g., let $f(t,v,\phi) = g(v,\phi)\exp(\lambda t)$; your original equation will reduce to the same equation you would need to solve to get $s(v,\phi)$.
4. Oct 19, 2012
### sith
Ah, yes, I see that.. =P But I think I actually have found a way to solve the problem. First to separate out the time dependence as you suggested. Then use the coordinate substitution $r(\phi, v)$ as I wrote in the first post, and simply put $s(\phi, v) = v$. Then of course $B\cos(\phi)$ has to be rewritten in terms of $r$ and $v$. I end up with an ODE that is solveable. The final result I got was
$$f(t, \phi, v) = f_0(r(\phi, v))\exp\left(\lambda\left[t \pm \int_0^v{\frac{d u}{\sqrt{B^2 - \left[A\left(\frac{u^3 - v^3}{3} - v_r(u - v)\right) + B\sin(\phi)\right]^2}}}\right]\right),$$
where $\lambda$ is a constant, $f_0$ is an arbitrary function of $r(\phi,v)$, and the $\pm$ depends on $\phi$ (minus if $|\phi| < \pi/2$). Well, the integral doesn't have an analytical solution as far as I know, and Mathematica didn't think so either, so I guess this is as far as one can get. And since the equation is linear one can add an arbitrary sum of these solutions as well. Anyway, thanks for the help Mute. :)
5. Oct 20, 2012
### Staff: Mentor
Maybe I'm missing something, but it seems to me this equation can be solved using the method of characteristics.
6. Oct 21, 2012
### sith
Yes, it was actually the method that I used to solve it (before I knew there was a name for the method =P). The substitution
$$r(\phi, v) = A\left(\frac{1}{3}v^3 - v_r^2 v\right) - B\sin(\phi)$$
is the characteristic of the $\phi, v$-differential equation. | 998 | 3,203 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2017-43 | longest | en | 0.86643 |
https://www.studypool.com/discuss/239912/solve-math-word-problem-please-with-step-by-step-answers-3?free | 1,508,313,354,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822822.66/warc/CC-MAIN-20171018070528-20171018090528-00406.warc.gz | 995,745,166 | 14,655 | Time remaining:
label Algebra
account_circle Unassigned
schedule 0 Hours
account_balance_wallet \$5
Oct 17th, 2017
4.
cos theta –sin theta =1
When theta = 0 or theta= 3pi/2
cos theta –sin theta =1
Hence 0, 3pi/2 are the roots.
Option b
5.
T (october) = 17 cos (pi/6 *10-7pi/6)+75= cos (3*pi/6)+75= cos (pi/2)+75=75
Option e
6.
(tan u+cot u)(sin u+cos u) =( sinu/cos u+cos u/sinu)(sin u +cosu)
= (sin2u+cos2u)/(sinu cosu)(sin u+cosu)= 1*(sin u+cos u)/(sinu cosu) = 1/cosu +1/sin u
= sec u+ csc u=csc u+sec u
Valid
7.
(sec 2u-1)/sec2 2u = tan22u* cos2u =sin2 2u/ cos2u * cos2u= sin2 2u
Valid
Sep 23rd, 2014
...
Oct 17th, 2017
...
Oct 17th, 2017
Oct 18th, 2017
check_circle | 308 | 691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-43 | latest | en | 0.515256 |
https://origin.geeksforgeeks.org/count-of-squads-of-positive-integers-such-that-a-b-c-d-n/ | 1,685,613,157,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647639.37/warc/CC-MAIN-20230601074606-20230601104606-00077.warc.gz | 508,554,880 | 43,482 | GFG App
Open App
Browser
Continue
# Count of squads of positive integers such that (A * B) + (C * D) = N
Given a positive integer N, the task is to find the count of squads of positive integers (A, B, C, D) such that (A * B) + (C * D) = N
Note: (A, B, C, D) is different from (A, D, B, C).
Example:
Input: 4
Output: 8
Explanation:
(A, B, C, D)=(1, 1, 1, 3)
(A, B, C, D)=(1, 1, 3, 1)
(A, B, C, D)=(1, 2, 1, 2)
(A, B, C, D)=(1, 2, 2, 1)
(A, B, C, D)=(1, 3, 1, 1)
(A, B, C, D)=(2, 1, 1, 2)
(A, B, C, D)=(2, 1, 2, 1)
(A, B, C, D)=(3, 1, 1, 1)
Naive Approach: The basic way to solve the problem is as follows:
In this implementation, we use four nested loops to iterate over all possible values of A, B, C, and D. We check if (AB) + (CD) equals N and that A is less than or equal to B, and C is less than or equal to D to avoid counting the same squad multiple times.
Below is the implementation of the above approach:
## C++
`// C++ code for the above approach` `#include ` `using` `namespace` `std;` `// Function to count number of squads` `int` `countSquads(``int` `N)` `{` ` ``int` `count = 0;` ` ``int` `i, j, k, l;` ` ``// Iterating 4 nested loop to find` ` ``// four varible i j k l` ` ``for` `(i = 1; i <= N; i++) {` ` ``for` `(j = 1; j <= N; j++) {` ` ``for` `(k = 1; k <= N; k++) {` ` ``for` `(l = 1; l <= N; l++) {` ` ``// If (a*b) + (c*d) == N` ` ``// then increase the` ` ``// counter by 1.` ` ``if` `((i * j) + (k * l) == N) {` ` ``count++;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// Return the total count` ` ``return` `count;` `}` `// Driver code` `int` `main()` `{` ` ``int` `N = 4;` ` ``// Function call` ` ``int` `ans = countSquads(N);` ` ``cout << ans << endl;` ` ``return` `0;` `}`
## Java
`// Java code for the above approach` `import` `java.io.*;` `class` `Main {` ` ``// Function to count number of squads` ` ``static` `int` `countSquads(``int` `N)` ` ``{` ` ``int` `count = ``0``;` ` ``int` `i, j, k, l;` ` ``// Iterating 4 nested loop to find` ` ``// four variable i j k l` ` ``for` `(i = ``1``; i <= N; i++) {` ` ``for` `(j = ``1``; j <= N; j++) {` ` ``for` `(k = ``1``; k <= N; k++) {` ` ``for` `(l = ``1``; l <= N; l++) {` ` ``// If (a*b) + (c*d) == N` ` ``// then increase the` ` ``// counter by 1.` ` ``if` `((i * j) + (k * l) == N) {` ` ``count++;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// Return the total count` ` ``return` `count;` ` ``}` ` ``// Driver code` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``int` `N = ``4``;` ` ``// Function call` ` ``int` `ans = countSquads(N);` ` ``System.out.println(ans);` ` ``}` `}` `// This code is contributed by Prajwal Kandekar`
## Python3
`# Function to count number of squads` `def` `countSquads(N):` ` ``count ``=` `0` ` ``# Iterating 4 nested loop to find` ` ``# four variable i j k l` ` ``for` `i ``in` `range``(``1``, N``+``1``):` ` ``for` `j ``in` `range``(``1``, N``+``1``):` ` ``for` `k ``in` `range``(``1``, N``+``1``):` ` ``for` `l ``in` `range``(``1``, N``+``1``):` ` ``# If (a*b) + (c*d) == N` ` ``# then increase the` ` ``# counter by 1.` ` ``if` `(i ``*` `j) ``+` `(k ``*` `l) ``=``=` `N:` ` ``count ``+``=` `1` ` ``# Return the total count` ` ``return` `count` `# Driver code` `N ``=` `4` `# Function call` `ans ``=` `countSquads(N)` `print``(ans)`
## C#
`// C# code for the above approach` `using` `System;` `public` `class` `GFG {` ` ``// Function to count number of squads` ` ``static` `int` `countSquads(``int` `N)` ` ``{` ` ``int` `count = 0;` ` ``int` `i, j, k, l;` ` ``// Iterating 4 nested loop to find four variable i j` ` ``// k l` ` ``for` `(i = 1; i <= N; i++) {` ` ``for` `(j = 1; j <= N; j++) {` ` ``for` `(k = 1; k <= N; k++) {` ` ``for` `(l = 1; l <= N; l++) {` ` ``// If (a*b) + (c*d) == N then` ` ``// increase the counter by 1.` ` ``if` `((i * j) + (k * l) == N) {` ` ``count++;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// Return the total count` ` ``return` `count;` ` ``}` ` ``static` `public` `void` `Main()` ` ``{` ` ``// Code` ` ``int` `N = 4;` ` ``// Function call` ` ``int` `ans = countSquads(N);` ` ``Console.WriteLine(ans);` ` ``}` `}` `// This code is contributed by karthik.`
## Javascript
`// Function to count number of squads` `function` `countSquads(N) {` ` ``let count = 0;` ` ``// Iterating 4 nested loops to find` ` ``// four variables i, j, k, l` ` ``for` `(let i = 1; i <= N; i++) {` ` ``for` `(let j = 1; j <= N; j++) {` ` ``for` `(let k = 1; k <= N; k++) {` ` ``for` `(let l = 1; l <= N; l++) {` ` ``// If (a*b) + (c*d) == N` ` ``// then increase the counter by 1.` ` ``if` `((i * j) + (k * l) === N) {` ` ``count++;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// Return the total count` ` ``return` `count;` `}` `// Driver code` `const N = 4;` `// Function call` `const ans = countSquads(N);` `console.log(ans);`
Output
`8`
Time Complexity: O(N4)
Auxilairy Space: O(1)
Efficient Approach: To solve the problem follow the below idea:
We can write the given number in the form X + Y and we have to find how many ways we express X and Y in the form of U×V. so we can find the total number of (X Y) combination by traversing from 1 to the number and we can find I and i-n. and in each operation, we have to find (U, V) for X and Y .for that we can find all divisors of the number. And it can be proved that number of divisors is equal to the number of (|U, v) combinations.So, we can find the C1 combination for X and C2 combination for Y . And in total we can get C1 * C2 combination in each iteration.
Steps involved in the implementation of code:
• Iterate a loop from 1 to the number N.
• Find all the combinations (A, B) such that A + B = N.
• Find two numbers A and B where A = i and B = N-i.
• By the countDivisor() function we are counting all the divisors of A and B.
• And by multiplying these we can get all possible combinations.
Below is the implementation of the above approach:
## C++
`// C++ code for the above approach` `#include ` `using` `namespace` `std;` `// Function for counting all divisors` `int` `countDivisors(``int` `n)` `{` ` ``int` `cnt = 0;` ` ``for` `(``int` `i = 1; i <= ``sqrt``(n); i++) {` ` ``if` `(n % i == 0) {` ` ``if` `(n / i == i)` ` ``cnt++;` ` ``else` ` ``cnt = cnt + 2;` ` ``}` ` ``}` ` ``// Return count` ` ``return` `cnt;` `}` `// Driver code` `int` `main()` `{` ` ``int` `n = 4;` ` ``int` `count = 0;` ` ``for` `(``int` `i = 1; i < n; i++) {` ` ``int` `X = i;` ` ``int` `Y = n - i;` ` ``// Function call` ` ``int` `a1 = countDivisors(Y);` ` ``// a1 is permutaion of (u, v) for X;` ` ``int` `a2 = countDivisors(X);` ` ``// a2 is permutaion of (u, v) for Y;` ` ``count += (a1 * a2);` ` ``// Total permutation for (X, Y)` ` ``}` ` ``cout << count << endl;` ` ``return` `0;` `}`
## Python3
`import` `math` `# Function for counting all divisors` `def` `countDivisors(n):` ` ``cnt ``=` `0` ` ``for` `i ``in` `range``(``1``, ``int``(math.sqrt(n)) ``+` `1``):` ` ``if` `n ``%` `i ``=``=` `0``:` ` ``if` `n ``/``/` `i ``=``=` `i:` ` ``cnt ``+``=` `1` ` ``else``:` ` ``cnt ``+``=` `2` ` ``# Return count` ` ``return` `cnt` `# Driver code` `def` `main():` ` ``n ``=` `4` ` ``count ``=` `0` ` ``for` `i ``in` `range``(``1``, n):` ` ``X ``=` `i` ` ``Y ``=` `n ``-` `i` ` ``# Function call` ` ``a1 ``=` `countDivisors(Y)` ` ``# a1 is permutation of (u, v) for X` ` ``a2 ``=` `countDivisors(X)` ` ``# a2 is permutation of (u, v) for Y` ` ``count ``+``=` `(a1 ``*` `a2)` ` ``# Total permutation for (X, Y)` ` ``print``(count)` `if` `__name__ ``=``=` `"__main__"``:` ` ``main()`
## Javascript
`// Function for counting all divisors` `function` `countDivisors(n) {` ` ``let cnt = 0;` ` ``for` `(let i = 1; i <= Math.sqrt(n); i++) {` ` ``if` `(n % i === 0) {` ` ``if` `(n / i === i) {` ` ``cnt++;` ` ``} ``else` `{` ` ``cnt += 2;` ` ``}` ` ``}` ` ``}` ` ``// Return count` ` ``return` `cnt;` `}` `// Driver code` `function` `main() {` ` ``let n = 4;` ` ``let count = 0;` ` ``for` `(let i = 1; i < n; i++) {` ` ``let X = i;` ` ``let Y = n - i;` ` ``// Function call` ` ``let a1 = countDivisors(Y);` ` ``// a1 is permutaion of (u, v) for X;` ` ``let a2 = countDivisors(X);` ` ``// a2 is permutaion of (u, v) for Y;` ` ``count += a1 * a2;` ` ``// Total permutation for (X, Y)` ` ``}` ` ``console.log(count);` `}` `main();` `// akashish__`
Output
`8`
Time Complexity: O(N3/2)
Auxiliary Space: O(1)
My Personal Notes arrow_drop_up | 3,735 | 10,073 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-23 | latest | en | 0.618902 |
https://curriculum.illustrativemathematics.org/HS/teachers/3/corrections.html | 1,726,287,616,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651548.18/warc/CC-MAIN-20240914025441-20240914055441-00696.warc.gz | 158,062,461 | 29,434 | ## Corrections
Early printings of the course guide did not include sample responses for each modeling prompt. To access these, visit the modeling prompt pages online (link).
Course Guide. In the Scope and Sequence, Unit 5 contains 13 days and no optional lessons. The total number of days in Algebra 2 is 124.
In the Course Guide, under Scope and Sequence, the Pacing Guide for Algebra 2 Unit 3 was edited to remove lesson 13 from the list of optional lessons.
Unit 1, Lesson 1, Practice Problem 1. The sample solution to the first question should use 20 instead of 25.
Unit 1, Lesson 5, Lesson Synthesis. The indexing for each of the explicit formulas are now corrected to use $$n-1$$ instead of $$n$$. For example, the first geometric sequence is now $$f(n) = 2 \boldcdot 3^{n-1}$$ instead of $$f(n) = 2 \boldcdot 3^n$$.
Unit 1, Lesson 6, Practice Problem 3. Choice 4 is updated to $$d(n) = d(n-1)+n$$ because the lesson uses $$n=0$$ to begin the sequence.
Unit 1, Lesson 9, Activity 3. In the solution for part 3 of the Are You Ready for More?, the fourth value is 1.61803406.
Unit 1, Lesson 10, Activity 2. The solution for the last equation of the second question is updated to apply for $$n \geq 0$$.
Unit 1, Lesson 11, Practice Problem 5. Updated solution to use $$T(n)$$.
Unit 1, End of Unit Assessment, Item 1. The solutions each gave a recursive formula for $$n \geq 1$$, but should refer to $$n \geq 2$$
Unit 1, End of Unit Assessment, Item 2. The recursive definition for $$f$$ in the statement should use $$n \ge 2$$.
Unit 2, Lesson 2, Cool-down. The last question is corrected to begin, "After the 4th year, \$200 is added to the account." Unit 2, Lesson 8, Activity 3 Synthesis. In the second list item, change "Only changing the exponent to something 6 or higher would. . ." to "Changing the exponent to 6 would . . ." Unit 2, Lesson 8, Lesson Summary. The value in the table for the row labeled -10 and the column $$\text{-}20x^2$$ should be negative. Unit 2, Lesson 8, Practice Problem 5. In the solution for b, the constant term is 12. Unit 2, Lesson 10, Activity 2. In the solution for 1, C is "...larger in the negative direction." and D is "...larger in the positive direction." Unit 2, Lesson 13, Activity 4. For question 4, the cubic in standard form is $$\boxed{\phantom{30}}x^3 + 11x^2 - 17x + 6$$ Unit 2, Lesson 14, Practice Problem 6. Change the $$\text-15x$$ inside the table to $$\text-5x$$, and in the answer, change $$\text-3x$$ to $$7x$$. Unit 2, Lesson 16, Activity 2. In the solution to 1, the height associated with radius 8 is 2.2. Unit 2, Lesson 18, Practice Problem 2. The solution uses the line $$y = \text{-}3$$. Unit 2, Lesson 19. Change $$p(x)$$ to $$q(x)$$ throughout the lesson. Specifically: • activity 2: narrative (2 places), task statement (2 places) • activity 3: narrative (1 place), task statement (1 place), student response (3 places), synthesis (3 places) • cool-down: task statement (1 place) Unit 2, Lesson 19, Warm-up. The statement used the value 2772 instead of 2775. Unit 2, Lesson 19, Practice Problem 4. Added to the problem statement: "(Note: Some of the answer choices are not used and some answer choices are used more than once.)" Unit 2, Lesson 19, Practice Problem 5. The statement is updated to use $$f(\text{-}3)=0$$ and $$f(1) = 0$$. In the solution, the last term is $$(x+3)$$. Unit 2, Lesson 19, Practice Problem 8. Updated one of the choices to "The value of the expression is 99." instead of getting closer and closer to the value. Unit 2, Lesson 21, Activity 3. The solution to 3 is about 23 ohms. Unit 2, Lesson 22, Warm-up. The statement incorrectly had a 1 in the numerator instead of a 3. The solution suggestions also mention $$x(x+2)$$ instead of $$x(x-2)$$. These have been corrected. Unit 2, Lesson 23, Practice Problem 2. In the solution, change $$x^2+2x+1$$ to $$2x^2+4x+2$$. Unit 2, Lesson 24, Practice Problem 6. Choice 4 is changed to $$2(3x^2 + 6x + 4)$$. Unit 3. Learning goals and learning targets updated from pilot versions. Unit 3, Lesson 8, Activity 4. In the solution for question 2, $$m = 8$$. Unit 3, Lesson 3, Activity 3. The image in the launch is updated to show +1. Unit 3, Lesson 13, Activity 3. The solutions to parts 2 and 3 of Are You Ready for More? are updated to $$\text-6+i-9j+8k$$ and $$\text-6+7i+9j+4k$$, respectively. Unit 3, Lesson 14, Activity 2. The Are You Ready for More? solution to 1f is updated to $$\text{-}4 + \text{-}4i$$ Unit 3, Lesson 18, Activity 3. Row 3 for partner B is updated to $$2z^2+6z=\text{-}19$$ so that the solutions match. Unit 3, Lesson 19, Activity 1. For the first function, $$f(x)=0$$ when $$x = 0, \text{-}2$$ Unit 3, Lesson 19, Practice Problem 2. Updated the image to label the axes correctly. Unit 4, Lesson 2, Cool down. The sample response for question 2 is corrected to correctly calculate the value as 182.25. Unit 4, Lesson 3, Practice Problem 1. C and D are included as correct answers. Unit 4, Lesson 6, Activity 2. Data Card 1 is updated to "The rate of increase from 0 to 1.5 is 337.5%." Unit 4, Lesson 8, Practice Problem 3. Corrected the solution from -5 to -4 as the first entry in the top row. Unit 4, Lesson 8, Practice Problem 7. The solution to the second question should refer to 0.09375 picograms instead of 0.9375 picograms. Unit 4, Lesson 10, Activity 1. For the explanation in the student response, the second bullet point had the 1 and 0 reversed. This is fixed. Unit 4, Lesson 14, Practice Problem 2. The solution is $$d = \text{-}2$$ instead of $$d = 0$$. Unit 4, Lesson 18, Practice Problem 4. The horizontal line $$y = 1,\!000$$ should be used instead of $$y = 100$$. Unit 4, Mid-Unit Assessment, Item 7. The solution for the last part of the question used the incorrect initial value for the equation. This has been corrected. Unit 4, End-of-Unit Assessment, Item 7. The equation should be $$A(d) = 100 \boldcdot e^{0.25d}$$. The question and solutions are corrected. Unit 5, Lesson 1, Lesson Summary. Corrected 32 to 36 in the sentence, "What did multiplying by 45 and adding 36 do to the graph?" Unit 5, Lesson 2, Practice Problem 2. Updated solutions for parts b and c to $$B(30)= 375$$ and $$B(t) = P(t)+25$$ respectively. Unit 5, Lesson 2, Practice Problem 3. Updated solution for part b to indicate that the graph of $$h$$ is shifted to the left. Unit 5, Lesson 2, Practice Problem 5. Updated the problem statements to better indicate which line is being translated (from given line to the line containing the points). Updated solution to part a to reflect this change (left instead of right). Unit 5, Lesson 3, Practice Problem 3. The correct point for Han is $$(1,85)$$. Unit 5, Lesson 5, Activity 3. The blackline master is updated so that card N has 2 in the $$y$$ column to match the graph. Unit 5, Lesson 5, Practice Problem 7. Adjusted the graph back an hour to begin at $$(0.5, 80)$$ so that the translation fits on the given axes. Unit 5, Lesson 7, Activity 1. The statement (and solution) had the last part of the table listed in the wrong order. This has been corrected to ask for translations first, then reflections. Unit 5, Lesson 7, Lesson Synthesis. The order of the transformations for the third function has been corrected to do translations before reflection. Unit 5, Lesson 7, Practice Problem 2. The first question should have $$g(x) = -e^x + 2.7$$ Unit 5, Lesson 7, Practice Problem 3. Correct answers should open down with a negative leading coefficient. A possible response is $$y = \text{-}(x-2)^2 - 3$$. Unit 5, Lesson 8, Practice Problem 5. Changed $$f(0)$$ to 0 in the table. Unit 5, Lesson 9, Practice Problem 1. The solution to b should be $$k = 0.625$$ and the solution to d should be $$k = \text-\frac{1}{2}$$. Unit 5, Lesson 10, Practice Problem 2. The solution is corrected to $$x = 0$$ and $$x = \text{-}2$$. Unit 5, Lesson 10, Practice Problem 10. The solution is corrected so that $$r(t)$$ starts at 212. Unit 5, Lesson 11, Practice Problem 6. The graph is shifted right first, then compressed. The graph is compressed horizontally by a factor of $$\frac{1}{3}$$. Unit 6, Lesson 1, Practice Problem 5. Option d is updated to $$k(x) = \frac{3x^2 - 16x + 12}{x-6}$$ and choice 6 is updated to, "The graph approaches $$y = 3x+2$$." Unit 6, Lesson 2, Practice Problem 1. Updated choices C and D. C: $$\sin(C) = \frac{6}{10}$$ and D: $$\cos(C) = \frac{8}{10}$$. Unit 6, Lesson 3, Practice Problem 7. Exchanged the names of side lengths $$d$$ and $$e$$ so that $$d$$ is across from angle $$D$$ and $$e$$ is across from angle $$E$$. Unit 6, Lesson 4, Practice Problem 5. Added "$$BC$$ is shorter than $$AC$$" to the prompt. Unit 6, Lesson 7, Practice Problem 2. The solutions should use $$100+85\sin(\theta)$$ for the height. This also updates the estimates. Unit 6, Lesson 9, Cool-down. The $$x$$-axis is corrected to show $$\frac{5\pi}{6}$$. Unit 6, Lesson 11, Practice Problem 5. The solution to part d now correctly has a 2 in the denominator instead of a 4. Unit 6, Lesson 12, Practice Problem 6. The second and third questions are updated to 12 and 24 hours respectively. Unit 6, Lesson 12, Practice Problem 7. The solution to c is corrected to, "...translated 6 units to the left." Unit 6, Lesson 17, Activity 1. In the activity synthesis, corrected the first question to, "...and then translate it left 1 and down 3..." Unit 6, Lesson 19, Practice Problem 2. Added D as a correct response. Unit 6, Lesson 19, Practice Problem 5. The solution points are S and T instead of U and V. Unit 7, Lesson 1, Practice Problem 6. Added D as a correct response. Unit 7, Lesson 3, Activity 3. The sample solutions are updated. 2a is 14.6 square meters and 2d is 8.2 square meters. Unit 7, Lesson 6, Lesson Synthesis. The 3rd bullet should say that 4.5 is the mean. Unit 7, Lesson 10, Learning Target. Corrected to "I know that a smaller margin of error means less variability, ..." Unit 7, Lesson 13, Practice Problem 4. Removed the instructions to round from the question. Unit 7, Lesson 14, Practice Problem 3. Question b corrected to ask about more extreme than 2.5. Solution to b updated to 0.0139 and solution to c updated to match these changes. Unit 7, Lesson 14, Practice Problem 4. Updated part b to read, "...difference at least as great as the difference in means between the control and treatment groups?" Added additional tick marks on the $$y$$-axis of the image. Updated solution to part b to $$\frac{46}{100}$$. Unit 7, Lesson 14, Practice Problem 6. Corrected mention of "mean lengths" to "mean weights." Unit 7, Lesson 14, Activity 3. Solutions are updated to use the correct area of 0.0084 for questions 2 through 4. Unit 7, Lesson 15, Practice Problem 3. The solution to part e is updated to 0.005. Unit 7, End of Unit Assessment, Problem 6. Corrected the Tier 1 response to show the correct decimal places in the multiplication. ## Lesson Numbering for Learning Targets In some printed copies of the student workbooks, we erroneously printed a lesson number instead of the unit and lesson number. This table provides a key to match the printed lesson number with the unit and lesson number. Lesson Number Unit and Lesson Lesson Title 1 1.1 A Towering Sequence 2 1.2 Introducing Geometric Sequences 3 1.3 Different Types of Sequences 4 1.4 Using Technology to Work with Sequences 5 1.5 Sequences are Functions 6 1.6 Representing Sequences 7 1.7 Representing More Sequences 8 1.8 The$n^{\text{th}}$Term 9 1.9 What’s the Equation? 10 1.10 Situations and Sequence Types 11 1.11 Adding Up 12 2.1 Let’s Make a Box 13 2.2 Funding the Future 14 2.3 Introducing Polynomials 15 2.4 Combining Polynomials 16 2.5 Connecting Factors and Zeros 17 2.6 Different Forms 18 2.7 Using Factors and Zeros 19 2.8 End Behavior (Part 1) 20 2.9 End Behavior (Part 2) 21 2.10 Multiplicity 22 2.11 Finding Intersections 23 2.12 Polynomial Division (Part 1) 24 2.13 Polynomial Division (Part 2) 25 2.14 What Do You Know About Polynomials? 26 2.15 The Remainder Theorem 27 2.16 Minimizing Surface Area 28 2.17 Graphs of Rational Functions (Part 1) 29 2.18 Graphs of Rational Functions (Part 2) 30 2.19 End Behavior of Rational Functions 31 2.20 Rational Equations (Part 1) 32 2.21 Rational Equations (Part 2) 33 2.22 Solving Rational Equations 34 2.23 Polynomial Identities (Part 1) 35 2.24 Polynomial Identities (Part 2) 36 2.25 Summing Up 37 2.26 Using the Sum 38 3.1 Properties of Exponents 39 3.2 Square Roots and Cube Roots 40 3.3 Exponents That Are Unit Fractions 41 3.4 Positive Rational Exponents 42 3.5 Negative Rational Exponents 43 3.6 Squares and Square Roots 44 3.7 Inequivalent Equations 45 3.8 Cubes and Cube Roots 46 3.9 Solving Radical Equations 47 3.10 A New Kind of Number 48 3.11 Introducing the Number$i$49 3.12 Arithmetic with Complex Numbers 50 3.13 Multiplying Complex Numbers 51 3.14 More Arithmetic with Complex Numbers 52 3.15 Working Backwards 53 3.16 Solving Quadratics 54 3.17 Completing the Square and Complex Solutions 55 3.18 The Quadratic Formula and Complex Solutions 56 3.19 Real and Non-Real Solutions 57 4.1 Growing and Shrinking 58 4.2 Representations of Growth and Decay 59 4.3 Understanding Rational Inputs 60 4.4 Representing Functions at Rational Inputs 61 4.5 Changes Over Rational Intervals 62 4.6 Writing Equations for Exponential Functions 63 4.7 Interpreting and Using Exponential Functions 64 4.8 Unknown Exponents 65 4.9 What is a Logarithm? 66 4.10 Interpreting and Writing Logarithmic Equations 67 4.11 Evaluating Logarithmic Expressions 68 4.12 The Number$e$69 4.13 Exponential Functions with Base$e$70 4.14 Solving Exponential Equations 71 4.15 Using Graphs and Logarithms to Solve Problems (Part 1) 72 4.16 Using Graphs and Logarithms to Solve Problems (Part 2) 73 4.17 Logarithmic Functions 74 4.18 Applications of Logarithmic Functions 75 5.1 Matching up to Data 76 5.2 Moving Functions 77 5.3 More Movement 78 5.4 Reflecting Functions 79 5.5 Some Functions Have Symmetry 80 5.6 Symmetry in Equations 81 5.7 Expressing Transformations of Functions Algebraically 82 5.8 Scaling the Outputs 83 5.9 Scaling the Inputs 84 5.10 Combining Functions 85 5.11 Making a Model for Data 86 6.1 Moving in Circles 87 6.2 Revisiting Right Triangles 88 6.3 The Unit Circle (Part 1) 89 6.4 The Unit Circle (Part 2) 90 6.5 The Pythagorean Identity (Part 1) 91 6.6 The Pythagorean Identity (Part 2) 92 6.7 Finding Unknown Coordinates on a Circle 93 6.8 Rising and Falling 94 6.9 Introduction to Trigonometric Functions 95 6.10 Beyond$2\pi\$
96 6.11 Extending the Domain of Trigonometric Functions
97 6.12 Tangent
98 6.13 Amplitude and Midline
99 6.14 Transforming Trigonometric Functions
100 6.15 Features of Trigonometric Graphs (Part 1)
101 6.16 Features of Trigonometric Graphs (Part 2)
102 6.17 Comparing Transformations
103 6.18 Modeling Circular Motion
104 6.19 Beyond Circles
105 7.1 Being Skeptical
106 7.2 Study Types
107 7.3 Randomness in Groups
108 7.4 Describing Distributions
109 7.5 Normal Distributions
110 7.6 Areas in Histograms
111 7.7 Areas under a Normal Curve
112 7.8 Not Always Ideal
113 7.9 Variability in Samples
114 7.10 Estimating Proportions from Samples
115 7.11 Reducing Margin of Error
116 7.12 Estimating a Population Mean
117 7.13 Experimenting
118 7.14 Using Normal Distributions for Experiment Analysis
119 7.15 Questioning Experimenting
120 7.16 Heart Rates | 4,627 | 15,343 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-38 | latest | en | 0.830947 |
null | null | null | null | null | null | apples walking checkup
For urgent matters, please send an email to
Glomerular Diseases
On this page:
Many diseases affect kidney function by attacking the glomeruli, the tiny units within the kidney where blood is cleaned. Glomerular diseases include many conditions with a variety of genetic and environmental causes, but they fall into two major categories:
• Glomerulonephritis (gloh-MEHR-yoo-loh-nef-RY-tis) describes the inflammation of the membrane tissue in the kidney that serves as a filter, separating wastes and extra fluid from the blood.
• Glomerulosclerosis (gloh-MEHR-yoo-loh-skleh-ROH-sis) describes the scarring or hardening of the tiny blood vessels within the kidney.
Although glomerulonephritis and glomerulosclerosis have different causes, they can both lead to kidney failure.
What are the kidneys and what do they do?
The two kidneys are bean-shaped organs located just below the rib cage, one on each side of the spine. Everyday, the two kidneys filter about 120 to 150 quarts of blood to produce about 1 to 2 quarts of urine, composed of wastes and extra fluid.
Blood enters the kidneys through arteries that branch inside the kidneys into tiny clusters of looping blood vessels. Each cluster is called a glomerulus, which comes from the Greek word meaning filter. The plural form of the word is glomeruli. There are approximately 1 million glomeruli, or filters, in each kidney. The glomerulus is attached to the opening of a small fluid-collecting tube called a tubule. Blood is filtered in the glomerulus, and extra fluid and wastes pass into the tubule and become urine. Eventually, the urine drains from the kidneys into the bladder through larger tubes called ureters.
Drawing of a kidney. Labels show where blood with wastes enters the kidney, clean blood leaves the kidney, and wastes-urine-are sent to the bladder. An inset shows a microscopic view of a nephron. Labels point to the glomerulus and the tubule.
In the nephron (left), tiny blood vessels intertwine with fluid-collecting tubes. Each kidney contains about 1 million nephrons.
Each glomerulus-and-tubule unit is called a nephron. Each kidney is composed of about 1 million nephrons. In healthy nephrons, the glomerular membrane that separates the blood vessel from the tubule allows waste products and extra water to pass into the tubule while keeping blood cells and protein in the bloodstream.
How do glomerular diseases interfere with kidney function?
Glomerular diseases damage the glomeruli, letting protein and sometimes red blood cells leak into the urine. Sometimes a glomerular disease also interferes with the clearance of waste products by the kidney, so they begin to build up in the blood. Furthermore, loss of blood proteins like albumin in the urine can result in a fall in their level in the bloodstream. In normal blood, albumin acts like a sponge, drawing extra fluid from the body into the bloodstream, where it remains until the kidneys remove it. But when albumin leaks into the urine, the blood loses its capacity to absorb extra fluid from the body. Fluid can accumulate outside the circulatory system in the face, hands, feet, or ankles and cause swelling.
What are the symptoms of glomerular disease?
The signs and symptoms of glomerular disease include
• albuminuria: large amounts of protein in the urine
• hematuria: blood in the urine
• reduced glomerular filtration rate: inefficient filtering of wastes from the blood
• hypoproteinemia: low blood protein
• edema: swelling in parts of the body
• Proteinuria may cause foamy urine.
How is glomerular disease diagnosed?
Patients with glomerular disease have significant amounts of protein in the urine, which may be referred to as "nephrotic range" if levels are very high. Red blood cells in the urine are a frequent finding as well, particularly in some forms of glomerular disease. Urinalysis provides information about kidney damage by indicating levels of protein and red blood cells in the urine. Blood tests measure the levels of waste products such as creatinine and urea nitrogen to determine whether the filtering capacity of the kidneys is impaired. If these lab tests indicate kidney damage, the doctor may recommend ultrasound or an x ray to see whether the shape or size of the kidneys is abnormal. These tests are called renal imaging. But since glomerular disease causes problems at the cellular level, the doctor will probably also recommend a kidney biopsy—a procedure in which a needle is used to extract small pieces of tissue for examination with different types of microscopes, each of which shows a different aspect of the tissue. A biopsy may be helpful in confirming glomerular disease and identifying the cause.
What causes glomerular disease?
A number of different diseases can result in glomerular disease. It may be the direct result of an infection or a drug toxic to the kidneys, or it may result from a disease that affects the entire body, like diabetes or lupus. Many different kinds of diseases can cause swelling or scarring of the nephron or glomerulus. Sometimes glomerular disease is idiopathic, meaning that it occurs without an apparent associated disease.
The categories presented below can overlap: that is, a disease might belong to two or more of the categories. For example, diabetic nephropathy is a form of glomerular disease that can be placed in two categories: systemic diseases, since diabetes itself is a systemic disease, and sclerotic diseases, because the specific damage done to the kidneys is associated with scarring.
Autoimmune Diseases
When the body's immune system functions properly, it creates protein-like substances called antibodies and immunoglobulins to protect the body against invading organisms. In an autoimmune disease, the immune system creates autoantibodies, which are antibodies or immunoglobulins that attack the body itself. Autoimmune diseases may be systemic and affect many parts of the body, or they may affect only specific organs or regions.
Systemic lupus erythematosus (SLE) affects many parts of the body: primarily the skin and joints, but also the kidneys. Because women are more likely to develop SLE than men, some researchers believe that a sex-linked genetic factor may play a part in making a person susceptible, although viral infection has also been implicated as a triggering factor. Lupus nephritis is the name given to the kidney disease caused by SLE, and it occurs when autoantibodies form or are deposited in the glomeruli, causing inflammation. Ultimately, the inflammation may create scars that keep the kidneys from functioning properly. Conventional treatment for lupus nephritis includes a combination of two drugs, cyclophosphamide, a cytotoxic agent that suppresses the immune system, and prednisolone, a corticosteroid used to reduce inflammation. A newer immunosuppressant, mychophenolate mofetil (MMF), has been used instead of cyclophosphamide. Preliminary studies indicate that MMF may be as effective as cyclophosphamide and has milder side effects.
Goodpasture's Syndrome involves an autoantibody that specifically targets the kidneys and the lungs. Often, the first indication that patients have the autoantibody is when they cough up blood. But lung damage in Goodpasture's Syndrome is usually superficial compared with progressive and permanent damage to the kidneys. Goodpasture's Syndrome is a rare condition that affects mostly young men but also occurs in women, children, and older adults. Treatments include immunosuppressive drugs and a blood-cleaning therapy called plasmapheresis that removes the autoantibodies.
IgA nephropathy is a form of glomerular disease that results when immunoglobulin A (IgA) forms deposits in the glomeruli, where it creates inflammation. IgA nephropathy was not recognized as a cause of glomerular disease until the late 1960s, when sophisticated biopsy techniques were developed that could identify IgA deposits in kidney tissue.
The most common symptom of IgA nephropathy is blood in the urine, but it is often a silent disease that may go undetected for many years. The silent nature of the disease makes it difficult to determine how many people are in the early stages of IgA nephropathy, when specific medical tests are the only way to detect it. This disease is estimated to be the most common cause of primary glomerulonephritis—that is, glomerular disease not caused by a systemic disease like lupus or diabetes mellitus. It appears to affect men more than women. Although IgA nephropathy is found in all age groups, young people rarely display signs of kidney failure because the disease usually takes several years to progress to the stage where it causes detectable complications.
No treatment is recommended for early or mild cases of IgA nephropathy when the patient has normal blood pressure and less than 1 gram of protein in a 24-hour urine output. When proteinuria exceeds 1 gram/day, treatment is aimed at protecting kidney function by reducing proteinuria and controlling blood pressure. Blood pressure medicines—angiotensin—converting enzyme inhibitors (ACE inhibitors) or angiotensin receptor blockers (ARBs)—that block a hormone called angiotensin are most effective at achieving those two goals simultaneously.
Hereditary Nephritis—Alport Syndrome
The primary indicator of Alport syndrome is a family history of chronic glomerular disease, although it may also involve hearing or vision impairment. This syndrome affects both men and women, but men are more likely to experience chronic kidney disease and sensory loss. Men with Alport syndrome usually first show evidence of renal insufficiency while in their twenties and reach total kidney failure by age 40. Women rarely have significant renal impairment, and hearing loss may be so slight that it can be detected only through testing with special equipment. Usually men can pass the disease only to their daughters. Women can transmit the disease to either their sons or their daughters. Treatment focuses on controlling blood pressure to maintain kidney function.
Infection-related Glomerular Disease
Glomerular disease sometimes develops rapidly after an infection in other parts of the body. Acute post-streptococcal glomerulonephritis (PSGN) can occur after an episode of strep throat or, in rare cases, impetigo (a skin infection). The Streptococcus bacteria do not attack the kidney directly, but an infection may stimulate the immune system to overproduce antibodies, which are circulated in the blood and finally deposited in the glomeruli, causing damage. PSGN can bring on sudden symptoms of swelling (edema), reduced urine output (oliguria), and blood in the urine (hematuria). Tests will show large amounts of protein in the urine and elevated levels of creatinine and urea nitrogen in the blood, thus indicating reduced kidney function. High blood pressure frequently accompanies reduced kidney function in this disease.
PSGN is most common in children between the ages of 3 and 7, although it can strike at any age, and it most often affects boys. It lasts only a brief time and usually allows the kidneys to recover. In a few cases, however, kidney damage may be permanent, requiring dialysis or transplantation to replace renal function.
Bacterial endocarditis, infection of the tissues inside the heart, is also associated with subsequent glomerular disease. Researchers are not sure whether the renal lesions that form after a heart infection are caused entirely by the immune response or whether some other disease mechanism contributes to kidney damage. Treating the heart infection is the most effective way of minimizing kidney damage. Endocarditis sometimes produces chronic kidney disease (CKD).
HIV, the virus that leads to AIDS, can also cause glomerular disease. Between 5 and 10 percent of people with HIV experience kidney failure, even before developing full-blown AIDS. HIV-associated nephropathy usually begins with heavy proteinuria and progresses rapidly (within a year of detection) to total kidney failure. Researchers are looking for therapies that can slow down or reverse this rapid deterioration of renal function, but some possible solutions involving immunosuppression are risky because of the patients' already compromised immune system.
Sclerotic Diseases
Glomerulosclerosis is scarring (sclerosis) of the glomeruli. In several sclerotic conditions, a systemic disease like lupus or diabetes is responsible. Glomerulosclerosis is caused by the activation of glomerular cells to produce scar material. This may be stimulated by molecules called growth factors, which may be made by glomerular cells themselves or may be brought to the glomerulus by the circulating blood that enters the glomerular filter.
Diabetic nephropathy is the leading cause of glomerular disease and of total kidney failure in the United States. Kidney disease is one of several problems caused by elevated levels of blood glucose, the central feature of diabetes. In addition to scarring the kidney, elevated glucose levels appear to increase the speed of blood flow into the kidney, putting a strain on the filtering glomeruli and raising blood pressure.
Diabetic nephropathy usually takes many years to develop. People with diabetes can slow down damage to their kidneys by controlling their blood glucose through healthy eating with moderate protein intake, physical activity, and medications. People with diabetes should also be careful to keep their blood pressure at a level below 140/90 mm Hg, if possible. Blood pressure medications called ACE inhibitors and ARBs are particularly effective at minimizing kidney damage and are now frequently prescribed to control blood pressure in patients with diabetes and in patients with many forms of kidney disease.
Focal segmental glomerulosclerosis (FSGS) describes scarring in scattered regions of the kidney, typically limited to one part of the glomerulus and to a minority of glomeruli in the affected region. FSGS may result from a systemic disorder or it may develop as an idiopathic kidney disease, without a known cause. Proteinuria is the most common symptom of FSGS, but, since proteinuria is associated with several other kidney conditions, the doctor cannot diagnose FSGS on the basis of proteinuria alone. Biopsy may confirm the presence of glomerular scarring if the tissue is taken from the affected section of the kidney. But finding the affected section is a matter of chance, especially early in the disease process, when lesions may be scattered.
Confirming a diagnosis of FSGS may require repeat kidney biopsies. Arriving at a diagnosis of idiopathic FSGS requires the identification of focal scarring and the elimination of possible systemic causes such as diabetes or an immune response to infection. Since idiopathic FSGS is, by definition, of unknown cause, it is difficult to treat. No universal remedy has been found, and most patients with FSGS progress to total kidney failure over 5 to 20 years. Some patients with an aggressive form of FSGS reach total kidney failure in 2 to 3 years. Treatments involving steroids or other immunosuppressive drugs appear to help some patients by decreasing proteinuria and improving kidney function. But these treatments are beneficial to only a minority of those in whom they are tried, and some patients experience even poorer kidney function as a result. ACE inhibitors and ARBs may also be used in FSGS to decrease proteinuria. Treatment should focus on controlling blood pressure and blood cholesterol levels, factors that may contribute to kidney scarring.
Other Glomerular Diseases
Membranous nephropathy, also called membranous glomerulopathy, is the second most common cause of the nephrotic syndrome (proteinuria, edema, high cholesterol) in U.S. adults after diabetic nephropathy. Diagnosis of membranous nephropathy requires a kidney biopsy, which reveals unusual deposits of immunoglobulin G and complement C3, substances created by the body's immune system. Fully 75 percent of cases are idiopathic, which means that the cause of the disease is unknown. The remaining 25 percent of cases are the result of other diseases like systemic lupus erythematosus, hepatitis B or C infection, or some forms of cancer. Drug therapies involving penicillamine, gold, or captopril have also been associated with membranous nephropathy. About 20 to 40 percent of patients with membranous nephropathy progress, usually over decades, to total kidney failure, but most patients experience either complete remission or continued symptoms without progressive kidney failure. Doctors disagree about how aggressively to treat this condition, since about 20 percent of patients recover without treatment. ACE inhibitors and ARBs are generally used to reduce proteinuria. Additional medication to control high blood pressure and edema is frequently required. Some patients benefit from steroids, but this treatment does not work for everyone. Additional immunosuppressive medications are helpful for some patients with progressive disease.
Minimal change disease (MCD) is the diagnosis given when a patient has the nephrotic syndrome and the kidney biopsy reveals little or no change to the structure of glomeruli or surrounding tissues when examined by a light microscope. Tiny drops of a fatty substance called a lipid may be present, but no scarring has taken place within the kidney. MCD may occur at any age, but it is most common in childhood. A small percentage of patients with idiopathic nephrotic syndrome do not respond to steroid therapy. For these patients, the doctor may recommend a low-sodium diet and prescribe a diuretic to control edema. The doctor may recommend the use of nonsteroidal anti-inflammatory drugs to reduce proteinuria. ACE inhibitors and ARBs have also been used to reduce proteinuria in patients with steroid-resistant MCD. These patients may respond to larger doses of steroids, more prolonged use of steroids, or steroids in combination with immunosuppressant drugs, such as chlorambucil, cyclophosphamide, or cyclosporine.
What are renal failure and end-stage renal disease?
Renal failure is any acute or chronic loss of kidney function and is the term used when some kidney function remains. Total kidney failure, sometimes called end-stage renal disease (ESRD), indicates permanent loss of kidney function. Depending on the form of glomerular disease, renal function may be lost in a matter of days or weeks or may deteriorate slowly and gradually over the course of decades.
Acute Renal Failure
A few forms of glomerular disease cause very rapid deterioration of kidney function. For example, PSGN can cause severe symptoms (hematuria, proteinuria, edema) within 2 to 3 weeks after a sore throat or skin infection develops. The patient may temporarily require dialysis to replace renal function. This rapid loss of kidney function is called acute renal failure (ARF). Although ARF can be life-threatening while it lasts, kidney function usually returns after the cause of the kidney failure has been treated. In many patients, ARF is not associated with any permanent damage. However, some patients may recover from ARF and subsequently develop CKD.
Chronic Kidney Disease
Most forms of glomerular disease develop gradually, often causing no symptoms for many years. CKD is the slow, gradual loss of kidney function. Some forms of CKD can be controlled or slowed down. For example, diabetic nephropathy can be delayed by tightly controlling blood glucose levels and using ACE inhibitors and ARBs to reduce proteinuria and control blood pressure. But CKD cannot be cured. Partial loss of renal function means that some portion of the patient's nephrons have been scarred, and scarred nephrons cannot be repaired. In many cases, CKD leads to total kidney failure.
Total Kidney Failure
To stay alive, a patient with total kidney failure must go on dialysis—hemodialysis or peritoneal dialysis—or receive a new kidney through transplantation. Patients with CKD who are approaching total kidney failure should learn as much about their treatment options as possible so they can make an informed decision when the time comes. With the help of dialysis or transplantation, many people continue to lead full, productive lives after reaching total kidney failure.
Points to Remember
• The kidneys filter waste and extra fluid from the blood.
• The filtering process takes place in the nephron, where microscopic blood vessel filters, called glomeruli, are attached to fluid-collecting tubules.
• A number of different disease processes can damage the glomeruli and thereby cause kidney failure. Glomerulonephritis and glomerulosclerosis are broad terms that include many forms of damage to the glomeruli.
• Some forms of kidney failure can be slowed down, but scarred glomeruli can never be repaired.
• Treatment for the early stages of kidney failure depends on the disease causing the damage.
• Early signs of kidney failure include blood or protein in the urine and swelling in the hands, feet, abdomen, or face. Kidney failure may be silent for many years.
The Nephrotic Syndrome
• The nephrotic syndrome is a condition marked by very high levels of protein in the urine; low levels of protein in the blood; swelling, especially around the eyes, feet, and hands; and high cholesterol.
• The nephrotic syndrome is a set of symptoms, not a disease in itself. It can occur with many diseases, so prevention relies on controlling the diseases that cause it.
• Treatment of the nephrotic syndrome focuses on identifying and treating the underlying cause, if possible, and reducing high cholesterol, blood pressure, and protein in the urine through diet, medication, or both.
• The nephrotic syndrome may go away once the underlying cause, if known, is treated. However, often a kidney disease is the underlying cause and cannot be cured. In these cases, the kidneys may gradually lose their ability to filter wastes and excess water from the blood. If kidney failure occurs, the patient will need to be on dialysis or have a kidney transplant.
Signs and Symptoms of Glomerulonephritis
edema (eh-DEE-muh): Swelling caused by the accumulation of fluid in cells and tissues. In kidney failure, fluid may collect in the feet, hands, abdomen, or face.
hematuria (HEE-muh-TOOR-ee-uh): Blood in the urine. Blood may turn the urine pink or cola-colored.
hypoproteinemia (HY-po-PRO-teen-EE-mee-uh): Reduced levels of protein in the blood.
proteinuria (PRO-tee-NOOR-ee-uh): Large amounts of protein in the urine.
uremia (yoo-REE-mee-uh): Accumulation of urea and other wastes in the blood. These wastes, which become toxic in large amounts, are normally eliminated through urination.
Diseases and Conditions
autoimmune (AW-toh-ih-MYOON) disease: A disease in which the body's own disease-fighting cells attack the body itself.
hypertension (HY-per-TEN-shun): High blood pressure, a condition that can cause kidney damage or be caused by kidney disease.
idiopathic (id-ee-o-PATH-ik) disease: A disease that occurs without a known cause.
nephrotoxic (NEF-ro-TOKS-ik): Damaging to the kidneys.
sclerotic (skleh-ROT-ik) disease: A disease in which tissues become hardened or scarred.
systemic (sis-TEM-ik) disease: A disease that affects multiple parts of the body, often as a result of substances circulating in the blood.
Treatments and Procedures
biopsy (BY-op-see): A procedure in which a needle is used to obtain small pieces of tissue from an organ for examination under different types of microscopes, each of which shows a different aspect of the tissue.
dialysis (dy-AL-ih-sis): A medical treatment that removes wastes and extra fluid from the blood after the kidneys have stopped working.
immunosuppressant (im-YOON-oh-suh-PRESS-unt): A medicine given to block the body's immune system.
plasmapheresis (PLAZ-muh-fer-EE-sis): A medical treatment in which the blood is treated outside the body to remove harmful antibodies, and then returned to the patient.
Kidney Parts and Organic Substances
antibody (AN-tee-BOD-ee): A molecule that protects the body against disease by attacking foreign tissues or organisms. Antibodies are also called immunoglobulins.
antigen (AN-tih-jen): A substance that triggers a response from the body's immune system.
autoantibody (AW-toh-AN-tee-bod-ee): An antibody that attacks the body itself.
creatinine (kree-AT-ih-nin): A waste product in the blood that results from the normal breakdown of muscle. Healthy kidneys filter creatinine from the blood.
glomerulus (gloh-MEHR-yoo-lus): The tiny cluster of looping blood vessels in the nephron, where wastes are filtered from the blood.
lipid (LIP-id): One of several fatty substances used in cells. Excess lipids in the blood may result in harmful deposits in blood vessels.
nephron (NEF-rahn): One of a million tiny filtering units in each kidney. Each nephron is made up of both a glomerulus and a fluid-collecting tubule that processes extra water and wastes.
protein (PRO-teen): A substance found in food and used by the body to grow, repair tissue, and fight disease.
urea (yoo-REE-uh): A waste material found in blood after protein has been broken down. Healthy kidneys remove urea from the blood. Damaged kidneys may allow urea to accumulate in the blood, thus causing uremia.
For More Information
American Association of Kidney Patients
2701 North Rocky Point Drive, Suite 150
Tampa, FL 33607
Fax: 813–636–8122
Internet: leaving site icon
American Kidney Fund
11921 Rockville Pike, Suite 300
Rockville, MD 20852
Phone: 1–800–638–8299
Internet: leaving site icon
Medical Education Institute, Inc.
414 D'Onofrio Drive, Suite 200
Madison, WI 53719
Phone: 1–800–468–7777 or 608–833–8033
Fax: 608–833–8366
Internet: leaving site icon leaving site icon leaving site icon leaving site icon
National Kidney Foundation
30 East 33rd Street
New York, NY 10016–5337
Fax: 212–689–9261
Internet: leaving site icon
The NephCure Foundation
15 Waterloo Avenue
Berwyn, PA 19312
Phone: 1–866–NephCure (1–866–637–4287)
Internet: leaving site icon
National Kidney and Urologic Diseases Information Clearinghouse
3 Information Way
Bethesda, MD 20892–3580
Phone: 1–800–891–5390
TTY: 1–866–569–1162
Fax: 703–738–4929
NIH Publication No. 14–4358
February 2014
Page last updated April 2, 2014
National Kidney and Urologic Diseases Information Clearinghouse
3 Information Way
Bethesda, MD 20892–3580
Phone: 1–800–891–5390
TTY: 1–866–569–1162
Fax: 703–738–4929
NIH...Turning Discovery Into Health ®
| null | null | null | null | null | null | null | null | null |
https://gradeup.co/important-short-tricks-to-solve-compound-interest-questions-i-fdcdba43-e058-11e5-ae9c-5384be3d1625 | 1,624,556,739,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556482.89/warc/CC-MAIN-20210624171713-20210624201713-00082.warc.gz | 265,276,940 | 60,360 | # Important Short Tricks to solve Compound Interest Questions
By Sandeep Baliyan|Updated : April 11th, 2017
Numerical ability section is considered to be one of the toughest subjects of SSC Exams but it can be scored off well if prepared well. Compound Interest is one of the toughest chapter which leaves candidates a bit confused and most of the aspirants leave these questions untouched.
To make the chapter easy for you all, we are providing you all some Important Short Tricks to solve Compound Interest Questions which will surely make the chapter easy for you all.
## Important Short Tricks to solve Compound Interest Questions
Compound Interest :- Sometimes it so happens that the borrower and the lender agree to fix up a certain unit of time, say yearly or half-yearly or quarterly to settle the precious account.
In such cases, the amount after first unit of times becomes the principal for the second unit the amount after second unit becomes the principle for the third unit and so on.
After a specified period, the difference between the amount and the money borrowed is called the Compound Interest (abbreviated as C.I.) for that period.
### Important Facts & Formulas on Compound Interest
Case 1: Let principle = P, time = n years and rate = r% per annum and let A be the total amount at the end of n years, then
Example: Albert invested an amount of Rs.8000 in a fixed deposit scheme for 2 years at compound interest rate 5 p.c.p.a. how much amount will Albert get on maturity of the fixed deposit.
Solution:
Amount = Rs.
= Rs.
Case 2: When compound interest is reckoned half-yearly.
If the annual rate is r% per annum and is to be calculated for n years, then in this case, rate = (n/2%) half-yearly and time = (2n) half-yearly.
Form the above we get
Example: Sam investment Rs.15,000 @ 10% per annum for one year. If the interest is compounded half-yearly, then the amount received by Sam at the end of the year will be.
Solution:
P = Rs. 15000; R = 10% p.a = 5% half-year, T = 1 year = 2 half year
Amount = Rs
= Rs.16537.50
Case 3: When compound interest is reckoned quarterly.
In this case, rate = (r/4%) quarterly and time = (4n) quarter years.
As before,
Example:
Find the compound interest on Rs. 15,625 for 9 months at 16% per annum compounded quarterly.
Solution:
P = Rs. 15625, n= 9 months = 3 quarters, R = 16% p.a. = 4% per quarter.
Amount = Rs.
= Rs.17576
C.I = Rs. (17576 – 15625 ) = Rs. 1951.
Note: The difference between the compound interest and the simple interest over two years is given by
or
Case 4: When interest is compounded annually but time is in fraction, say years.
Amount =
Example:
What is the difference between the compound interest on Rs. 5000 for at 4% per annum compounded yearly and half-yearly?
Solutions:
C.I. when interest is compounded yearly
= Rs.
= Rs.5304
C.I. when interest is compounded half-yearly
Difference = Rs.(5306.04 – 5304 ) = Rs.2.04.
Case 5: Present worth of Rs.x due n years hence is given by:
Present Worth =
Example:
The principle that amounts to Rs.4913 in 3 years at per annum compound interest compounded annually, is :
Solution:
Principle = Rs.
We hope you all like the post. Don't forget to share your feedback with us in the comment box.
More from Us:
GK Digest: Physics Capsule for SSC Exams
Advanced Maths: Geometry Digest for SSC Exams
GK Digest: Chemistry Capsule for SSC Exams
Thank You!
Posted by:
Member since Oct 2015
Currently working as a Senior Community Manager at Gradeup. He has been a Professional Educator with diverse experience of more than 6 years in the E-learning industry. He has worked in Bankersadda as a content manager and build the foundation of SSC Adda also. He has been a pioneer in providing excellent study material to the students. He has cracked many government exams like SSC, NDA and various police departmental exams. His email address is sandeep@gradeup.co.
write a comment
Mix And MatchFeb 4, 2019
How 9 monts is 3 quarter plz someone explain....im nt able to understand
Asif KhanFeb 11, 2019
Hi
Priya AgrawalMar 2, 2019
Case number 4 I am not able to understand. ...please explain
Shivani KanojiyaMay 13, 2019
Sir... Case 4 & 5 k solutions smjh ni aa rhe. Can u plz explain ?
VilvinJun 12, 2019
How do you get interest rate?
Himanshu SinghNov 16, 2019
In case 5, rate is taken 25/4 .How? Please explain.
GayathriDec 9, 2019
Can someone explain the case 4 example .
Apachy NightDec 11, 2019
Dear Sir, Its not trick, This is traditional way to solve this question.
Subrat Kumar SahooJan 12, 2021
case 4&5 question is incomplete
Prudhvi GoenkaApr 6, 2021
case 4 rate 25/4...??? how..
GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 support@gradeup.co | 1,262 | 4,810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-25 | latest | en | 0.941343 |
https://amara.org/en/subtitles/1HTyU7qeVNx7/en/5/download/Another%20Common%20Factor%20-%20Visualizing%20Algebra.en.srt | 1,632,541,675,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00690.warc.gz | 155,967,374 | 1,124 | 1 00:00:00,012 --> 00:00:04,943 We can factor an a from ax and ay so we have a times x plus y. We can also 2 00:00:04,943 --> 00:00:10,345 factor a 2 from 2x and 2y. So we'll have plus 2 times x plus y. And notice if we 3 00:00:10,345 --> 00:00:15,688 were to just distribute we'd get back with what we started with. But in this 4 00:00:15,688 --> 00:00:21,339 line there's something incredible happening. This would be one term made up of 5 00:00:21,339 --> 00:00:26,854 two factors a times x plus y. This would be another term made up of two factors, 6 00:00:26,854 --> 00:00:31,742 2 and x plus y. So do you see it? What's the common factor for each of these 7 00:00:31,742 --> 00:00:33,880 terms? Write your answer here. | 263 | 724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-39 | latest | en | 0.96776 |
http://mathhelpforum.com/calculus/9166-relative-maximum-newton-s-method-integrate.html | 1,480,744,365,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540839.46/warc/CC-MAIN-20161202170900-00201-ip-10-31-129-80.ec2.internal.warc.gz | 174,057,605 | 12,278 | 1. ## relative maximum,Newton’s Method,Integrate.......
please try to solve these questions.
2. Originally Posted by m777
please try to solve these questions.
Question # 1
The derivative of the continuous function is given. Find all critical points and determines whether a relative maximum, relative minimum or neither occurs there
$f'(x)=2\,\sin^3(x) - \sin^2(x),\ \ \ 0
The critical points are solutions of:
$f'(x)=2\,\sin^3(x) - \sin^2(x)=0,\ \ \ 0
which occur either when $sin(x)=0$ or $2\,\sin(x)-1=0$, the roots of these are:
$\pi$ for the first and $\pi/6$ and $5 \pi/6$ for the second.
Now these correspond to local maxima when $f''(x)<0$, minima when $f''(x)>0$ and (stationary) points of inflection when $f''(x)=0$.
$f''(x)=2\,\sin(x)\cos(x)[3\, \sin(x)-1]$
So the first of the roots gives $f''(x)=0$, the second $f''(x)>0$ and the third $f''(x)<0$
RonL
3. Originally Posted by m777
please try to solve these questions.
Question # 2
Solve it with the help of Newton’s Method up to four terms
$sin(x)=x^2$
We want to find the roots of: $f(x)=\sin(x)-x^2$ using Newton's method. This is the ittereation:
$x_{n+1}=x_n-f(x_n)/f'(x_n)$
and here: $f'(x)=\cos(x)-2x$.
This can be set up in Excel (see attachment) from which we see that there is a root at x~=0.8767
4. Hello, m777!
1) The derivative of a continuous function is given.
Find all critical points and determine if a rel.max., rel.min. or neither occurs there.
$f'(x)\:=\:2\sin^3x - \sin^2x,\quad 0 < x <2\pi$
Solve $f'(x) = 0$
. . $2\sin^3x - \sin^2x \:= \:0\quad\Rightarrow\quad \sin^2x\left(2\sin x - 1\right)\:=\:0$
We have two equations to solve:
. . $\sin^2x \,= \,0\quad \Rightarrow\quad \sin x \,= \,0\quad \Rightarrow \quad \boxed{ x \:=\:\pi}$
. . $2\sin x - 1 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:\frac{1}{2} \quad\Rightarrow\quad \boxed{ x \:=\:\frac{\pi}{6},\,\frac{5\pi}{6}}$
Second derivative: . $f''(x) \:=\:6\sin^2x\cos x - 2\sin x\cos x\:=\:2\sin x\cos x\left(3\sin x - 1\right)$
$f''(\pi) \:=\:2\!\cdot\!\sin\pi\!\cdot\!\cos\pi(3\sin\pi - 1) \:=\:2\cdot0\cdot(\text{-}1)\,[3\cdot0 - 1]$ $\:=\:0\quad\hdots\quad\boxed{\text{ neither at }x = \pi}$
$f''\left(\frac{\pi}{6}\right) \:=\:2\!\cdot\!\sin\frac{\pi}{6}\!\cdot\cos\frac{\ pi}{6}\left(3\sin\frac{\pi}{6} - 1\right) \:=\:2\!\left(\frac{1}{2}\right)\!\left(\frac{\sqr t{3}}{2}\right)\!\left(3\!\cdot\!\frac{1}{2} - 1\right)\:>\:0$
. . $f(x)$ is concave up: $\cup\quad\hdots\quad\boxed{\text{relative minimum at }x = \frac{\pi}{6}}$
$f''(x)\left(\frac{5\pi}{6}\right) \:=\:2\cdot\sin\!\frac{5\pi}{6}\!\cdot\cos\!\frac{ 5\pi}{6}\left(3\sin\!\frac{5\pi}{6} - 1\right)$ $\:=\:2\!\left(\frac{1}{2}\right)\!\left(\text{-}\frac{\sqrt{3}}{2}\right)\!\left(3\!\cdot\!\frac{ 1}{2} - 1\right) \:<\:0$
. . $f(x)$ is concave down: $\cap\quad\hdots\quad\boxed{\text{relative maximim at }x = \frac{5\pi}{6}}$
5. #3 Integrate it with respect to “y”
(i) $\int (2+y^2)^2dy$
Integrate it with respect to “x”
(ii) $\int \frac{1}{x^6} dx$
Hello,
to (i). Expand the bracket first and then integrate:
$\int (2+y^2)^2dy=\int(4+4y^2+y^4)dy=4y+\frac{4}{3}y^3+\ frac{1}{5}y^5+C$
to (ii):
$\int \frac{1}{x^6} dx=\int x^{-6}dx=-\frac{1}{5}x^{-5}+C$
EB
Merry Christmas and a happy New Year!
6. ## once again thanks.
Hello,
Merry christmas and happy new year to all of you.specially to soroban, earboth and captain black.
from
m777 | 1,333 | 3,378 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 37, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2016-50 | longest | en | 0.699842 |
https://www.teacherspayteachers.com/Product/Long-Division-Task-Cards-The-Box-or-Area-Method-3580874 | 1,542,398,451,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743184.39/warc/CC-MAIN-20181116194306-20181116220306-00251.warc.gz | 1,032,915,493 | 21,398 | # Long Division Task Cards: The Box or Area Method
Subject
Resource Type
Common Core Standards
Product Rating
4.0
9 Ratings
File Type
PDF (Acrobat) Document File
2 MB|17 pages
Share
Also included in:
1. Are you working on multi-digit division in your classroom? This is one of the most challenging concepts that we face as teachers. Luckily, there are several very effective strategies for teaching this concept! This is a bundle of nine sets of long division task cards. Each set of task cards provid
\$29.50
\$20.97
Save \$8.53
2. This is a discounted bundle of long division resources. Please view the bundle contents to see exactly what is included.
\$47.49
\$36.99
Save \$10.50
Product Description
These task cards are also included in a Long Division Big Bundle. This includes nine sets of task cards for a variety of Long Division alternatives and strategies. See it HERE.
Are you looking for even more support with teaching long division in your classroom? You might be interested in this self-paced, student-centered Long Division Station that will allow your students to move through all of these strategies and approaches at their own pace. That station can be found HERE.
***********************************************************************
Are you working on multi-digit division in your classroom? This is one of the most challenging concepts that we face as teachers. Luckily, there are several very effective strategies for teaching this concept!
The box method - also known as the area method - is an effective alternative for traditional long division. It is a mental math based approach in which a division equation is solved in manageable parts in order to find the final quotient.
The box method involves multiplying and then subtracting parts until you get to 0, or as close to 0 as possible. If you cannot get down to 0, then there is a remainder.
This is a fantastic way to lead students into the partial quotients strategy - a "must-teach" alternative for long division.
This resource is a fantastic supplement to those teachers currently using the Long Division Station in their classroom. It can be used as extra practice for the Box Method level when students get to that point in the station.
This resource includes:
- detailed explanations and examples of the box method for division
- a link to a video explaining the box method
- a box method strategy poster to hang in the classroom for easy reference
- 24 task cards that will have students practicing the box method in a variety of different ways to enhance understanding
- recording sheets to keep students organized
- answer keys to make self-checking a breeze
You may also be interested in:
- The Long Division Station
- Long Division Task Cards: The Big Bundle
Total Pages
17 pages
Included
Teaching Duration
N/A
Report this Resource
\$3.50 | 587 | 2,846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-47 | latest | en | 0.937614 |
http://mathhelpforum.com/differential-geometry/90994-inequalities.html | 1,481,203,539,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542588.29/warc/CC-MAIN-20161202170902-00044-ip-10-31-129-80.ec2.internal.warc.gz | 174,590,649 | 11,536 | 1. ## Inequalities
1. Prove that $n! > \left(\frac{n}{e} \right)^{n}$ for $n \geq 1$.
Proof. We use induction on $n$. For $n=1$, $1 > \frac{1}{e}$. Now suppose that for some $k > 1$, $k! > \left(\frac{k}{e} \right)^{k}$. So we want to show that $(k+1)! > \left(\frac{k+1}{e} \right)^{k+1}$. Now $(k+1)! > (k+1) \cdot \left(\frac{k}{e} \right)^{k}$. Can we somehow use the fact that $\left(1+ \frac{1}{n}\right)^{n} < e$ for all $n$? We could flip it so that $\frac{1}{e} < \frac{1}{\left(1+\frac{1}{n} \right)^{n}}$ for all $n$. And so $\left(\frac{n}{e} \right)^{n} < \frac{n^n}{\left(1+\frac{1}{n} \right)^{n^2}}$. Or another approach may be to show that $\frac{k^{k} \cdot (k+1)}{e^k} > \frac{(k+1)^{k+1}}{e^{k+1}}$? The denominator of the second term is larger, indicating that it would probably be smaller.
2. Show that $n! < \left(\frac{n+1}{n} \right)^{n}$ for $n>1$, and deduce that $n! < e \left(\frac{n}{2} \right)^{n}$.
For this one you would use the AM-GM inequality? Because the LHS is $n(n-1)(n-2) \cdots 1$. So taking $\sqrt[n]{n!}$ we can use the AM-GM inequality?
2. Originally Posted by Sampras
1. Prove that $n! > \left(\frac{n}{e} \right)^{n}$ for $n \geq 1$.
Proof. We use induction on $n$. For $n=1$, $1 > \frac{1}{e}$. Now suppose that for some $k > 1$, $k! > \left(\frac{k}{e} \right)^{k}$. So we want to show that $(k+1)! > \left(\frac{k+1}{e} \right)^{k+1}$. Now $(k+1)! > (k+1) \cdot \left(\frac{k}{e} \right)^{k}$. Can we somehow use the fact that $\left(1+ \frac{1}{n}\right)^{n} < e$ for all $n$? We could flip it so that $\frac{1}{e} < \frac{1}{\left(1+\frac{1}{n} \right)^{n}}$ for all $n$. And so $\left(\frac{n}{e} \right)^{n} < \frac{n}{\left(1+\frac{1}{n} \right)^{n^2}}$. Or another approach may be to show that $\frac{k^{k} \cdot (k+1)}{e^k} > \frac{(k+1)^{k+1}}{e^{k+1}}$? The denominator of the second term is larger, indicating that it would probably be smaller.
2. Show that $n! < \left(\frac{n+1}{n} \right)^{n}$ for $n>1$, and deduce that $n! < e \left(\frac{n}{2} \right)^{n}$.
For this one you would use the AM-GM inequality? Because the LHS is $n(n-1)(n-2) \cdots 1$. So taking $\sqrt[n]{n!}$ we can use the AM-GM inequality?
For (1), $k^{k} \cdot (k+1) \not > (k+1)^{k+1}$. So the inequality depends on the denominators $e^{k}$ and $e^{k+1}$. Is this where we use the fact that $\left(1+\frac{1}{n} \right)^{n} < e$ for all $n$?
3. Originally Posted by Sampras
1. Prove that $n! > \left(\frac{n}{e} \right)^{n}$ for $n \geq 1$.
Proof. We use induction on $n$. For $n=1$, $1 > \frac{1}{e}$. Now suppose that for some $k > 1$, $k! > \left(\frac{k}{e} \right)^{k}$. So we want to show that $(k+1)! > \left(\frac{k+1}{e} \right)^{k+1}$. Now $(k+1)! > (k+1) \cdot \left(\frac{k}{e} \right)^{k}$.
so we just need to prove that $(k+1)\left(\frac{k}{e} \right)^k > \left(\frac{k+1}{e} \right)^{k+1},$ which after simplifying gives us this equivalent inequality: $\left(1 + \frac{1}{k} \right)^k < e,$ which we know is true.
2. Show that $n! < \left(\frac{n+1}{n} \right)^{n}$ for $n>1.$
this is false! check what you wrote again!
4. Originally Posted by NonCommAlg
so we just need to prove that $(k+1)\left(\frac{k}{e} \right)^k > \left(\frac{k+1}{e} \right)^{k+1},$ which after simplifying gives us this equivalent inequality: $\left(1 + \frac{1}{k} \right)^k < e,$ which we know is true.
this is false! check what you wrote again!
It should be $n! < \left(\frac{n+1}{2} \right)^{n}$. So basically you simplified the given inequality into an inequality we know?
5. Originally Posted by Sampras
It should be $n! < \left(\frac{n+1}{2} \right)^{n}$.
this is a very straightforward application of AM-GM: $\frac{n+1}{2}=\frac{\frac{n(n+1)}{2}}{n}=\frac{1+2 + \cdots + n}{n} > \sqrt[n]{1 \cdot 2 \cdot \cdots n}=\sqrt[n]{n!}.$
6. Originally Posted by NonCommAlg
this is a very straightforward application of AM-GM: $\frac{n+1}{2}=\frac{\frac{n(n+1)}{2}}{n}=\frac{1+2 + \cdots + n}{n} > \sqrt[n]{1 \cdot 2 \cdot \cdots n}=\sqrt[n]{n!}.$
And to deduce that $n! < e \left(\frac{n}{2} \right)^{n}$ we know that $n! < \left(\frac{n+1}{2} \right)^{n}$. So maybe simplify the first inequality to get the second one? We can do the following: $n! < \left[\left(1+\frac{1}{n} \right) \cdot \frac{n}{2} \right]^{n}$? We then get the second inequality.
Is this correct?
7. Originally Posted by Sampras
And to deduce that $n! < e \left(\frac{n}{2} \right)^{n}$ we know that $n! < \left(\frac{n+1}{2} \right)^{n}$. So maybe simplify the first inequality to get the second one? We can do the following: $n! < \left[\left(1+\frac{1}{n} \right) \cdot \frac{n}{2} \right]^{n}$? We then get the second inequality.
Is this correct?
you just need to show that $\left(\frac{n+1}{2} \right)^n < e \left(\frac{n}{2} \right)^n,$ which, again, after simplifying gives you this equivalent and true inequality: $\left( 1+ \frac{1}{n} \right)^n < e.$
8. Originally Posted by NonCommAlg
you just need to show that $\left(\frac{n+1}{2} \right)^n < e \left(\frac{n}{2} \right)^n,$ which, again, after simplifying gives you this equivalent and true inequality: $\left( 1+ \frac{1}{n} \right)^n < e.$
But my simplification of the current inequality to the inequality we proved is incorrect? | 1,973 | 5,211 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 74, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2016-50 | longest | en | 0.630545 |
https://www.univerkov.com/iron-weighing-14-g-was-fused-with-sulfur-weighing-4-8-g-and-hcl-was-added-to-the-resulting-mixture/ | 1,708,683,684,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474377.60/warc/CC-MAIN-20240223085439-20240223115439-00040.warc.gz | 1,085,609,839 | 6,482 | # Iron weighing 14 g was fused with sulfur weighing 4.8 g, and hcl was added to the resulting mixture.
Iron weighing 14 g was fused with sulfur weighing 4.8 g, and hcl was added to the resulting mixture. What gases are released and what are their volumes?
We write down the equation of reactions.
Fe + S = FeS.
We find the amount of the substance iron and sulfur. We use the following formula.
n = m / M.
We calculate the molar mass of iron and sulfur.
M (Fe) = 56 g / mol.
M (S) = 32 g / mol.
We find the amount of substance.
n (Fe) = 14 g / 56 g / mol = 0.25 mol.
n (S) = 4.8 g / 32 g / mol = 0.15 mol.
This means that iron is in excess, which means we will count by sulfur.
We write down the uranium of reactions.
0.15 mol S = 0.15 mol FeS.
FeS + 2 HCL = FeCl 2 + H2S.
Next, we find the amount of hydrogen sulfide substance by the reaction equation.
0.15 mol FeS – х mol H2S
1 mol FeS – 1 mol H2S
Hence, n (H2S) = 0.15 mol.
Next, we find the volume.
n = V / Vm.
V = n × Vm = 0.15 mol × 22.4 L / mol = 3.36 L.
Answer: hydrogen sulfide H2S gas with a volume of 3.36 liters will be released.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 416 | 1,404 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-10 | latest | en | 0.929448 |
http://www.brainbashers.com/showabcview.asp?date=0914&which=1 | 1,505,976,752,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687702.11/warc/CC-MAIN-20170921063415-20170921083415-00425.warc.gz | 396,744,382 | 5,321 | [OK] Privacy Policy - Terms & Conditions - See DetailsWe use cookies to personalise content and ads. We also share information about your use of our site with our advertising partners who may combine it with other information you've provided to them or they've collected from your use of their services.
Daily ABC View
Solve a random Easy - Medium - Hard Japanese-type logic puzzle chosen from today's collection.
September 14 - Puzzle 1
Show mistakes when checking
Warn when leaving page
Auto-pencil marks
Highlight clues
P = auto-pencil marks (or click the check box)
SHIFT + 1 = highlight squares with A
SHIFT + 2 = highlight squares with B
SHIFT + 3 = highlight squares with C
SHIFT + 0 = clear highlighted squares
(Only when there are pencil marks)
Today's ABC View Puzzles
Notes
• Objective / Rules
- Each row and column requires exactly one A, B and C (and two blank squares).
- The clues around the edge tell you which letter appears first in that row or column.
• Help
Read the help/walkthrough page on ABC View puzzles for a more detailed explanation and a walkthrough.
• Keyboard Usage
For pencil marks simply enter multiple letters per square.
X is treated as a blank square.
P = auto-pencil marks (updates A, B and C only).
CTRL + arrows = move around the grid.
SHIFT + 1 = highlight all squares containing A.
SHIFT + 2 = highlight all squares containing B.
SHIFT + 3 = highlight all squares containing C.
SHIFT + 0 = clear highlighted squares.
• Checking
If you click 'Check' the system will check for incorrect squares. If 'Show mistakes when checking' is checked they will be marked in red. The system only checks squares that contain one large A, B C, or X, and ignores all others. If 'Highlight clues' is checked any clue where A, B or C appears first in the row/column will be highlighted - this does not mean that your answer is correct.
• Uniqueness
Each puzzle has exactly one solution, which can be found using logic alone and no guesses are ever required. If you think you've found another solution, then please double check the rules.
• JavaScript
This game requires JavaScript to be enabled.
[Unique Reference = ABCView-0914-1-358099] | 493 | 2,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-39 | longest | en | 0.85631 |
http://sst2011-s105maths.blogspot.com/2011/09/s1-viva-voce-by-nicholas-lee-incomplete.html | 1,532,104,054,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591718.31/warc/CC-MAIN-20180720154756-20180720174756-00443.warc.gz | 340,488,508 | 15,133 | ## Sunday, September 11, 2011
### S1 (Viva Voce) By Nicholas Lee (Incomplete)
S1 (Viva Voce) Part A Question 1:
Steps to solve question 1:
Firstly, you need to list down all known variables;
In this case, AB = (2x+3)cm, AD = (4x-30)cm and DC = (3x-7)cm
According to the diagram, AB = DC which means (2x+3)cm = (3x-7)cm
(i) x would be equal to 3+7=10
(ii) To find the Area of Rectangle ABCD, you must first find the value of the breath of the rectangle, AD.
Since 1x is 10cm, 4x-30cm is equal to 40cm - 30cm=10cm | 188 | 517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-30 | latest | en | 0.914877 |
https://debasishg.blogspot.com/2015/06/baking-can-teach-you-bit-of.html | 1,726,663,996,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651895.12/warc/CC-MAIN-20240918100941-20240918130941-00338.warc.gz | 176,761,361 | 19,504 | Saturday, June 13, 2015
Baking a π can teach you a bit of Parametricity
Even though I got my copy of Prof. Eugenia Cheng's awesome How to Bake π a couple of weeks back, I started reading it only over this weekend. I am only on page 19 enjoying all the stuff regarding cookies that Prof. Cheng is using to explain abstraction. This is a beautiful piece of explanation and if you are a programmer you may get an extra mile out of the concepts that she explains here. Let's see if we can unravel a few of them ..
She starts with a real life situation such as:
If Grandma gives you five cookies and Grandpa gives you five cookies, how many cookies will you have ?
Let's model this as box of cookies that you get from your Grandma and Grandpa and you need to count them and find the total. Let's model this in Scala and we may have something like the following ..
case class CookieBox(count: Int)
and we can define a function that gives you a CookieBox containing the total number of cookies from the 2 boxes that we pass to the function ..
def howManyCookies(gm: CookieBox, gp: CookieBox) = {
}
and we use howManyCookies to find the count ..
scala> val gm = CookieBox(5)
scala> val gp = CookieBox(5)
res5: CookieBox = CookieBox(10)
.. so we have 10 cookies from our Grandma & Grandpa .. Perfect!
The problem is .. the child answers: "None, because I'll eat them all". To model this let's add a function eat to our CookieBox abstraction ..
case class CookieBox(count: Int) {
// let's assume n < count for simplicity
def eat(n: Int): CookieBox = CookieBox(count - n)
}
So instead of the correct way to answer the question, the child cheats and implements howManyCookies as ..
def howManyCookies(gm: CookieBox, gp: CookieBox) = {
}
and we get the following ..
scala> howManyCookies(gm, gf)
res6: CookieBox = CookieBox(0)
Prof. Cheng continues ..
The trouble here is that cookies do not obey the rules of logic, so using math to study them doesn't quite work. .. We could impose an extra rule on the situation by adding "... and you're not allowed to eat the cookies". If you're not allowed to eat them, what's the point of them being cookies ?
This is profound indeed. When we are asked to count some stuff, it really doesn't matter if they are cookies or stones or pastries. The only property we need here is to be able to add together the 2 stuff that we are handed over. The fact that we have implemented howManyCookies in terms of CookieBox gives the little child the opportunity to cheat by using the eat function. More information is actually hurting us here, being concrete with data types is actually creating more avenues for incorrect implementation.
Prof. Cheng is succinct here when she explains ..
We could treat the cookies as just things rather than cookies. We lose some resemblance to reality, but we gain scope and with it efficiency. The point of numbers is that we can reason about "things" without having to change the reasoning depending on what "thing" we are thinking about.
Yes, she is talking about generalization, being polymorphic over what we count. We just need the ability to add 2 "things", be it cookies, monkeys or anchovies. In programming we model this with parametric polymorphism, and use a universal quantification over the set of types for which we implement the behavior.
def howMany[A](gm: A, gp: A) = //..
We have made the implementation parametric and got rid of the concrete data type CookieBox. But how do we add the capability to sum the 2 objects and get the result ? You got it right - we already have an abstraction that makes this algebra available to a generic data type. Monoids FTW .. and it doesn't get simpler than this ..
trait Monoid[T] {
def zero: T
def append(t1: T, t2: T): T
}
zero is the identity function and append is a binary associative function over 2 objects of the type. So given a monoid instance for our data type, we can model howMany in a completely generic way irrespective of whether A is a CookieBox or Monkey.
def howMany[A : Monoid](gm: A, gp: A): A = gm append gp
Implementing a monoid for CookieBox is also simple ..
object CookieBox {
implicit val CookieBoxMonoid = new Monoid[CookieBox] {
val zero = CookieBox(0)
}
}
With the above implementation of howMany, the little child will not be able to cheat. By providing a simpler data type we have made the implementation more robust and reusable across multiple data types.
Next time someone wants me to explain parametricity, I will point them to Page 19 of How to Bake π.
narayan iyer said...
FYI, for those of us who use only eBooks (especially with re-flowable content), the section on parametricity is "Cookies - How things that are too real don't obey mathematics".
Unknown said...
Good catch .. I am still an old fashioned dude who loves to read the dead tree version :-). Thanks for clarifying - indeed that's the section I was referring to ..
narayan iyer said...
:). BTW, nice post (as always). Thanks!
esfand said...
That was profound indeed. Thank you very much for the great post which was short, sweet but very profound.
Anonymous said...
Functional newbie here, so please be kind =)
Your final CookieBox lacks the "eat" behaviour that allowed the child to write a "cheat" version of howManyCookies. If "eat" is added to your final implementation isn't it then easy for the child to write a cheat version of "append"?
Industrial Computers said...
Very profound indeed! Another great read thanks very much!
Perevesti said...
This is a very useful article and desperately useful
JJ said...
Great article | 1,275 | 5,589 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-38 | latest | en | 0.89045 |
null | null | null | null | null | null | You are here: Home > Opinion > Current Article
Merger of Our Two Hospitals
By / May 28, 2013 / Comments
(Reprinted here with permission of the author)
by Hope Campbell
So, you know that state law required our Hospital Authority to hold a public hearing last spring before awarding the management contract for Palmyra. (That very hospital OUR Authority bought for $195,000,000.00, out of their operating funds of, well, zero.) But did you know that our Authority advertised the hearing exactly one time? Do you think that they could possibly have wanted to limit attendance at the hearing…… OUR Authority? Surely NOT!
Even if you didn’t attend, you must have heard, the turnout was HUGE! There was overwhelming opposition to the proposed plan to award Phoebe Putney a 40-year lease for the Palmyra facility for the amazing sum of $1.00 per year. Those few who did speak out in favor of the lease were in some way tied to Phoebe, hardly a resounding endorsement. You would probably think that the members of the Authority would actually consider whether such a merger was in the best interests of our community.
Person after person arose to point out very compelling reasons to find a different lessor. How could it not matter that Phoebe has engaged in anticompetitive practices for so many years with impunity? How could it not matter that the FTC had asked the US Supreme Court to rule on the legality of the Phoebe-Palmyra debacle? What was the rush to sign a lease? Shouldn’t we at least wait until the Supreme Court decided whether or not to hear the case? What about this one? Phoebe Putney, the one that offers “world-class medicine with a hometown commitment,” has a less than exemplary record when it comes to quality of care! Or this one. It actually costs MORE THAN AVERAGE to avail yourself of the LESS THAN AVERAGE care provided by Phoebe.
Isn’t it reasonable to think a hospital that has sought to squelch competition through possibly illegal means has an agenda that isn’t in the best interest of the community?
As I stated earlier, the only Phoebe proponents who attended the hearing were in some way tied to Phoebe. And I do not mean that they were successfully treated at Phoebe! There were NO satisfied consumers there, not a single one! There were Phoebe doctors, sons of those doctors, board members and employees. (All of the pro-Phoebe spokesmen were in the Phoebe pocket, so to speak.) Are we so ignorant in their estimation that they thought we would believe their support was spontaneous? And did they think for a minute that their motivation was not transparent?
If you attended the hearing, as the proceedings evolved, you were probably one of many who noted with something approaching despair , that even though MANY potentially negative outcomes to the Phoebe-Palmyra merger were outlined every member of our Hospital Authority sat with bored far-away looks and not so much as a pretense of interest in the seriousness of the implications for the citizens whose interests they should represent. And when they quickly voted UNANIMOUSLY to proceed with the Phoebe lease at the earliest possible date, you surely realized that the hearing was a perfunctory exercise.
Why did they not even consider the possibility that another hospital management company might entertain the idea of a forty-year lease at $1.00 per year?
Then, this spring, when the Supreme Court of the United States of America decided unanimously against Phoebe, you must have thought our Hospital Authority would finally take its responsibility seriously and pursue courses of actions other than granting an unbid contract to Phoebe….
Well, my friend you were wrong. Our Hospital Authority has failed us as they have sat indifferent to the fact that Phoebe has sought to circumvent the Supreme Court by changing Georgia law to allow SOME, but not ALL, county hospital authorities to enter into a monopolistic leasing agreement with a hospital management company. Do you see that the proposed law was tailor made to fit the Phoebe situation? How could our Authority sit back and allow Phoebe Putney to do that? The members of our Hospital Authority apparently either misunderstand their job, or they don’t care that Phoebe, in its arrogance has decided to thumb its collective nose at the FTC, the US Supreme Court and the citizens of our community who will be so adversely affected by their takeover of Palmyra.
Perhaps, those on the Authority with integrity should take this opportunity to resign. As to the others , we should recall every one of them. We should require our county commission, which is complicit in this mess, to install a new Authority which is independent of Phoebe’s powerful influence, who understands the implications of selling out our interests to Phoebe Putney and who will oversee the proper delivery of health care at a reasonable price to our citizens.
Writte by Hope H. Campbell, a lifelong resident of Albany. She was educated in Albany Schools and owned a business in Albany for many years
Print Email
You might also like...
Punish the deed not the breed.
Read More →
SEO Powered by Platinum SEO from Techblissonline | null | null | null | null | null | null | null | null | null |
http://www.jiskha.com/display.cgi?id=1270427255 | 1,498,616,478,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128322275.28/warc/CC-MAIN-20170628014207-20170628034207-00651.warc.gz | 553,050,419 | 3,753 | # trig
posted by .
which of the following sets gives the x coordinates where the parabola y=x^2+4X-50 and the line y=7x-10 intersect when drawn in the coordinate plane
• trig -
7x-10 = x^2 + 4 x - 50
x^2 -3x -40 = 0
(x-8)(x+5) = 0
x = 8 or x = -5
(-5,-45)
(8,46) | 110 | 267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-26 | latest | en | 0.776417 |
http://www.slideserve.com/vangie/warm-up-4-16 | 1,508,368,546,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823153.58/warc/CC-MAIN-20171018214541-20171018234541-00401.warc.gz | 571,409,221 | 12,562 | Warm Up 4/16
1 / 15
# Warm Up 4/16 - PowerPoint PPT Presentation
Warm Up 4/16. Lesson 13: Populations and Samples. Objective: Students differentiate between a population and a sample. Students investigate statistical questions that involve generalizing from a sample to a larger population. Populations vs. Samples.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
## PowerPoint Slideshow about ' Warm Up 4/16' - vangie
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Lesson 13: Populations and Samples
Objective:
• Students differentiate between a population and a sample.
• Students investigate statistical questions that involve generalizing from a sample to a larger population.
Populations vs. Samples
• Population - A population is the entire set of objects (people, animals, plants, etc.) from which data might be collected.
• Sample - A sample is a subset of the population.
Fact
There are many different ways to select a sample from a population. To make valid inferences about a population, you must choose a random sample very carefully so that it accurately represents the population.
Fact
• The results of an unbiased sample are proportional to the results of the population. So, you can use unbiased samples to make predictions about the population.
• Biased samples are not representative of the population. So, you should not use them to make predictions about the population because the predictions may not be valid. | 425 | 1,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-43 | latest | en | 0.905233 |
http://mathhelpforum.com/discrete-math/115399-solved-application-pigeonhole-principle.html | 1,506,096,417,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688997.70/warc/CC-MAIN-20170922145724-20170922165724-00694.warc.gz | 228,062,546 | 11,642 | # Thread: [SOLVED] Application of the Pigeonhole Principle
1. ## [SOLVED] Application of the Pigeonhole Principle
I have some problems relating to the pigeonhole principle. Here they are:
Prove that at a party where there are at least two people, there are two people who know the same number of other people there.
The other one is,
Let n1, n2, ... , nt be positive integers. Show that if n1+n2+...+nt - t + 1 objects are placed into t boxes, then for some i, i = 1, 2, ... , t, the ith box contains at least ni objects.
Now, I was given the hint that the pigeonhole principle can help me here, but I don't know how to apply it. Are there any other hints anyone could think of to help me prove this? Thanks.
2. Originally Posted by Oijl
Prove that at a party where there are at least two people, there are two people who know the same number of other people there.
We must make some assumptions here.
First “If A knows B then B knows A”.
Second “A knows at most $n-1$ other people at the party".
This problem is equivalent to the theorem: In a simple graph at two vertices have the same degree.
For the pigeonholes, there $n$ of them: $0,1,\cdots,n-1$.
Anyone at the party can know from $0$ to $n-1$ other people.
Suppose that none of those pigeonholes has more than one person in it.
Convince yourself that the $[n-1]^{th}$ hole would be empty.
3. Not to be rude, but I think simply applying the fact that we have $n-1$ holes as you described, and n people in the party, the pidgeonhole principle directly gives us that there must be two people in at least one hole.
4. Originally Posted by Defunkt
Not to be rude, but I think simply applying the fact that we have $n-1$ holes as you described, and n people in the party, the pidgeonhole principle directly gives us that there must be two people in at least one hole.
That is not being rude. It is being unable to count.
There are $n$ holes: $0,1,\cdots,n-1$.
5. Oops, didn't see the 0. Sorry.
6. I see! All the pigeonholes cannot be filled, because that is a contradiction, because the hole representing zero cannot be filled while the hole representing knowing n - 1 is filled... and since there are the same number of elements as holes, and all the holes cannot be filled, there must be some hole with more than one element in it.
,
,
,
### let n1, n2,....nt be positive integer show that if n1 n2 ..nt-t.v., the objects are placed into t boxes, then for some I, i=1,2,....t, the I th box contains at least n:objects
Click on a term to search for related topics. | 664 | 2,536 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-39 | longest | en | 0.966009 |
https://www.coursehero.com/file/6626920/Homework-3-Solutions/ | 1,500,555,052,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423183.57/warc/CC-MAIN-20170720121902-20170720141902-00525.warc.gz | 774,776,142 | 23,434 | Homework 3 Solutions
# Homework 3 Solutions - HW 3 SOLUTIONS Chapters 4 and 5(from...
This preview shows pages 1–2. Sign up to view the full content.
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.
Unformatted text preview: HW 3 SOLUTIONS Chapters 4 and 5 (from 3 rd ed. of text) 1. 5.23. A certain assay for serum alanine aminotransferase (ALT) is rather imprecise. The results of repeated assays of a single specimen follow a normal distribution with mean equal to the ALT concentration for that specimen and standard deviation equal to 4 U/Li (as in Exercise 4.40). Suppose a hospital lab measures many specimens every day, and specimens with reported ALT values of 40 or more are flagged as “unusually high.” If a patient's true ALT concentration is 35 U/Li find the probability that his specimen will be flagged as “unusually high” a. if the reported value is the result of a single assay Correct: The result of a single assay is like a random observation Y from the population of assays. A value Y ≥ 40 will be flagged as “unusually high”. TI-84 normalcdf(40, E99, 35, 4) = 0.1056 Thus, Pr{specim en will be flagged as “unusually high”} = 0.1056. b. Find the 99th percentile of this distribution. TI-84 invNorm(.99, 35, 4) = 44.31 U/Li c. if the reported value is the mean of three independent assays of the same specimen The reported value is the mean of three independent assays, which is like the mean Y of a sample of size n = 3 from the population of assays. A value ¡ ≥ 40 will be flagged as “unusually high.” We are concerned with thehigh....
View Full Document
### Page1 / 3
Homework 3 Solutions - HW 3 SOLUTIONS Chapters 4 and 5(from...
This preview shows document pages 1 - 2. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 496 | 1,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-30 | latest | en | 0.900927 |
null | null | null | null | null | null | About Francesco Azzola
Android ListView Pull-to-Refresh
In this post I want to describe how to create a custom ListView widget that behaves like Gmail list, so that when you slide down and you are at the top of the list it will be refreshed. This new UI pattern is knows as Pull-to-refresh. There are several implementation of it this is my own version.
To have this behavior we have to create a custom component so that we can have a full control. So we have to:
• Create a custom component that extends the standard ListView widget
• Create a custom adapter (optional)
• Create an interface (listener)
Creating custom component
The first thing we have to do is creating a custom widget that extends the ListView widget and implements the pull down to refresh functionality. We have to mind that we have to preserve all the behaviors in the standard ListView. The first thing we have to consider is when the list view items refresh takes place. There are two conditions that must be satisfied:
• We are at the top of the list view, in other word the first element is visible
• User has to touch and move down his finger with a path length longer than a specified threshold
To achieve these requirements we have to override the onTouchEvent and in this method we have to verify if those conditions are satisfied.
public boolean onTouchEvent(MotionEvent event) {
//System.out.println("First ["+this.getFirstVisiblePosition()+"]");
float y = event.getY();
switch (event.getAction()) {
case MotionEvent.ACTION_MOVE: {
if ( ( y - startY) > THRESHOLD && STATE_REFRESH_ENABLED && !STATE_REFRESHING ) {
case MotionEvent.ACTION_DOWN: {
startY = y;
STATE_REFRESH_ENABLED = getFirstVisiblePosition() == 0; // We are on the first element so we can enable refresh
case MotionEvent.ACTION_UP: {
STATE_REFRESHING = false;
return super.onTouchEvent(event);
In particular, when user touches the screen the first time (MotionEvent.ACTION_DOWN) at line 13-15 we have to verify which item is visible using getFirstVisiblePosition(). If the result of this method is 0 then we can enable the refresh status and check if the user moves down his finger. At the same time we store the y coord on the touch screen. Then we verify if the user moves down his finger catching the MotionEvent.ACTION_MOVE event (line 8-9). While user moves down his finger and the path length is longer than the a pre-configured threshold then we can trigger the refresh event.
Implement the refresh event
Once all the conditions are satisfied we can trigger the refresh event. To notify to the caller that the user wants to refresh the list we can use a simple interface with only one method, so the if the caller implements this interface and register itself as a listener it will be notified when a refresh should take place.
public interface OnRefreshListListener {
public void onRefresh();
One thing we can improve is when a refresh status is already triggered and we don’t want to trigger it again. To do it we simply use an internal boolean attribute. Another important thing is to adapt the threshold to the real screen size and dpi. Of course this isn’t a complete library ready-to-use but it can give some hints.
Reference: Android ListView Pull-to-Refresh from our JCG partner Francesco Azzola at the Surviving w/ Android blog.
Subscribe to our newsletter to start Rocking right now!
JPA Mini Book
JVM Troubleshooting Guide
Given email address is already subscribed, thank you!
Oops. Something went wrong. Please try again later.
Please provide a valid email address.
Please complete the CAPTCHA.
Please fill in the required fields.
Leave a Reply
+ 7 = twelve
Java Rockstar?
Subscribe to our newsletter to start Rocking right now!
Get ready to Rock!
You can download the complementary eBooks using the links below: | null | null | null | null | null | null | null | null | null |
https://www.slideserve.com/ellema/using-time-to-explore-angles | 1,513,467,725,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948589512.63/warc/CC-MAIN-20171216220904-20171217002904-00665.warc.gz | 804,538,244 | 12,490 | 1 / 21
# Using Time to Explore Angles - PowerPoint PPT Presentation
Using Time to Explore Angles. A. endpoints. Look in your book on page 440. Geometry Terms to Know. Line - a straight path between two points that goes on forever in both directions. Point - an exact place in space.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
## PowerPoint Slideshow about 'Using Time to Explore Angles' - ellema
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
### Using Time to Explore Angles
endpoints
Look in your book on page 440.
### Geometry Terms to Know
Line - a straight path between two points that goes on forever in both directions.
Point - an exact place in space.
Line Segment - part of a line; has a distinct starting point and ending point.
### Geometry Terms to Know
Ray - part of a line; has a definite starting point, but no ending point; goes on forever in one direction.
Endpoint
### Geometry Terms to Know
Angle - formed by two lines, rays, or line segments that have the same endpoint.
### Geometry Terms to Know
Angle - formed by two lines, rays, or line segments that have the same endpoint. This end point is called the vertex.
### Geometry Terms to Know
We measure angles by measuring the distance between the two sides of the angle.
### Geometry Terms to Know
We measure this distance in a unit called degrees.
The symbol for degrees is a small superscript circle.
Degrees = 45°
### Types of Angles
Let’s use our clocks to discover the names of these angles.
There are essentially three different types of angles.
### Right Angles
The hands of the clock form a right angle.
### Right Angles
The hands of the clock form a right angle.
### Right Angles
Right angles measure
exactly 90 degrees.
### Acute Angles
The hands of the clocks form acute angles.
### Acute Angles
The hands of the clocks form acute angles.
### Acute Angles
Acute angles measure less than 90 degrees.
### Straight Angles
The hands of the clock form a straight angle.
### Straight Angles
Straight angles measure exactly 180 degrees.
### Obtuse Angles
The hands of the clocks form obtuse angles. | 603 | 2,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-51 | latest | en | 0.888958 |
http://gmatclub.com/forum/the-product-of-the-first-twelve-positive-integers-is-44758.html | 1,436,071,242,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375097204.8/warc/CC-MAIN-20150627031817-00027-ip-10-179-60-89.ec2.internal.warc.gz | 116,045,506 | 46,284 | Find all School-related info fast with the new School-Specific MBA Forum
It is currently 04 Jul 2015, 20:40
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# The product of the first twelve positive integers is
Author Message
TAGS:
Manager
Joined: 20 Nov 2006
Posts: 215
Followers: 1
Kudos [?]: 6 [0], given: 0
The product of the first twelve positive integers is [#permalink] 20 Apr 2007, 09:23
1
This post was
BOOKMARKED
00:00
Difficulty:
(N/A)
Question Stats:
100% (01:51) correct 0% (00:00) wrong based on 7 sessions
The product of the first twelve positive integers is divisible by all of the following EXCEPT
(A) 210
(B) 88
(C) 75
(D) 60
(E) 34
VP
Joined: 08 Jun 2005
Posts: 1146
Followers: 6
Kudos [?]: 136 [0], given: 0
since 1*2*3*4*5*6*7*8*9*10*11*12 =
only 34 has prime factors grater then 12 = 17*2
that cannot be divided !!
Similar topics Replies Last post
Similar
Topics:
2 If P represents the product of the first 13 positive integers, which 3 04 Dec 2014, 06:39
5 The product of the fir
st eight positive even integers is 4 25 Feb 2012, 16:01
5 The product of first twelve positive integers is divisible 5 05 Sep 2010, 07:51
If n is a positive integer and the product of all the 3 31 Mar 2008, 05:43
If n is a positive integer and the product of all the 1 10 Dec 2005, 15:16
Display posts from previous: Sort by | 552 | 1,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2015-27 | longest | en | 0.873645 |
https://www.physicsforums.com/threads/system-of-linear-equations.348413/ | 1,510,989,448,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804666.54/warc/CC-MAIN-20171118055757-20171118075757-00681.warc.gz | 874,838,855 | 14,666 | # System of linear equations
1. Oct 23, 2009
### IniquiTrance
1. The problem statement, all variables and given/known data
Find coefficients a, b, c, d so that the circle with the following 3 points satisfies the equation:
$$ax^{2} + ay^{2} + bx + cy + d = 0$$
Points:
(-4, 5)
(4, -3)
(-2, 7)
2. Relevant equations
3. The attempt at a solution
I'm wondering if since I can only construct 3 equations from the 3 points, if I will have to make one unknown a parameter - probably d.
Is there a way to construct a 4 th equation which I'm missing?
Thanks!
2. Oct 23, 2009
### Dick
The parameters a,b,c and d are not independent if you are given that it's a circle. Write the equation of a circle in the form (x-a)^2+(y-b)^2=r^2. Now you only have three parameters. And you have three points.
Last edited: Oct 23, 2009
3. Oct 23, 2009
### IniquiTrance
What if I used Gauss Jordan elimination to find a,b and c in terms of parameter d, would that sufficiently answer the question?
4. Oct 23, 2009
### Dick
Sure, I suppose. The 'fourth parameter' is really that you can divide your whole equation by any one of the four parameters that is nonzero and eliminate it. It was never really there to begin with. I.e. x^2+y^2+bx+cy+d=0 is also just as good.
Last edited: Oct 23, 2009 | 376 | 1,289 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-47 | longest | en | 0.914408 |
https://apps.apple.com/us/app/math-jungle-grade-1/id1073980208?at=11ltdS&uo=4&mt=8 | 1,725,708,073,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00534.warc.gz | 80,762,352 | 152,616 | # Math Jungle : Grade 1 4+
• \$0.99
## Description
Your child will love practicing their grade 1 math skills with this fun and engaging game. This game is aligned with the Common Core State Standards.
Studies show that practice is the most important factor in developing strong math skills. Our play based approach to learning keeps your child eager to practice. We play test our apps to make sure they are fun enough to play repeatedly. The focus is on mastering skills through practice. Hints are available to help your child learn different techniques for solving problems.
SKILLS
Children will learn and master the following skills
- Add and subtract within 20 (1.OA.A.1)
- Count to 120, starting at any number less than 120 (1.NBT.A.1)
- Understand multiples of 10 (1.NBT.C.5)
- Compare two two-digit numbers (1.NBT.B.3)
- Add within 100 (1.NBT.C.4)
- Find 10 more or 10 less than the number (1.NBT.C.5)
- Subtract multiples of 10 from multiples of 10 (1.NBT.C.6)
- Relate counting to addition and subtraction (1.OA.C.5)
- Add three whole numbers whose sum is less than or equal to 20 (1.OA.A.2)
- Apply properties of operations as strategies to add and subtract (1.OA.B.3)
- Understand subtraction as an unknown-addend problem (1.OA.B.4)
- Determine the unknown whole number in an addition or subtraction equation (1.OA.D.8)
- Use objects, drawings, and equations with a symbol for the unknown number (1.OA.A.1, 1.OA.A.2)
- Understand the meaning of the equal sign (1.OA.D.7)
- Understand that the two digits of a two-digit number represent amounts of tens and ones. (1.NBT.B.2)
- Understand that 10 can be thought of as a bundle of ten ones (1.NBT.B.2.A)
- Understand that the numbers from 11 to 19 are composed of a ten and ones (1.NBT.B.2.B)
PLAY
Your child will help a monkey collect bananas in a jungle. Most bananas are collected by solving math problems. Other bananas are collected through simple game play. The game includes a high score feature. This provides motivation for your child to improve. Higher scores are achieved by answering correctly. This provides incentive to focus on getting the correct answer.
HINTS
Educational hints are always available. They include different techniques for solving problems.
PRIVACY
- No third party advertising
- No external links
- No in-app purchases
- No push notifications
- No personal information collected
We take your privacy seriously
CONNECT
Let us know if you need help, have a question, have a suggestion, or just want to say hello.
info@littlebigthinkers.com
facebook.com/LittleBigThinkers
twitter.com/LilBigThinkers
## What’s New
Version 1.0
This app has been updated by Apple to display the Apple Watch app icon.
## Ratings and Reviews
5.0 out of 5
1 Rating
1 Rating
Lhamoland ,
### Fun way to practice grade appropriate math skills
My first Grader is enjoying this app which gets her to practice the kinds of problems she is doing in her Go Math program at school. I like how it gets her out of skill and drill worksheet environment into a playful landscape with a monkey. Nice job tying to CCSS.
## App Privacy
The developer, Little Big Thinkers, has not provided details about its privacy practices and handling of data to Apple. For more information, see the developer’s privacy policy.
### No Details Provided
The developer will be required to provide privacy details when they submit their next app update. | 815 | 3,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-38 | latest | en | 0.896509 |
https://www.physicsforums.com/threads/will-2-particles-with-paths-crossed-collide.801693/ | 1,508,439,208,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823360.1/warc/CC-MAIN-20171019175016-20171019195016-00078.warc.gz | 1,002,690,271 | 17,541 | # Will 2 particles with paths crossed collide?
1. Mar 6, 2015
### jefer
1. The problem statement, all variables and given/known data
The first particle A is starting at origin and it has a velocity vector with magnitude 10km/h and moves at an angle of 30 degrees relative to y axis. The second particle B is starting at (0,5) and it has velocity vector with magnitude 8km/h and moves at an agle of 60 degrees relative to y axis. Will these 2 particles colide, and if not, what is the minimal distance they will be at?
2. Relevant equations
3. The attempt at a solution
I found the velocity vectors with given angles/magnitudes. I'm not sure what to do now.
2. Mar 6, 2015
### azizlwl
The particles will collide at time where both at x and y displacement are equal if time t exist.
3. Mar 6, 2015
### jefer
Yeah I got position vector rA and rB from formula r(t)=v*t, where r and v are vectors. They have this form rA=x*t*i + y*t*j where i and j are unit vectors. What do I get from equatting positions, how do I calculate time from that?
The correct answer is that they don't collide and minimal distance is 1.8km. So does it really make sense to equate postitions if we're not 100% sure they'll collide? Im really confused :/
Last edited: Mar 6, 2015
4. Mar 6, 2015
### azizlwl
Find time in y direction where both displacements are equal. Then on x direction or component. If both time are equal then they meet.
5. Mar 7, 2015
### TSny
jefer,
Have you studied the concept of relative velocity? If so, what is the velocity of B relative to A?
6. Mar 8, 2015
### vela
Staff Emeritus
When trying to solve the problem, you can assume they'll collide and see what consequences arise from that assumption. If the consequences don't make sense, then you can conclude that the assumption was wrong. It's like a proof by contradiction in math, if you're familiar with those. | 497 | 1,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2017-43 | longest | en | 0.921479 |
https://mathematica.stackexchange.com/questions/187974/help-solving-complicated-polynomial-equation | 1,563,876,756,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529175.83/warc/CC-MAIN-20190723085031-20190723111031-00416.warc.gz | 459,889,219 | 35,940 | # Help solving complicated polynomial equation
I am trying to solve the following two equations, but I am having troubles doing so.
Equation 1 :
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3) ;
Solve[eq1[n] == 0, n]
If I can find an "n", I would then plug it into the following equation and solve for V.
eq2[n_] := 1024/5*n + 7133.17 n^3 -
0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[nFound] == 0, V]
Sorry, I am not very sure what you want to ask. Maybe you want to substitute the solution of the first equation into the second equation? Then you can do this.
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3);
ansList = Values@Solve[eq1[n] == 0, n]
eq2[n_] :=
1024/5*n + 7133.17 n^3 - 0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[#] == 0, V] & /@ ansList
You also can use the following code to get the value of n.
sol=Solve[eq1[n] == 0, n]
nList=n/.sol
That's all. Please enjoy the fun of Mathematica.
• Thanks ! This was exactly what I was looking for. – james Dec 16 '18 at 9:13
• Never thought about using the Values function in this context. Thanks for the new idea! – Thies Heidecke Dec 16 '18 at 14:02
Solve can handle simultaneous equations. Solve is an exact solver so use Rationalze to provide exact numbers as input. Solve will work without doing this but will provide a warning that it did it internally. Or use NSolve.
eq1[n_] :=
Evaluate[2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3) //
Rationalize[#, 0] & // Simplify];
eq2[n_] := Evaluate[1024/5*n + 7133.17 n^3 -
0.06287*V^2*2.21355*10^(-16)/((1 - n)^2) // Rationalize[#, 0] &];
Solve[{eq1[n] == 0, eq2[n] == 0}, {n, V}] // N
(* {{n -> 0.00634203 + 0.0986751 I,
V -> -7.58581*10^8 - 6.22509*10^8 I}, {n -> 0.00634203 + 0.0986751 I,
V -> 7.58581*10^8 + 6.22509*10^8 I}, {n -> 0.00634203 - 0.0986751 I,
V -> 7.58581*10^8 - 6.22509*10^8 I}, {n -> 0.00634203 - 0.0986751 I,
V -> -7.58581*10^8 + 6.22509*10^8 I}, {n -> 0.587316,
V -> -4.37685*10^9}, {n -> 0.587316, V -> 4.37685*10^9}} *)
For real solutions
Solve[{eq1[n] == 0, eq2[n] == 0}, {n, V}, Reals] // N
(* {{n -> 0.587316, V -> -4.37685*10^9}, {n -> 0.587316, V -> 4.37685*10^9}} *)
For real, positive solutions
Solve[{eq1[n] == 0, eq2[n] == 0, V > 0}, {n, V}] // N
(* {{n -> 0.587316, V -> 4.37685*10^9}} *)
• This is a great answer ! Thank you very much ! – james Dec 17 '18 at 14:24
• How would you do it with Rationalize ? – james Dec 17 '18 at 14:42
• Look at the definition of eq1 above. At the end you will see that Rationalize and Simplify are used. Since SetDelayed ( := ) is used in the definition rather than Set ( = ), Evaluate is used to have these evaluated at the time of definition rather than for each use of the function eq1. Similarly for eq2 except Simplify wasn't needed. Read the documentation for each of these functions. – Bob Hanlon Dec 17 '18 at 15:52 | 1,216 | 3,055 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-30 | latest | en | 0.709152 |
https://princetoneclub.org/what-is-the-least-common-multiple-of-5-and-7/ | 1,656,318,048,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103329963.19/warc/CC-MAIN-20220627073417-20220627103417-00783.warc.gz | 524,165,435 | 5,023 | LCM of 5 and also 7 is the smallest number among all typical multiples the 5 and 7. The first couple of multiples that 5 and also 7 space (5, 10, 15, 20, 25, 30, 35, . . . ) and (7, 14, 21, 28, 35, . . . ) respectively. There are 3 frequently used methods to discover LCM that 5 and 7 - by department method, by element factorization, and by listing multiples.
You are watching: What is the least common multiple of 5 and 7
1 LCM the 5 and also 7 2 List that Methods 3 Solved Examples 4 FAQs
Answer: LCM of 5 and 7 is 35.
Explanation:
The LCM of 2 non-zero integers, x(5) and also y(7), is the smallest confident integer m(35) that is divisible by both x(5) and also y(7) without any remainder.
Let's look in ~ the different methods because that finding the LCM the 5 and 7.
By department MethodBy Listing MultiplesBy prime Factorization Method
### LCM that 5 and 7 by division Method
To calculation the LCM of 5 and also 7 by the division method, we will certainly divide the numbers(5, 7) by their prime components (preferably common). The product of this divisors provides the LCM the 5 and 7.
Step 3: continue the steps until only 1s space left in the last row.
The LCM that 5 and 7 is the product of all prime number on the left, i.e. LCM(5, 7) by division method = 5 × 7 = 35.
### LCM the 5 and 7 through Listing Multiples
To calculate the LCM of 5 and also 7 through listing out the typical multiples, we deserve to follow the given listed below steps:
Step 1: list a few multiples of 5 (5, 10, 15, 20, 25, 30, 35, . . . ) and 7 (7, 14, 21, 28, 35, . . . . )Step 2: The usual multiples native the multiples that 5 and also 7 room 35, 70, . . .Step 3: The smallest usual multiple of 5 and 7 is 35.
∴ The least typical multiple of 5 and also 7 = 35.
### LCM the 5 and also 7 by element Factorization
Prime administrate of 5 and 7 is (5) = 51 and (7) = 71 respectively. LCM the 5 and 7 can be derived by multiply prime determinants raised to your respective highest power, i.e. 51 × 71 = 35.Hence, the LCM of 5 and 7 by prime factorization is 35.
## FAQs ~ above LCM the 5 and 7
### What is the LCM of 5 and 7?
The LCM the 5 and also 7 is 35. To find the LCM (least typical multiple) that 5 and also 7, we need to uncover the multiples the 5 and also 7 (multiples of 5 = 5, 10, 15, 20 . . . . 35; multiples that 7 = 7, 14, 21, 28 . . . . 35) and also choose the the smallest multiple that is precisely divisible through 5 and also 7, i.e., 35.
### If the LCM of 7 and also 5 is 35, uncover its GCF.
LCM(7, 5) × GCF(7, 5) = 7 × 5Since the LCM the 7 and 5 = 35⇒ 35 × GCF(7, 5) = 35Therefore, the greatest common factor (GCF) = 35/35 = 1.
### What room the techniques to uncover LCM of 5 and also 7?
The typically used techniques to uncover the LCM of 5 and also 7 are:
Division MethodPrime administer MethodListing Multiples
### What is the least Perfect Square Divisible by 5 and 7?
The the very least number divisible through 5 and 7 = LCM(5, 7)LCM of 5 and also 7 = 5 × 7 ⇒ least perfect square divisible by every 5 and also 7 = LCM(5, 7) × 5 × 7 = 1225 Therefore, 1225 is the forced number.
See more: A Temporary Group Of Employees Responsible For Bringing About A Particular Change Is A
### How to uncover the LCM of 5 and 7 by prime Factorization?
To uncover the LCM the 5 and 7 using prime factorization, we will find the element factors, (5 = 5) and (7 = 7). LCM that 5 and 7 is the product of prime components raised to your respective greatest exponent among the numbers 5 and also 7.⇒ LCM of 5, 7 = 51 × 71 = 35. | 1,143 | 3,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2022-27 | latest | en | 0.944942 |
http://openstudy.com/updates/4f8c58fee4b0935d57ea1693 | 1,519,482,354,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815812.83/warc/CC-MAIN-20180224132508-20180224152508-00053.warc.gz | 244,936,510 | 8,706 | • anonymous
Match each set of measures of a and b with the measure of c that gives a volume of 24 cubic inches. a = 2in and b = 8 in a = 4in and b = 2 in a = 3in and b = 4 in a = 6in and b = 4in options are A. c = 2 in B. c = 3 in C. c = 4 in D. c = 6 in
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. | 346 | 1,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-09 | latest | en | 0.341123 |
https://www.physicsforums.com/threads/physics-and-geometry.702253/ | 1,529,301,886,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860089.11/warc/CC-MAIN-20180618051104-20180618071104-00424.warc.gz | 892,722,870 | 17,824 | # Homework Help: Physics and Geometry
1. Jul 20, 2013
### Bashyboy
1. The problem statement, all variables and given/known data
A certain quaternary star system consists of three stars, each of mass m, moving in the same circular orbit of radius r about a central star of mass M. The stars orbit in the same sense and are positioned one-third of a revolution apart from one another. Show that the period of each of the three stars is given by
2. Relevant equations
3. The attempt at a solution
What I am having difficulty is with the geometry of this problem. I attached a diagram that the answer key provides. How am I to know that the three planets form a equilateral triangle, what betokens this. Likewise, why is the angle between two sides of the triangle 60 degrees?
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
#### Attached Files:
• ###### Capture.PNG
File size:
5.7 KB
Views:
266
2. Jul 20, 2013
### phinds
is that not clear?
3. Jul 20, 2013
### 462chevelle
think about it.
360/3=120
so if everything is equal. wherever each point is has to equal said number
and a triangle =180 and divided by 3 equals......
4. Jul 20, 2013
### tiny-tim
Hi Bashyboy!
If they move uniformly in a circle, the total force on one of of the outer planets from the other two must point towards the centre …
isn't it obvious then that the position must be symmetic?
(and the angles of an equilateral triangle must be 60° because the angles of any triangle must add to 180°)
5. Jul 20, 2013
### Bashyboy
Well, I had already determined that a 120 degree angle was was maintained between adjacent planets. However, I was not sure if that related to anything.
Why must the angles of any triangle sum to 180 degrees, does this follow from some definition?
6. Jul 20, 2013
### tiny-tim
erm … you should be able to prove this in about 17 different ways!
(eg divide the triangle into two right-angled triangles)
you need to study an elementary geometry book!!
7. Jul 20, 2013
### Bashyboy
I have another question, would the force of one planet on another produce some tangential acceleration?
8. Jul 20, 2013
### haruspex
Yes, but by symmetry the tangential affects of each pair on the third cancel.
9. Jul 20, 2013
### Bashyboy
Oh, I see. There are two tangential forces acting on each planet, each of which is equal and opposite to each other, is this correct?
10. Jul 20, 2013
### tiny-tim
yes, but you're analysing this too much …
isn't it obvious that, if you have two planets of the same mass at the same distance, then the total force will be toward their midpoint?
11. Jul 20, 2013
### Bashyboy
No, it is not immediately evident; however, after having analyzed the problem, I can see that. I don't think I am analyzing the problem too much, I want to understand every detail of every problem I solve.
12. Jul 20, 2013
### Bashyboy
Tiny Tim, is what you say always true?
13. Jul 21, 2013
### tiny-tim
yes, because of symmetry
if you reflect it in a mirror (through the midpoint), you'll have exactly the same …
so the force in the reflected situation must be same as the original force, in other words it must be its own reflection, in other words it must be in the mirror itself, ie towards (or away from) the midpoint
it's this concept of symmetry that you're missing …
many physics exam problems are deliberately constructed with a symmetry in, to help you and to save you time
it is perfectly acceptable in an exam to say "from symmetry, it is obvious that …"
you need to think about symmetry a lot (sorry, but it isn't really a subject you can look up in books), until you're used to spotting it, and using it!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook
Have something to add?
Draft saved Draft deleted | 960 | 3,857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-26 | latest | en | 0.92759 |
https://www.education.com/lesson-plans/fifth-grade/el/math/CCSS-Math-Content-5-NBT-A/ | 1,623,710,406,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00427.warc.gz | 671,328,464 | 24,560 | # Search Our Content Library
11 filtered results
11 filtered results
English Learner (EL)
Math
5.NBT.A
Sort by
Powers of 10
Lesson Plan
Powers of 10
Teach your students to recognize patterns of zeros when multiplying by powers of 10 with exponents.
Math
Lesson Plan
Many, Many Millions
Lesson Plan
Many, Many Millions
Teach your students to read, write, and understand place value in numbers up to the millions place.
Math
Lesson Plan
Menu Math: Rounding Decimals
Lesson Plan
Menu Math: Rounding Decimals
Are your students hungry for math? In this lesson, students pretend to order their favorite takeout foods with their classmates all while practicing rounding decimals so that they know what to expect when the bill comes!
Math
Lesson Plan
Decimals, Decimals, Decimals!
Lesson Plan
Decimals, Decimals, Decimals!
This lesson combines three essential skills in one entertaining game! Students will need to add, subtract, and understand place value with decimals.
Math
Lesson Plan
Dividing Decimals: Estimation
Lesson Plan
Dividing Decimals: Estimation
Teach your students to round decimals to whole numbers to estimate a quotient. In this lesson, students will have so much fun playing the Estimation Station game, they might forget they are learning!
Math
Lesson Plan
Decimal Placement with Division
Lesson Plan
Decimal Placement with Division
Encourage learners to discuss decimal point placement with division and apply their understanding of standard algorithm decimal division. Use this lesson on its own or use it as support to the lesson Dividing Decimals Dash.
Math
Lesson Plan
Expanded Form to Millions
Lesson Plan
Expanded Form to Millions
Saying lengthy numbers doesn't have to be a challenge! Relate the digits in numbers to their place value to help students say the expanded form. Use this lesson on its own or use it as support to the lesson Many, Many Millions.
Math
Lesson Plan
Divide by Powers of Ten
Lesson Plan
Divide by Powers of Ten
Build on students' understanding of the power of ten within multiplication to understand its effect within division. Use this as a standalone lesson or as support to the lesson Dividing Decimals by Powers of 10.
Math
Lesson Plan
Decimal Comparisons
Lesson Plan
Decimal Comparisons
Make decimal comparisons! Your students will focus on comparing decimals and using necessary language to say their comparisons. Use this lesson by itself or use it as support for the Decimals, Decimals, Decimals lesson.
Math
Lesson Plan
Dissecting Decimals
Lesson Plan
Dissecting Decimals
Challenge students to think about decimals while identifying numbers to the hundredths and thousandths. They'll discuss decimal attributes too. Use this lesson on its own or as support to the lesson Dividing Decimals: Estimation. | 594 | 2,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2021-25 | latest | en | 0.884699 |
http://www.cut-the-knot.org/Curriculum/Geometry/CircleCentersOnRadicalAxes.shtml | 1,500,652,480,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423785.29/warc/CC-MAIN-20170721142410-20170721162410-00669.warc.gz | 409,891,361 | 17,315 | # Circle Centers on Radical Axes
## What is this about? A Mathematical Droodle
(Three circles. Drag by any boundary point. Change radius by dragging the center.)
31 March 2016, Created with GeoGebra
Explanation
The applet attempts to suggest the following statement:
Let there be three circles (P), (Q), (R) centered at P, Q, and R, respectively. Assume that P lies on the radical axes of (Q) and (R); Q, on that of (P) and (R). Then R lies on the radical axis of (P) and (Q).
### Proof
The radical axis of two circles is perpendicular to their centerline. Let's denote the radical axes of circles centered at points U and V as LUV. If, as stipulated in the problem, P lies LQR and Q lies on LPR, then the two lines serve as altitudes (from P and Q) of ΔPQR. Their intersection - the radical center O of (P), (Q), (R) - doubles as the orthocenter of the triangle. The third altitude (from R) is perpendicular to PQ and passes through the radical center O, but there is only one perpendicular to PQ through O and this is the radical axis LPQ. It follows that R ∈ LPQ.
Note: this problem generalizes another one where the three circle have been assume to be concurrent. The proof is only a slight modification of one of the proofs of the latter. | 311 | 1,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-30 | longest | en | 0.93954 |
null | null | null | null | null | null | Re: Real World (Re: Mixing OO and DB)
From: rpost <>
Date: Mon, 17 Mar 2008 20:15:24 +0100
Message-ID: <e97ac$47dec34c$839b4533$>
topmind wrote:
>I cannot speak for every domain, but in the domain I am most familiar
>with: custom biz apps, there is no "real world". It's mostly dealing
>with intellectual property (money, invoices, laws, etc.) Even a paper
>invoice is merely a representation and not THE "invoice" per se. Paper
>is simply an old-style implementation. Before that they maybe used
>rocks or animal bones.
>Thus, any operations used are artificial anyhow; all mental
OK, it's partly abstract and intangible, but in the end money buys you food, laws can make people go to jail, etc.
In the example I used (numbers of inhabitants of countries in the world) the operation of counting all people in a country is also "artificial": it doesn't really happen that way. Still, the numbers correspond to that operation. Most attributes in your biz apps are more abstract, but we can still give them concrete meaning in the same way. (This is not to say that we should write code to mimic such operations and consider that code to be a specification.)
>Now, I will agree that OOP generally models how a typical
>clerk might do it: one paper at a time and one operation at time via a
>cursor-oriented pencil. Manual labor has no real set operations.
>Manual labor is generally imperative and thus "navigational".
>However, that is NOT the same as "modeling the real world", but rather
>modeling the old fashioned *implementation*.
You have a point. I agree it would be terrible to define the meaning of, say, an attribute, in terms of how particular implementations of the operations that use it are executed in detail. But OO doesn't direct us to do that. It distinguishes between an operation and its implementation. E.g. we can abstract from the details of how collections and operations on them are implemented.
Received on Mon Mar 17 2008 - 14:15:24 CDT
Original text of this message | null | null | null | null | null | null | null | null | null |
https://math.stackexchange.com/questions/2006467/prove-this-xm1fmx-x | 1,643,219,504,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304959.80/warc/CC-MAIN-20220126162115-20220126192115-00455.warc.gz | 432,430,863 | 37,468 | # prove this $x^{m+1}|f^{(m)}(x)-x$
Let $f(x)$ is polynomial with complex coefficients,such $$x^2|f(x)-e^{\frac{2\pi i}{m}}\cdot x$$ where $m>1$ be give postive integers,and define $$f^{(1)}(x)=f(x),f^{(2)}(x)=f(f(x)),f^{(3)}(x)=f(f(f(x))),\cdots,f^{(m)}(x)=f(f^{(m-1)}(x))$$ show that $$x^{m+1}|f^{(m)}(x)-x$$
I try to use mathematical induction to prove.But there is no proof
• It's difficult to use induction, because you would have to take with you the condition that $x^2\mid f(x)-e^{\frac{2\pi i}{m}}\cdot x$ from one case to the next. The $m$ in there looks to me like it makes it very unpractical, perhaps impossible to prove the induction step. Nov 9 '16 at 11:57
• Interesting problem. May I ask where it comes from?
– dxiv
Nov 11 '16 at 5:10
• Experimentation with Mathematica shows that if $f(x)=w x+\sum_{k=2}^m a_kx^k$ then $f^{(m)}(x)=w^mx+\frac{1-w^m}{1-w}\cdot \sum_{k=2}^mP_kx^k+O(x^{m+1})$, where $P_k$ are polynomials in $w$ and $a_j$. Nov 19 '16 at 19:19
• @Inequality: I suggest to remove the acceptance mark of my answer, since a bounty question which is not already accepted might attract more user. Nov 20 '16 at 11:30
Consider the space $A = t\Bbb C[[t]] = \{a_1t + a_2t^2 + a_3t^3 + \ldots \mid a_i \in \Bbb C\}$.
$A$ can be equipped with the composition law $\circ$ which is linear in the first argument
( $f \circ h + g \circ h = (f+g) \circ h$ and $(\lambda f) \circ g = \lambda (f \circ g)$ for $\lambda \in \Bbb C$).
Thus for any $g \in A$, we get a linear endomorphism $\rho_g(f) = f \circ g$.
If $g = b_1t + b_2t^2 + b_3t^3 + \ldots$, then $\rho_g(t^k) = g(t)^k = b_1^k t^k + \ldots$.
This shows that the "matrix" of $\rho_g$ is triangular (in particular, $\rho_g$ is compatible with the $t$-adic topology) and the coefficients on the diagonal are the sequence $(b_1^n)$.
Restricting modulo $t^{n+1}$ gives you a linear map $\rho_g^{[n]} : A_n \to A_n$ (where $A_n = t\Bbb C_{n-1}[t]$ has dimension $n$) defined by $\rho_g^{[n]}(f) = \rho_g(f) \pmod {t^{n+1}}$ whose matrix is simply the $n \times n$ submatrix in the topleft corner of the infinite matrix of $\rho_g$. Its eigenvalues are the $b_1^k$ for $k=1 \ldots n$.
Now suppose you are looking at a $g$ whose $b_1$ is a primitive $n$th root of unity $\zeta_n$.
Then $\rho_g^{[n]}$ has eigenvalues $\zeta_n^k$ for $k=1 \ldots n$, and since they are all distinct this is diagonalisable, and since their $n$th power is $1$, $(\rho_g^{[n]})^n$ is the identity of $A_n$.
Going back to $A$, this proves that the topleft $n \times n$ block in the matrix of $\rho_g^n$ is $I_n$, and so that for any $f \in A$, $f \circ g^{\circ n} \equiv f \pmod {t^{n+1}}$
Applying this to $t \in A$ you get $g^{\circ n} \equiv t \pmod {t^{n+1}}$
• Instructive approach! Most of the creative work here is finding a proper framework to formulate the problem. If this is done the answer can be seen at a glance. Very nice! (+1) Nov 24 '16 at 7:27
• wonderful proof ! Nov 26 '16 at 15:33
I have offered a bounty for compensation in order to support a correct answer to OPs question.
Let $m>1$ be a positive integer and $f(x)$ a polynomial with complex coefficients and degree $n\geq m$. We denote with $\zeta_m$ the following $m$-th root of unity $$\zeta_m=\exp\left(\frac{2\pi i}{m}\right)$$
Claim: The following is valid for $m>1$
with \begin{align*} f^{(m)}(x)&:=f^{(m-1)}\left(f(x)\right)\qquad m>1\\ f^{(1)}(x)&:=f(x)\\ f^{(0)}(x)&:=x\\ \end{align*}
We introduce some more settings for convenience. Since $x^m|f(x)-\zeta_m x$ there is a polynomial $$q(x)=\sum_{j=0}^{n-m}a_jx^j$$ of degree $n-m$ with \begin{align*} x^mq(x)&=f(x)-\zeta_m x\\ \text{resp.}\qquad\qquad\\ f(x)&=x^mq(x)+\zeta_m x\tag{2} \end{align*}
Approach: The idea is to repeatedly apply (2) and so reduce $m$ in $f^{(m)}$ until we see that (1) is valid. In order to keep the calculation manageable we will consequently simplify expressions modulo $x^{m+1}$.
We do not go the shortest way, but add some intermediate steps to easier see what is going on and to motivate the claim (9) which is central for proving the answer.
Step: $m\rightarrow (m-1)$
We obtain \begin{align*} f^{(m)}(x)-x&=f^{(m-1)}\left(f(x)\right)-x\\ &=f^{(m-1)}\left(x^mq(x)+\zeta_m x\right)-x\tag{3}\\ &\equiv f^{(m-1)}\left(x^ma_0+\zeta_m x\right)-x&\pmod{x^{m+1}}\tag{4}\\ \end{align*}
Comment:
• In (3) we substitute the RHS of (2) for $f(x)$.
• In (4) we note that \begin{align*} f(x)&=x^mq(x)+\zeta_m x\\ &=x^m\left(a_0+a_1x+\cdots+a_{n-m}x^{n-m}\right)+\zeta_m\\ &\equiv x^m(a_0)+\zeta_m&\pmod{x^{m+1}} \end{align*} This behaviour is also valid, when we consider compositions of $f$. This will become more obvious when we calculate the next steps.
Step: $(m-1)\rightarrow (m-2)$
We obtain from (4)
\begin{align*} f^{(m)}&(x)-x\\ &\equiv f^{(m-2)}\left(f\left(x^ma_0+\zeta_m x\right)\right)-x&\pmod{x^{m+1}}\\ &\equiv f^{(m-2)}\left(\left(x^ma_0+\zeta_m x\right)^mq\left(x^ma_0+\zeta_m x\right)\right.\\ &\qquad\qquad\quad +\left.\zeta_m\left(x^ma_0+\zeta_m x\right)\right)-x&\pmod{x^{m+1}}\tag{5}\\ &\equiv f^{(m-2)}\left(x^m q\left(x^m a_0+\zeta_m x\right)+\zeta_m x^m _0+\zeta_m^2 x\right)-x&\pmod{x^{m+1}}\tag{6}\\ &\equiv f^{(m-2)}\left(x^m a_0+\zeta_m x^m a_0+\zeta_m^2 x\right)-x&\pmod{x^{m+1}}\tag{7}\\ \end{align*}
Comment:
• In (5) we note the only contribution of $\left(x^ma_0+\zeta_m x\right)^m\pmod{x^{m+1}}$ is $\zeta_m^m x^m$ and since $\zeta_m^m=1$ we get $x^m$.
• In (6) we note the only contribution of $x^mq\left(x^m a_0+\zeta_m x\right)\pmod{x^{m+1}}$ is the constant part $a_0$ of $q$ multiplied with $x^m$.
Observation: When looking at (4) and (7) we might see a pattern, but to be sure we add one step more.
Step: $(m-2)\rightarrow (m-3)$
We see from (4),(7) and (8) a pattern which we will prove next. In fact we could start the answer with the next step.
Step: $(m-k)\rightarrow (m-k-1)$
We show the following is valid for $1\leq k \leq m-1$: \begin{align*} f^{(m-k)}&\left(a_0 x^m\sum_{j=0}^{k-1} \zeta_m^j+\zeta_m^k x\right)\\ &\equiv f^{(m-k-1)}\left(a_0 x^m\sum_{j=0}^{k} \zeta_m^j+\zeta_m^{k+1} x\right)&\pmod{x^{m+1}}\tag{9} \end{align*}
We obtain \begin{align*} f^{(m-k)}&\left(a_0 x^m\sum_{j=0}^{k-1} \zeta_m^j+\zeta_m^k x\right)\\ &\equiv f^{(m-k-1)}\left(f\left(a_0 x^m\sum_{j=0}^{k-1} \zeta_m^j+\zeta_m^k x\right)\right)&\pmod{x^{m+1}}\\ &\equiv f^{(m-k-1)}\left(\left(a_0 x^m\sum_{j=0}^{k-1} \zeta_m^j+\zeta_m^k x\right)^mq\left(xa_0 x^m\sum_{j=0}^{k-1} \zeta_m^j+\zeta_m^k x\right)\right.\\ &\qquad\qquad\quad +\left.\zeta_m\left(a_0 x^m\sum_{j=0}^{k-1} \zeta_m^j+\zeta_m^k x\right)\right)&\pmod{x^{m+1}}\\ &\equiv f^{(m-k-1)}\left(x^mq\left(xa_0 x^m\sum_{j=0}^{k-1} \zeta_m^j+\zeta_m^k x\right)\right.\\ &\qquad\qquad\quad+\left(\zeta_m\left(a_0 x^m\sum_{j=0}^{k-1} \zeta_m^j+\zeta_m^k x\right)\right)&\pmod{x^{m+1}}\\ &\equiv f^{(m-k-1)}\left(x^ma_0 +\zeta_m\left(a_0 x^m\sum_{j=0}^{k-1} \zeta_m^j+\zeta_m^k x\right)\right)&\pmod{x^{m+1}}\\ &\equiv f^{(m-k-1)}\left(a_0 x^m\sum_{j=0}^{k} \zeta_m^j+\zeta_m^{k+1} x\right)&\pmod{x^{m+1}}\\ \end{align*} and the claim follows.
Putting all together
With the help of (9) we can show OPs claim (1).
Comment:
• In (10) we apply the result (4) of the first step to derive $f^{(m-1)}$ from $f^{(m)}$.
• In (11) we apply the main result (9) $m-2$ times to reduce $f^{(m-1)}$ to $f^{(1)}=f$.
• In (12) we apply (2), the representation $f(x)=x^mq(x)+\zeta_m x$.
• In (13) to (14) we do simplifications $\pmod{x^{m+1}}$ similarly to the steps before.
• In (15) we note $\zeta_m^m=1$ so that $\zeta_m^{m} x-x=0$ and we also use \begin{align*} \sum_{j=0}^{m-1} \zeta_m^j=\frac{1-\zeta_m^{m}}{1-\zeta_m}=0 \end{align*}
• Do I understand it correctly that you did prove the claim only for functions of the type $f(x)=x^mq(x)+\zeta_m x$ and in fact this answer is only partial? Nov 20 '16 at 6:53
• @inequality: Many thanks for accepting my answer and granting the bounty! :-) Nov 20 '16 at 6:56
• Please can you answer my question? Nov 20 '16 at 7:00
• Yes, but OPs condition was that $x^2|f(x)-\zeta_m x$ not $x^m|f(x)-\zeta_m x$. Nov 20 '16 at 7:13
• @Nemo: Oh! Yes, you're right! I've misread the claim. Thanks for pointing at it. I've offered a bounty for compensation. Nov 20 '16 at 8:18 | 3,342 | 8,222 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2022-05 | latest | en | 0.793731 |
https://whatsupwhimsy.com/find-the-volume-in-the-first-octant-bounded-by-84 | 1,675,004,833,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00115.warc.gz | 642,239,171 | 6,005 | ## Determine the volume of the solid in the first octant bounded
How do you find the volume of the solid in the first octant, which is bounded by the coordinate planes, the cylinder x2 + y2 = 9, and the plane x+z=9? Calculus Using Integrals to Find Areas and Volumes Calculating Volume using
## Volume of region in the first octant bounded by coordinate
Deal with math problem
Solve mathematic equations
Do math problems
Work on the task that is interesting to you
Get math help online
## 1. Find the volume of the region in the first octant bounded by
• Reach support from expert tutors
• Figure out math problems
## Clients said
It's in any language, and so easy to use. It's helped me get homework done in about an hour when it usually takes up to 2-4 hours, it's also very aesthetic ✨.
Ryan Bearden
There are very good step by step instructions. Often I try not too read to far into the solution just using the first few steps to get me going on a problem. It's a 5/5 experience. Not only does it pick up any problem in a complete mess of notes and in any writing format, but it also gives amazing and in-depth explanations of how to solve your problem.
Kurt Townsley
Thanks a bunch for your app, so much helpful in difficult math problems. When it gives the answeres it also gives you the steps, and the process on how to get that solution to that problem. As of this point I have been only been using this app for a few weeks now and it has been great.
Hung Flood
## Volume in the first octant bounded by the coordinate planes
Top Specialists
More than just an app
If you want to save time, do your research and plan ahead.
Get arithmetic support online
I can't believe I have to scan my math problem just to get it checked.
Passing Rate
Solving math problems can be a fun and rewarding experience.
## Find the volume of the solid in the first octant bounded by
Viewed 4k times 0 Find the volume of the solid in the first octant bounded by the cylinder z = 9 − y 2 and the plane x = 2 Can I solve this problem using triple integrals in the
We have the best specialists in the business.
Solve mathematic problems
Solving math equations can be challenging, but it's also a great way to improve your problem-solving skills.
Always on Time
There are many ways to improve your writing skills. | 538 | 2,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-06 | longest | en | 0.944566 |
https://forum.allaboutcircuits.com/threads/another-math-problem-fun.141975/ | 1,579,296,770,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250591234.15/warc/CC-MAIN-20200117205732-20200117233732-00365.warc.gz | 456,531,345 | 24,397 | # Another math problem (fun)
#### MrChips
Joined Oct 2, 2009
19,728
Three friends were in a restaurant and at the end of the meal the bill comes up to $27 including gratuities. They each give$10 to the waiter. The waiter goes to the cashier who notices that the waiter had made a mistake in tallying the tab. The correct total is $25. The cashier gives the waiter five$1 coins in change (they don't print $1 bills anymore). On the way back to the table, the waiter ponders on how he is going to split$5 three ways.
No problem, he thinks. "They think the total is $27. I will give them each$1 in change and I'll keep the remaining $2. Problem solved." Here is the math question. Each customer paid$9, times three equals $27. The waiter kept$2, equals $29. What happened to the missing$1?
(This one works best after your friends have had a few drinks.)
#### WBahn
Joined Mar 31, 2012
24,966
Each customer paid $9 totaling$27. The waiter kept $2 of the$27 dollars, leaving $25 to pay the bill. #### MaxHeadRoom Joined Jul 18, 2013 19,387 That one has been around since the 1960's to my knowledge. Max. #### strantor Joined Oct 3, 2010 5,177 This one got me good when I was a kid. I remember trying to work it out with 30 scraps of paper. Thread Starter #### MrChips Joined Oct 2, 2009 19,728 That one has been around since the 1960's to my knowledge. Max. Must be. That was when it was told to us by our math teacher. The mathematically challenged in the class couldn't figure it out no matter how hard you tried. #### strantor Joined Oct 3, 2010 5,177 Must be. That was when it was told to us by our math teacher. The mathematically challenged in the class couldn't figure it out no matter how hard you tried. Mathematically challenged? Maybe I'm just trying to give myself some consolation, but I think that problem would stump all but the brightest people on the first go-round (and probably the 2nd, 3rd) #### WBahn Joined Mar 31, 2012 24,966 Mathematically challenged? Maybe I'm just trying to give myself some consolation, but I think that problem would stump all but the brightest people on the first go-round (and probably the 2nd, 3rd) It's not so much a difficulty with math as it is letting the person that asks a question bias you with a false assumption. If the "math question" has simply been to account for the entire$30 originally given to the waiter, would anyone be stumped? No. What happens to every dollar was clearly stated. But the person that asked the question explicitly adds $2 to$27 even though there is NOTHING in the problem that indicates any justification for doing so. But by incorporating it into the question, the person asking the question leads the person being asked to defer to authority and blindly accept that performing this summation is a reasonable thing to do when accounting for the money.
My guess is that if you tested the people that are given this problem you would find that those that really struggle with it tend to be highly field dependent, because field dependence is highly correlated with deferring to authority while field independence is highly correlated with questioning authority (i.e., not accepting things at face value just because it is presented by someone whose credentials aren't suspect).
#### GopherT
Joined Nov 23, 2012
8,012
Just order an extra beer and it all comes out even.
#### shortbus
Joined Sep 30, 2009
7,219
This was going around the room at our Thanksgiving gathering.
#### WBahn
Joined Mar 31, 2012
24,966
Even though I've had plenty of experience with people whose reasoning and logic skills are at least as bad as her's purports to be, I take all of these videos with a huge grain of salt because so many of them are nothing but staged fakes. But whether it is or not, the interviewer is doing a horrible job because he just keeps asking the same question over and over and over (until the very end) when it is clear she doesn't understand the question and it is equally clear how she is interpreting it. If it isn't staged, then why should he be surprised that someone answers the same question the same way time after time, and if it is staged, it isn't a very good script because what is so noteworthy about someone giving the same wrong answer to the same question repeatedly?
#### Raymond Genovese
Joined Mar 5, 2016
1,658
Even though I've had plenty of experience with people whose reasoning and logic skills are at least as bad as her's purports to be,/---/
No argument from me as to the possibility of "staged entertainment". If it is genuine, however, I would argue that her responding has little to do with reasoning or logic skills, it has much more to do with listening deficits and comprehension bias. Those, in my view, are distinct, but not necessarily completely independent, from reasoning and logic.
She insists from the start on hearing the question, "How old were you five years ago?" and not "How old are you if you were born five years ago?" Possibly primed by previous addition/subtraction problems. She has a well designed filter at work - a predisposition. We all have those I suppose and this is just an umm, "robust" example. Not unlike the old children's riddle, "What is heavier, a pound of feathers or a pound of lead?"
In defense of the recipients of many a "Duh?".... if only they changed the question to "What feels heavier, a pound of feathers or a pound of lead?"
Last edited:
#### GopherT
Joined Nov 23, 2012
8,012
She has a whole series of these idiot videos...
Here is another "interview"
Joined Jul 18, 2013
19,387
The Local paper just published a list from a local travel agent that consisted of questions potential travelers ask of her over the years.
The one I particularly liked was a person planning a trip to the UK, she wanted to know if they spoke English there!.
Max.
#### Raymond Genovese
Joined Mar 5, 2016
1,658
She has a whole series of these idiot videos...
Here is another "interview"
I just watched several of this "series". I don't know, but I am convinced that they are staged (possibly including some natural talent). Phrases like "hot mess", "male fantasy" and "Bimbo appeal" come to mind. Incredibly large number of views for many of them. Quite successful if that was the intent.
#### MrChips
Joined Oct 2, 2009
19,728
The Local paper just published a list from a local travel agent that consisted of questions potential travelers ask of her over the years.
The one I particularly liked was a person planning a trip to the UK, she wanted to know if they spoke English there!.
Max.
Is Scouse English? I can't understand a word of it!
My inlaws are all scousers!
Joined Jul 18, 2013
19,387
Scouse is Liverpool/Merseyside, (The Beatles Origin).
Most are of Irish descent so may have some influence on it.!
Max.
#### sarathd
Joined Mar 1, 2018
1
Three friends were in a restaurant and at the end of the meal the bill comes up to $27 including gratuities. They each give$10 to the waiter. The waiter goes to the cashier who notices that the waiter had made a mistake in tallying the tab. The correct total is $25. The cashier gives the waiter five$1 coins in change (they don't print $1 bills anymore). On the way back to the table, the waiter ponders on how he is going to split$5 three ways.
No problem, he thinks. "They think the total is $27. I will give them each$1 in change and I'll keep the remaining $2. Problem solved." Here is the math question. Each customer paid$9, times three equals $27. The waiter kept$2, equals $29. What happened to the missing$1?
(This one works best after your friends have had a few drinks.)
Forget the \$ 27.00 value that is the wrong value and recalculate in and out numbers. | 1,843 | 7,685 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-05 | latest | en | 0.969916 |
https://www.queryhome.com/puzzle/12075/if-3-4-3-4-4-35-5-4-4-6-4-12-7-4-60-then-8-4 | 1,542,088,054,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741219.9/warc/CC-MAIN-20181113041552-20181113063552-00308.warc.gz | 970,446,273 | 31,146 | # If 3+4=3; 4+4=35; 5+4=4; 6+4=12; 7+4=60 then 8+4=?
2,945 views
If
3+4=3
4+4=35
5+4=4
6+4=12
7+4=60
Then
8+4=?
posted Jan 10, 2016
Ans will be 5
Look
3+4=3 in this
Multiply these 3 and 4 and when u get the two digit ans, square the both, higher digit square will be substracted by lower digit square
Now 3*4=12
2^ - 1^ = 4-1=3
Here ^ is used by me as square
4+4=35
4*4=16
6^ - 1^=36-1=35
Solve all like this and finally
8+4
8*4=32
3^ - 2^=9-4=5
And 5 is ans
Ans given by 8601084829
Krishna P. Sharma
answer Aug 13, 2016 by anonymous
Wow wonderful, touch one to crack
Similar Puzzles
If
2+3=8,
3+7=27,
4+5=32,
5+8=60,
6+7=72
then
7+8=??
–1 vote
If 2+3=3, 4+4=5, 4+5=7, 5+7=10, 5+8=12, 6+5=8
then (3+7)+2 = ? | 374 | 716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-47 | latest | en | 0.568094 |
http://forum.onlineconversion.com/showthread.php?p=72888 | 1,398,337,064,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00119-ip-10-147-4-33.ec2.internal.warc.gz | 124,945,532 | 9,981 | Welcome to OnlineConversion.com Forums
OnlineConversion Forums Inches to Square foot
User Name Remember Me? Password
[ Home ] [ Forum Home ] Register FAQ Calendar Search Today's Posts Mark Forums Read
Convert and Calculate Post any conversion related questions and discussions here. If you're having trouble converting something, this is where you should post.* Guest Posting is allowed.
Thread Tools Display Modes
#41
08-25-2010, 07:27 AM
eleni Guest Posts: n/a
Re: Inches to Square foot
i have 18x18 tiles i need to know how many square feet are the 62 tiles 18x18
#42
08-25-2010, 11:08 AM
JohnS Moderator Join Date: Dec 2007 Location: SE Michigan, USA Posts: 9,547 Rep Power: 19
Re: Inches to Square foot
Quote:
Originally Posted by eleni i have 18x18 tiles i need to know how many square feet are the 62 tiles 18x18
1 ft = 12 in, so
18" = 1.5'
18 " x 18" = 1.5' x 1.5' = 2.25 ft²
62 tiles is 139.5 ft²
(If you place tiles with a slight gap, filled with grout, it will be a little more, depending on the spacing)
#43
08-29-2010, 06:50 AM
Unregistered Guest Posts: n/a
Re: Inches to Square foot
my house roof is 51'.2" X 51'. How do I convert it to square feet?
#44
08-29-2010, 09:25 AM
JohnS Moderator Join Date: Dec 2007 Location: SE Michigan, USA Posts: 9,547 Rep Power: 19
Re: Inches to Square foot
Quote:
Originally Posted by Unregistered my house roof is 51'.2" X 51'. How do I convert it to square feet?
2" is 1/6 of a foot 0.167) so 51 x 51.167 = 2609.5 ft²
#45
03-22-2011, 01:26 AM
Unregistered Guest Posts: n/a
Re: Inches to Square foot
Quote:
Originally Posted by Unregistered I have two pictures 1) measuring 46 inches x 150 inches. Need to know how much would that be in square feet 2) the other is measuring 248 inches x 168 inches. How much would that be in square feet
How much square feet would be (a) 32 inches X 11 inches ? (b) 32 inches X 7 inches ?
#46
03-22-2011, 02:48 AM
JohnS Moderator Join Date: Dec 2007 Location: SE Michigan, USA Posts: 9,547 Rep Power: 19
Re: Inches to Square foot
Quote:
Originally Posted by Unregistered How much square feet would be (a) 32 inches X 11 inches ? (b) 32 inches X 7 inches ?
32 in x 11 in x 1 ft²/144 in² = 2.44 ft²
You can do the other one the same way
#47
04-21-2011, 06:29 PM
Unregistered Guest Posts: n/a
Re: Inches to Square feet
bought a raw material and the size is 2mm x 60" x80"
what is the conversion in square feet, please help
#48
04-22-2011, 02:57 AM
JohnS Moderator Join Date: Dec 2007 Location: SE Michigan, USA Posts: 9,547 Rep Power: 19
Re: Inches to Square feet
Quote:
Originally Posted by Unregistered bought a raw material and the size is 2mm x 60" x80" what is the conversion in square feet, please help
I assume you want the area of the flat 60"x80" sheet converted to square meters
60" x 80" x (0.0254 m/ft)² = 4.13 m²
(or you could just measure it in meters and multiply )
#49
04-25-2011, 11:18 AM
Unregistered Guest Posts: n/a
Re: Inches to Square foot
Need to know how to convert inches into sq. ft for example 11" x 86" what is the sq. ft for this
#50
04-25-2011, 02:51 PM
JohnS Moderator Join Date: Dec 2007 Location: SE Michigan, USA Posts: 9,547 Rep Power: 19
Re: Inches to Square foot
Quote:
Originally Posted by Unregistered Need to know how to convert inches into sq. ft for example 11" x 86" what is the sq. ft for this
There's 12 inches in a foot, so 144 in² in 1 ft²
11 in x 86 in x 1 ft²/144 in² = 6.57 ft²
Thread Tools Display Modes Linear Mode
Posting Rules You may post new threads You may 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 Off Forum Rules
Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Main Forums Convert and Calculate Resources General Chat
All times are GMT -8. The time now is 02:57 AM.
Contact Us - OnlineConversion.com - Archive - Privacy Statement - Top
Powered by vBulletin® Version 3.8.7
Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
Copyright ©1997-2005, 2007, OnlineConversion.com | 1,244 | 4,105 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2014-15 | latest | en | 0.839218 |
https://iq.opengenus.org/remove-k-digits-to-make-smallest-number/ | 1,726,717,302,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00308.warc.gz | 275,644,617 | 20,566 | ×
Search anything:
# Remove K digits to make smallest number
#### Algorithms Monotonic Stack Stack Greedy Algorithms
Open-Source Internship opportunity by OpenGenus for programmers. Apply now.
In this article, we will be solving the problem of removing K digits from a given number to form the smallest number possible without changing the order of the original number. We will use the idea of Monotonic Stack.
Table of content:
1. Understanding the problem
2. Algorithm & Pseudocode
3. Implementations
4. Explanation / Code walk-through
5. Time and Space Complexity
# Understanding the problem
In this problem, the user provides us with two input numbers NUM and K, and we have to form the smallest number possible from the input number NUM by removing K digits from it without changing the order of the original number.
Suppose the user has provided us with the number NUM = 9154 and K = 2.
What exactly is K here? K is the number of times that we are allowed to remove a digit from the given number in order for us to find the smallest number. So in the number 9154, we first remove the digit 9, which is a peak digit.
What exactly is a peak digit?
Any digit in the number that is greater than the digit present ahead of it in the number is a peak digit. We have to remove as many peak digits from the number as we can to find the smallest number.
After removing 9, K becomes 1 and the number becomes 154. After that, the digit 5 is removed since it is a peak digit, K becomes 0 and the number becomes 14.
Since K has become 0, we cannot remove any more digits from the number, and hence the resultant number that we have is our final answer, which in this case, is 14.
You might be wondering as to what will happen if the input number provided by the user is something like 11111, where we cannot locate a peak digit. What do we do in such cases?
Well, all the corner cases will be handled as well in this article, so keep on reading!
This problem can be solved in many ways, however, we will be solving it by making use of a greedy algorithm along with the stack data structure.
# Algorithm & Pseudocode
In short, we use the idea of Monotonic Stack to solve this problem in linear time.
The idea is to insert each digit into a Stack provided the top of Stack is greater than current element. If top of stack is less than current element, then pop the top of stack and insert current element on checking the condition again. Note the pop operation can be done only K times as per the problem (removing K digits).
The intuitive form of the algorithm is as follows:-
1. Convert the input number into a type where each digit of the number can be iterated properly
2. Iterate through each digit of the input number
2.1 If the digit is lesser than the digit present at the top of the stack, delete the element at the top of the stack and keep track of K
2.2 Add the digit of the input number to the stack
3. Perform a check for corner cases
4. Return the stack in the form of an integer as our final answer
A more elaborate form of the Algorithm / Pseudocode is as follows:-
``````def smallestNumber(num, K) as:
1. If K == 0 do:
1.1 return num
2. If K == length of num do:
2.1 return 0
3. Convert num into a type so that it will be easier to traverse through each digit
4. Declare an empty stack with an appropriate data type, if required
5. For each digit in num do:
5.1 While K != 0 and length of stack > 0 and top element of stack > digit do:
5.1.2 Decrement K by 1 and remove the top element from the stack
5.2 Add digit to the stack
6. If K > 0 do:
6.1 Replace stack with stack itself by reducing the number of elements present in it by K times from the right
7. Return the stack in the form of an integer as our final answer
``````
# Implementations
Implementation in Python:
``````def smallestNumber(num, K):
if(K == 0):
return num
if(K == len(str(num))):
return 0
l, stack = [int(x) for x in str(num)], []
for i in l:
while(K>0 and len(stack)>0 and stack[-1]>i):
K -= 1
stack.pop()
stack.append(i)
if K > 0:
stack = stack[:-K]
return str("".join([str(n) for n in stack]))
num, K = 95163, 3
print(smallestNumber(num, K))
``````
Implementation in Java:
``````import java.util.Stack;
public class Sky {
static int smallestNumber(int num, int K) {
String temp1 = String.valueOf(num);
StringBuilder temp2 = new StringBuilder();
int numLength = temp1.length();
if(K == numLength) {
return 0;
}
if(K == 0) {
return num;
}
int index = 0;
Stack<Integer> stack = new Stack<Integer>();
while(index < numLength) {
while(K > 0 && !stack.isEmpty() && stack.peek() > Character.getNumericValue(temp1.charAt(index))) {
K--;
stack.pop();
}
stack.push(Integer.parseInt(String.valueOf(temp1.charAt(index))));
index++;
}
while(K > 0) {
stack.pop();
K--;
}
while(!stack.isEmpty()) {
temp2.append(stack.remove(0));
}
return Integer.parseInt(String.valueOf(temp2));
}
public static void main(String args[]) {
System.out.println(smallestNumber(95163, 3));
}
}
``````
# Explanation / Code walk-through
Let's go through the code line by line and understand as to how the algorithm works in detail.
• In this algorithm, we will go through each digit present in the input number and insert it or delete it from the stack depending upon the results of some checks/conditions.
• The input number is 95163. It is in the integer format and so in order for us to traverse/iterate over each digit in the number, we will have to first convert it into a list.
Converting it into a list, we get:
l = [9, 5, 1, 6, 3]
• An empty stack is also declared, which will be used to perform various operations, comparisons and also store our final answer.
• We also check for corner cases such as if the value of K is 0 or equal to the length of the input number, in which cases we will simply return the number itself and 0 respectively.
• {Entering the first loop} In this loop, we will be iterating over each digit that is present in our list l. We also have another loop inside this loop, that will only run if the three conditions that are given below are true:-
(1) The value of K should be greater than 0.
(2) The total number of elements in the stack should be greater than one.
(3) If the stack is not empty, then the element present at the top of the stack should be greater than the current digit from the list.
• When we come out of the inner loop, then the current digit is added to the stack and then these steps are repeated again until we reach the end of our list.
• Let's perform these steps on our list and visualize as to how the algorithm works on the back-end as well.
• We have l = [9, 5, 1, 6, 3]
For our first iteration, we have i = 9, stack = [], K = 3.
Since two of the conditions for the while loop to run are not true, i.e., length of stack is not greater than zero and the top element of the stack is not greater than i, the loop won't run. Hence, the digit 9 is added to the stack.
• For the second iteration, i = 5, stack = [9], K = 3.
Since all the conditions for the while loop are true, i.e., K > 0, length of stack is greater than zero (1) and the top element of the stack is greater than i (9 > 5), we go inside the while loop.
(1) K becomes 2 as it is decremented by 1.
(2) The element 9 is removed from the stack.
(3) Finally, when we come out of the while loop, the digit 5 is added to the stack.
• For the third iteration, i = 1, stack = [5], K = 2.
Since all the conditions for the while loop are true, i.e., K > 0, length of stack is greater than zero (1) and the top element of the stack is greater than i (5 > 1), we go inside the while loop.
(1) K becomes 1 as it is decremented by 1.
(2) The element 5 is removed from the stack.
(3) Finally, outside of the while loop, the digit 1 is added to the stack.
• For the the fourth iteration, i = 6, stack = [1], K = 1.
Since one of the condition for the while loop is not true, i.e., the top element of the stack is not greater than i (1 !> 6), the while loop will be skipped and i (6) is directly added to the stack.
• For the fifth iteration, i = 3, stack = [1, 6], K = 1.
Since all the conditions for the while loop are true, i.e., K > 0, length of stack is greater than zero (2) and the top element of the stack is greater than i (6 > 3), we go inside the while loop.
(1) K becomes 0 as it is decremented by 1.
(2) The element 6 is removed from the stack.
(3) Finally, when we come out of the while loop, the digit 3 is added to the stack.
• We have now completed the main loop of our algorithm and our stack = [1, 3].
If we combine these digits from the starting of the stack (0th index till nth index), we will get our final answer, i.e., 13. We can return this as our final answer.
• You might have noticed that we perform one more check after our main loop that checks if the value of K > 0.
We perform this check for corner cases for numbers such as 11111, 33333, etc. where there is no peak digit. If this condition is true, then we will simply remove K elements from the number and return it as our final answer.
# Time and Space Complexity
Time complexity:
We have a main loop in the algorithm that runs over n times, where n is the number of digits present in the input number. We also have another conditional loop inside the main loop that will run K times.
We know that each element is going to be processed at most twice with respect to the operations performed (1 push and 1 pop) and hence there will be 2n operations at most. This makes our complexity (2N).
Let us assume that a constant amount of work C is being done each time inside each loop (different operations such as comparison, insertion, removal, etc.), which makes our total complexity ((2N) + C).
We can ignore the constants.
Hence, the overall complexity becomes: `O(N)`.
Space complexity:
The length/size of the stack depends upon the input from the user, i.e., n.
Hence, the space complexity will be `O(N)`.
With this article at OpenGenus, you must have the complete idea of solving this problem of removing K digits to make smallest number using the idea of Monotonic Stack.
Remove K digits to make smallest number | 2,532 | 10,105 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-38 | latest | en | 0.935926 |
http://nathankraft.blogspot.com/2015/ | 1,539,980,098,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512434.71/warc/CC-MAIN-20181019191802-20181019213302-00128.warc.gz | 245,052,101 | 23,305 | ## Thursday, October 29, 2015
### Can you remember more than 7 digits?
The other day, I came across this website that tests your ability to remember digits.
I thought it was interesting that, according to the website, the average person can remember 7 numbers at once. I've heard this before. This is supposedly the reason why telephone numbers are 7 digits long. At this point, I'm sure you're wondering if you are an "average" person. So, go try it...http://www.humanbenchmark.com/tests/number-memory.
Did you do it? I did it a few times myself and the farthest I got was 12 digits (my worst was 10). This probably means that I'm a superhuman or I have evolved past the rest of you. I'm sorry, but your days are numbered. (Numbered! Get it? No, of course you don't.)
I was still curious about this 7 digit claim, so I posed the problem to my students. Can the average person really only remember 7 numbers?
I had all of my students load the website and play along. After everyone was finished, I recorded the results and made a line plot with the data.
I asked the students to talk to their neighbors about whether or not this data confirms that the average person can remember 7 digits. Overwhelmingly, they felt pretty good about it, especially since the median of the data was 7. (I should note that sixth grade standards are all about analyzing distributions.) They were also able to see that more than half of the students were able to remember at least 7 digits, but less than half could remember 8 or more. Another reason to believe the claim that the average person could remember 7 digits.
We then discussed strategies for memorizing the numbers. Some students mentioned that they chunked the data...remembering 62 as "sixty-two" instead of "six-two". Some of them would practice typing them to build the motor memory.
I also shared a couple of my own strategies...sometimes I could associate a number with something. For instance, once I saw a 53 and, for whatever bizarre reason, I remember that as Bobby Abreu's jersey number. Once I had that image of Bobby Abreu in my head, I stopped worrying about remembering 53. For the longer sets of digits, I would repeat the second half of digits over and over again while staring at the first half of digits. This way, I was relying on both my visual and auditory memory.
Now that the students had some new strategies, I gave them another chance to increase their digits.
As you can see, the data changed, but there really wasn't much improvement. Many students did worse while a few did marginally better. We couldn't make much sense of it, though we suspected that some of these strategies need to be practiced before we could see some results.
At this point, it would have been nice to keep practicing to see if we could improve, but my period is only 37 minutes long. I also had a couple of situations where students figured out they could copy and paste their answers. Cheating would be difficult to monitor.
Side note: Some of my students with IEPs could only remember three digits. This was consistent each time they made an attempt. This was eye-opening for me...when short-term memory is so weak, learning anything must be a huge struggle.
## Saturday, October 24, 2015
### Why, Common Core? Why?
The other day, I was checking students' work on mean, median, and mode. One of the problems involved finding out what grade you would need to get on a fourth test to have an average of 85 for the class. It's basically a mean problem in reverse, and for students who have never solved this problem, it can be challenging.
One of my students was struggling with this and wrote in her notebook, "WHY COMMON CORE WHY". I laughed and assured her that this problem has been around a lot longer than Common Core. What I really found amusing was that, in terms of content, this sixth grader really hasn't been exposed to some of the more unique things about Common Core. Most of that is happening in elementary school and Pennsylvania only switched over last year, when she was in fifth grade.
In all likelihood, this girl's hatred towards Common Core probably stems from something she overheard her parents say. And now, every time I present her with a challenge, a little voice in the back of her head is going to tell her that this problem is Common Core and it's not really important for her to figure it out. And that's all she needs...another reason to give up.
## Tuesday, September 29, 2015
### Warm-Ups with a Purpose
Warm-ups last year:
I would display four or five review problems on the Smartboard for students to work through as I took attendance. I would then walk around the classroom to see how students were progressing, but would often struggle to help very many of them, nor would I have a good sense of how the class did as a whole. We would then review every problem which was time consuming and not always helpful. The next day, I would create a few more warm-up exercises but I never had a clear picture of what my students were still struggling with or why.
Warm-ups this year:
I was asked to move into a new classroom where every student would have his or her own computer. Over the summer, I looked at several websites that would help me use formative assessment on a daily basis. I was happy to find Socrative (which is FREE!) and I use it everyday for my warm-ups. Students can quickly log in and start working on the exercises. I can create multiple choice, true/false, or short answer questions, and as students are answering them, I can see their responses live! It looks something like this...
This is kind of a big deal. As soon as a student gets something right or wrong, I know. And there's a lot I can do with that information. During those exercises, you'll routinely hear me say things like...
"Mary, awesome job on that last one. Everyone's having trouble with it."
"Almost everybody's getting #1 wrong. Make sure you read it carefully!"
"Sheri, that last one...how are you supposed to set up an addition problem with decimals?"
"Fawn, you seem to be having trouble with greatest common factor. Can I see your work for that last problem?"
"Hey, Andrew. Where's your notebook? Stop trying to do the work in your head. You're not Rain Man!"
After the students finish the exercises, I share the results with them and I let them tell me which ones we need to review (and which ones we don't). We look at commonly selected wrong answers and think about what mistakes students were making.
At the end of the day, I can throw this data onto a spreadsheet (shown below) and decide which topics/skills students have a firm grasp and which need further review. I can see how students progress in some skills over time and share that as a model of learning.
I love that students are getting instant feedback. I love that I have evidence of their growth. I love that we can review results as a class and, rather than students only focusing on their own mistakes, we can ask ourselves, what are we, as a class, doing wrong? What are we, as a class, doing right?
## Sunday, September 20, 2015
A few weeks ago, I blogged about how I was going to stop putting grades on quizzes. This decision was heavily influenced by Dylan Wiliam's ideas from his book, Embedded Formative Assessment. I also need to mention that Ashli Black has very helpful explaining how she does comments-only grading and pushing me to design a system of grading that works.
This past week, I was finally able to test-run this idea after the students took a quiz on the Order of Operations. I explained my reasoning to the students and, for the most part, they seemed to be okay with it. I told them that this creates a better working environment where students can feel less embarrassed about their performance and work together to identify and correct their mistakes, no matter how well they did. I marked the quizzes by circling the problem number for every wrong solution and then color-coding three problems that I wanted the student to correct. If a problem had a pink mark, they had to identify their error. If there was a purple mark, they had to rework the problem. If a student did not get anything wrong, I gave them a more challenging problem to solve. Finally, while grades were not written on the quizzes, they were calculated and recorded into the online gradebook so parents and students could see them at home.
Overall, I thought it went really well. The students had about 10 minutes to work alone or together on their mistakes and handed the quizzes back to me. Those who did not finish had extra time overnight to do so.
The next day, I used socrative (an online quizzing tool) to ask my students how they felt about my "no grade" policy. The good news is that 70% of my students either liked it or didn't care. More students liked it than didn't like it. However, there is still 30% of my students that didn't like it. While it was not obvious in their responses, I believe that this frustration comes from not having that instant gratification of knowing what your grade is. This impatience isn't unexpected. Many times students will ask me if I graded their quiz ten minutes after handing it in.
In the end, I think the benefit of students revisiting their work and working together to fix mistakes outweighs the annoyance of not getting their grades right away. I'm hoping that, over time, students will begin to also see that benefit.
As a side note, I should say that I'm not really doing "comments-only grading". I had considered writing out comments, but it occurred to me that most of what I'd be writing could later be discovered by the student upon more reflection or figured out with help from a classmate. I believe that writing comments on every wrong answer would have been extremely time consuming and would have deprived my students from discovering their own mistakes.
Update 2/15/15:
Carolina Vila (@MsVila on twitter) asked me if I have kept up with this system. As with anything I experiment with, I look for more efficient ways to do things. (Okay, maybe I just got lazier.)
I mentioned that I color-coded problems in the beginning of the year and that these colors would tell students how I wanted them to reflect on each problem (identify the error/explain what they did wrong or rework the problem). After doing this a few times, it just seemed to make more sense to have students do both things. On a separate piece of paper, they would have to tell me which three problems they chose to rework, tell me (in sentence form) what they did wrong, and finally, rework the problem.
For students that got everything right, I backed away from trying to give them a more challenging problem, and instead, asked them to help other students make their corrections.
Students would turn in their corrections along with their quiz, I would check to see that it was done, AND THEN, I would write their grade on the quiz to give back to them the next day. When I first started taking grades off of the quizzes, I had hoped that I could just put their grades online for them to check, but I ran into too many issues where students and parents couldn't check the grades online because they lost their passwords or didn't have internet access at home. By finally putting the grades on the quizzes, students complained less and respected the correction process more.
On the student side, one of the biggest misconceptions was that making quiz corrections would improve their quiz grade. I explained that they would get credit for making the corrections (similar to a homework grade), but that their quiz grade would remain the same. The only way their grade would improve would be to retake the quiz, and the only way a student would be allowed to retake a quiz is if he or she made the corrections on the first quiz. Altogether, there is plenty of incentive to make these corrections.
## Monday, August 3, 2015
### Spaced Practice and Repercussions for Teaching
I've been reading John Hattie's book, Visible Learning, in which he ranks the effect sizes of different strategies that help student achievement. One of the strategies that is pretty high on the list is that it is better to give students spaced (or distributed) practice as opposed to mass practice. In other words, rather than having a student practice something over and over again in one day, it is much better to spread that practice out over multiple days or weeks. (You can read one of these studies here.) The main benefit is that spaced practice helps with long-term retention.
While this research certainly gives some justification for providing students with multiple opportunities to revisit older topics, I am left to wonder if this should change how I structure my lessons and assessments. I, like many others, teach by units. My students might spend a month on fractions followed by a test. They then get a month of algebra followed by another test. We, as teachers, create this span of time when all learning about a particular topic must happen. We don't always give students the time to practice these ideas, particularly the more challenging ones that almost always happen at the end of the unit and right before the test.
Based on what I've read about spaced practice, I would propose that teachers shouldn't give tests at the end of a unit. Perhaps students need time to practice these skills over several weeks before you should assess them. This is something I'm going to explore this year with some of the concepts that were challenging for my students last year.
Note: This is probably not an original idea and I'm sure someone else out there has probably explored it. If you have any resources to share on the subject, I'd greatly appreciate it!
Another note: I do allow my students to retake quizzes which I had hoped would send the message that learning doesn't stop after the quiz is taken. However, very few of my students have taken advantage of this in the past. I am hoping to correct that this year with some ideas from Dylan Wiliam, Ashli Black, and others.
Update: Henri Piccioto has written about this and calls it "lagging homework". He also reinforces the idea that quizzing should happen much later then when the material was taught. Thanks to Mary Bourassa and Chris Robinson for helping me find his work!
## Sunday, August 2, 2015
### Movie Popcorn
I ordered a small popcorn at the movie theater and the cashier asked me if I'd like the large size for only \$1 more. I knew that this had to be the better deal, so I took it. I mean, what if I had gotten the small popcorn and ran out during the movie? That would be unacceptable.
However, as I left the theater, I noticed that I didn't actually eat all of the popcorn. There was about two and a half inches of popcorn left at the bottom of the bucket. I could take it home with me, but stale popcorn doesn't sound too appetizing and I decide to throw it away. Did I just get ripped off? Should I have just bought the small popcorn?
There's a couple of ways of modifying this task to address the needs of different grade levels. It all depends on what information is given to the students. If you can just give the students the number of cups of popcorn in each bucket, then this is a fairly simple unit price problem. If you just give dimensions of the buckets, you will need to derive and use formulas. It would also be extremely helpful to use a spreadsheet.
Info required...
Questions to explore...
What is the unit price for each size?
What is the percent change in size, price, unit price?
What is the least amount of popcorn from the large container (in cups) you would need to eat so that you don't get ripped off? (This is not as interesting a question as the 8th grade version because you can't usually tell how many cups of popcorn are left in a bucket.)
Info required...
Volume of a truncated cone:
You will notice that there is a little bit of popcorn above the rim of each bucket. There is also a small gap on the bottom of each bucket. I assumed that the added and subtracted volumes of this popcorn would more or less cancel each other out. I could be wrong about this!!!
Questions to explore...
What is the capacity of each size?
What is the unit price for each size?
What is the percent change in size, price, unit price?
How many inches of popcorn would be left in the large bucket if you eat just as much as the small bucket?
What is the least amount of popcorn from the large container you would need to eat so that you don't get ripped off? In other words, how many inches of popcorn can I leave at the bottom of the bucket?
I'm not leaving my full solution here because I'm curious to see how others might solve it. Basically, I used a spreadsheet to test different heights of popcorn eaten to determine where the unit price of the large matches the unit price of the small. If you think about it, this is further complicated because as you eat popcorn, the height AND top radius changes. You will have to come up with a formula that calculates the top radius based on the height.
I determined that you get ripped off if you leave more than two inches of popcorn at the bottom of the bucket.
## Sunday, July 26, 2015
### My Grudge with "Grudge"
I'm flying home from Twitter Math Camp near Los Angeles, and after successfully figuring out how to steal the airplane's wifi, I decided to write another post. This is what I do. I go to a conference, get inspired to contribute to the MTBoS community, and write a blog post. You must understand that once I get home, all motivation to do such a thing will be lost. That's what Netflix would like me to believe anyway.
There is one contribution I've made to the online community that has received a lot of good feedback from students and other teachers. This is a game called Grudge. I gave a survey to my students at the end of this year and asked them what were their favorite things were from my class. Grudge was near the top of the list. ("Mr. Kraft" was at the very top of the list, of course.)
There is no question in my mind that it is a review game that engages almost all of my students almost all of the time. I also feel that I present it in such a way that students seriously consider their answers and are eager to understand their mistakes. But there is a problem with the game. On occasion, students will team up on other students, and while it is not always expressed, I do believe that feelings can be hurt. As Matt Vaudrey once expressed in a tweet, it hurts the class culture. It promotes competition instead of collaboration.
I've learned that any activity I use in my class should not only be engaging and promote academic growth, but should also encourage students to be respectful to one another.
## Sunday, April 19, 2015
### What the hell is mean absolute deviation?
When I first started looking at the Common Core standards for sixth grade a couple of years ago, admittedly, there was one standard I had to do a double-take on:
6.SP.B.5.C: Giving quantitative measures of center (median and/or mean) and variability (interquartile range and/or mean absolute deviation), as well as describing any overall pattern and any striking deviations from the overall pattern with reference to the context in which the data were gathered.
And, like many of my colleagues, I thought, "What the hell is mean absolute deviation?" My horror was confirmed when I googled it and saw how complicated it would likely be for my students.
Looking in some textbooks and online resources, I was continually left wondering why my students would even care about mean absolute deviation. I mean, you do all of these steps, you get a number, and then what? What does mean absolute deviation tell you?
I figured that the only way my students are going to have any access to this would be to compare different data sets, make a quick judgement about which one has more variability (which can be very subjective) and find some way of quantifying that variability. On top of that, I wanted my students to create their own data where the goal would be to have the least amount of variability.
I then remembered the "Best Triangle" activity I did with Dan Meyer. In this activity, Dan asked four teachers to draw their best equilateral triangles. (Notice that Andrew and I have points in our nostrils.)
Rather than having the students evaluate the teachers' triangles, I had them create their own. I started the lesson off by asking the students to draw, what was in their mind, the perfect triangle. Immediately, there were several hands that shot up from students who wanted some clarification, but I told them to just do what they thought was best. After a quick walk-around and throwing some random triangles up on the document camera, it seemed that almost everyone was trying to draw an equilateral triangle. A few students argued that a right triangle could be considered a perfect triangle and I admitted that my instructions were very vague and their interpretations were justified.
We then brainstormed all the things we should look for in the perfect equilateral triangle. Students agreed that we needed three equal sides and three equal angles. They then made a second attempt on the whiteboards to draw perfect equilateral triangles. I asked everyone to make a quick judgement about which triangles they thought were the best, but soon ran out of time for the day. After the students left, I quickly took pictures of their triangles and took measurements in millimeters. (Admittedly, this is something I would have preferred having the students do on their own, but my class time is unbelievably short...37 minutes.)
The next day, I told my students that I took those measurements and found a way to rank all of the triangles from all of my classes. Next, I showed them the five triangles which represent the minimum (best), first quartile, second quartile, third quartile, and maximum (worst) of the data (in order below). This was a nice way to show a sample of the triangles as my students had just finished learning about box-and-whisker plots.
When I first showed them these triangles, I asked them to figure out which triangle represented the maximum and the third quartile. The other three triangles were not easily identified, however, we noticed that if you reorient the triangles so that one of the other two sides was on the bottom, the inferior triangles no longer looked equilateral (leaning to the left or right).
I explained that ranking these five different triangles didn't provide too much difficulty, but I was confused how to rank triangles that looked very similar. I gave the three following triangles as an example and had students vote on which one they believed looked the best:
In each class, there was a lot of disagreement about which triangle was the best, and more often than not, the majority picked the wrong one. I then provided the side lengths of each triangle (above in millimeters) and asked the students, "how can we use these measurements to rank these three triangles?"
After a few unproductive guesses, someone usually asked to find the differences between the measurements, which lead to someone else asking to find the sum of those differences or the range. They notice that the ranges for each triangle are all 20 mm. Someone usually calls me out for doing this intentionally...which I did.
Next, somebody will ask about the mean of the numbers. I act dumb, as I do with every suggestion, and we explore that possibility. We find the means, and it would seem that we have again hit a dead end.
I have say that at this point, some classes were completely stuck, and some kept going with it. For those that were stuck, I told them that to me, the mean (157 mm for the first triangle) represented the side length that the triangle drawer had intended for each side, but sometimes he or she fell a little short of that goal (149 mm), or overshot it (169 mm). I then asked them to compare each drawn side to "the perfect side length". We found the differences of each length and the mean, and soon after, someone suggested finding the sum of those differences.
At this point, most of my classes were satisfied that we found a method of comparing the triangles. We just had to look at the sum of the differences from the mean. The best triangle was the triangle that had the lowest sum. A couple of classes even went one step further to find the mean of those differences. In reality, there was nothing wrong with either of those methods. However, the second method WAS THE MEAN ABSOLUTE DEVIATION!!! When I first started planning this lesson, never did I think my students would intuitively come up with this concept.
This was the first time I've taught this lesson and I realize that there was a lot more I could have done with it. Given more time, I could have had students work in groups to come up with their own methods for determining the best triangle (similar to Dan's lesson plan) and we could have compared the methods later.
Side note: Dan says that "the best solution is to use the fact that an equilateral triangle is the triangle that encloses the most area for a given perimeter". Sixth graders are not at a point yet where they can find the area of a triangle just given the side lengths, so some other solution was necessary. Technically, my method is flawed because it favors smaller triangles. If you double or triple the size of a triangle, it doubles or triples the mean absolute deviation. This is noticeable in the data as smaller triangles were preferred. A better method would have been to compute the percent differences from the mean, but this would have greatly complicated an idea I was just trying to introduce for the first time. | 5,436 | 25,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-43 | latest | en | 0.981325 |
https://brainmass.com/computer-science/c/medians-and-order-statistics-71589 | 1,529,811,441,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866191.78/warc/CC-MAIN-20180624024705-20180624044705-00178.warc.gz | 559,950,201 | 18,490 | Explore BrainMass
Share
# Medians and Order statistics
Please review problem and verify the solution.
problem
---------
In the algorithm SELECT, the input elements are divided into groups of 5. Will the algorithm work in linear time if they are divided into groups of 7? Argue that SELECT does not run in linear time if groups of 3 are used.
solution
---------
Use groups of k for the analysis. The worst case SELECT will be called recursively on at most n - (n/4 - k) = 3n/4 + k elements.
The recurrence is T(n) <= T(ceiling(n/k)) + T(3n/4 + k) + O(n)
Solve the recurrence by substitution
I believe the solution is the following:
T(n) <= T(ceiling(n/k)) + T(3n/4 + k) + O(n)
<= c(n/k + 1) + 3cn/k + c(k+1) + O(n)
<= cn(1/k + 3/4) + c(k + 1) + O(n) [ this only holds for k > 4 so we have proved it works for any group size of 4 or more]
<= cn
© BrainMass Inc. brainmass.com June 23, 2018, 11:37 pm ad1c9bdddf
#### Solution Summary
Medians and Order statistics are evaluated.
\$2.19 | 302 | 991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-26 | latest | en | 0.869154 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-3rd-edition/chapter-2-limits-2-2-limits-a-numerical-and-graphical-approach-exercises-page-55/62 | 1,695,622,015,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506686.80/warc/CC-MAIN-20230925051501-20230925081501-00601.warc.gz | 858,657,243 | 12,806 | # Chapter 2 - Limits - 2.2 Limits: A Numerical and Graphical Approach - Exercises - Page 55: 62
$L(n)=\dfrac{n-1}{2}$, $n$ integer
#### Work Step by Step
We have to determine the limit: $L(n)=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{n}{1-x^n}-\dfrac{1}{1-x}\right)$ Determine $L(n)$ for $n=1,2,3$: $L(1)=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{1}{1-x^1}-\dfrac{1}{1-x}\right)=0$ $L(2)=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{2}{1-x^2}-\dfrac{1}{1-x}\right)$ $=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{2-1-x}{1-x^2}\right)$ $=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{1-x}{(1-x)(1+x)}\right)$ $=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{1}{1+x}\right)$ $=\dfrac{1}{2}$ $L(3)=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{3}{1-x^3}-\dfrac{1}{1-x}\right)$ $=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{3}{(1-x)(1+x+x^2)}-\dfrac{1}{1-x}\right)$ $=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{3-1-x-x^2}{(1-x)(1+x)}\right)$ $=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{-x^2-x+2}{(1-x)(1+x+x^2)}\right)$ $=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{x+2}{1+x+x^2}\right)$ $=1$ $L(n)=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{n}{1-x^n}-\dfrac{1}{1-x}\right)$ $=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{n}{(1-x)(1+x+x^2+...+x^{n-1})}-\dfrac{1}{1-x}\right)$ $=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{n-1-x-x^2-...-x^{n-1}}{(1-x)(1+x+x^2+...+x^{n-1})}\right)$ $=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{(1-x)+(1-x^2)+...+(1-x^{n-1})}{(1-x)(1+x+x^2+...+x^{n-1})}\right)$ $=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{(1-x)(1+(1+x)+...+(1+x+...+x^{n-2})}{(1-x)(1+x+x^2+...+x^{n-1})}\right)$ $=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{1+(1+x)+...+(1+x+...+x^{n-2})}{1+x+x^2+...+x^{n-1}}\right)$ $=\dfrac{1+2+3+...+(n-1)}{1+1+...+1}$ $=\dfrac{\dfrac{(n-1)n}{2}}{n}$ $=\dfrac{n-1}{2}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 896 | 2,032 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2023-40 | latest | en | 0.263286 |
null | null | null | null | null | null | Wednesday, December 12, 2012
How I taught our boys to read: The informal version
I am frequently asked how I taught our boys to read. I think my friends are curious for two reasons: (1) our boys learned early, and (2) I've been fairly laid-back about it. Or at least not terribly formal about it. I haven't used a curriculum, and no flashcards around here.
I'm writing this post because it's pretty hard to answer in two or three sentences.
So I'll try to keep it to under 50. Or 100. :) I realize long posts are against blog etiquette or blog wisdom, but I think keeping everything together would be simpler.
This will not be a post that systematically explains how I took my kids through the steps of learning to read. Maybe I'll post that in the future, but this post will (hopefully) give lots of ideas to parents who want to prepare their children for a time when they or someone else will take them through a formal curriculum. It will be full of things that, when scattered through their days over a number of years, teach a giant chunk of what they'll need to know. The pieces probably won't put themselves together in a finished whole (kids decoding words), but I bet they'll be very close.
As for learning to read, here's my M.O.: Capitalize on the words, phrases, and books around them that they're already interested in, and also create interest in additional words and phrases around them. Beginning around their second birthday, gradually, gradually, gradually teach them to spell and recognize those words, dissect them, play with them, contort them, and play tricks with them until they have a nice arsenal of phonetic skills and sight words. And then teach them to use that arsenal to attack new words. All the while, teach them that they can do this reading thing and they shouldn't be intimidated by it.
I'll tell you what that's looked like practically around our house:
Environmental print: This is a phrase used in the teaching world to describe words in kids' environments, such as "Colgate" or "Home Depot." Some of these words just aren't worth pursuing (my boys call it "toothpaste," not "Colgate"), but others certainly are.
When your child asks you which faucet is for hot water, say, "This one. See how it has an "H" on it? H-h-h-hot. So this one is for cold. See? C-c-c-cold."
Point out stop signs when you see them. They're everywhere, of course, and somewhere around two or three, kids seem to think it's pretty fun to shout that they've seen them at every single intersection. So take advantage of it. Point out that "STOP" is spelled S-T-O-P. (And "stop" is only spelled S-T-O-P, not P-O-T-S!) Shout it out every time you see one: "Stop! S-T-O-P!" You can do the same thing with "Open" signs if you like. They're everywhere.
Just teaching them that print goes from left-to-right can take a two-year-old months, but they can learn it, and if it's just done with the word on the shopping cart handle or on their lunchbox, it can be done without any skin off your back.
You can do these things even before your kids know their letters and sounds; really, you can do this to teach them their letters and sounds. Leap Frog toys, TV shows, and placemats can be nice ways to teach letters and sounds to kids, too, but, really, just spell things everywhere you go.
Pay attention for additional environmental print that just happens to be everywhere and in many different contexts: My husband and I are proud Texas Longhorns and own a fair amount of beautiful burnt orange clothing. Between those shirts, plus driving past Texas State Bank and Don't Mess with Texas signs and the many other narcissistic ways that Texans seem to plaster the word "Texas" everywhere, well, the word "Texas" is everywhere. And that is very, very useful because it comes in many different fonts. Don't fool yourself thinking your child knows how to read "McDonalds;" they know how to recognize a particular logo. But if they can recognize the word "Texas" in many different fonts, well, that means they know how to recognize "Texas."
Pay attention for unexpected words in your environment that appear in different kinds of places.
Artificial environmental print that they must interact with: Create environmental print. Preschool teachers are often encouraged to label everything in their classroom so that kids have a chance to learn words. you think most kids really pay attention to the label "door" over the door or "table" on the table? Neither do I. But you can create print they must interact with.
Even if your child can't read yet, their chore chart can be text only. No picture cues. If you daily help them figure out that "F-f-f-fish" is their next chore, they'll catch on after a few weeks and go grab the fish food on their own. If more than one chore starts with the letter "P," you can gradually teach them to distinguish between "plants" and "pajamas" by the second letter.
Create other situations where kids have to recognize a particular word to know how to proceed. Label drawers in your toy storage with words like "Animals" and "Toy food." Write each kid's name on a different party favor and have them find their own and their brother's. There are probably ways in your world that you answer the question of which is which for your child; well, label those things and teach them to figure it out themselves.
Their own name and the names of family and friends: There's a fair chance your child's name is embroidered on their lunchbox or plastered on their wall or written inside their backpack.
Take huge advantage of your child's love for their own name. :) They should know how to spell their own name early, and then you can begin to help them understand the sounds in their name. Some names are much trickier than others, of course, but do what you can. Do the same with the names of their siblings or best friends.
Our sons' real names (not these fakes that I use on the blog) are actually names that occur very frequently in books. It's kind of amazing how much we stumble across them. That's been fun for both of them. Even when they were just little guys, they each knew their own name and their brother's, so they were pretty stoked to come across them in real books. They felt like they were really reading something (even though it was just the same word every time). Great! I want them to feel very empowered to read.
Most kids' names won't fall into that category, of course; Olivia's real name doesn't show up in many books, so she won't have that advantage. Thankfully we have some nice grandmothers who have scouted out books that do have characters with her name. Maybe you can find one or two for your own children.
Have them begin to recognize lots of family members' names. Maybe you can tape pictures with names written underneath at their eye level on the refrigerator. (Do this for some of their favorite things, too, like Elmo or trucks.) Maybe they can be "Santa" at Christmas and pass out presents. Maybe they can help you find your sister's name in your speed dial. Look for chances.
Read, and have them believe they can read: You were expecting this one. Maybe you read to your kids a lot, or maybe you don't and feel guilty about it. Actually, I don't read to my kids a whole, whole lot.
But I do read to my kids some, and I try to take advantage of that time. Not too much advantage, mind you--I want them to enjoy reading, and there are many other things to be gotten out of reading a book together other than learning how to decode words. But when I am able to sneak in some "work" without them minding, or minding much, I do so. Some examples:
Have them read anything they possible can. If you are reading Dr. Seuss' ABCs (an absolutely fabulous book for teaching reading), have them read each of the letters as they come up. Sure, that's just the letter F, but they're participating, and that's golden.
Along the same lines, look for rebus books. We have an old, out-of-print one that we use around here (and looks like this):
so I haven't really researched to find out what else is out there to buy now. I do know Big Backyard magazine includes a rebus every month. Rebuses (rebi?) are great because it gives kids a sense of how print works (left-to-right, and then continuing on the next line) and because they get kids used to piping up with anything they are able to contribute.
Other ideas:
Have them "read" the main character's name every time it appears. In Where the Wild Things Are, "Max" is on every other page or so, plus it's written on the side of his boat. That's frequent enough for the kid to get the hang of what they're supposed to say without slowing everything way down.
In Arnie the Doughnut, Jake's job was first to "read" the name "Arnie" every time it appeared. Then I asked him to start reading "Mr. Bing." Of course he just said "Arnie" for a while at first; he thought his job was to say "Arnie" every time I paused and waited expectantly. So we had to work on him being able to distinguish between the print of "Arnie" and of "Mr. Bing"--which of course look very different, so that didn't take all that long. It was a bit of a game for him.
In Happy Dog, Sad Dog, my kids had to read the word "dog" on each page. In any of the Biscuit books, the kids can be responsible for each of the Woof! Woofs!
Have them read favorite lines. Jake loved to shout, "Arnie! Arnie! Wait up!" at the appropriate place every time we got to it. It's not that he knew how to recognize those words--at first. He just recognized the crazy picture and enormous and distinctive font. But he thought it was funny to shout it when we got there. Then I started pointing at each word as he said it, and soon it was his job to do so. Then he noticed that "up" was also on the movie Up!...and in Ten Apples Up on Top... and so on.
Have them "read" repeating sections. In (the fantastic, marvelous masterpiece) The Cat in the Hat, there's a page that says: "So all we could do was to sit!...sit!...sit!...sit! And we did not like it, not one little bit." Read the first "sit" and have them read the rest. They won't know what to do at first, of course, but say, "Look! That's the same word. See? S-I-T, S-I-T. So it's 'sit' again." Read them the page, pointing at all the "sits" as you go, and then have them do it. They'll feel like they're reading, right? If you do that every time you read The Cat in the Hat over the course of a year or two, well, you get the idea. Inch by inch.
If you're reading The Foot Book, look at the front cover before you began and say, "Where does it say 'book?' No, not there. B-b-b-book. Do you hear how 'book' starts with a 'buh' sound (de-emphasizing the 'uh')? Then it has to be this word because it begins with B and B says 'buh.' That word is 'foot' and it starts with an F. F-f-foot. But this says "book.'" And then read the book without making them do anything else with it, if you like. What you did with the front cover was a good lesson for a three-year-old.
You can also watch for books that have onomatopoeias or other "words" that will lend themselves to emphasizing how letters translate to sound. In Snuggle Puppy, the author repeats an "Oooooooooooooooooo" on many pages. Run your finger under the letters as you make a big deal out of the sound. You can also do the same thing with characters who sleep ("Zzzzzzzz") or draw out a particular word ("I caaaaaaan't!" or "Noooooooo!").
As they get older and begin a math book or any kind of other curriculum, expect them to read as much of the directions as they're able to (even if it's just the words "the" and "and"). Have them help you read recipes or maps.
Once they get some words down, try to trick them. If they've been reading "Max" and you see the word "mat" somewhere, ask them what it says. They'll likely say "Max" at first and you'll have to show them why it doesn't. Same with "Sam" and "sat," or "car" and "cart," or "pig" and "pigs." (As an aside, The Three Little Pigs is fantastic because your child will have to navigate back and forth between "pig" and "pigs.") Make a game out of the "tricking" and they'll think it's fun.
Show them how the words they already know (like "all" or "in") are in other words ("tall," "into," or "win").
And, just as you expect them to read...
Expect them to write: I've got another post about our makeshift handwriting,, those words are too generous...handwriting doings around here, but in addition to that, if your child can write letters, have them do so in real situations. If you're wrapping a gift for a birthday party, have your child write the card to their friend: "To Isaac. From Michael, Jake, and Olivia."
That wouldn't be where you start, of course. You might start with them writing their name only, or the first letter of each name, or tracing it all on a paper you printed out. But give them opportunities to do some genuine, purposeful writing. Approach it very casually as, "Here write this. Of course you can write; we all write." And provide them a way, however modified, to do it.
And the next thing you know, you might have kids teaming up to make "tickets to Hippo's house" (for some make-belief circus) or asking how to spell "Creature Power Suit" for the Wild Kratt's makeshift-costume they're pasting together...
I teach a class at a Classical Conversations group once a week, and one year my class was full of very young four-year-olds. We had to learn locations on a map every week, and so I began the year expecting them to label their own maps. It was only the first letter of each word, which I give them of course, and I soon began encouraging them to write the whole word if they were able. All along, I stood at the front saying, "A-a-athens. See right there? And see it on your paper? Write 'A' for A-a-athens right there." And you'd be surprised--they really did!
Reading and writing are two sides of the same coin. There's a giant, positive feedback loop between reading and writing. Encourage them to write anything they'll write. Anything! Make them figure out what letter the word they want to write starts with...and then what comes next...and next...etc. Around here, a big Magnadoodle has been pretty key in this.
Rhyme, etc.: Maybe this one should have been listed at the beginning. You can't read unless you can manipulate sounds. This is called phonemic awareness, and it's different that phonological awareness (phonics). Phonics involves print. Phonemic awareness can be done in the dark or on a desert island. It's just working with what the kid can hear, and it's every bit as important as knowing letters and sounds.
Pull out nursery rhymes book, listen to lots of songs, and play silly games like "Jake, Jake, bo Bake, banana fana fo fake, me my mo Make, Jake." Rhyme words of food while you're at the grocery store and of their favorite stuffed animals while you're at home.
Kids also need to be able to manipulate sounds in other ways, like by identifying what sound a word starts with, and later, what it ends with. Sometimes when they're walking around the house, just ask what sound "duck" or "house" starts with. Once they learn to do that, ask them what letter "duck" or "house" starts with. Once they've got that skill down, ask them to figure out the next sound (and letter) in the word. (Oh, now I'm mixing phonological awareness and phonemic awareness in together...oh, wait, that's the way it's supposed to end up.) Work your way up gradually.
Once they've got that, have them make up silly alliterative sentences with you: "Michael makes mudpies on Mondays." Have them help you expand the sentence and make it even sillier. Buy this great song by Raffi or this great (but tricky!) song called "Gumball" by Dr. Jean. At dinner, start things with the same sound just for fun: "Pake, pill poo pass pe pacaroni please?"
Give them sight words: "Sight words" is a dirty word (term) in some reading circles, but I am all for the right way. Yes, I teach my kids sight words at first--lots of them--such as the names of their family members, stop, Texas, all those things I mentioned--but so that they can learn to mess with them, break them down, twist them, and learn phonics and new words through them. My goal is phonemic and phonological awareness through the words. They will be broken down at some point. Most will not remain merely "sight words."
So. Teach some words like "the" and "and" and "could" and "should." If your child can get these basic words:
...down, your child will have something to carry them through longer sentences without feeling completely defeated. They can work on decoding the title The Cat in the Hat, because, well, they're only decoding two words.
Mostly you can work on sight words as you read together. Just pick one or two at first and expect them to read it every time it pops up.
We also have these lovely magnets for our fridge:
Or you can tape a word or two or three to the bathroom mirror and talk about them while they brush their teeth.
Okay, that's the end of my informal ideas for helping your child be ready to read.
And last but not least, watch SuperWhy! and do what they do!
1. This is great! Thanks for the many awesome tips! We're not homeschooling (as you know) and have no set curriculum. Learning to read and write is just happening organically for V. He love, love, loves to write his name and finds "his" letters everywhere in the environment. One question though: How did you progress from capital to lower case letters?
2. Thank you, thank you fortune fun and practical ways you have taught the boys to read. I appreciate you taking the time to give so many examples.
3. Agnes, I bet your style and my style are pretty similar (except that we're only handling one language!). I've thought of the word "organic" before, too.
As far as lowercase letters, as long as they've recognized all of them and known that "b" = "B," I haven't been too worried about when they begin writing with them. I've taught both boys to write in all caps first, which is what the occupational therapist at my old school recommended because they're so much more developmentally appropriate. Many kindergarten teachers (so here, for five- to six-year-olds) get pretty frustrated with kids who come in writing in all caps because it's a hard habit to break and also because they spend a lot of time teaching kids the mechanics of sentences--and if they're writing in all caps, obviously you can't tell if they've capitalized the proper noun, etc. But since our boys both began writing at three, I thought it was much more important that they had letters they were actually able to form correctly (and were willing to attempt, even), so all caps it's been around here. Michael gradually started writing in lowercase letters on his own...maybe around his fifth birthday? If Jake doesn't do it on his own, then I'll give him either tracing pages of sentences he likes, or maybe copywork of sentences he likes, so that he'll have to start doing so. I think that's still a number of months off, though. He still has trouble enough writing some of his capital letters with the strokes in the correct order, so we'll hold off on "e."
4. Lindsey - first of all, thanks for commenting on my blog so I could find YOUR blog. It's absolutely wonderful, and I can't wait to read (and learn from) more of your posts.
This one instantly caught my eye because I've encouraged my children to read at an early age as well. With my oldest I used the famous Teach Your Child to Read in 100 Easy Lessons. It worked like a dream. Now I'm using it with my next son. We're on Lesson 42, but I feel like we're losing some steam and I should focus more on just making it fun and a part of our every day life. So I really, really love your ideas in this post. I think he has a great foundation, and now maybe I need to just relax and enjoy words and phrases and sentences together without going about it in a terribly structured way. | null | null | null | null | null | null | null | null | null |
https://www.answers.com/Q/How_many_gallons_equals_24_ounces | 1,603,616,352,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107888402.81/warc/CC-MAIN-20201025070924-20201025100924-00368.warc.gz | 598,409,347 | 34,036 | Units of Measure
# How many gallons equals 24 ounces?
123
###### 2008-08-17 23:56:37
24 ounces = 3/16 of a gallon 24 ounces / 128 ounces to the gallon = 3/16 gallon
๐
0
๐คจ
0
๐ฎ
0
๐
0
## Related Questions
(24 gallons) x (128 fluid ounces per gallon) = 3,072fluid ounces
If 16 ounces are in 24 gallons then there are 1.5 ounces in a gallon 16 ounces don't equal 24 gallons. 128 U.S. Fluid ounces equal 1 U.S. liquid gallon. There would be 3072 fluid ounces in 24 gallons.
(24 x 8 fluid ounces) / (128 fluid ounces per gallon) = 1.5 gallons
24 US fluid ounces is about 709.76 mL.
There is 16 ounces in every pound. Therefore, 24 pounds equals to 384 ounces.
There are 16 ounces in one pound. Therefore, 1.5 pounds equals 24 ounces.
Since there are 16 ounces in a pound, 24 ounces equals 1 lb 8 oz
24 fluid ounces equals about 7.1dL
Not sure of your question....Here are the two possible answers from what I get from your question. 1. 6 ounces of A for one gallon of B....how many ounces of A are needed for 24 ounces of B. Answer is 1.125 ounces 2. 6 ounces of A for one gallon of B....24 ounces of A will be used for how many gallons of B. Answer is 4 gallons
1 U.S. fluid ounce equals 29.5735 milliliters. 24 ounces equals 709.765 milliliters.
24 fluid ounces are equal to:0.70976471 liters (L)709.7647095 milliliters (mL)0.1875 gallons (gal)
8 ounces in a cup, so three cups in 24 ounces.
###### Math and ArithmeticUnits of MeasureNutritional MeasurementLabor and Employment LawCooking Measurements
Copyright ยฉ 2020 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. | 505 | 1,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-45 | latest | en | 0.797852 |
https://socratic.org/questions/how-do-you-solve-1-2-x-40#146588 | 1,670,483,821,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711278.74/warc/CC-MAIN-20221208050236-20221208080236-00527.warc.gz | 567,623,839 | 5,785 | # How do you solve 1/2 x = 40?
Consider that you can move $\frac{1}{2}$ to the right of your expression (but inverting it) as:
$x = \left(\frac{2}{1}\right) 40$
So that $x = 40 \cdot 2 = 80$
Remember that your equation is like a balance; everything you move from one side to the other (of the equal sign) must change sign. In this case on the left you had a division by $2$...it goes to the right as a multiplication by $2$. | 127 | 425 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-49 | latest | en | 0.92232 |
http://mathhelpforum.com/pre-calculus/100092-cubic-print.html | 1,513,226,748,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948539745.30/warc/CC-MAIN-20171214035620-20171214055620-00766.warc.gz | 178,831,792 | 4,058 | # a cubic
• Sep 1st 2009, 08:28 AM
pacman
a cubic
Factor a^3 + b^3 + c^3 - 3abc , where to start with this?
• Sep 1st 2009, 09:13 AM
Taluivren
Hello, I got
$a^3+b^3+c^3-3abc = a(a^2-bc)+b(b^2-ac)+c(c^2-ab)=$
$=a(a^2-bc -ac -ab +b^2 +c^2)-a(-ac -ab +b^2 +c^2)+$
$+b(b^2-ac-bc-ab+a^2+c^2)-b(-bc-ab+a^2+c^2)+$
$+c(c^2-ab-ac-bc+a^2+b^2)-c(-ac-bc+a^2+b^2)=$
$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$+a^2c+a^2b-ab^2-ac^2 +b^2c+ab^2-a^2b-bc^2 +ac^2+bc^2-a^2c-b^2c=$
$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
• Sep 1st 2009, 09:14 AM
ynj
Quote:
Originally Posted by pacman
Factor a^3 + b^3 + c^3 - 3abc , where to start with this?
You only have to remember it:
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$
• Sep 1st 2009, 10:38 AM
RobLikesBrunch
Quote:
Originally Posted by ynj
You only have to remember it:
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$
Simply memorizing mathematics without understanding is absurd and idiotic. It is much better to not memorize a certain formula and derive it each time...even if it takes you longer.
• Sep 1st 2009, 12:27 PM
Taluivren
Quote:
Originally Posted by RobLikesBrunch
Simply memorizing mathematics without understanding is absurd and idiotic.
(Giggle) agree with you, but some things are good to know. I don't remember this formula, but what initially helped me to factor the expression was the formula $x^3+y^3 = (x+y)(x^2+y^2 -xy)$.
So here is the other way how to factorize, and pacman you'll find another ways if you try.
$a^3+b^3+c^3-3abc = a^3+b^3+c^3-3abc +3ab(a+b) - 3ab(a+b)=$
$= (a+b)^3 +c^3 - 3abc -3ab(a+b) = (a+b)^3 +c^3 -3ab(a+b+c) =$
[now we make use of $x^3+y^3 = (x+y)(x^2+y^2 -xy)$]
$=(a+b+c)((a+b)^2+c^2-(a+b)c) -3ab(a+b+c) =$
$=(a+b+c)((a+b)^2+c^2-(a+b)c -3ab) =(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$
• Sep 1st 2009, 07:11 PM
pacman
I appreciate all you guys, but the hard thing about this is that how did you figure out the sequence of steps that leads to the identity . . . . that one alone bogged me down. Thanks for the efforts to help. (Clapping)(Clapping)(Clapping)(Clapping)(Clapping)
• Sep 2nd 2009, 01:17 AM
Taluivren
Quote:
Originally Posted by pacman
I appreciate all you guys, but the hard thing about this is that how did you figure out the sequence of steps that leads to the identity . . . . that one alone bogged me down.
for me, it is about messing around with symbols using identities i already know and recognizing patterns and symmetries. Maybe we should start to learn from computers, i suppose they are better in this than humans. Can anybody explain how computers do it? (i mean how they factorize (Wink))
• Sep 2nd 2009, 03:03 AM
pacman
i found this from the web, now i wonder why it is difficult to factor
The identity would probably be known to Lagrange from his extensive study of algebraic equations If w is a primitive cubic root of unity then a^3 + b^3 + c^3 - 3abc is the constant term of the polynomial satisfied by a +bw +cww
a^3 + b^3 + c^3 - 3abc = (a+b+c)(a +bw +cww)(a +bww +cw).
This and other similar identities occur when symmetrical functions of the roots of polynomial
equations are calculated I know Newton studied symmetric functions of roots. He may have been aware of this identity.
The (a +bw +cww) is a special form of Eisenstein cubic integrers and a^3 + b^3 + c^3 - 3abc ist its norm. thus, a^3 + b^3 + c^3 - 3abc can also be referrred to a a termary cubic form. | 1,228 | 3,371 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-51 | longest | en | 0.696264 |
null | null | null | null | null | null | Can you break down this song musically for me?
March 22, 2013 6:00 AM Subscribe
Can someone tell me what is going on musically in this song? I'm mostly interested in chord progressions and key changes, and especially what happens in the breakdown starting around 3:20 and continuing for the next minute or so-- does it just go up an octave or is there a key change, and why is that so emotionally affective? I'm not interested in production techniques or the percussion(which I already know), only the melody of the lead and interplay with the bass and pads in the background.
posted by empath to Media & Arts (6 answers total) 4 users marked this as a favorite
The chords and key don't actually change at all - at 3:06 the drums drop out (bar the snare/handclap sound) then at about 3:26 the synth melody, which has previously been a very repetitive pattern, starts becoming a higher in pitch bit-by-bit, and then stays up there for a while, still in quite a randomized pattern but overall higher in pitch than for the rest of the song so far.
The loss of drums makes you suddenly aware of all the pads and stuff in the background so that will create a change of mood, and the introduction of a higher melody is an oft-used device to "add sparkle" and bring in another layer, often in a chorus section but in this style of music, in the later part of the song to hold the listener's interest.
posted by greenish at 6:15 AM on March 22, 2013 [1 favorite]
I agree with greenish, as well as noting that the background ghostly/wooshy sound gets louder. Combined with the percussion dropping out and the melody using the higher notes, this (to me) creates a feeling of tension.
I agree that the melody doesn't really change much, it sounds sort of like an "improv" break in a jazz or blues type of song. The artist is choosing notes in the same key, but choosing different, higher ones that are further away from the root of whatever key the melody is in. Again adding tension.
posted by gjc at 6:26 AM on March 22, 2013
Agreeing with the above that it's one key throughout and one chord droning throughout. What I'm hearing is that leading into the break, those ghostly notes in the background are providing a consistent and predictable tension build and release as they cycle through 8-7-6-7-8 (or Do-Ti-La-Ti-Do) repeatedly. With 8 (Do) being the root of the song's key (and chord), that's where the tension is resolved. As it wanders away the harmonic tension builds, and as it returns the tension resolves.
At the break, that predictable pattern drops out completely and those ghostly notes spend more time on unresolved 2s (Re) and 6s (La). At the same time the main melodic part jumps up into a higher register, as you heard in the octave lift. From that point on, it's a very slow resolution to the end of the song, as those ghostly notes start to land a little more regularly on more harmonically resolved 3s (Mi) and eventually end up back in the original, predictable, and familiar pattern of 8-7-6-7-8.
posted by Balonious Assault at 6:58 AM on March 22, 2013
I disagree with others who say the chords don't change. The notes don't change much, but the bassline implies a very strong IV-V-iv harmonic progression.
Having said that, I don't think there is a satisfactory "music theoretic" answer to why the song is evocative. To me it's mostly those killer synth "howls" and the interesting arrangement of parts and sounds (and not necessarily the specific "notes" being played).
posted by grog at 12:00 PM on March 22, 2013
Grog you're totally right about the chords, but I meant they don't change at that point in the song. The IV-V-VI progression is constant throughout, if I remember rightly. The OP was asking what changed at the breakdown :)
posted by greenish at 4:27 PM on March 22, 2013
I'll stand by the assertion that in the traditional sense of chords this song has one, and it drones in the background from start to finish. But I will also agree that it has pronounced implications of chord progression, and that's probably what carries the bulk of the song's emotional weight.
posted by Balonious Assault at 5:41 PM on March 22, 2013
« Older I have a job interview! (proje... | Please help me find a song I h... Newer »
This thread is closed to new comments. | null | null | null | null | null | null | null | null | null |
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-5-section-5-5-multiplying-polynomials-exercise-set-page-379/65 | 1,537,933,324,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267163146.90/warc/CC-MAIN-20180926022052-20180926042452-00109.warc.gz | 757,166,550 | 14,356 | ## Algebra: A Combined Approach (4th Edition)
Published by Pearson
# Chapter 5 - Section 5.5 - Multiplying Polynomials - Exercise Set - Page 379: 65
#### Answer
$x^{2}-\frac{2}{7}x-\frac{3}{49}$
#### Work Step by Step
$(x+\frac{1}{7})(x-\frac{3}{7})=x(x-\frac{3}{7})+\frac{1}{7}(x-\frac{3}{7})=(x)(x)+(x)(-\frac{3}{7})+(\frac{1}{7})(x)+(\frac{1}{7})(-\frac{3}{7})=x^{2}+\frac{1}{7}x-\frac{3}{7}x-\frac{3}{49}=x^{2}-\frac{2}{7}x-\frac{3}{49}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 231 | 608 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-39 | latest | en | 0.599468 |
https://thefloatingschoolid.org/and-pdf/3254-square-and-cube-of-1-to-30-pdf-931-686.php | 1,653,204,442,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545090.44/warc/CC-MAIN-20220522063657-20220522093657-00488.warc.gz | 653,574,495 | 11,205 | # Square And Cube Of 1 To 30 Pdf
File Name: square and cube of 1 to 30 .zip
Size: 22653Kb
Published: 13.05.2021
In arithmetic and algebra , the cube of a number n is its third power , that is, the result of multiplying three instances of n together. The cube is also the number multiplied by its square :. It is an odd function , as. The volume of a geometric cube is the cube of its side length, giving rise to the name.
## Square Root Table 1 1000 Pdf 21
All questions and answers from the Mathematics Solutions Book of Class 8 Math Chapter 2 are provided here for you for free. Hence, the perfect squares between 1 and are 1, 4, 9, 16, 25, 36, 49, 64, 81, , , , , , , , , , , , and Write 3-digit numbers ending with 0, 1, 4, 5, 6, 9, one for each digit, but none of them is a perfect square. Find the th and th triangular numbers, and find their sum. Verify the Statement 8 for this sum. Therefore, when a perfect number is divided by 5, then the remainder will be 0, 1, 4, 0, 1 or 4.
## Cubes and Cube Roots
To cube a number, just use it in a multiplication 3 times Note: we write "3 Cubed" as 3 3 the little 3 means the number appears three times in multiplying. The cube root of a number is The cube root of 27 is This is the special symbol that means "cube root", it is the "radical" symbol used for square roots with a little three to mean cube root. You can use it like this: we say "the cube root of 27 equals 3".
A square root 74 of a number is a number that when multiplied by itself yields the original number. Every positive real number has two square roots, one positive and one negative. If the radicand 77 , the number inside the radical sign, is nonzero and can be factored as the square of another nonzero number, then the square root of the number is apparent. In this case, we have the following property:. The radicand may not always be a perfect square. If a positive integer is not a perfect square, then its square root will be irrational.
Add standard and customized parametric components - like flange beams, lumbers, piping, stairs and more - to your Sketchup model with the Engineering ToolBox - SketchUp Extension - enabled for use with the amazing, fun and free SketchUp Make and SketchUp Pro. We don't collect information from our users. Only emails and answers are saved in our archive. Cookies are only used in the browser to improve user experience. Some of our calculators and applications let you save application data to your local computer.
Squares and cubes form an important part of Quantitative Aptitude section. Earlier we discussed an article on how to increase your calculation speed during this lockdown. Such questions involving squares and cubes can always be asked in PO, Clerk and also in SSC exams so in this article we are explaining some quick tips and tricks to use squares and cubes along with letting the aspirants know the squares and cubes of numbers from 1 to Knowing squares and cubes is an important mathematical operation by which aspirants can actually make their calculations fast and moreover, there will be less probability of error. It is essential for the candidates to maintain speed and accuracy in the bank exams or for that matter in any such competitive exam.
Use this table to find the squares and square roots of numbers from 1 to You can also use this table to estimate the square roots of larger numbers. For instance, if you want to find the square root of , look in the middle column until you find the number that is closest to You don't need to use prime factorization or other methods every time you find the square roots.
### Service Unavailable in EU region
Сьюзан спустилась по лестнице на несколько ступенек. Горячий воздух снизу задувал под юбку. Ступеньки оказались очень скользкими, влажными из-за конденсации пара. Она присела на решетчатой площадке. - Коммандер. Стратмор даже не повернулся. Он по-прежнему смотрел вниз, словно впав в транс и не отдавая себе отчета в происходящем.
Я не могу, - повторила. - Я не могу выйти за тебя замуж. - Она отвернулась.
Дэвид Беккер умрет. Халохот поднимался вверх с пистолетом в руке, прижимаясь вплотную к стене на тот случай, если Беккер попытается напасть на него сверху. Железные подсвечники, установленные на каждой площадке, стали бы хорошим оружием, если бы Беккер решил ими воспользоваться. Но если держать дистанцию, можно заметить его вовремя. У пистолета куда большая дальность действия, чем у полутораметрового подсвечника.
number system notes for ssc cgl pdf, number system questions pdf, number system tricks for ssc, number system in 1 Squares of numbers from 51 to Square values up to number 30 and cube values up to number 16 are given below.
#### Practice Mock Tests
- Откроем пачку тофу. - Нет, спасибо. - Сьюзан шумно выдохнула и повернулась к. - Я думаю, - начала она, -что я только… -но слова застряли у нее в горле. Она побледнела. - Что с тобой? - удивленно спросил Хейл. Сьюзан встретилась с ним взглядом и прикусила губу.
Между шифровалкой и стоянкой для машин не менее дюжины вооруженных охранников. - Я не такой дурак, как вы думаете, - бросил Хейл. - Я воспользуюсь вашим лифтом. Сьюзан пойдет со. А вы останетесь. - Мне неприятно тебе это говорить, - сказал Стратмор, - но лифт без электричества - это не лифт.
По сути, это был самый настоящий шантаж. Он предоставил АНБ выбор: либо рассказать миру о ТРАНСТЕКСТЕ, либо лишиться главного банка данных. Сьюзан в ужасе смотрела на экран. Внизу угрожающе мигала команда: ВВЕДИТЕ КЛЮЧ Вглядываясь в пульсирующую надпись, она поняла. Вирус, ключ, кольцо Танкадо, изощренный шантаж… Этот ключ не имеет к алгоритму никакого отношения, это противоядие. Ключ блокирует вирус.
Вижу, - сказал Бринкерхофф, стараясь сосредоточиться на документе. - Это данные о сегодняшней производительности. Взгляни на число дешифровок.
Уже на середине комнаты она основательно разогналась. За полтора метра до стеклянной двери Сьюзан отпрянула в сторону и зажмурилась. Раздался страшный треск, и стеклянная панель обдала ее дождем осколков. Звуки шифровалки впервые за всю историю этого здания ворвались в помещение Третьего узла.
Ему было не привыкать работать допоздна даже по уикэндам; именно эти сравнительно спокойные часы в АНБ, как правило, были единственным временем, когда он мог заниматься обслуживанием компьютерной техники. Просунув раскаленный паяльник сквозь проволочный лабиринт у себя над головой, он действовал с величайшей осмотрительностью: опалить защитную оболочку провода значило вывести аппарат из строя. Еще несколько сантиметров, подумал Джабба. Работа заняла намного больше времени, чем он рассчитывал. Когда он поднес раскаленный конец паяльника к последнему контакту, раздался резкий звонок мобильного телефона.
Но глаза… твои глаза, - сказал Беккер, чувствуя себя круглым дураком.
Это очень и очень плохо. - Спокойствие, - потребовал Фонтейн. - На какие же параметры нацелен этот червь. На военную информацию. Тайные операции.
Соши заливалась слезами. - Джабба, - спросил Фонтейн, - много они похитили. - Совсем мало, - сказал Джабба, посмотрев на монитор.
Не важно, сколько посетителей стоят в очереди, - секретарь всегда бросит все дела и поспешит поднять трубку. Беккер отбил шестизначный номер. Еще пара секунд, и его соединили с больничным офисом. Наверняка сегодня к ним поступил только один канадец со сломанным запястьем и сотрясением мозга, и его карточку нетрудно будет найти.
В подавленном настроении Сьюзан приняла ванну.
### Related Posts
5 Response
1. Nieriapayda1960
Professional guide to diseases pdf computer and information security handbook 3rd edition pdf
2. Geoffrey C.
cube n3 square root number n square n2 cube n3 square root. 1. 1. 1. 41
3. Eudoxia S.
Hello Friends, Today we are sharing Square table from 1 to PDF. This is very helpful for various competitive exams and improve solve.
4. Arridano P. | 2,258 | 7,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2022-21 | latest | en | 0.892767 |
http://mathhelpforum.com/calculus/8526-parametric-equation.html | 1,526,863,039,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863811.3/warc/CC-MAIN-20180520224904-20180521004904-00588.warc.gz | 187,238,707 | 10,452 | 1. ## parametric equation
Question
If Lsub1 has parametric equation x=1+3t, y=1+t, z=4-t, and Lsub2 has parametric equation x=7-6t, y=-2t, z=3+2t, then Lsub1 and Lsub2 are parallel.
Please show me why it is true. Thank you very much.
2. Originally Posted by Jenny20
Question
If Lsub1 has parametric equation x=1+3t, y=1+t, z=4-t, and Lsub2 has parametric equation x=7-6t, y=-2t, z=3+2t, then Lsub1 and Lsub2 are parallel.
Please show me why it is true. Thank you very much.
Hello Jenny,
your equations describe two straight lines in $\displaystyle \mathbb{R}^3$
$\displaystyle L_1: [x,y,z]=\underbrace{[1,1,4]}_{\text{fixed point}}+t\cdot \underbrace{[3,1,-1]}_{\text{direction}}$
$\displaystyle L_2: [x,y,z]=\underbrace{[7,0,3]}_{\text{fixed point}}+t\cdot \underbrace{[-6, -2, 2]}_{\text{direction}}$
By comparison you can see, that
$\displaystyle [-6, -2, 2]=(-2) \cdot [3,1,-1]$. That means the direction vectors are collinear: They have the same direction but different length.
Therefore $\displaystyle L_1$ and $\displaystyle L_2$ are at least parallel. To proof if they are actually the same you have to show that the fixed point of $\displaystyle L_1$ belongs to $\displaystyle L_2$ too.
EB
3. Thank you very much , earboth!
4. Hello, Jenny!
If $\displaystyle L_1$ has parametric equations: .$\displaystyle \begin{Bmatrix}x\:= & 1+3t\\ y\:= & 1+t \\ z\:= & 4-t\end{Bmatrix}$
and $\displaystyle L_2$ has parametric equations: .$\displaystyle \begin{Bmatrix} x\:= & 7-6t \\ y\:= & -2t \\ z\:= & 3+2t\end{Bmatrix}$
then $\displaystyle L_1$ and $\displaystyle L_2$ are parallel.
Two lines are parallel if their direction vectors are parallel.
. . (They do not have to be collinear.)
$\displaystyle L_1$ has direction vector: $\displaystyle \vec{u}\:=\:\langle 3,1,\text{-}1\rangle$
$\displaystyle L_2$ has direction vector: $\displaystyle \vec{v}\:=\:\langle\text{-}6,\text{-}2,2\rangle \:=\:\text{-}2\langle3,1,\text{-}1\rangle$
Since $\displaystyle \vec{v} = -2\vec{u}\!:\;\;\vec{u} \parallel \vec{v}$ . . . . Q.E.D.
5. Two lines are parallel if their direction vectors are parallel.
. . (They do not have to be collinear.)
L_1 has direction vector: \vec{u}\:=\:\langle 3,1,\text{-}1\rangle
L_2 has direction vector: \vec{v}\:=\:\langle\text{-}6,\text{-}2,2\rangle \:=\:\text{-}2\langle3,1,\text{-}1\rangle
Since \vec{v} = -2\vec{u}\!:\;\;\vec{u} \parallel \vec{v} . . . . Q.E.D.
Hi Soroban,
Thank you very much! | 863 | 2,440 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-22 | latest | en | 0.711181 |
https://math.stackexchange.com/questions/2899147/gradient-descent-on-non-linear-function-with-linear-constraints | 1,579,515,885,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250598217.23/warc/CC-MAIN-20200120081337-20200120105337-00432.warc.gz | 544,573,521 | 30,712 | # Gradient descent on non-linear function with linear constraints
Here is an optimization problem I'm trying to solve:
Objective function to be minimized:
$$f(x) = -\sum_{i=1}^{n}(x_{i}+a_{i})\bigg[1-\exp\bigg(-\frac{x_ib_i}{x_i+a_i}\bigg)\bigg]$$ where the constants: $$a_{i},b_{i} \ge 0$$ are positive for all $i$.
Constraints: $$x_{i} \ge 0$$ $$\sum_{x=1}^{n}x_{i} \le A$$ (A is a positive constant)
I have tried solving this in python and matlab using built-in solvers. It works well, however the problem is that the constants order of magnitude is very large (also largely different among them, so can't normalize) and the solvers do not output correct results for these values. I am trying to solve this in a programming language using gradient descent, which I read it is suitable for such occasions (even with the risk of stopping into a local minima if the function is not convex). I've read the theory about gradient descent which makes sense.
However I also read that since I have constraints, I have to project my space on them. I am not sure how exactly I can do this. Can anyone give any suggestions and advice? Thanks!
You can add a slack variable $x_{n+1}\geq 0$ such that $x_1+\dots +x_{n+1} = A$. Then you can apply the projected gradient method $$x^{k+1} = P_C(x^k - \alpha \nabla f(x^k)),$$ where in every iteration you need to project onto the set $C = \{x\in\mathbb{R}^{n+1}_{+}\colon x_1+\dots + x_{n+1} = A\}$. The set $C$ is called the simplex and the projection onto it is more or less explicit: it needs only sorting of the coordinates, and thus requires $O(n\log n)$ operations. There are many versions of such algorithms, here is one of them Fast Projection onto the Simplex and the $l_1$ Ball by L. Condat.
Since $C$ is a very important set in applications, it has been already implemented for various languages. | 513 | 1,849 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-05 | latest | en | 0.927412 |
http://www.papercamp.com/essay/145396/1-5-Constructions-With-Tech | 1,496,135,455,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463614620.98/warc/CC-MAIN-20170530085905-20170530105905-00182.warc.gz | 782,580,903 | 6,349 | # 1.5 Constructions with Tech
Submitted by: Submitted by
Views: 96
Words: 329
Pages: 2
Category: Other Topics
Date Submitted: 01/27/2015 05:37 AM
Report This Essay
Part:2
1. Constructing a Equilateral Triangle Inscribed in a Circle: The way I constructed the equilateral triangle inscribed in the circle using Geogebra, is I used the compass tool to make point A and B. Then I used the Line tool to draw a line to connect and create line AB. Then I used the intersecting tool to mark the points of intersection between circle A and AB. Then used the compass tool to make another circle with a Radius of AD. I marked the points intersected between circle A and Circle D. I labeled these points E and F. Lastly I drew segments BE, BF, and EF which makes the polygon BEF a equilateral Triangle.
2. Constructing Parallel Lines: Using Geogebra I used the line tool and created line AB first. The I used point tool to make point C not on the line but above it. Used point tool again to make point D under line AB and connected C and D to create line CD. Used intersect tool next between line AB and CD. I label this point E. Then I used the compass tool to create circle E with a radius DE. Then I used the compass tool again to create circle C with a radius equal in length to DE. Then I marked all the intersection points. I used the compass tool again to make another circle and labeled the intersection points. Then I drew the lines to make them parallel.
Part:3
1. I like using the constructions with technology because its more easy to use and I have better control then using a compass and straight edge.
2. The benefits are that you have more control, easier use, and more accurate measure.
3. The limitations are that you don’t do as many steps as you do with a compass and straight edge.
4. The longer the segment the bigger the circle. If it’s a shorter segment the circle is smaller.
### Related Essays
Feasibility Of Concurrent In The Nigerian Construction Industry
AND CONSTRUCTION 19 2.8.1 NATURE OF THE OPERATION 19 2.9 CHARACTERISTICS OF THE NIGERIAN CONSTRUCTION INDUSTRY 21 2.10 THE CURRENT USE OF ICT IN THE CONSTRUCTION | 506 | 2,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-22 | longest | en | 0.873201 |
https://www.geeksforgeeks.org/travelling-salesman-problem-tsp-using-reduced-matrix-method/ | 1,716,441,167,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00777.warc.gz | 691,184,903 | 54,574 | # Travelling Salesman Problem (TSP) using Reduced Matrix Method
Last Updated : 21 Apr, 2024
Given a set of cities and the distance between every pair of cities, the problem is to find the shortest possible route that visits every city exactly once and returns to the starting point.Â
Examples:
Input:Â
Example of connections of cities
Output: 80
Explanation: An optimal path is 1 – 2 – 4 – 3 – 1.
Dynamic Programming Approach: This approach is already discussed in Set-1 of this article.
Branch and Bound Approach: The branch and bound approach is already discussed in this article.
Reduced Matrix: This approach is similar to the Branch and Bound approach. The difference here is that the cost of the path and the bound is decided based on the method of matrix reduction. The following are the assumptions for a reduced matrix:
• A row or column of the cost adjacency matrix is said to be reduced if and only if it contains at least one zero element and all remaining entries in that row or column ≥ 0.
• If all rows and columns are reduced then only the matrix is reduced matrix.Â
• Tour length (new) = Tour length (old) – Total value reduced.
• We first rewrite the original cost adjacency matrix by replacing all diagonal elements from 0 to InfinityÂ
The basic idea behind solving the problem is:
• The cost to reduce the matrix initially is the minimum possible cost for the travelling salesman problem.
• Now in each step, we need to decide the minimum possible cost if that path is taken i.e., a path from vertex u to v is followed.Â
• We can do that by replacing uth row and vth column cost by infinity and then further reducing the matrix and adding the extra cost for reduction and cost of edge (u, v) with the already calculated minimum path cost.
• Once at least one path cost is found, that is then used as upper bound of cost to apply the branch and bound method on the other paths and the upper bound is updated accordingly when a path with lower cost is found.
Following are the steps to implement the above procedure:
1. Step1: Create a class (Node) that can store the reduced matrix, cost, current city number, level (number of cities visited so far), and path visited till now.Â
2. Step2: Create a priority queue to store the live nodes with the minimum cost at the top.Â
3. Step3: Initialize the start index with level = 0 and reduce the matrix. Calculate the cost of the given matrix by reducing the row and then the column. The cost is calculated in the following way:
• Row reduction – Find the min value for each row and store it. After finding the min element from each row, subtract it from all the elements in that specific row.Â
• Column reduction – Find the min value for each column and store it. After finding the min element from each column, subtract it from all the elements in that specific column. Now the matrix is reduced.
• Now add all the minimum elements in the row and column found earlier to get the cost.Â
4. Step4: Push the element with all information required by Node into the Priority Queue.Â
5. Step5: Now, perform the following operations till the priority queue gets empty.Â
• Pop the element with the min value from the priority queue.
• For each pop operation check whether the level of the current node is equal to the number of nodes/cities or not.Â
• If yes then print the path and return the minimum cost.
• If No then, for each and every child node of the current node calculate the cost by using the formula-Â
Child->Cost = parent_matrix_cost + cost_from_parentTochild + Child_reducedMatrix_cost.Â
• The cost of a reduced Matrix can be calculated by converting all the values of its rows and column to infinity and also making the index Matrix[Col][row] = infinity.Â
• Then again push the current node into the priority queue.Â
6. Step6: Repeat Step5 till we don’t reach the level = Number of nodes – 1.
Follow the illustration below for a better understanding.
Illustration:
Consider the connections as shown in the graph:
Example of connections
Initially the cost matrix looks like:
row/colno 1 2 3 4 1 – 10 15 20 2 10 – 35 25 3 15 35 – 30 4 20 25 30 –
After row and column reduction the matrix will be:
row/colno 1 2 3 4 1 – 0 5 10 2 0 – 25 15 3 0 20 – 15 4 0 5 10 –
and row minimums are 10, 10, 15 and 20.
row/colno 1 2 3 4 1 – 0 0 0 2 0 – 20 5 3 0 20 – 5 4 0 5 5 –
and the column minimums are 0, 0, 5 and 10.
So the cost reduction of the matrix is (10 + 10 + 15 + 20 + 5 + 10) = 70
Now let us consider movement from 1 to 2: Initially after substituting the 1st row and 2nd column to infinity, the matrix will be:
row/colno 1 2 3 4 1 – – – – 2 – – 20 5 3 0 – – 5 4 0 – 5 –
• After the matrix is reduced the row minimums will be 5, 0, 0Â
row/colno 1 2 3 4 1 – – – – 2 – – 15 0 3 0 – – 5 4 0 – 5 –
• and the column minimum will be  0, 5, 0
row/colno 1 2 3 4 1 – – – – 2 – – 10 0 3 0 – – 5 4 0 – 0 –
• So the cost will be 70 + cost (1, 2) + 5 + 5 = 70 + 0 + 5 + 5 = 80.
Continue this process till the traversal is complete and find the minimum cost.
Given below the structure of the recursion tree along with the bounds:
The recursion diagram with bounds
Below is the implementation of the above approach.
C++ ```// C++ code to implement the approach #include <bits/stdc++.h> using namespace std; // N is the number of cities/Node given #define N 4 #define INF INT_MAX // Structure to store all the necessary information // to form state space tree struct Node { // Helps in tracing the path when the answer is found // stores the edges of the path // completed till current visited node vector<pair<int, int> > path; // Stores the reduced matrix int reducedMatrix[N][N]; // Stores the lower bound int cost; // Stores the current city number int vertex; // Stores the total number of cities visited int level; }; // Formation of edges and assigning // all the necessary information for new node Node* newNode(int parentMatrix[N][N], vector<pair<int, int> > const& path, int level, int i, int j) { Node* node = new Node; // Stores parent edges of the state-space tree node->path = path; // Skip for the root node if (level != 0) { // Add a current edge to the path node->path.push_back(make_pair(i, j)); } // Copy data from the parent node to the current node memcpy(node->reducedMatrix, parentMatrix, sizeof node->reducedMatrix); // Change all entries of row i and column j to INF // skip for the root node for (int k = 0; level != 0 && k < N; k++) { // Set outgoing edges for the city i to INF node->reducedMatrix[i][k] = INF; // Set incoming edges to city j to INF node->reducedMatrix[k][j] = INF; } // Set (j, 0) to INF // here start node is 0 node->reducedMatrix[j][0] = INF; // Set number of cities visited so far node->level = level; // Assign current city number node->vertex = j; // Return node return node; } // Function to reduce each row so that // there must be at least one zero in each row int rowReduction(int reducedMatrix[N][N], int row[N]) { // Initialize row array to INF fill_n(row, N, INF); // row[i] contains minimum in row i for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { if (reducedMatrix[i][j] < row[i]) { row[i] = reducedMatrix[i][j]; } } } // Reduce the minimum value from each element // in each row for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { if (reducedMatrix[i][j] != INF && row[i] != INF) { reducedMatrix[i][j] -= row[i]; } } } return 0; } // Function to reduce each column so that // there must be at least one zero in each column int columnReduction(int reducedMatrix[N][N], int col[N]) { // Initialize all elements of array col with INF fill_n(col, N, INF); // col[j] contains minimum in col j for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { if (reducedMatrix[i][j] < col[j]) { col[j] = reducedMatrix[i][j]; } } } // Reduce the minimum value from each element // in each column for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { if (reducedMatrix[i][j] != INF && col[j] != INF) { reducedMatrix[i][j] -= col[j]; } } } return 0; } // Function to get the lower bound on the path // starting at the current minimum node int calculateCost(int reducedMatrix[N][N]) { // Initialize cost to 0 int cost = 0; // Row Reduction int row[N]; rowReduction(reducedMatrix, row); // Column Reduction int col[N]; columnReduction(reducedMatrix, col); // The total expected cost is // the sum of all reductions for (int i = 0; i < N; i++) { cost += (row[i] != INT_MAX) ? row[i] : 0, cost += (col[i] != INT_MAX) ? col[i] : 0; } return cost; } // Function to print list of cities // visited following least cost void TSPPAthPrint(vector<pair<int, int> > const& list) { for (int i = 0; i < list.size(); i++) { cout << list[i].first + 1 << " -> " << list[i].second + 1 << "\n"; } } // Comparison object to be used to order the heap struct Min_Heap { bool operator()(const Node* lhs, const Node* rhs) const { return lhs->cost > rhs->cost; } }; // Function to solve the traveling salesman problem // using Branch and Bound int solve(int CostGraphMatrix[N][N]) { // Create a priority queue to store live nodes // of the search tree priority_queue<Node*, vector<Node*>, Min_Heap> pq; vector<pair<int, int> > v; // Create a root node and calculate its cost. // The TSP starts from the first city, i.e., node 0 Node* root = newNode(CostGraphMatrix, v, 0, -1, 0); // Get the lower bound of the path // starting at node 0 root->cost = calculateCost(root->reducedMatrix); // Add root to the list of live nodes pq.push(root); // Finds a live node with the least cost, // adds its children to the list of live nodes, // and finally deletes it from the list while (!pq.empty()) { // Find a live node with // the least estimated cost Node* min = pq.top(); // The found node is deleted from // the list of live nodes pq.pop(); // i stores the current city number int i = min->vertex; // If all cities are visited if (min->level == N - 1) { // Return to starting city min->path.push_back(make_pair(i, 0)); // Print list of cities visited TSPPAthPrint(min->path); // Return optimal cost return min->cost; } // Do for each child of min // (i, j) forms an edge in a space tree for (int j = 0; j < N; j++) { if (min->reducedMatrix[i][j] != INF) { // Create a child node and // calculate its cost Node* child = newNode(min->reducedMatrix, min->path, min->level + 1, i, j); child->cost = min->cost + min->reducedMatrix[i][j] + calculateCost(child->reducedMatrix); // Add a child to the list of live nodes pq.push(child); } } // Free node as we have already stored edges (i, j) // in vector. So no need for a parent node while // printing the solution. delete min; } return 0; } // Driver code int main() { int CostGraphMatrix[N][N] = { { INF, 10, 15, 20 }, { 10, INF, 35, 25 }, { 15, 35, INF, 30 }, { 20, 25, 30, INF } }; // Function call cout << "Total cost is " << solve(CostGraphMatrix); return 0; } ``` Java ```import java.util.*; public class Main { // Define the number of vertices and infinity value static final int N = 4; static final int INF = Integer.MAX_VALUE; // Node class to store each node along with the cost, level, and vertex static class Node { ArrayList<int[]> path = new ArrayList<>(); int[][] reducedMatrix = new int[N][N]; int cost; int vertex; int level; } public static void main(String[] args) { // Define the cost matrix int[][] CostGraphMatrix = { { INF, 10, 15, 20 }, { 10, INF, 35, 25 }, { 15, 35, INF, 30 }, { 20, 25, 30, INF } }; // Print the total cost of the tour System.out.println("Total cost is " + solve(CostGraphMatrix)); } // Function to allocate a new node static Node newNode(int[][] parentMatrix, ArrayList<int[]> path, int level, int i, int j) { Node node = new Node(); node.path = (ArrayList<int[]>)path.clone(); // Add this edge to the path if (level != 0) { node.path.add(new int[]{i, j}); } // Copy data from parent matrix to current matrix for (int r = 0; r < N; r++) { node.reducedMatrix[r] = parentMatrix[r].clone(); } // Change all entries of row i and column j to infinity // Also change the entry for vertex k to infinity if (level != 0) { for (int k = 0; k < N; k++) { node.reducedMatrix[i][k] = INF; node.reducedMatrix[k][j] = INF; } node.reducedMatrix[j][0] = INF; } // Update the level of node node.level = level; // Update the vertex number node.vertex = j; return node; } // Function to perform row reduction static int rowReduction(int[][] reducedMatrix, int[] row) { // Initialize row array to INF Arrays.fill(row, INF); // Row[i] contains minimum in row i for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { if (reducedMatrix[i][j] < row[i]) { row[i] = reducedMatrix[i][j]; } } } // Reduce the minimum value from each element in each row for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { if (reducedMatrix[i][j] != INF && row[i] != INF) { reducedMatrix[i][j] -= row[i]; } } } return 0; } // Function to perform column reduction static int columnReduction(int[][] reducedMatrix, int[] col) { // Initialize col array to INF Arrays.fill(col, INF); // Col[j] contains minimum in col j for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { if (reducedMatrix[i][j] < col[j]) { col[j] = reducedMatrix[i][j]; } } } // Reduce the minimum value from each element in each column for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { if (reducedMatrix[i][j] != INF && col[j] != INF) { reducedMatrix[i][j] -= col[j]; } } } return 0; } // Function to calculate the cost of the path static int calculateCost(int[][] reducedMatrix) { int cost = 0; int[] row = new int[N]; rowReduction(reducedMatrix, row); int[] col = new int[N]; columnReduction(reducedMatrix, col); // Calculate the cost by adding the reduction values for (int i = 0; i < N; i++) { cost += (row[i] != INF) ? row[i] : 0; cost += (col[i] != INF) ? col[i] : 0; } return cost; } // Function to print the path static void printPath(ArrayList<int[]> list) { for (int[] path : list) { System.out.println((path[0] + 1) + " -> " + (path[1] + 1)); } } // Function to solve the TSP problem static int solve(int[][] CostGraphMatrix) { // Create a priority queue to store live nodes of the search tree PriorityQueue<Node> pq = new PriorityQueue<>(Comparator.comparingInt(node -> node.cost)); ArrayList<int[]> v = new ArrayList<>(); // Create a root node and calculate its cost Node root = newNode(CostGraphMatrix, v, 0, -1, 0); root.cost = calculateCost(root.reducedMatrix); // Add root to the list of live nodes pq.add(root); // Continue until the priority queue becomes empty while (!pq.isEmpty()) { // Find a live node with the least estimated cost Node min = pq.poll(); // Get the vertex number int i = min.vertex; // If all the cities have been visited if (min.level == N - 1) { min.path.add(new int[]{i, 0}); printPath(min.path); return min.cost; } // Generate all the children of min for (int j = 0; j < N; j++) { if (min.reducedMatrix[i][j] != INF) { Node child = newNode(min.reducedMatrix, min.path, min.level + 1, i, j); child.cost = min.cost + min.reducedMatrix[i][j] + calculateCost(child.reducedMatrix); pq.add(child); } } } return 0; } } ``` Python3 ```import sys from queue import PriorityQueue # Define the number of vertices and infinity value N = 4 INF = sys.maxsize # Node class to store each node along with the cost, level, and vertex class Node: def __init__(self, parentMatrix, path, level, i, j): self.path = path.copy() self.reducedMatrix = [row.copy() for row in parentMatrix] self.cost = 0 self.vertex = j self.level = level # Add this edge to the path if level != 0: self.path.append((i, j)) # Change all entries of row i and column j to infinity # Also change the entry for vertex k to infinity if level != 0: for k in range(N): self.reducedMatrix[i][k] = INF self.reducedMatrix[k][j] = INF self.reducedMatrix[j][0] = INF def __lt__(self, other): return self.cost < other.cost # Function to perform row reduction def rowReduction(reducedMatrix): row = [INF]*N # Row[i] contains minimum in row i for i in range(N): for j in range(N): if reducedMatrix[i][j] < row[i]: row[i] = reducedMatrix[i][j] # Reduce the minimum value from each element in each row for i in range(N): for j in range(N): if reducedMatrix[i][j] != INF and row[i] != INF: reducedMatrix[i][j] -= row[i] return row # Function to perform column reduction def columnReduction(reducedMatrix): col = [INF]*N # Col[j] contains minimum in col j for i in range(N): for j in range(N): if reducedMatrix[i][j] < col[j]: col[j] = reducedMatrix[i][j] # Reduce the minimum value from each element in each column for i in range(N): for j in range(N): if reducedMatrix[i][j] != INF and col[j] != INF: reducedMatrix[i][j] -= col[j] return col # Function to calculate the cost of the path def calculateCost(reducedMatrix): cost = 0 row = rowReduction(reducedMatrix) col = columnReduction(reducedMatrix) # Calculate the cost by adding the reduction values for i in range(N): cost += (row[i] if row[i] != INF else 0) cost += (col[i] if col[i] != INF else 0) return cost # Function to print the path def printPath(path): for pair in path: print(f"{pair[0] + 1} -> {pair[1] + 1}") # Function to solve the TSP problem def solve(CostGraphMatrix): # Create a priority queue to store live nodes of the search tree pq = PriorityQueue() # Create a root node and calculate its cost root = Node(CostGraphMatrix, [], 0, -1, 0) root.cost = calculateCost(root.reducedMatrix) # Add root to the list of live nodes pq.put((root.cost, root)) # Continue until the priority queue becomes empty while not pq.empty(): # Find a live node with the least estimated cost min = pq.get()[1] # Get the vertex number i = min.vertex # If all the cities have been visited if min.level == N - 1: min.path.append((i, 0)) printPath(min.path) return min.cost # Generate all the children of min for j in range(N): if min.reducedMatrix[i][j] != INF: child = Node(min.reducedMatrix, min.path, min.level + 1, i, j) child.cost = min.cost + min.reducedMatrix[i][j] + calculateCost(child.reducedMatrix) pq.put((child.cost, child)) return 0 # Define the cost matrix CostGraphMatrix = [ [ INF, 10, 15, 20 ], [ 10, INF, 35, 25 ], [ 15, 35, INF, 30 ], [ 20, 25, 30, INF ] ] # Print the total cost of the tour print("Total cost is", solve(CostGraphMatrix)) ``` JavaScript ```class Node { constructor() { this.path = []; this.reducedMatrix = Array.from({ length: N }, () => Array(N).fill(0)); this.cost = 0; this.vertex = 0; this.level = 0; } } const INF = Number.MAX_SAFE_INTEGER; const N = 4; function newNode(parentMatrix, path, level, i, j) { const node = new Node(); node.path = [...path]; if (level !== 0) { node.path.push([i, j]); } for (let r = 0; r < N; r++) { node.reducedMatrix[r] = parentMatrix[r].slice(); } if (level !== 0) { for (let k = 0; k < N; k++) { node.reducedMatrix[i][k] = INF; node.reducedMatrix[k][j] = INF; } node.reducedMatrix[j][0] = INF; } node.level = level; node.vertex = j; return node; } function rowReduction(reducedMatrix, row) { row.fill(INF); for (let i = 0; i < N; i++) { for (let j = 0; j < N; j++) { if (reducedMatrix[i][j] < row[i]) { row[i] = reducedMatrix[i][j]; } } } for (let i = 0; i < N; i++) { for (let j = 0; j < N; j++) { if (reducedMatrix[i][j] !== INF && row[i] !== INF) { reducedMatrix[i][j] -= row[i]; } } } return 0; } function columnReduction(reducedMatrix, col) { col.fill(INF); for (let i = 0; i < N; i++) { for (let j = 0; j < N; j++) { if (reducedMatrix[i][j] < col[j]) { col[j] = reducedMatrix[i][j]; } } } for (let i = 0; i < N; i++) { for (let j = 0; j < N; j++) { if (reducedMatrix[i][j] !== INF && col[j] !== INF) { reducedMatrix[i][j] -= col[j]; } } } return 0; } function calculateCost(reducedMatrix) { let cost = 0; const row = Array(N).fill(0); const col = Array(N).fill(0); rowReduction(reducedMatrix, row); columnReduction(reducedMatrix, col); for (let i = 0; i < N; i++) { cost += row[i] !== INF ? row[i] : 0; cost += col[i] !== INF ? col[i] : 0; } return cost; } function printPath(list) { for (const path of list) { console.log(`\${path[0] + 1} -> \${path[1] + 1}`); } } function solve(CostGraphMatrix) { const pq = new PriorityQueue((a, b) => a.cost - b.cost); const v = []; const root = newNode(CostGraphMatrix, v, 0, -1, 0); root.cost = calculateCost(root.reducedMatrix); pq.enqueue(root); while (!pq.isEmpty()) { const min = pq.dequeue(); const i = min.vertex; if (min.level === N - 1) { min.path.push([i, 0]); printPath(min.path); return min.cost; } for (let j = 0; j < N; j++) { if (min.reducedMatrix[i][j] !== INF) { const child = newNode(min.reducedMatrix, min.path, min.level + 1, i, j); child.cost = min.cost + min.reducedMatrix[i][j] + calculateCost(child.reducedMatrix); pq.enqueue(child); } } } return 0; } class PriorityQueue { constructor(comparator) { this.heap = []; this.comparator = comparator || ((a, b) => a - b); } enqueue(element) { this.heap.push(element); this.bubbleUp(); } dequeue() { const min = this.heap[0]; const last = this.heap.pop(); if (this.heap.length > 0) { this.heap[0] = last; this.bubbleDown(); } return min; } isEmpty() { return this.heap.length === 0; } bubbleUp() { let index = this.heap.length - 1; while (index > 0) { const parentIndex = Math.floor((index - 1) / 2); if (this.comparator(this.heap[index], this.heap[parentIndex]) >= 0) { break; } [this.heap[parentIndex], this.heap[index]] = [this.heap[index], this.heap[parentIndex]]; index = parentIndex; } } bubbleDown() { let index = 0; while (index < this.heap.length) { const left = 2 * index + 1; const right = 2 * index + 2; let smallest = index; if (left < this.heap.length && this.comparator(this.heap[left], this.heap[smallest]) < 0) { smallest = left; } if (right < this.heap.length && this.comparator(this.heap[right], this.heap[smallest]) < 0) { smallest = right; } if (smallest === index) { break; } [this.heap[index], this.heap[smallest]] = [this.heap[smallest], this.heap[index]]; index = smallest; } } } const CostGraphMatrix = [ [INF, 10, 15, 20], [10, INF, 35, 25], [15, 35, INF, 30], [20, 25, 30, INF] ]; console.log("Total cost is " + solve(CostGraphMatrix)); ```
Output
```1 -> 3
3 -> 4
4 -> 2
2 -> 1
Total cost is 80```
Time Complexity: O(2N * N2) where N = number of node/ cities.
Space Complexity: O(N2)
Share your thoughts in the comments | 6,372 | 22,140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-22 | latest | en | 0.937387 |
null | null | null | null | null | null | Sunday, January 8, 2012
The Outdoor Kitchen
It was some time in 1905 or 1906 and John who was always known as Jack and his wife Rachel who hated her name and went by the catchall Sis, had been married about six or seven months. They were living with Sis’ parents on their property "Rose Valley Farm" in Texas, Queensland. Jack decided he should get to and build an outdoor kitchen for Sis and so he set to work. Over a number of weeks the job progressed until finally it was finished and Jack felt quite proud of what he had achieved. It had all the requirements of a modern kitchen with an outstanding oven, the stack of which proudly pointed into the air high enough to carry all the smoke away.
Its not known how long after but it was probably a year or so and Sis was working in the kitchen with the oven fire burning away making its usual whooshing sound as the smoke rose through the chimney when the disaster struck. That stack of which Jack had been so proud, suddenly collapsed and came down over Sis, setting fire to her sleeve and severely burning her. She was saved by her sister who was able to throw a bucket of water over her. She was months in care and when she finally recovered her arm was fused to her body so that she couldn’t lift her arm over her head. That was why about fifty years later, although I didn’t know it then, I used to peg the clothes on the clothesline for her which led me to having many enjoyable talks with Nanna.
My cousin Margarette says that John never forgave himself for, as he saw it, not having built that stack correctly. She told me that at about 90 years of age, not long before he left us and with Rachel herself some years gone, he confided in her that he still felt guilty about it.
My grandfather John who I was named after, was a truly wonderful man. What a shame it was that he felt he had to carry that guilt for the whole of his life.
1. How sad! I wonder whether there is a Justice Department 'fire inquest' file at Queensland State Archives. My Web site (look under 'Tips for Queensland Research') has brief advice about inquest records.
2. My grandmother didn't die in the fire so do you think there would be an "inquest" for this incident? | null | null | null | null | null | null | null | null | null |
Subsets and Splits