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+0 +1 54 2 +109 Let $$f(x) = \left\{ \begin{array}{cl} ax+3, &\text{ if }x>2, \\ x-5 &\text{ if } -2 \le x \le 2, \\ 2x-b &\text{ if } x <-2. \end{array} \right.$$ Find a+b if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper). Aug 24, 2020 #1 0 a + b = 2 + 8 = 10. Aug 24, 2020 #2 +109 -1,+0 # HELP ASAP 0 103 1 Let $$f(x) = \left\{ \begin{array}{cl} ax+3, &\text{ if }x>2, \\ x-5 &\text{ if } -2 \le x \le 2, \\ 2x-b &\text{ if } x <-2. \end{array} \right.$$ Find $$a+b$$ if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper). Aug 22, 2022 #1 +2540 0 Because the graph is continuous when x = 2, $$x- 5 = ax+ 3$$. Subbing in what we know for x, we get $$a = -3$$ Now, do the same thing when x = -2. Aug 22, 2022,+0 # Help ASAP +2 128 4 +475 Let $$$f(x) = \left\{ \begin{array}{cl} ax+3, &\text{ if }x>2, \\ x-5 &\text{ if } -2 \le x \le 2, \\ 2x-b &\text{ if } x <-2. \end{array} \right.$$$ Find a+b if the piecewise function is continuous. (which means that its graph can be drawn without lifting your pencil from the paper). Jul 22, 2020 #1 +31506 +1 If the curve is continuous then we must have: At x = 2: a*2 + 3 = 2 - 5 or 2a = -6 so a = -3 At x = -2: 2*(-2) - b = -2 - 5 or -4 - b = -7 so b = 3 Jul 22, 2020 #2 +475 +1 How do we find the equation of a continuous piecewise function though? Thank you very much! Jul 22, 2020 #3 +31506 +1 Well, given a and b we can write Alan Jul 22, 2020 #4 +475 +1 Thank you! Jul 22, 2020,+0 # help with piecewise defined functions pls 0 165 3 +229 Let $f(x) = \left\{ \begin{array}{cl} ax+3, &\text{ if }x>2, \\ x-5 &\text{ if } -2 \le x \le 2, \\ 2x-b &\text{ if } x <-2. \end{array} \right.$ Find $a+b$ if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper). Mar 5, 2021 #1 +34464 +2 The line of the graph needs to be continuous so at x = 2 ax+3 has to equal x-5 ax+3 = x-5 ax = x - 8 at x= 2 a(2) = 2-8 a = -3 and at -2 x-5 has to equal 2x-b x-5 = 2x-b b+5 = x for x = -2 b+5 = -2 b = -7 Mar 5, 2021 #3 +34464 +1 *** corrected *** Had an incorrect '+' sign: and at -2 x-5 has to equal 2x-b x-5 = 2x-b b - 5 = x for x = -2 b - 5 = -2 b = 3 ElectricPavlov Mar 5, 2021 #2 +229 0 thats wrong but thanks Mar 5, 2021,+0 # piecewise defined function 0 116 1 Let f(x) = 2x^2 - 4 if x <= 2, ax + 5 if x > 2. Find a if the graph of y = f(x) is continuous (which means the graph can be drawn without lifting your pencil from the paper). Oct 14, 2020 #1 +10820 +1 Let f(x) = 2x^2 - 4 if x <= 2 and ax+5 if x > 2 Find a if the graph of y = f(x) is continuous. (which means the graph can be drawn without lifting your pencil from the paper). Hello Guest! $$\color{BrickRed}f(x)=2x^2-4=ax+5\\ x=2\\ 2\cdot 2^2-4=2a+5\\ a=(2\cdot 2^2-4-5)/2$$ $$\large a=-\frac{1}{2}$$ ! Oct 14, 2020,### Home > INT1 > Chapter 11 > Lesson 11.2.2 > Problem11-66 11-66. Graph the piecewise function $f ( x ) = \left\{ \begin{array} { l l } { 2 ^ { x } } & { \text { if } x \geq 0 } \\ { - x } & { \text { if } x < 0 } \end{array} \right.$. Use the eTool below to solve the problem. Click the link at the right to view the full version of the eTool: Int1 11-66 HW eTool.,### Home > PC3 > Chapter 3 > Lesson 3.2.4 > Problem3-140 3-140. Graph the piecewise-defined function at right and determine if the function is continuous at $x=2$. Explain your reasoning. $f ( x ) = \left\{ \begin{array} { l l } { x ^ { 2 } + x + 1 } & { \text { for } x \leq 2 } \\ { - \frac { 1 } { 2 } x + 7 } & { \text { for } x > 2 } \end{array} \right.$ Create two tables. One for $x\le2$ and one for $x>2$.,Home > PC > Chapter 4 > Lesson 4.1.3 > Problem4-44 4-44. Sketch the graph of $f ( x ) = \left\{ \begin{array} { c l } { - x ^ { 2 } + 4 } & { \text { for } x < 0 } \\ { x ^ { 2 } - 1 } & { \text { for } x \geq 0 } \end{array} \right.$. Is $f(x)$ continuous? Explain. 1. Find the height at $0$ for both parts of the piece-wise function. 2. If they are the same, it is continuous. 3. Plot the points found above and continue with the appropriate graph on either side.,+0 0 134 1 For real numbers $x$, let $f(x) = \left\{ \begin{array}{cl} x+2 &\text{ if }x>3, \\ 2x+a &\text{ if }x\le 3. \end{array} \right.$What must the value of $a$ be to make the piecewise function continuous (which means that its graph can be drawn without lifting your pencil from the paper)? Aug 23, 2018 #1 +98130 +2 $$f(x) = \left\{ \begin{array}{cl} x+2 &\text{ if }x>3, \\ 2x+a &\text{ if }x\le 3. \end{array} \right.$$ Let x = 3 and set the functions equal.... we have 3 + 2 = 2(3) + a 5 = 6 + a 5 - 6 = a - 1 = a See the graph here that shows that this value of a makes the function continuous : https://www.desmos.com/calculator/ux7xrnhaxi Aug 23, 2018,+0 # help pls 0 44 1 +229 For real numbers $x$, let $f(x) = \left\{ \begin{array}{cl} x+2 &\text{ if }x>3, \\ 2x+a &\text{ if }x\le 3. \end{array} \right.$ What must the value of $a$ be to make the piecewise function continuous (which means that its graph can be drawn without lifting your pencil from the paper)? Mar 6, 2021,+0 # Piecewise Functions 0 56 1 Let $$f(x) = \left\{ \begin{array}{cl} -x + 3 & \text{if } x \le 0, \\ 2x - 5 & \text{if } x > 0. \end{array} \right.$$ How many solutions does the equation $$f(f(x))=4$$ have? May 18, 2020,+0 # Piecewise Functions 0 144 1 Let $$f(x) = \left\{ \begin{array}{cl} -x + 3 & \text{if } x \le 0, \\ 2x - 5 & \text{if } x > 0. \end{array} \right.$$ How many solutions does the equation $$f(f(x))=4$$ have? May 18, 2020,### Home > GB8I > Chapter 11 Unit 12 > Lesson INT1: 11.2.2 > Problem11-66 11-66. Graph the piecewise function $f ( x ) = \left\{ \begin{array} { l l } { 2 ^ { x } } & { \text { if } x \geq 0 } \\ { - x } & { \text { if } x < 0 } \end{array} \right.$. Use the eTool below to solve the problem. Click the link at the right to view the full version of the eTool: Int1 11-66 HW eTool.,Math Help - Need a function for this graph 1. Need a function for this graph 2. Originally Posted by sanghoon93 Have you considered a piecewise function? 3. $f(x)=\left\{\begin{array}{cc} x,&\mbox{ if } 0 \leq x \leq 1\\2-x, & \mbox{ if } 1 < x \leq 2\\0, & \mbox{ if } x > 2\end{array}\right.$ Should work...,# Math Help - Need a function for this graph 1. ## Need a function for this graph 2. Originally Posted by sanghoon93 Have you considered a piecewise function? 3. $f(x)=\left\{\begin{array}{cc} x,&\mbox{ if } 0 \leq x \leq 1\\2-x, & \mbox{ if } 1 < x \leq 2\\0, & \mbox{ if } x > 2\end{array}\right.$ Should work...,### Home > CCA2 > Chapter C > Lesson C.1.3 > ProblemC-59 C-59. Graph the piecewise function $f ( x ) = \left\{ \begin{array} { l l } { 2 ^ { x } } & { \text { if } x \geq 0 } \\ { | x | } & { \text { if } x < 0 } \end{array} \right.$. Start by making a table of points for each of the two parts of the function.,$$(2)(-6)+8=-4$$. We don’t even care about the $$\boldsymbol{{x}^{2}}$$! It’s that easy. You can also see that we did this correctly by using the graph above. Now try $$x=4$$. We look at the right first, and see that our $$x$$ is greater than –2, so we plug it in the $${{x}^{2}}$$. (We can just ignore the $$2x+8$$ this time.) So $$f(x)$$ or $$y$$ is $${{4}^{2}}=16$$. # Graphing Piecewise Functions You’ll probably be asked to graph piecewise functions. Sometimes the graphs will contain functions that are non-continuous or discontinuous, meaning that you have to pick up your pencil in the middle of the graph when you are drawing it (like a jump!). Continuous functions means that you never have to pick up your pencil if you were to draw them from left to right. And remember that the graphs are true functions only if they pass the Vertical Line Test. Let’s draw these piecewise functions and determine if they are continuous or non-continuous. Note how we draw each function as if it were the only one, and then “erase” the parts that aren’t needed. We’ll also get the Domain and Range like we did here in the Algebraic Functions section. Piecewise Function Graph $$\displaystyle f\left( x \right)=\left\{ \begin{array}{l}-2x+8\,\,\,\,\,\,\text{if }x\le 4\\\frac{1}{2}x-2\,\,\,\,\,\,\,\,\,\text{if }x>4\end{array} \right.$$ Continuous Domain: $$\mathbb{R},\,\,\,\text{or}\,\,\left( {-\infty ,\infty } \right)$$ Range:,+0 # Please help asap 0 81 1 For real numbers $x$, let $f(x) = \left\{ \begin{array}{cl} x+2 &\text{ if }x>3, \\ 2x+a &\text{ if }x\le 3. \end{array} \right.$What must the value of $a$ be to make the piecewise function continuous (which means that its graph can be drawn without lifting your pencil from the paper)? Guest Aug 23, 2018 #1 +91186 +2 $$f(x) = \left\{ \begin{array}{cl} x+2 &\text{ if }x>3, \\ 2x+a &\text{ if }x\le 3. \end{array} \right.$$ Let x = 3 and set the functions equal.... we have 3 + 2 = 2(3) + a 5 = 6 + a 5 - 6 = a - 1 = a See the graph here that shows that this value of a makes the function continuous : https://www.desmos.com/calculator/ux7xrnhaxi CPhill Aug 23, 2018 ### New Privacy Policy We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website. For more information: our cookie policy and privacy policy.,+0 # Piecewise-Defined Functions +1 132 1 +147 For real numbers $x$, let $f(x) = \left\{ \begin{array}{cl} x+2 &\text{ if }x>3, \\ 2x+a &\text{ if }x\le 3. \end{array} \right.$ What must the value of $a$ be to make the piecewise function continuous (which means that its graph can be drawn without lifting your pencil from the paper)? Feb 24, 2021 ### 1+0 Answers #1 +602 +1 At $3$, it must be continuous. So $2*3+a=3+2\implies \boxed{a=-1}$. Feb 24, 2021,# Evaluate the piecewise defined function at the indicated values.a^m inHf(x)=left{begin{array}{11}{5}&{text{if}xleq2}{2x-3}&{text{if}x>2}end{array}righta^m inHf(-3),f(0),f(2),f(3),f(5) Question Functions Evaluate the piecewise defined function at the indicated values. $$a^m inH$$ $$f(x)= \begin{array}{11}{5}&\text{if}\ x \leq2 \ 2x-3& \text{if}\ x>2\end{array}$$ $$a^m inH$$ f(-3),f(0),f(2),f(3),f(5) 2021-01-14 1: $$f(-3)=5$$ $$f=5$$ 2: $$f(0)=5$$ $$f=5$$ 3: $$f(2)=5$$ $$f=5$$ 4: $$f(3)=2 \cdot (3)-3=3$$ $$f=3$$ 5: $$f(5)=2 \cdot (5)-3=7$$ $$f=7$$ ### Relevant Questions For what value of the constant c is the function f continuous on $$\displaystyle{\left(−∞,+∞\right)}?$$ $$\displaystyle{f{{\left({x}\right)}}}={\left\lbrace{\left({c}{x}^{{2}}\right)}+{4}{x},{\left({x}^{{3}}\right)}-{c}{x}\right\rbrace}{\quad\text{if}\quad}{x}{<}{5},{\quad\text{if}\quad}{x}\Rightarrow{5}$$ c= Use the piecewise-defined function to fill in the bla . $$\displaystyle{f{{\left({x}\right)}}}={\left\lbrace{4},-{4}{<}{x}{<}-{2},{2}{x}-{4},-{1}{<}{x}{<}{2},{3}{x},{2}\le{x}{<}{5}\right\rbrace}$$ a. The domain ____ is used when graphing the function $$f(x)=2x-4$$. b. The equation ____ is used to find $$f(4)=12.$$ Use exponential regression to find a function that models the data. $$\begin{array}{|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 14 & 7.1 & 3.4 & 1.8 & 0.8 \\ \hline \end{array}$$ Find the absolute max and min values at the indicated interval $$f(x)=2x^{3}-x^{2}-4x+10, [-1,0]$$ find the values of b such that the function has the given maximum or minimum value. $$f(x) = -x^2+bx-75$$, Maximum value: 25 Two random variables X and Y with joint density function given by: $$f(x,y)=\begin{cases}\frac{1}{3}(2x+4y)& 0\leq x,\leq 1\\0 & elsewhere\end{cases}$$ Find the marginal density of X. Two random variables X and Y with joint density function given by: $$f(x,y)=\begin{cases}\frac{1}{3}(2x+4y)&(...TRUNCATED)

+0 # help with piecewise defined functions pls 0 165 3 +229 Let $f(x) = \left\{ \begin{array}{cl} ax+3, &\text{ if }x>2, \\ x-5 &\text{ if } -2 \le x \le 2, \\ 2x-b &\text{ if } x <-2. \end{array} \right.$ Find $a+b$ if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper). Mar 5, 2021 #1 +34464 +2 The line of the graph needs to be continuous so at x = 2 ax+3 has to equal x-5 ax+3 = x-5 ax = x - 8 at x= 2 a(2) = 2-8 a = -3 and at -2 x-5 has to equal 2x-b x-5 = 2x-b b+5 = x for x = -2 b+5 = -2 b = -7 Mar 5, 2021 #3 +34464 +1 *** corrected *** Had an incorrect '+' sign: and at -2 x-5 has to equal 2x-b x-5 = 2x-b b - 5 = x for x = -2 b - 5 = -2 b = 3 ElectricPavlov Mar 5, 2021 #2 +229 0 thats wrong but thanks Mar 5, 2021,ax^2-bx+3 = a(2)^2-b(2)+3$$ Since the function whose limit we are trying to find is continuous everywhere (since it’s a polynomial) we can just plug in 2 for x to find this limit. $$a(2)^2-b(2)+3=a(2)^2-b(2)+3$$ $$4a-2b+3=4a-2b+3$$ Notice both sides of this equation are the same. Because of this, this equation will actually be true no matter what we put in for a and b. This doesn’t really help us at all in this case, but it was important to test it out and see what it told us. So we know that f will be continuous at $$x=2$$ as long as $$1=4a-2b.$$ At this point we have found a set of a and b values that make f continuous at $$x=2$$. But we need to find just one a and one b that will accomplish this for all x values. But this does not tell us anything about whether it would also be continuous at $$x=3$$. Checking that may give us another relationship between a and b that we can use to find the single unique solution for the two constants that will make f continuous everywhere. Continuity at $$x=3$$ Making sure that f is continuous at $$x=3$$ will be an extremely similar process to what we just did around $$x=2$$. Similar to above, we will need to make sure,# Math Help - Piecewise Differentiability 1. ## Piecewise Differentiability Let f be the piecewise defined function given by f(x)=ax+b x≤2. 2-x^2 2<x. Find a and b so that f is differentiable at x=2. 2. Originally Posted by Velvet Love Let f be the piecewise defined function given by f(x)=ax+b x≤2. 2-x^2 2<x. Find a and b so that f is differentiable at x=2. This will work if the two pieces are equal at $x=2$ (so that $f$ is continuous) and if their derivatives are equal at $x=2$. Spoiler: We need to solve these two equations: 1.) $a\cdot 2+b=2-2^2 \implies 2a+b=-2$ 2.) $a=-2\cdot 2 \implies a=-4$ Plug $a=-4$ back into the first equation to find that $b=6$. So $f(x)=\left\{\begin{array}{lr}-4x+6&:x\leq 2\\-x^2+2&:x>2\end{array}\right\}$ 3. I actually solved this problem a couple of minutes ago. Now i'm having trouble with this: Find the line which is perpendicular to the graph of f(x)=x^2+1 at x=4. You will need to use the definition to compute the derivative of f. 4. Originally Posted by Velvet Love Let f be the piecewise defined function given by f(x)=ax+b x≤2. 2-x^2 2<x. Find a and b so that f is differentiable at x=2. first, remember that if a function is differentiable at $x = c$ , then it is continuous at $x = c$. secondly ... $f'(c) =,ax^2-bx+3 = a(2)^2-b(2)+3$$ Since the function whose limit we are trying to find is continuous everywhere (since it’s a polynomial) we can just plug in 2 for x to find this limit. $$a(2)^2-b(2)+3=a(2)^2-b(2)+3$$ $$4a-2b+3=4a-2b+3$$ Notice both sides of this equation are the same. Because of this, this equation will actually be true no matter what we put in for a and b. This doesn’t really help us at all in this case, but it was important to test it out and see what it told us. So we know that f will be continuous at $x=2$ as long as $$1=4a-2b.$$ At this point we have found a set of a and b values that make f continuous at $x=2$. But we need to find just one a and one b that will accomplish this for all x values. But this does not tell us anything about whether it would also be continuous at $x=3$. Checking that may give us another relationship between a and b that we can use to find the single unique solution for the two constants that will make f continuous everywhere. Continuity at $x=3$ Making sure that f is continuous at $x=3$ will be an extremely similar process to what we just did around $x=2$. Similar to above, we will need to make sure,6 \end{array} \right. $$Show solution \blacktriangleright As with the last example, we seek to find the constants that make each piece of the piecewise function equal to one another at the interval break points. Let's work left-to-right and see what we can discern: set the left-most function definition equal to the middle function definition.$$7 = x^2 - 2ax + b$$Furthermore, we want these two parts of the function definition to be equivalent specifically at x=-2, where the piecewise function "changes over" from the first definition to the second. So let's let x be -2.$$7 = 4 + 4a + b$$This equation is indeterminate, having 2 variables in it. Let's pause on it and do the same thing for the other cutoff point of the piecewise function - set the two pieces equal.$$x^2 - 2ax + b = ax + b$$At the place that we want these two definitions to be equal, x is 6. Let's plug in that value and see what remains.$$36 - 12a + \cancel{b} = 6a + \cancel{b}36 - 18a = 0\boxed{a = 2}$$Now that we know a=2, we can use that result in the first equation we had set up but had to put on pause:$$7 = 4 + 4a + b7 = 4 + 4(2) + b\boxed{b = -5}$$Therefore, b = -5, and,Given the piecewise function 1-x,x<-1, (x^2)-x,-1<=x<=6, x-7,x>6, is it continuous at x=-1 and -6? Dec 22, 2015 It is continuous at $x = - 1$ and $x = - 6$, but not at $x = 6$. Explanation: *As a disclaimer, I assumed the $- 6$ at the very end of the question was supposed to be a $6$ and solved accordingly. * To solve, you simply plug in the bordering $x$-value and see if the two are the same. For $x = - 1$: $1 - \left(- 1\right)$ $1 + 1$ $2$ ${\left(- 1\right)}^{2} - \left(- 1\right)$ $1 + 1$ $2$ So, $f \left(x\right)$ is continuous at $x = - 1$ For $x = - 6$, the function is continuous because it is not a border of the piecewise. For $x = 6$: ${\left(6\right)}^{2} - \left(6\right)$ $36 - 6$ $30$ $\left(6\right) - 7$ $- 1$ So, $f \left(x\right)$ is not continuous at $x = 6$,yes Also, i don't know if it matters, but just looking at it i can tell you its a piecewise. Its linear for the first four points, then it turns curvelinear from (1,1) to (3,-1) no, we cannot tell from this. we just have discrete points here. descriptions like "piece-wise" do not apply. if these points are just a few points given for a non-discrete function, then we cannot tell whether the function is piece-wise or not here. 5. Ah! I think i get it! Tell me if im on the right track. $y=ax+b$So, if your going to add $ax+b$ to itself you would get $2ax+2b$. Hence, $2f(x) or, 2y?$ 6. Originally Posted by EyesForEars Ah! I think i get it! Tell me if im on the right track. $y=ax+b$So, if your going to add $ax+b$ to itself you would get $2ax+2b$. Hence, $2f(x) or, 2y?$ Exactly! Operations on functions may look bad, but they're really pretty simplistic. The "(f + f)(x)" means nothing more than "f(x) + f(x)", or "2f(x)". So if f(x) = mx + b, then (f + f)(x) = 2f(x) = 2(mx + b) = 2mx + 2b. Good work!,+0 # Help ASAP +2 128 4 +475 Let $$$f(x) = \left\{ \begin{array}{cl} ax+3, &\text{ if }x>2, \\ x-5 &\text{ if } -2 \le x \le 2, \\ 2x-b &\text{ if } x <-2. \end{array} \right.$$$ Find a+b if the piecewise function is continuous. (which means that its graph can be drawn without lifting your pencil from the paper). Jul 22, 2020 #1 +31506 +1 If the curve is continuous then we must have: At x = 2: a*2 + 3 = 2 - 5 or 2a = -6 so a = -3 At x = -2: 2*(-2) - b = -2 - 5 or -4 - b = -7 so b = 3 Jul 22, 2020 #2 +475 +1 How do we find the equation of a continuous piecewise function though? Thank you very much! Jul 22, 2020 #3 +31506 +1 Well, given a and b we can write Alan Jul 22, 2020 #4 +475 +1 Thank you! Jul 22, 2020,values slightly larger than 2. For x values slightly larger than 2, but infinitely close to 2, we would use the $$y=ax^2-bx+3$$ piece to define f. We already found f(2) above, so putting these two facts together we see $$\lim_{x \to 2^{+}} ax^2-bx+3 = a(2)^2-b(2)+3$$ Since the function whose limit we are trying to find is continuous everywhere (since it’s a polynomial) we can just plug in 2 for x to find this limit. $$a(2)^2-b(2)+3=a(2)^2-b(2)+3$$ $$4a-2b+3=4a-2b+3$$ Notice both sides of this equation are the same. Because of this, this equation will actually be true no matter what we put in for a and b. This doesn’t really help us at all in this case, but it was important to test it out and see what it told us. So we know that f will be continuous at $$x=2$$ as long as $$1=4a-2b.$$ At this point we have found a set of a and b values that make f continuous at $$x=2$$. But we need to find just one a and one b that will accomplish this for all x values. But this does not tell us anything about whether it would also be continuous at $$x=3$$. Checking that may give us another relationship between a and b that we can use to find the single unique solution for,and $b$ so that \begin{equation*} f(x) = \begin{cases} 5-2x, & x \le -1, \\ ax+b, & -1 \lt x \lt 2, \\ 2x-5, & x \ge 2, \end{cases} \end{equation*} is continuous at $x=-1$ and at $x=2\text{.}$,+0 I have trouble with piecewise functions. 0 102 1 If$f(x) = \begin{cases} 2x-5 &\quad \text{if } x \ge 3, \\ -x + 5 &\quad \text{if } x < 3, \end{cases}$then for how many values of $x$ is $f(f(x)) = 3$? Dec 31, 2021 f(f(1)) = f(4) = 3. The only solution is x = 1, so the number of solutions is $\boxed{1}$.,+0 # Piecewise-Defined Functions +1 132 1 +147 For real numbers $x$, let $f(x) = \left\{ \begin{array}{cl} x+2 &\text{ if }x>3, \\ 2x+a &\text{ if }x\le 3. \end{array} \right.$ What must the value of $a$ be to make the piecewise function continuous (which means that its graph can be drawn without lifting your pencil from the paper)? Feb 24, 2021 ### 1+0 Answers #1 +602 +1 At $3$, it must be continuous. So $2*3+a=3+2\implies \boxed{a=-1}$. Feb 24, 2021,look at.For $g(f(x))$:$$g\left(f(x)\right) = \left(\sqrt{x^2-2x-15}\right)^2 -4$$$$=x^2-2x-19$$This result is a polynomial function, and we know that polynomial functions are differentiable everywhere. However, like continuity, we aren't allowed to include any $x$ values in the domain that we couldn't include in the original functions. Therefore, since $g(x)$ isn't defined for $x \in (-3,5)$, $g(f(x))$ isn't either and is therefore also not differentiable. It is however differentiable at $-3$ and $5$ because the derivative of $g(f(x))$ is defined at those points. ## Unknown Coefficients Problems Teachers love to test your knowledge of differentiability with piecewise functions. One common way they do this is to ask you to find the unknown coefficients in a piecewise function such that the function is differentiable at the endpoints of the piecewise interval. Example 3Find $a$ and $b$ such that the following function $f(x)$ is continuous and differentiable.$\blacktriangleright$ From Algebra, we should know that having two unknowns in our problem implies that we need to solve two simultaneous equations. First, we need this function to be continuous. In order for that to be true, the left-side function near $x=2$ must equal the right-side function near $x=2$.$$ax^2 - bx + 1 = ax - b - 1$$but since we need this true at $x=2$, replace $x$ with $2$:$$4a - 2b + 1 = 2a - b -,## zmudz one year ago Let $$f$$ be the piecewise function such that $$f(x) = \begin{cases} x^2 - 5x - 64, & x \le 0 \\ x^2 + 3x - 38, & x > 0 \end{cases}$$ At how many points $$x$$ does $$f(x)=2$$? 1. jim_thompson5910 set each piece equal to 2 and solve for x x^2 - 5x - 64 = 2 leads to x = ?? or x = ?? x^2 + 3x - 38 = 2 leads to x = ?? or x = ?? 2. zmudz but the answer isn't 4 (I got for the first equation - -6, 11 and for the second equation 5, -8) 3. freckles when you solve f(x)=2 you have to make sure the x's fit in the inequality thingy x^2-5x-64=2 when x<=0 so you your x=-6 will only work for that one because 11 is greater than 0 so one solution so far... x^2+3x-38=2 when x>0 so your x=5 will only work for that one because -8 is less than 0 so 1+1=2 you have 2 solutions,# Thread: Find a and b so that the given piecewise function is continuous 1. ## Find a and b so that the given piecewise function is continuous f(x)={14, x<-3 {ax2+b, -3<x<2 {18-x, x>2 2. ## Re: Find a and b so that the given piecewise function is continuous Originally Posted by hplitz f(x)={14, x<-3 {ax2+b, -3<x<2 {18-x, x>2 Solve for $a~\&~b$ so that ${\lim _{x \to - {3^ + }}}f(x) = 14~\&~{\lim _{x \to {2^ - }}}f(x) = 16$. 3. ## Re: Find a and b so that the given piecewise function is continuous I got the answer a=-2/5 and b=88/5 Does that seem right? 4. ## Re: Find a and b so that the given piecewise function is continuous Originally Posted by hplitz I got the answer a=-2/5 and b=88/5 Does that seem right? Yes it does. At least a is correct. , , , , # find a and b on piecewise functions Click on a term to search for related topics.,+0 # Piecewise-Defined Functions +1 49 1 +147 For real numbers $x$, let $f(x) = \left\{ \begin{array}{cl} x+2 &\text{ if }x>3, \\ 2x+a &\text{ if }x\le 3. \end{array} \right.$ What must the value of $a$ be to make the piecewise function continuous (which means that its graph can be drawn without lifting your pencil from the paper)? Feb 24, 2021 At $3$, it must be continuous. So $2*3+a=3+2\implies \boxed{a=-1}$.,+0 # Piecewise-Defined Function -1 54 3 +194 If $f(x) = \begin{cases} 2x-5 &\quad \text{if } x \ge 3, \\ -x + 5 &\quad \text{if } x < 3, \end{cases}$ then for how many values of $x$ is $f(f(x)) = 3$? Feb 4, 2021 #1 0 The equation f(f(x)) = 3 has 1 solution. Feb 4, 2021 #3 0 Are you sure? How did you arrive at this answer? Guest Feb 4, 2021 #2 +1 This problem seems overwhelming, but let's take this problem one piece at a time. Instead of trying to wrap our heads around $$f(f(x))=3$$, we should instead ponder a far easier question. For what values of x does $$f(x) = 3$$? Well, let's investigate, shall we? $$f(x)$$ is defined as a piecewise function, so we must more or less treat it as two separate functions and ensure that our answers lie within the restrictions. $$2x-5=3\\ 2x=8\\ x_1=4\\ 4\geq 3 \checkmark$$ $$-x+5=3\\ -x=-2\\ x_2=2\\ 2<3\checkmark$$ I placed checkmarks around both proposed solutions because they respect the boundaries given by the original piecewise function. $$f(x_1)=3\text{ and } f(x_2)=3$$. Let and solve$$f(x) = x_1\text{ or } f(x) = x_2$$to determine the number of solutions to $$f(f(x)) = 3$$. $$f(f(x))=3\\ f(x_1)=3\\ f(x)=x_1\\ f(x)=4$$ $$f(f(x))=3\\ f(x_2)=3\\ f(x)=x_2\\ f(x)=2$$ Let's find all the values for x that satisfy these two,two values – either $$3$$ or $$-2$$. It’s a good practice to always check your answers if they satisfy the equation too. In this case, both solutions satisfy the equation $$|2x-1|=5$$ ## Concept Check Questions 1. Evaluate $$|-7(2)+3|$$ 2. Sketch the graph of $$y=|2x-1|$$ Label all important features 3. Sketch the graph of $$y=|(x-2)^2-3|-4$$ 4. Express $$f(x)=|x-5|$$ as a piecewise function 5. Express $$f(x)=-|5-x|$$ as a piecewise function 6. Express $$f(x)=|2x+4|-3$$ as a piecewise function 7. Solve $$|2x-5|=35$$ 8. Solve $$|x+2|-5=10$$ 1. 11 2. $$f(x) = \begin{cases} 5-x & ;x<5 \\ x-5 & ;x\geq5 \end{cases}$$ 3. $$f(x) = \begin{cases} x-5 & ;x<5 \\ 5-x & ;x\geq5 \end{cases}$$ 4. $$f(x) = \begin{cases} -2x-7 & ;x<-2 \\ 2x+1 & ;x\geq-2 \end{cases}$$ 5. $$2x-5=35$$ or $$2x-5= -35$$ $$∴x=20$$ or $$-15$$ 6. $$|x+2|-5=10$$ $$|x+2|=15$$ $$x+2=15$$ or $$x+2=-15$$ $$x=13$$ or $$-17$$,$x=5$ $x^2 + bx + a$ if $x<5$ • October 14th 2007, 07:49 PM ffezz • October 15th 2007, 04:19 AM topsquark Quote: Originally Posted by ffezz hehe alright =) i missed a questions though somehow =\ Find constants a and b such that the function is continuous for all of x $f(x) = ax+3,$ if $x>5$ $8$ if $x=5$ $x^2 + bx + a$ if $x<5$ New questions should go in new threads. It looks to me like you simply need to find an a and b such that $ax + 3 = x^2 + bx + a = 8$ when x = 5. -Dan • October 15th 2007, 04:21 AM topsquark Quote: Originally Posted by ffezz,+0 # Piecewise-Defined Functions -1 68 2 +204 If $f(x) = \begin{cases} x^2-4 &\quad \text{if } x \ge -4, \\ x + 3 &\quad \text{otherwise}, \end{cases}$ then for how many values of $x$ is $f(f(x)) = 5$? Feb 21, 2021 #1 +74 -1 I believe it's 5. Feb 21, 2021 #2 +112966 +2 $$f(x) = \begin{cases} x^2-4 &\quad \text{if } x \ge -4, \\ x + 3 &\quad \text{otherwise}, \end{cases}$$ then for how many values of $$x$$ is $$f(f(x)) = 5$$ ? --------- You need to look at 2 different cases. first assume f(x) is less than -4 solve 5=x+3 x=-2 -2 is not less than -4 so no solutions here. now assume f(x) >= -4 $$5=x^2-4\\ 9=x^2\\ x=\pm3$$ Both of these are bigger than -4 so they both could have answers. That is, for this problem f(x) could be +3 or -3 Now you need to look at each of these for each of the scenarios x<-4 and x>=-4 and see how many answers you get that are valid. Feb 22, 2021(...TRUNCATED)

+0 # help with piecewise defined functions pls 0 165 3 +229 Let $f(x) = \left\{ \begin{array}{cl} ax+3, &\text{ if }x>2, \\ x-5 &\text{ if } -2 \le x \le 2, \\ 2x-b &\text{ if } x <-2. \end{array} \right.$ Find $a+b$ if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper). Mar 5, 2021 #1 +34464 +2 The line of the graph needs to be continuous so at x = 2 ax+3 has to equal x-5 ax+3 = x-5 ax = x - 8 at x= 2 a(2) = 2-8 a = -3 and at -2 x-5 has to equal 2x-b x-5 = 2x-b b+5 = x for x = -2 b+5 = -2 b = -7 Mar 5, 2021 #3 +34464 +1 *** corrected *** Had an incorrect '+' sign: and at -2 x-5 has to equal 2x-b x-5 = 2x-b b - 5 = x for x = -2 b - 5 = -2 b = 3 ElectricPavlov Mar 5, 2021 #2 +229 0 thats wrong but thanks Mar 5, 2021,+0 +1 54 2 +109 Let $$f(x) = \left\{ \begin{array}{cl} ax+3, &\text{ if }x>2, \\ x-5 &\text{ if } -2 \le x \le 2, \\ 2x-b &\text{ if } x <-2. \end{array} \right.$$ Find a+b if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper). Aug 24, 2020 #1 0 a + b = 2 + 8 = 10. Aug 24, 2020 #2 +109 -1,+0 # HELP ASAP 0 103 1 Let $$f(x) = \left\{ \begin{array}{cl} ax+3, &\text{ if }x>2, \\ x-5 &\text{ if } -2 \le x \le 2, \\ 2x-b &\text{ if } x <-2. \end{array} \right.$$ Find $$a+b$$ if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper). Aug 22, 2022 #1 +2540 0 Because the graph is continuous when x = 2, $$x- 5 = ax+ 3$$. Subbing in what we know for x, we get $$a = -3$$ Now, do the same thing when x = -2. Aug 22, 2022,+0 # Help ASAP +2 128 4 +475 Let $$$f(x) = \left\{ \begin{array}{cl} ax+3, &\text{ if }x>2, \\ x-5 &\text{ if } -2 \le x \le 2, \\ 2x-b &\text{ if } x <-2. \end{array} \right.$$$ Find a+b if the piecewise function is continuous. (which means that its graph can be drawn without lifting your pencil from the paper). Jul 22, 2020 #1 +31506 +1 If the curve is continuous then we must have: At x = 2: a*2 + 3 = 2 - 5 or 2a = -6 so a = -3 At x = -2: 2*(-2) - b = -2 - 5 or -4 - b = -7 so b = 3 Jul 22, 2020 #2 +475 +1 How do we find the equation of a continuous piecewise function though? Thank you very much! Jul 22, 2020 #3 +31506 +1 Well, given a and b we can write Alan Jul 22, 2020 #4 +475 +1 Thank you! Jul 22, 2020,+0 # Help ASAP! +2 50 4 +59 Let $$f(x) = \left\{ \begin{array}{cl} 2x + 7 & \text{if } x < -2, \\ -x^2 - x + 1 & \text{if } x \ge -2. \end{array} \right.$$ Find the sum of all values of x such that f(x) = -5 Mar 24, 2021 #1 +721 +6 grinding some piecewise-defined functions today, aren't we? we have $2x + 7 = -5$, so $x = -6.$ we also have $-x^2 - x + 1 = -5$, and solving for $x$ gives us solutions of $-3$ and $2.$ As $-3$ is not greater or equal than $-2$, we only consider the $2.$ So, the values of $x$ are $-6$ and $2.$ the sum of these is $\boxed{-4}.$ nice Mar 24, 2021 #1 +721 +6 grinding some piecewise-defined functions today, aren't we? we have $2x + 7 = -5$, so $x = -6.$ we also have $-x^2 - x + 1 = -5$, and solving for $x$ gives us solutions of $-3$ and $2.$ As $-3$ is not greater or equal than $-2$, we only consider the $2.$ So, the values of $x$ are $-6$ and $2.$ the sum of these is $\boxed{-4}.$ nice CentsLord Mar 24, 2021 #2 +59 0 D**n, I hate it when I get exposed Mar 24, 2021 #3,+0 # Piecewise-Defined Functions +1 132 1 +147 For real numbers $x$, let $f(x) = \left\{ \begin{array}{cl} x+2 &\text{ if }x>3, \\ 2x+a &\text{ if }x\le 3. \end{array} \right.$ What must the value of $a$ be to make the piecewise function continuous (which means that its graph can be drawn without lifting your pencil from the paper)? Feb 24, 2021 ### 1+0 Answers #1 +602 +1 At $3$, it must be continuous. So $2*3+a=3+2\implies \boxed{a=-1}$. Feb 24, 2021,+0 0 134 1 For real numbers $x$, let $f(x) = \left\{ \begin{array}{cl} x+2 &\text{ if }x>3, \\ 2x+a &\text{ if }x\le 3. \end{array} \right.$What must the value of $a$ be to make the piecewise function continuous (which means that its graph can be drawn without lifting your pencil from the paper)? Aug 23, 2018 #1 +98130 +2 $$f(x) = \left\{ \begin{array}{cl} x+2 &\text{ if }x>3, \\ 2x+a &\text{ if }x\le 3. \end{array} \right.$$ Let x = 3 and set the functions equal.... we have 3 + 2 = 2(3) + a 5 = 6 + a 5 - 6 = a - 1 = a See the graph here that shows that this value of a makes the function continuous : https://www.desmos.com/calculator/ux7xrnhaxi Aug 23, 2018,6 \end{array} \right. $$Show solution \blacktriangleright As with the last example, we seek to find the constants that make each piece of the piecewise function equal to one another at the interval break points. Let's work left-to-right and see what we can discern: set the left-most function definition equal to the middle function definition.$$7 = x^2 - 2ax + b$$Furthermore, we want these two parts of the function definition to be equivalent specifically at x=-2, where the piecewise function "changes over" from the first definition to the second. So let's let x be -2.$$7 = 4 + 4a + b$$This equation is indeterminate, having 2 variables in it. Let's pause on it and do the same thing for the other cutoff point of the piecewise function - set the two pieces equal.$$x^2 - 2ax + b = ax + b$$At the place that we want these two definitions to be equal, x is 6. Let's plug in that value and see what remains.$$36 - 12a + \cancel{b} = 6a + \cancel{b}36 - 18a = 0\boxed{a = 2}$$Now that we know a=2, we can use that result in the first equation we had set up but had to put on pause:$$7 = 4 + 4a + b7 = 4 + 4(2) + b\boxed{b = -5}$$Therefore, b = -5, and,### Home > PC > Chapter 3 > Lesson 3.3.1 > Problem3-113 3-113. Given $f ( x ) = \left\{ \begin{array} { l l } { a x ^ { 2 } + b \:\text { for } } & \:\:\:{ x < 0 } \\ { 2 a x + 5\: \text { for } } & { 0 \leq x < 1 } \\ { 3 x - b } \:\:\:\:\text { for } & \:\:\:{x \geq 1 } \end{array} \right.$, find the values of $a$ and $b$ so that $f\left(x\right)$ is continuous at both $x = 0$ and $x = 1$ $ax^2+b=2ax+5\text{ for }x=0$ $2ax+5=3x-b\text{ for }x=1$ Use the eTool below to visualize this problem. Click on the link to the right to view the full version of the eTool: 3-113 eTool,+0 # HELP THANKS!! 0 83 4 For real numbers $$x$$, let$$f(x) = \left\{ \begin{array}{cl} x+2 &\text{ if }x>3, \\ 2x+a &\text{ if }x\le 3. \end{array} \right.$$What must the value of $$a$$ be to make the piecewise function continuous (which means that its graph can be drawn without lifting your pencil from the paper)? Apr 20, 2021 #1 0 The value of a is 3. Apr 20, 2021 #2 0 Incorrect, Sorry. Guest Apr 20, 2021 #3 +33803 +1 at x = 3 both pieces must have the same value to be continuous x+2 = 2x + a when x = 3 3+2 = 2(3) + a 5 = 6 + a a = -1 Apr 20, 2021 #4 0 Correct $$a=-1$$ Guest Apr 20, 2021,+0 # Piecewise-Defined Functions +1 49 1 +147 For real numbers $x$, let $f(x) = \left\{ \begin{array}{cl} x+2 &\text{ if }x>3, \\ 2x+a &\text{ if }x\le 3. \end{array} \right.$ What must the value of $a$ be to make the piecewise function continuous (which means that its graph can be drawn without lifting your pencil from the paper)? Feb 24, 2021 At $3$, it must be continuous. So $2*3+a=3+2\implies \boxed{a=-1}$.,+0 # Piecewise-Defined Functions 0 75 1 +184 For some constants $a$ and $b$ let $f(x) = \left\{ \begin{array}{cl} 9 - 2x & \text{if } x \le 3, \\ ax + b & \text{if } x > 3. \end{array} \right.$ The function $f$ has the property that $f(f(x)) = x$ for all $x$. What is $a+b$? Apr 17, 2021 #1 +420 0 if $$f(f(x)) = x$$ for all x, then f(x) is an involution. The properties of an involution include the fact that it is reflexive across the line y=x, which means that y=ax+b must be a reflected version of y=9-2x through the line y=x. To reflect that line, just find the inverse of that line by changing y for x and solving for y: $$x=9-2y\\2y=9-x\\y=4.5-0.5x$$ a = -0.5, b = 4.5, so a+b $$\boxed{4}$$ Apr 17, 2021,+0 # Piecewise function 0 162 1 Let f(x) = -x-3 if x <= -7 f(x) = 2x + 1 if x > -7 Find the sum of all values of $x$ such that f(x) = 0. May 12, 2022 #1 +9462 0 Suppose x <= -7. When f(x) = 0, -x - 3 = 0. This means x = -3, which is not <= -7. This root is rejected. Suppose x > -7. When f(x) = 0, 2x + 1 = 0. This means $$x = -\dfrac{\boxed{\phantom{\text{aaaaa}}}}{\boxed{\phantom{\text{aaaaa}}}}$$. Therefore, the sum of all values of x such that f(x) = 0 is $$-\dfrac{\boxed{\phantom{\text{aaaaa}}}}{\boxed{\phantom{\text{aaaaa}}}}$$, as it is the only solution to f(x) = 0.,Given the piecewise function 1-x,x<-1, (x^2)-x,-1<=x<=6, x-7,x>6, is it continuous at x=-1 and -6? Dec 22, 2015 It is continuous at $x = - 1$ and $x = - 6$, but not at $x = 6$. Explanation: *As a disclaimer, I assumed the $- 6$ at the very end of the question was supposed to be a $6$ and solved accordingly. * To solve, you simply plug in the bordering $x$-value and see if the two are the same. For $x = - 1$: $1 - \left(- 1\right)$ $1 + 1$ $2$ ${\left(- 1\right)}^{2} - \left(- 1\right)$ $1 + 1$ $2$ So, $f \left(x\right)$ is continuous at $x = - 1$ For $x = - 6$, the function is continuous because it is not a border of the piecewise. For $x = 6$: ${\left(6\right)}^{2} - \left(6\right)$ $36 - 6$ $30$ $\left(6\right) - 7$ $- 1$ So, $f \left(x\right)$ is not continuous at $x = 6$,### Home > PC3 > Chapter 3 > Lesson 3.2.4 > Problem3-140 3-140. Graph the piecewise-defined function at right and determine if the function is continuous at $x=2$. Explain your reasoning. $f ( x ) = \left\{ \begin{array} { l l } { x ^ { 2 } + x + 1 } & { \text { for } x \leq 2 } \\ { - \frac { 1 } { 2 } x + 7 } & { \text { for } x > 2 } \end{array} \right.$ Create two tables. One for $x\le2$ and one for $x>2$.,+0 # Piecewise Functions 0 56 1 Let $$f(x) = \left\{ \begin{array}{cl} -x + 3 & \text{if } x \le 0, \\ 2x - 5 & \text{if } x > 0. \end{array} \right.$$ How many solutions does the equation $$f(f(x))=4$$ have? May 18, 2020,− 2 to make it positive. This is indicated by placing −(x − 2) above the line to the left of 2. • To the right of 2, the expression x − 2 is positive (note the plus sign (+) below the line), so |x − 2| = x − 2. That is, we simply remove the absolute value bars because the quantity inside is already positive. This is indicated by placing x − 2 above the line to the right of 2 (see the number line above). We can use this last number line summary to construct a piecewise definition of the expression |x − 2|. $|x-2|=\left\{\begin{array}{ll}{-(x-2),} & {\text { if } x<2,} \\ {x-2,} & {\text { if } x \geq 2}\end{array}=\left\{\begin{array}{ll}{-x+2,} & {\text { if } x<2} \\ {x-2,} & {\text { if } x \geq 2}\end{array}\right.\right.$ Our number line and piecewise definition agree: |x − 2| = −(x − 2) to the left of 2 and |x − 2| = x − 2 to the right of 2. Further, note how we’ve included the critical value of 2 “on the right” in our piecewise definition. Let’s summarize the method we followed to construct the piecewise function above. Constructing a Piecewise Definition for Absolute Value When presented with the absolute value of an algebraic expression,,Home > PC3 > Chapter 9 > Lesson 9.1.2 > Problem9-37 9-37. Given $f ( x ) = \left\{ \begin{array} { l l } { a x ^ { 2 } + b \:\text { for } } & \:\:\:{ x < 0 } \\ { 2 a x + 5\: \text { for } } & { 0 \leq x < 1 } \\ { 3 x - b } \:\:\:\:\text { for } & \:\:\:{x \geq 1 } \end{array} \right.$, determine the values of $a$ and $b$ so that $f$ is a continuous function $ax^2+b=2ax+5\text{ for }x=0$ $2ax+5=3x-b\text{ for }x=1$,+0 # Piecewise Functions 0 144 1 Let $$f(x) = \left\{ \begin{array}{cl} -x + 3 & \text{if } x \le 0, \\ 2x - 5 & \text{if } x > 0. \end{array} \right.$$ How many solutions does the equation $$f(f(x))=4$$ have? May 18, 2020,+0 # Please help asap 0 81 1 For real numbers $x$, let $f(x) = \left\{ \begin{array}{cl} x+2 &\text{ if }x>3, \\ 2x+a &\text{ if }x\le 3. \end{array} \right.$What must the value of $a$ be to make the piecewise function continuous (which means that its graph can be drawn without lifting your pencil from the paper)? Guest Aug 23, 2018 #1 +91186 +2 $$f(x) = \left\{ \begin{array}{cl} x+2 &\text{ if }x>3, \\ 2x+a &\text{ if }x\le 3. \end{array} \right.$$ Let x = 3 and set the functions equal.... we have 3 + 2 = 2(3) + a 5 = 6 + a 5 - 6 = a - 1 = a See the graph here that shows that this value of a makes the function continuous : https://www.desmos.com/calculator/ux7xrnhaxi CPhill Aug 23, 2018 ### New Privacy Policy We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website. 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Difference between revisions of "2005 AIME I Problems/Problem 4" Problem The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are $5$ members left over. The director realizes that if he arranges the group in a formation with $7$ more rows than columns, there are no members left over. Find the maximum number of members this band can have. Solution Solution 1 If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$. If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$. Thus, if the number of columns is $n$, the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$. In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$, so this number works and no larger number can. Thus, the answer is $\boxed{294}$. Solution 2 Define,# Difference between revisions of "2005 AIME I Problems/Problem 4" ## Problem The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have. ## Solution If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$. If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$. Thus, if the number of columns is $n$, the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$. In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$, so this number works and no larger number can. Thus, the answer is 294.,Difference between revisions of "2005 AIME I Problems/Problem 4" Problem The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have. Solution Solution 1 If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$. If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$. Thus, if the number of columns is $n$, the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$. In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$, so this number works and no larger number can. Thus, the answer is $\boxed{294}$. Solution 2 Define,An army contingent of 616 members is to march behind an army band of 32 members in a parade Question. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? Solution: 616 and 32 $616=32 \times 19+8$ $32=4 \times 8$ $\operatorname{HCF}$ of $(616,32)=8$ Editor,# Difference between revisions of "2005 AIME II Problems" ## Problem 1 Six circles form a ring with with each circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle $C$ with radius $30$. Let $K$ be the area of the region inside circle $C$ and outside of the six circles in the ring. Find $\lfloor K \rfloor$. ## Problem 2 For each positive integer k, let $S_k$ denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is k. For example, $S_3$ is the squence $1,4,7,10 ...$. For how many values of k does $S_k$ contain the term 2005? ## Problem 3 How many positive integers have exactly three proper divisors, each of which is less than 50? ## Problem 4 The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have. ## Problem 5 Robert has 4 indistinguishable gold coins and 4,band has 36 members . they are arrange into 6 equal rows . How many band members are in each row? A band has 36 members . they are arrange into 6 equal rows . How many band members are in each row?... ### What evidence does the author provide to support that money used for the space program should be used to address poverty (article included) A. She acknowledges that exploring other worlds is inspiring for children B. She presents facts to demonstrate that space program spending is excessive C. She explains the importance of space research and exploration D. She shows why reducing the nation's debt should be a priority What evidence does the author provide to support that money used for the space program should be used to address poverty (article included) A. She acknowledges that exploring other worlds is inspiring for children B. She presents facts to demonstrate that space program spending is excessive C. She e...,each row? Math can look so pretty, all nicely formatted in the textbook. Square Formation is a Technology and infantry Formation in Empire: Total War. Below are two diagrams showing symmetrical hollow square formations. Hint 1: Different sizes of squares can be made with some or all of the pieces.However, a square can be made that uses all five pieces, side by side, with none of the pieces on top or over lapping another piece. The square wave is a special case of a pulse wave which allows arbitrary durations at minimum and … This figure is then multiplied by a half. Another way to prevent getting this page in the future is to use Privacy Pass. Squaring a number means to raise it to the second power. To make HST place two equal size squares of fabric Right Sides Together (RST). In an ideal square wave, the transitions between minimum and maximum are instantaneous. A complex number is a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i 2 = −1.For example, 2 + 3i is a complex number. • If you cluster pairs of rows by the number of ones in each column they fall into 19 equivalence classes. If you are at an,8 ------------- 9 9 ------------- 8 12 ----------- 6 18 ----------- 4 24 ----------- 3 Total of 8 different formations. How did you find out that the number of people in each row is the same ?? Senior Manager Joined: 13 Oct 2016 Posts: 359 GPA: 3.98 Re: If a marching band has 72 members that always march in fo [#permalink] Show Tags 16 Dec 2016, 05:55 1 NYC5648 wrote: If a marching band has 72 members that always march in formations of at least three rows and at least 3 members in each row, how many different formations can they march in? OA: 8 Many thanks guys! We have $$X$$ rows and $$Y$$ columns (members in each row) $$X*Y = 72$$ We need to find out the number of ways 72 can be expressed as a product of 2 factors. $$72 = 2^3*3^2$$ # of ways = $$\frac{(3+1)*(2+1)}{2} = 6$$ But we can’t use this answer because in our question ORDER MATTERS. Rows are different from the columns (number of people in a row). $$3*24$$ here is different from $$24*3$$. Hence we should not divide by 2 and we get 12 total possibilities. Next step: we need to take into consideration additional restriction: $$X ≥ 3$$, $$Y ≥ 3$$. We need to deduct following 4 cases from,# An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march ? ## Solution : To find the maximum number of columns, we have to find the H.C.F. of 616 and 32 i.e. 616 = 32 $$\times$$ 19 + 8 and 32 = 8 $$\times$$ 4 + 0 $$\therefore$$ H.C.F of 616 and 32 is 8. Hence, maximum number of columns is 8.,Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too! # An army contingent of 616 members is to march behind an army band of 32 members in a parade. Question: An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? Solution: We are given that an army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. We need to find the maximum number of columns in which they can march. Members in army = 616 Members in band = 32. Therefore, Maximum number of columns = H.C.F of 616 and 32. By applying Euclid’s division lemma $616=32 \times 19+8$ $32=8 \times 4+0$ Therefore, H.C.F. = 8 Hence, the maximum number of columns in which they can march is 8 . #### Leave a comment None Free Study Material,# Band (mathematics) (Redirected from Right-zero band) In mathematics, a band (also called idempotent semigroup) is a semigroup in which every element is idempotent (in other words equal to its own square). Bands were first studied and named by A. H. Clifford (1954); the lattice of varieties of bands was described independently in the early 1970s by Biryukov, Fennemore and Gerhard.[1] Semilattices, left-zero bands, right-zero bands, rectangular bands, normal bands, left-regular bands, right-regular bands and regular bands, specific subclasses of bands which lie near the bottom of this lattice, are of particular interest and are briefly described below. ## Varieties of bands A class of bands forms a variety if it is closed under formation of subsemigroups, homomorphic images and direct product. Each variety of bands can be defined by a single defining identity.[2] ### Semilattices Semilattices are exactly commutative bands; that is, they are the bands satisfying the equation • xy = yx for all x and y. ### Zero bands A left zero band is a band satisfying the equation • xy = x, whence its Cayley table has constant rows. Symmetrically, a right zero band is one satisfying • xy = y, so that the Cayley table has constant columns. ### Rectangular bands A rectangular band is a band S that satisfies • xyx =,of nonempty bands is equivalent to $\mathrm{Set}_{\ne \emptyset} \times \mathrm{Set}_{\ne \emptyset}$, where$\mathrm{Set}_{\ne \emptyset}$ is the category with nonempty sets as objects and functions as morphisms. This implies that not only that every nonempty rectangular band is isomorphic to one coming from a pair of sets, but also these sets are uniquely determined up to a canonical isomorphism, and all homomorphisms between bands come from pairs of functions between sets I thought these facts were restricted to rectangular bands, not the representation of general bands. In this thread Yemon Choi says: rectangular bands can serve as basic building blocks in a structure theory for general bands. … The structure theorem is due to Petrich, and is explained in Section 4.3 of Howie’s An Introduction To Semigroup Theory. A quick google turns up Andreas Distler - “Bands of order at most 10” that says: The building blocks for bands are two types of bands with additional properties, semilattices and rectangular bands. A classic result by Clifford states that every band is a semilattice of rectangular bands. A. H. Clifford. Bands of semigroups. Proc. Amer. Math. Soc., 5:499-504, 1954. None of this general band stuff appears in the WIkipedia article. Posted by: RodMcGuire on June 20, 2015 7:17 PM | Permalink | Reply to this ### Re: Semigroup Puzzles Yes, this,several ways to organize band members into rectangular arrays on the field for a performance.Should the instructor have 89 members or 99 members on the field? Explain. Answer: The Instructor should have 99 members on the field so that he can arrange the band members into rectangular arrays. Explanation: Given that a band instructor wants to organize band members into rectangular arrays on the field for a performance. It is also given that the instructor wants to arrange into an array of whether 89 members or 99 members Now, Factors of 89 are: 1, 89 Factors of 99 are: 1, 3, 9, 11, 33, 99 Hence, from the above, Since the 99 members can be arranged into different arrays, We can conclude that the instructor can arrange the participants into an array of 99 members. Question 16. DIG DEEPER! A paramedic is arranging bandages into 4 bins. An equal number of bandages are in each bin. Did the paramedic arrange a prime number or a composite number of bandages? Explain. Answer: The paramedic has to arrange a composite number of bandages. Explanation: Gien that a paramedic is arranging bandages into 4 bins. It is also given that there is an equal number of bandages in each bin. Now, Given there are 4 bins and 4 is a Composite,# Band (mathematics) (Redirected from Band (algebra)) In mathematics, a band (also called idempotent semigroup) is a semigroup in which every element is idempotent (in other words equal to its own square). Bands were first studied and named by A. H. Clifford (1954); the lattice of varieties of bands was described independently in the early 1970s by Biryukov, Fennemore and Gerhard.[1] Semilattices, left-zero bands, right-zero bands, rectangular bands, normal bands, and regular bands, specific subclasses of bands which lie near the bottom of this lattice, are of particular interest and are briefly described below. ## Varieties of bands A class of bands forms a variety if it is closed under formation of subsemigroups, homomorphic images and direct product. Each variety of bands can be defined by a single defining identity.[2] ### Semilattices Semilattices are exactly commutative bands; that is, they are the bands satisfying the equation • xy = yx for all x and y. ### Zero bands A left zero band is a band satisfying the equation • xy = x, whence its Cayley table has constant rows. Symmetrically, a right zero band is one satisfying • xy = y, so that the Cayley table has constant columns. ### Rectangular bands A rectangular band is a band S which satisfies • xyx = x for all xy,∈ S. Equivalently, • xyz = xz. For example, given arbitrary non-empty sets I and J one can define a semigroup operation on I × J by setting $(i, j) \cdot (k, \ell) = (i, \ell) \,$ The resulting semigroup is a rectangular band because 1. for any pair (ij) we have (ij) · (ij) = (ij) 2. for any two pairs (ixjx), (iyjy) we have $(i_x, j_x) \cdot (i_y, j_y) \cdot (i_x, j_x) = (i_x, j_x)$ In fact, any rectangular band is isomorphic to one of the above form. Left zero and right zero bands are rectangular bands, and in fact every rectangular band is isomorphic to a direct product of a left zero band and a right zero band. All rectangular bands of prime order are zero bands, either left or right. A rectangular band is said to be purely rectangular if it is not a left or right zero band.[3] ### Normal bands A normal band is a band S satisfying • xyzx = xzyx for all x, y, and z ∈ S. This is the same equation used to define medial magmas, and so a normal band may also be called a medial band, and normal bands are examples of medial magmas.[3] ### Regular bands A regular band is a band S satisfying •,Kattis # Rock Band Every day after school, you and your friends get together and play in a band. Over the past couple of months, the band has been rehearsing a large number of songs. Now it’s time to go out and perform in front of a crowd for the first time. In order to do so, a set list for the concert needs to be determined. As it turns out, every band member has a different taste in music. (Who would have thought?) Everybody is very picky: a band member doesn’t want to play any particular song $X$ unless he also gets to play all songs he likes better than song $X$. This holds for every band member and for every song $X$. Furthermore, obviously at least one song should be performed. The organisers of the concert do not want you to play too many songs, so a selection needs to be made that is as small as possible. As the unofficial leader of the band, you have taken it upon yourself to find a minimum length set list that meets the requirements. ## Input The first line contains two integers $M$ and $S$, satisfying $M \geq 1$ and $S \geq 1$ as well as $M\cdot S \leq 10^6$. These denote the total number of band members,# Organize musicians into rows, with weight and population restrictions So I just completed this project for my java programming class and was wondering if there was any way I could streamline it to really impress my professor. Here were the instructions, followed by my code. The University of Miami's "Band of the Hour" needs a program to organize where the musicians will stand when they play at away games. Each away stadium is different, so when they arrive the conductor gets the following information from the local organizer: The number of rows they have to stand on. The maximum number of rows is 10. The rows are labelled with capital letters, 'A', 'B', 'C', etc. For each row, the number of positions in the row. The maximum number of positions is 8. The positions are numbered with integers, 1, 2, 3, etc. The conductor then starts assigning people to positions, but is constrained by weight limits: Musicians, fully clothed and holding their instruments, weigh from 45kg to 200kg, and the total weight of a row may not exceed 100kg per position (e.g., a row with 5 positions may not have more than 500kg of musicians on it). The conductor wants a program that allows musicians to be added and removed from positions, while ensuring the constraints are,in each row. There is also a row of 6 people who carry flags. How many people are in the marching band in all? Multiplication equation: Addition equation: There are _______ people in the marching band in all. Answer: A marching band has 7 rows with 8 musicians in each row. 1 row – 8 musicians 7 rows – 7 × 8 musicians = 56 musicians There is also a row of 6 people who carry flags. 56 + 6 = 62 people Therefore there are 62 people in the marching band in all. Show and Grow Question 17. A table has 3 rows with 8 prizes in each row. There is also a row of 4 prizes on the floor. How many prizes are there in all? Answer: 28 prizes Explanation: A table has 3 rows with 8 prizes in each row. 1 row – 8 prizes 3 rows – 8 × 3 = 24 prizes There is also a row of 4 prizes on the floor 24 + 4 = 28 prizes Thus there are 28 prizes in all. Question 18. DIG DEEPER! One section of a parking lot has 2 rows of 8 cars. Another section of the parking lot has 8 rows of 6 cars. How many cars are in the parking lot in,because $(A\times \emptyset)\cong (B\times \emptyset)$ for all sets $A$ and $B$. I think the characterization of rectangular bands requires a focus on non-empty bands and non-empty sets. Correct me if I am wrong. This is in contrast to the situation with involutive rectangular bands that I mentioned below. The monadicity of that adjunction has been analyzed by Francois Metayer here. Posted by: Sam Staton on June 16, 2015 11:14 AM | Permalink | Reply to this ### Re: Semigroup Puzzles Sam wrote: $(A\times \emptyset)\cong (B\times \emptyset)$ for all sets $A$ and $B$. I think the characterization of rectangular bands requires a focus on non-empty bands and non-empty sets. Oh! You’re right, $A \times \emptyset$ and $B \times \emptyset$ are always isomorphic as rectangular bands. So my hoped-for equivalence of the category of rectangular bands with $Set \times Set$, despite being “well known in the semigroup community”, is actually false. But the category of nonempty rectangular bands is equivalent to the category $NonemptySet \times NonemptySet$. Right? Posted by: John Baez on June 16, 2015 11:44 AM | Permalink | Reply to this ### Re: Semigroup Puzzles You are right of course. In semigroup theory we often forget the empty semigroup. You need nonempty rectangular bands. The category of non empty rectangular bands is isomorphic to the product of two,how many different formations can they march in? OA: 8 Many thanks guys! We have $$X$$ rows and $$Y$$ columns (members in each row) $$X*Y = 72$$ We need to find out the number of ways 72 can be expressed as a product of 2 factors. $$72 = 2^3*3^2$$ # of ways = $$\frac{(3+1)*(2+1)}{2} = 6$$ But we can’t use this answer because in our question ORDER MATTERS. Rows are different from the columns (number of people in a row). $$3*24$$ here is different from $$24*3$$. Hence we should not divide by 2 and we get 12 total possibilities. Next step: we need to take into consideration additional restriction: $$X ≥ 3$$, $$Y ≥ 3$$. We need to deduct following 4 cases from our total set: $$1*72$$, $$2*36$$ and $$72*1$$, $$36*2$$ Final answer: $$12 – 4 = 8$$ Non-Human User Joined: 09 Sep 2013 Posts: 8859 Re: If a marching band has 72 members that always march in fo [#permalink] ### Show Tags 24 Oct 2018, 06:21 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(...TRUNCATED)

of members the band could have? Level 5 Algebra Let $x$ be the number of band members in each row for the original formation, when two are left over. Then we can write two equations from the given information: $$rx+2=m$$ $$(r-2)(x+1)=m$$ Setting these equal, we find: $$rx+2=(r-2)(x+1)=rx-2x+r-2$$ $$2=-2x+r-2$$ $$4=r-2x$$ We know that the band has less than 100 members. Based on the first equation, we must have $rx$ less than 98. We can guess and check some values of $r$ and $x$ in the last equation. If $r=18$, then $x=7$, and $rx=126$ which is too big. If $r=16$, then $x=6$, and $rx=96$, which is less than 98. Checking back in the second formation, we see that $(16-2)(6+1)=14\cdot 7=98$ as it should. This is the best we can do, so the largest number of members the band could have is $\boxed{98}$. What is the degree of the polynomial $(4 +5x^3 +100 +2\pi x^4 + \sqrt{10}x^4 +9)$? Level 3 Algebra This polynomial is not written in standard form. However, we don't need to write it in standard form, nor do we need to pay attention to the coefficients. We just look for the exponents on $x$. We have an $x^4$ term and no other term of higher degree, so $\boxed{4}$ is the degree of the polynomial. Evaluate $\left\lceil3\left(6-\frac12\right)\right\rceil$. Level 3,Difference between revisions of "2005 AIME I Problems/Problem 4" Problem The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are $5$ members left over. The director realizes that if he arranges the group in a formation with $7$ more rows than columns, there are no members left over. Find the maximum number of members this band can have. Solution Solution 1 If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$. If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$. Thus, if the number of columns is $n$, the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$. In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$, so this number works and no larger number can. Thus, the answer is $\boxed{294}$. Solution 2 Define,Difference between revisions of "2005 AIME I Problems/Problem 4" Problem The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have. Solution Solution 1 If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$. If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$. Thus, if the number of columns is $n$, the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$. In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$, so this number works and no larger number can. Thus, the answer is $\boxed{294}$. Solution 2 Define,# Difference between revisions of "2005 AIME I Problems/Problem 4" ## Problem The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have. ## Solution If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$. If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$. Thus, if the number of columns is $n$, the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$. In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$, so this number works and no larger number can. Thus, the answer is 294.,8 ------------- 9 9 ------------- 8 12 ----------- 6 18 ----------- 4 24 ----------- 3 Total of 8 different formations. How did you find out that the number of people in each row is the same ?? Senior Manager Joined: 13 Oct 2016 Posts: 359 GPA: 3.98 Re: If a marching band has 72 members that always march in fo [#permalink] Show Tags 16 Dec 2016, 05:55 1 NYC5648 wrote: If a marching band has 72 members that always march in formations of at least three rows and at least 3 members in each row, how many different formations can they march in? OA: 8 Many thanks guys! We have $$X$$ rows and $$Y$$ columns (members in each row) $$X*Y = 72$$ We need to find out the number of ways 72 can be expressed as a product of 2 factors. $$72 = 2^3*3^2$$ # of ways = $$\frac{(3+1)*(2+1)}{2} = 6$$ But we can’t use this answer because in our question ORDER MATTERS. Rows are different from the columns (number of people in a row). $$3*24$$ here is different from $$24*3$$. Hence we should not divide by 2 and we get 12 total possibilities. Next step: we need to take into consideration additional restriction: $$X ≥ 3$$, $$Y ≥ 3$$. We need to deduct following 4 cases from,how many different formations can they march in? OA: 8 Many thanks guys! We have $$X$$ rows and $$Y$$ columns (members in each row) $$X*Y = 72$$ We need to find out the number of ways 72 can be expressed as a product of 2 factors. $$72 = 2^3*3^2$$ # of ways = $$\frac{(3+1)*(2+1)}{2} = 6$$ But we can’t use this answer because in our question ORDER MATTERS. Rows are different from the columns (number of people in a row). $$3*24$$ here is different from $$24*3$$. Hence we should not divide by 2 and we get 12 total possibilities. Next step: we need to take into consideration additional restriction: $$X ≥ 3$$, $$Y ≥ 3$$. We need to deduct following 4 cases from our total set: $$1*72$$, $$2*36$$ and $$72*1$$, $$36*2$$ Final answer: $$12 – 4 = 8$$ Non-Human User Joined: 09 Sep 2013 Posts: 8859 Re: If a marching band has 72 members that always march in fo [#permalink] ### Show Tags 24 Oct 2018, 06:21 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,+0 # HELLLLP! QUADRATIC INEQUALITIES! WORD PROBLEMS!! 0 717 1 +118 A band is marching in a rectangular formation with dimensions $n-2$ and $n + 8$. In the second stage of their performance, they re-arrange to form a different rectangle with dimensions $n$ and $2n - 3$, excluding all the drummers. If there are at least 4 drummers, then find the sum of all possible values of $n$. Mar 14, 2019 #1 +6196 +2 $$\text{There are }M=(n-2)(n+8) \text{ total marchers}\\ \text{let }D\text{ be the number of drummers},~D\geq 4\\ M-D = n(2n-3)$$ $$n^2+6n-16 = 2n^2 - 3n+D\\ n^2 -9n+(16+D)=0\\ n = \dfrac{9\pm \sqrt{81-4(16+D)}}{2} \in \mathbb{N}$$ $$81 - 4(16+D) = k^2,~k \in \mathbb{N}\\ \text{The only possible value for }D \text{ is 4}\\ n = \dfrac{9 \pm 1}{2} = 5,4\\ 4+5=9$$ . Mar 14, 2019 #1 +6196 +2 $$\text{There are }M=(n-2)(n+8) \text{ total marchers}\\ \text{let }D\text{ be the number of drummers},~D\geq 4\\ M-D = n(2n-3)$$ $$n^2+6n-16 = 2n^2 - 3n+D\\ n^2 -9n+(16+D)=0\\ n = \dfrac{9\pm \sqrt{81-4(16+D)}}{2} \in \mathbb{N}$$ $$81 - 4(16+D) = k^2,~k \in \mathbb{N}\\ \text{The only possible value for }D \text{ is 4}\\ n = \dfrac{9 \pm 1}{2} = 5,4\\ 4+5=9$$ Rom Mar 14, 2019,= 657x + 936 (-15). Hence, 9 = 657x —14445 9 + 14445 = 657x 14454 = 657x $x = \frac{ 14454 }{ 657 }$ Now, solve the above equation, ⇒ x = 22. Q.7: An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to in the same number of columns. What is the maximum number of columns in which they can march? Sol. It is given that an army contingent of 616 members is to march behind an army band of 32 members in a parade. Also, the two groups are to march in the same number of columns. We need to find the maximum number of columns in which they can march. So, Members in army = 616 Members in band = 32. Therefore, the maximum number of columns = H.C.F of 616 and 32. By applying Euclid’s division lemma 616 = 32 x 19 + 8 32 = 8 x 4 + 0. So, H.C.F. = 8 ∴ The maximum number of columns in which they can march is 8 Q.8: A merchant has 120 liters of oil of one kind, 180 liters of another and 240 liters of the third kind. He wants to sell the oil by filling the,all the members march in 4 equal rows? • Use square tiles to represent the problem. Then draw an array to show your work. • Can 26 band members march in 4 equal rows? Use factors or divisibility to explain. 4 is not the factor of 26. They cannot march in 4 equal rows. Explanation: There are 26 members in the marching band. All the members march in 4 equal rows. => Multiples of 26: 1 × 26 = 26. 2 × 13 = 26. Use a visual model to answer and explain. Question 3. Is 6 a factor of 32? No, 32 is not a factor of 6. Explanation: Multiples of 32: 1 × 32 = 32. 2 × 16 = 32. 4 × 8 = 32. Question 4. Is 5 a factor of 25? Yes, 5 a factor of 25. Explanation: Multiples of 25: 1 × 25 = 25. 5 × 5 = 25. Question 5. Use Repeated Reasoning Explain whether or not the following statement is true: Since 89 is not divisible by 2, then 89 is not divisible by any multiple of 2. Yes, the following statement is true because 89 cannot be divided by any factor of 2 as 89 is not an even number. Explanation: 89 is not divisible by 2, then,compared. Looking at the dimensions, you would have been able to see that the tapestry would not fit. Let's look at another example: Example The band leader has arranged 56 musicians that will be participating in a parade. The number of players he placed in each row was 10 more than the number of rows. How many rows were there? To approach this problem, we need to assume that the musicians are arranged in equal rows. Then we know we are going to have to look for factors of 56 which fit these qualifications. We can work to “guess and check” for numbers that work. The guessing and checking strategy uses $7 \times 8 = 56$, but that doesn’t work. $28 \times 2 = 56$, but that doesn’t work either. $14 \times 4 = 56$ and that works! If there are 14 musicians in each row, there will be 10 more than the number of rows. Guessing and checking isn’t the only way to solve this problem. You might also choose to draw a picture or make an organized list. ## Real Life Example Completed The Four Thousand Footers Here is the original problem once again. Reread it and underline any important information. While at Galehead Hut, Kelly found a book on the different mountains in the Presidential,Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too! # An army contingent of 616 members is to march behind an army band of 32 members in a parade. Question: An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? Solution: We are given that an army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. We need to find the maximum number of columns in which they can march. Members in army = 616 Members in band = 32. Therefore, Maximum number of columns = H.C.F of 616 and 32. By applying Euclid’s division lemma $616=32 \times 19+8$ $32=8 \times 4+0$ Therefore, H.C.F. = 8 Hence, the maximum number of columns in which they can march is 8 . #### Leave a comment None Free Study Material,we set up the equation? 4. Originally Posted by kohane hi thanks for your help. But at least how many people must attend? how would we set up the equation? Let $x \in \mathbb{N}$ denote the number of listeners. Collect all results and solve for x: $6 \frac{\}{person} x \geq \ 550 + \dfrac{\ 1200 - \550}{0.15}~\implies~x \geq 814 \ persons$ 5. kohane, here's one approach for progressing from a verbal rendering of the situation, to the algebraic inequality earboth gave you... The question can be though of as, How many tonedeaf (I've heard Mary's band) people must attend, so that • The amount that the band will receive, based on the agreement, • will be at least • \$1,200? One at a time, and letting x denote the number of listeners, sticking with earboth's notation... • The amount the band will receive, at 6 bucks a head, is 550 + 0.15(6x - 550) • "will be at least" can be rendered as "greater than or equal to", aka $\geq$ • 1,200 is, well, 1,200 Stringing the three components together, the question can thus be expressed as 550 + 0.15(6x - 550) $\geq$ 1,200 ...and then solving for x will get you to earboth's solution. I hope that helped a bit. Best of luck.,# An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march ? ## Solution : To find the maximum number of columns, we have to find the H.C.F. of 616 and 32 i.e. 616 = 32 $$\times$$ 19 + 8 and 32 = 8 $$\times$$ 4 + 0 $$\therefore$$ H.C.F of 616 and 32 is 8. Hence, maximum number of columns is 8.,x + y = 22 x^2 + y = 402 (X1,Y1)= (20,2) for (X2,Y2)= (-19,41) So there was either 41 or 2 band acts in 2016? so for 2017 2500x + 5500y = 90000 (X1,X2)=(22/5,158/11) or (25,5) So there was either 158/11 or 5 band acts in 2017? xy = 50 + 3x for for 2018 x + y = 30 133x + 400y^2 = 59994 (X1,Y1) = (18,12) or (16667/400,4667,400) For all of my results I will use the positive and whole numbers as said by you " so you must ignore any solutions you get for x or y that are not positive, Whole numbers." Putting (18,12) into 2625x + zy = 120000 so Z=-2937.5. Not to sure if I've done any of this correct? #### Otis ##### Senior Member … For all of my results I will use the positive and whole numbers … Okay, but you don't seem to be doing that. … so for 2016 … (X1,Y1)=(20,2) … (X2,Y2)=(-19,41) So there was either 41 or 2 band acts in 2016? No. We need to ignore the solution (-19,41) because x=-19 is not positive. There were two band acts in 2016. … so for 2017 … (22/5,158/11) or (25,5) So there was either 158/11 or 5 band acts in 2017? … No. We need,men. (i) Calculate the number of different groups of 4 people that have exactly 3 women. (ii) Calculate the number of different groups of at most 4 people where the number of women is the same as the number of men. 9. (CIE $0606 / 2019 / \mathrm{m} / 22 / \mathrm{q} 1)$ A band can play 25 different pieces of music. From these pieces of music, 8 are to be selected for a concert. (i) Find the number of different ways this can be done. The 8 pieces of music are then arranged in order. (ii) Find the number of different arrangements possible. The band has 15 members, Three members are chosen at random to be the treasurer, secretary and agent. (iii) Find the number of ways in which this can be done.[1] 10. (CIE $0606 / 2019 / \mathrm{w} / 22 / \mathrm{q} 3)$ A 5 -digit code is formed using the following characters. No character can be repeated in a code. Find the number of possible codes if (i) there are no restrictions,$[2]$ (ii) the code starts with a symbol followed by two letters and then two numbers,$[2]$ (iii) the first two characters are numbers, and no other numbers appear in the code. 11. (CIE $0606 / 2020 / \mathrm{w} / 11 / \mathrm{q} 5)$,An army contingent of 616 members is to march behind an army band of 32 members in a parade Question. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? Solution: 616 and 32 $616=32 \times 19+8$ $32=4 \times 8$ $\operatorname{HCF}$ of $(616,32)=8$ Editor,vertical bands in this figure — the first three look the same (although the second one appears to be reflected), and the last two also look the same, although reflections of each other. The first band is all ternary expansions in this new set beginning with 0.10. How do these relate to the whole set? Well, 1/9 of the set consists of expansions beginning with 0.001… or 0.002…, and then adding digits different from those previous. Adding 1/3 therefore gives all expansions beginning with 0.101… or 0.102…, and then adding different digits. This implies that the self-similarity describing the first vertical band is $y=\dfrac 19x+\dfrac13.$ The second band consists of those expansions in our strings beginning with 0.12. But if x is an expansion in our set beginning with 0.10, then 1 – x must be an expansion in our set beginning with 0.12, since we may write 1 as .22222…, repeating. Therefore, the second band is represented by the transformation $y=1-\left(\dfrac 19x+\dfrac13\right)=\dfrac23-\dfrac19x.$ We can think of the third band just as we did the first — except that this band consists of number beginning with 0.20 (rather than 0.10). So this band is represented by the transformation $y=\dfrac 19x+\dfrac23.$ The last two bands consist of those expansions beginning with 0.21. Here, we break into the two cases,### N000ughty Thoughts How many noughts are at the end of these giant numbers? ### DOTS Division Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}. ### Mod 3 Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3. # Indivisible ##### Age 14 to 16 ShortChallenge Level Some students (fewer than $100$) are having trouble lining up for a school production. When they line up in $3$s, two people are left over. When they line up in $4$s, three people are left over. When they line up in $5$s, four people are left over. When they line up in $6$s, five people are left over. How many students are there in the group? This problem is taken from the UKMT Mathematical Challenges. You can find more short problems, arranged by curriculum topic, in our short problems collection.,## bprice26 2 years ago Auditions for a new, upcoming band in Chicago were held with 102 men and women trying out for the lead singer position. If twice the number of men less three times the number of women is equal to -6, how many women auditioned? 1. PaxPolaris Using two variables let m be the number of men, ... & w the number of women $\Large m+w=102$ $\Large 2m-3w=-6$ 2. bprice26 i already had an equation written out ; i just don't know what to do from there .. 3. PaxPolaris multiply the first eq by 2 4. PaxPolaris then subtract the second equation 5. bprice26 What ? How would i do that .. Do i multiply 102 by 2 ??? I am soooo confused , i have been reading all my lessons and doing all my work i just don't get it, it makes no sense to me ... 6. PaxPolaris Well actually there are to methods, the use of simultaneous equations that i used above and using substitution... 7. bprice26 Okay well how does the substituion method work ? 8. PaxPolaris *2 methods Using substitution: $\large m+w= 102$$\large \implies w=102-m$ put that in 2nd equation 9. PaxPolaris $\large 2m-3\left( 102-m \right)=-6$ 10. bprice26 Okay ; now i get it ! Thanks (: 11. PaxPolaris,the number of rows/columns of the square formation as $s$, and the number of rows of the rectangular formation $r$ (so there are $r - 7$ columns). Thus, $s^2 + 5 = r(r-7) \Longrightarrow r^2 - 7r - s^2 - 5 = 0$. The quadratic formula yields $r = \frac{7 \pm \sqrt{49 - 4(1)(-s^2 - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}$. $\sqrt{4s^2 + 69}$ must be an integer, say $x$. Then $4s^2 + 69 = x^2$ and $(x + 2s)(x - 2s) = 69$. The factors of $69$ are $(1,69), (3,23)$; $x$ is maximized for the first case. Thus, $x = \frac{69 + 1}{2} = 35$, and $r = \frac{7 \pm 35}{2} = 21, -14$. The latter obviously can be discarded, so there are $21$ rows and $21 - 7 = 14$ columns, making the answer $294$. Solution 3 The number of members is $m^2+5=n(n+7)$ for some $n$ and $m$. Multiply both sides by $4$ and complete the square to get $4m^2+69=(2n+7)^2$. Thus, we have $69=((2n+7)+2m)((2n+7)-2m)$. Since we want to maximize $n$, set the first factor equal to $69$ and the second equal to $1$. Solving gives $n=14$, so the answer is $14\cdot21=294$. Solution 4 Partially completing the square Geometrically: Split up the formation of $n + 7$ rows and $n$ columns into a square of $n$(...TRUNCATED)

of members the band could have? Level 5 Algebra Let $x$ be the number of band members in each row for the original formation, when two are left over. Then we can write two equations from the given information: $$rx+2=m$$ $$(r-2)(x+1)=m$$ Setting these equal, we find: $$rx+2=(r-2)(x+1)=rx-2x+r-2$$ $$2=-2x+r-2$$ $$4=r-2x$$ We know that the band has less than 100 members. Based on the first equation, we must have $rx$ less than 98. We can guess and check some values of $r$ and $x$ in the last equation. If $r=18$, then $x=7$, and $rx=126$ which is too big. If $r=16$, then $x=6$, and $rx=96$, which is less than 98. Checking back in the second formation, we see that $(16-2)(6+1)=14\cdot 7=98$ as it should. This is the best we can do, so the largest number of members the band could have is $\boxed{98}$. What is the degree of the polynomial $(4 +5x^3 +100 +2\pi x^4 + \sqrt{10}x^4 +9)$? Level 3 Algebra This polynomial is not written in standard form. However, we don't need to write it in standard form, nor do we need to pay attention to the coefficients. We just look for the exponents on $x$. We have an $x^4$ term and no other term of higher degree, so $\boxed{4}$ is the degree of the polynomial. Evaluate $\left\lceil3\left(6-\frac12\right)\right\rceil$. Level 3,Difference between revisions of "2005 AIME I Problems/Problem 4" Problem The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are $5$ members left over. The director realizes that if he arranges the group in a formation with $7$ more rows than columns, there are no members left over. Find the maximum number of members this band can have. Solution Solution 1 If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$. If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$. Thus, if the number of columns is $n$, the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$. In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$, so this number works and no larger number can. Thus, the answer is $\boxed{294}$. Solution 2 Define,Difference between revisions of "2005 AIME I Problems/Problem 4" Problem The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have. Solution Solution 1 If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$. If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$. Thus, if the number of columns is $n$, the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$. In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$, so this number works and no larger number can. Thus, the answer is $\boxed{294}$. Solution 2 Define,# Difference between revisions of "2005 AIME I Problems/Problem 4" ## Problem The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have. ## Solution If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$. If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$. Thus, if the number of columns is $n$, the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$. In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$, so this number works and no larger number can. Thus, the answer is 294.,# Datasets:competition_math Languages: en Multilinguality: monolingual Size Categories: 10K<n<100K Language Creators: expert-generated Annotations Creators: expert-generated Source Datasets: original Dataset Preview problem (string)level (string)type (string)solution (string) Let $f(x) = \left\{ \begin{array}{cl} ax+3, &\text{ if }x>2, \\ x-5 &\text{ if } -2 \le x \le 2, \\ 2x-b &\text{ if } x <-2. \end{array} \right.$Find $a+b$ if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper). Level 5 Algebra For the piecewise function to be continuous, the cases must "meet" at $2$ and $-2$. For example, $ax+3$ and $x-5$ must be equal when $x=2$. This implies $a(2)+3=2-5$, which we solve to get $2a=-6 \Rightarrow a=-3$. Similarly, $x-5$ and $2x-b$ must be equal when $x=-2$. Substituting, we get $-2-5=2(-2)-b$, which implies $b=3$. So $a+b=-3+3=\boxed{0}$. A rectangular band formation is a formation with $m$ band members in each of $r$ rows, where $m$ and $r$ are integers. A particular band has less than 100 band members. The director arranges them in a rectangular formation and finds that he has two members left over. If he increases the number of members in each row by 1 and reduces the number of rows by 2, there are exactly enough places in the new formation for each band member. What is the largest number,8 ------------- 9 9 ------------- 8 12 ----------- 6 18 ----------- 4 24 ----------- 3 Total of 8 different formations. How did you find out that the number of people in each row is the same ?? Senior Manager Joined: 13 Oct 2016 Posts: 359 GPA: 3.98 Re: If a marching band has 72 members that always march in fo [#permalink] Show Tags 16 Dec 2016, 05:55 1 NYC5648 wrote: If a marching band has 72 members that always march in formations of at least three rows and at least 3 members in each row, how many different formations can they march in? OA: 8 Many thanks guys! We have $$X$$ rows and $$Y$$ columns (members in each row) $$X*Y = 72$$ We need to find out the number of ways 72 can be expressed as a product of 2 factors. $$72 = 2^3*3^2$$ # of ways = $$\frac{(3+1)*(2+1)}{2} = 6$$ But we can’t use this answer because in our question ORDER MATTERS. Rows are different from the columns (number of people in a row). $$3*24$$ here is different from $$24*3$$. Hence we should not divide by 2 and we get 12 total possibilities. Next step: we need to take into consideration additional restriction: $$X ≥ 3$$, $$Y ≥ 3$$. We need to deduct following 4 cases from,how many different formations can they march in? OA: 8 Many thanks guys! We have $$X$$ rows and $$Y$$ columns (members in each row) $$X*Y = 72$$ We need to find out the number of ways 72 can be expressed as a product of 2 factors. $$72 = 2^3*3^2$$ # of ways = $$\frac{(3+1)*(2+1)}{2} = 6$$ But we can’t use this answer because in our question ORDER MATTERS. Rows are different from the columns (number of people in a row). $$3*24$$ here is different from $$24*3$$. Hence we should not divide by 2 and we get 12 total possibilities. Next step: we need to take into consideration additional restriction: $$X ≥ 3$$, $$Y ≥ 3$$. We need to deduct following 4 cases from our total set: $$1*72$$, $$2*36$$ and $$72*1$$, $$36*2$$ Final answer: $$12 – 4 = 8$$ Non-Human User Joined: 09 Sep 2013 Posts: 8859 Re: If a marching band has 72 members that always march in fo [#permalink] ### Show Tags 24 Oct 2018, 06:21 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,so apply division lemma on divisor 36 and remainder 9 36 = 9 x 4 + 0. Therefore, H.C.F. = 9. Given H.C.F = 657x + 936 (-15). Therefore, 9 = 657x —14445 9 + 14445 = 657x 14454 = 657x $x = \frac{ 14454 }{ 657 }$ On solving the above, we have, x = 22. Hence obtained. Q.7: An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to in the same number of columns. What is the maximum number of columns in which they can march? Sol. We are given that an army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. We need to fund the maximum number of columns in which they can march. Members in army = 616 Members in band = 32. Therefore, Maximum number of columns = H.C.F of 616 and 32. By applying Euclid’s division lemma 616 = 32 x 19 + 8 32 = 8 x 4 + 0. Therefore, H.C.F. = 8 Hence, the maximum number of columns in which they can march is 8 Q.8: A merchant has 120 liters of oil of,rows and $n$ columns and a separate rectangle of the dimensions $7$ rows by $n$ columns. We want to take the rows from the rectangle and add them to the square to get another square and $5$ left over. If we attach exactly $2$ rows on the top and exactly $2$ rows on the side of the $n$ x $n$ square, then we have an $(n + 2)$ x $(n + 2)$ square that's missing a $2$ x $2$ corner. For the remaining $3n$ to fill this square plus the $5$ extra members, $n$ must be $3$. If we instead plaster exactly $3$ rows from the $7$ x $n$ formation to two adjacent sides of the $n$ x $n$ square, we have an $(n + 3)$ x $(n + 3)$ formation that's missing a $3$ x $3$ corner. For the remaining row of length $n$ to fill this plus five, $n = 14$. Plugging these in, we find $n = 14$ has a much higher count of members: $(n + 7)n; n = 14 --> 21(14) = 294$ Algebraically: We have $n^2 + 7n = m$, where $m$ is the number of members in the band and $n$ is a positive integer. We partially complete the square for $n$ to get $n^2 + 7n = (n + 1)^2,Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too! # An army contingent of 616 members is to march behind an army band of 32 members in a parade. Question: An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? Solution: We are given that an army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. We need to find the maximum number of columns in which they can march. Members in army = 616 Members in band = 32. Therefore, Maximum number of columns = H.C.F of 616 and 32. By applying Euclid’s division lemma $616=32 \times 19+8$ $32=8 \times 4+0$ Therefore, H.C.F. = 8 Hence, the maximum number of columns in which they can march is 8 . #### Leave a comment None Free Study Material,band has 36 members . they are arrange into 6 equal rows . How many band members are in each row? A band has 36 members . they are arrange into 6 equal rows . How many band members are in each row?... ### What evidence does the author provide to support that money used for the space program should be used to address poverty (article included) A. She acknowledges that exploring other worlds is inspiring for children B. She presents facts to demonstrate that space program spending is excessive C. She explains the importance of space research and exploration D. She shows why reducing the nation's debt should be a priority What evidence does the author provide to support that money used for the space program should be used to address poverty (article included) A. She acknowledges that exploring other worlds is inspiring for children B. She presents facts to demonstrate that space program spending is excessive C. She e...,in which they can march? ### Solution What is known? We are told that there is an army contingent of $$616$$ members and an army band of $$32$$ members. The two groups are to march in the same number of columns What is unknown? The maximum number of columns in which they can march. Reasoning: Here, we have to pay attention to the point that the army band members and army contingent members have to march in the same number of columns and that the number of columns must be the maximum possible. The definition of HCF states – HCF of two positive integers $$a$$ and $$b$$ is the largest positive integer $$d$$ that divides both $$a$$ and $$b$$. In other words, HCF of two numbers is the highest number (maximum) that divides both the numbers. Thus, we have the find the HCF of the members in the army band and the army contingent. Steps: HCF ($$616, 32$$) will give the maximum number of columns in which they can march. We use Euclid’s algorithm to find the H.C.F \begin{align} 616&=(32\,\times \,19)\,+\,8 \\32&=(8\,\times \,4)+0\end{align} The HCF ($$616, 32$$) is $$8.$$ Therefore, they can march in $$8$$ columns each. ## Chapter 1 Ex.1.1 Question 4 Use Euclid’s division lemma to show that the square of any positive integer is either,K. (36) Answered on 18/07/2019 Learn CBSE/Class 10/Mathematics/UNIT I: Number systems/Real numbers/NCERT Solutions/Exercise 1.1 An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two... ...more An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? Here a maximum number of columns=H.C.F. between 32 and 616. By division method 616=32x19+8 32=8x4+0, so 8 is H.C, F. between616 and 32.Hence the maximum number of columns=8 Dislike Bookmark Answered on 18/07/2019 Learn CBSE/Class 10/Mathematics/UNIT I: Number systems/Real numbers/NCERT Solutions/Exercise 1.1 Let a and b are two integers, where a>b, then according to Euclid division lemma a=bq+r where 0<=r<b, Here let b=6, then a=6q+r so 0<=r<6 When r=0, a=6q, when r=1,a=6q+1, for r=2, a=6q+2, for r=3, a=6q+3, for r=4, a=6q+4, for r=5, a=6q+5.here 6q,6q+2 and 6q+4 are divisible by 2, so even. 6q+1,6q+3... ...more Let a and b are two integers, where a>b, then according to Euclid division lemma a=bq+r where 0<=r<b, Here let b=6, then a=6q+r so 0<=r<6 When r=0, a=6q, when r=1,a=6q+1, for r=2, a=6q+2, for r=3, a=6q+3, for r=4, a=6q+4, for r=5, a=6q+5.here 6q,6q+2 and,6 or Six. Total number of seats = Number of rows of seats Students set up for a music concert × Number of seats in each row Students set up for a music concert = 6 × 6 = 36 or Thirty Six. Essential Question How Can You Break Apart Arrays to Multiply? Answer: You can break apart Arrays to multiply by using the Distributive Property. Explanation: Yes, you can Break Apart Arrays to Multiply. It says that you can break apart a multiplication fact into the sum of two other multiplication facts. Visual Learning Bridge The members of the band march in 6 equal rows. There are 8 band members in each row. How many are in the band? What You Show Find 6 × 8. Use 5s facts and 1s facts. Make an array for each multiplication sentence. What You Think 6 × 8 is 6 rows of 8. That is 5 eights plus 1 more eight. 5 eights are 40. 8 more is 48. 40 + 8 = 48 So, 6 × 8 = 48. The band has 48 members. Convince Me! Use Structure Use a 5s fact and a 1s fact to find 6 x 9. Draw two arrays. Explain your drawings. Another‌ ‌Example‌ ‌!‌ ‌ Find‌ ‌7‌ ‌×‌ ‌8.‌ ‌ Use‌ ‌5s‌ ‌facts‌,QUESTION # An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? Hint – Here we will proceed from the integer which is larger and then apply Euclid’s division lemma to both the integers. Then we will repeat the algorithm up to the time we get remainder as zero. Hence we will get the desired result. According to Euclid’s division lemma, if we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition $a = b \times q + r$ where $0 < = r < = b$. $\Rightarrow 616 > 32$ $\Rightarrow 616 = 32 \times 19 + 8$ $\Rightarrow 32 = 8 \times 4 + 0$,An army contingent of 616 members is to march behind an army band of 32 members in a parade Question. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? Solution: 616 and 32 $616=32 \times 19+8$ $32=4 \times 8$ $\operatorname{HCF}$ of $(616,32)=8$ Editor,3m + 2. Now, we have to prove that the cube of these can be rewritten in the form 9q, 9q + 1 or 9q + 8. Now, $$(3m)^3$$ = $$27m^3$$ = $$9(m^3)$$ = 9q, where … ## Use Euclid’s Division Lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. Solution : By Euclid’s Division Algorithm, we have a = bq + r …………..(i) On putting b = 3 in (1), we get a = 3q + r, [0 $$\le$$ r < 3] If r = 0 a = 3q $$\implies$$ $$a^2$$ = $$9q^2$$ … ## An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march ? Solution : To find the maximum number of columns, we have to find the H.C.F. of 616 and 32 i.e. 616 = 32 $$\times$$ 19 + 8 and 32 = 8 $$\times$$ 4 + 0 $$\therefore$$ H.C.F of 616 and 32 is 8. Hence, maximum number of columns is 8.,compared. Looking at the dimensions, you would have been able to see that the tapestry would not fit. Let's look at another example: Example The band leader has arranged 56 musicians that will be participating in a parade. The number of players he placed in each row was 10 more than the number of rows. How many rows were there? To approach this problem, we need to assume that the musicians are arranged in equal rows. Then we know we are going to have to look for factors of 56 which fit these qualifications. We can work to “guess and check” for numbers that work. The guessing and checking strategy uses $7 \times 8 = 56$, but that doesn’t work. $28 \times 2 = 56$, but that doesn’t work either. $14 \times 4 = 56$ and that works! If there are 14 musicians in each row, there will be 10 more than the number of rows. Guessing and checking isn’t the only way to solve this problem. You might also choose to draw a picture or make an organized list. ## Real Life Example Completed The Four Thousand Footers Here is the original problem once again. Reread it and underline any important information. While at Galehead Hut, Kelly found a book on the different mountains in the Presidential,# An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march ? ## Solution : To find the maximum number of columns, we have to find the H.C.F. of 616 and 32 i.e. 616 = 32 $$\times$$ 19 + 8 and 32 = 8 $$\times$$ 4 + 0 $$\therefore$$ H.C.F of 616 and 32 is 8. Hence, maximum number of columns is 8.,– 15, find x. We need to find x if the H.C.F of 657 and 963 is expressible in the form. Given integers are 657 and 963. By applying Euclid’s division lemma, we get Since the remainder, so apply division lemma on divisor 657 and remainder 306 Since the remainder, so apply division lemma on divisor 306 and remainder 45 Since the remainder, so apply division lemma on divisor 45 and remainder 36 Since the remainder, so apply division lemma on divisor 36 and remainder 9 Therefore, H.C.F. = 9. Given H.C.F =. Therefore, #### Question 7: An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? We are given that an army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. We need to find the maximum number of columns in which they can march. Members in army = 616 Members in band = 32. Therefore, Maximum number of columns = H.C.F of 616 and 32. By applying Euclid’s division lemma Therefore, H.C.F. =(...TRUNCATED)

$$\mathbb{Q}$$ are $$d_1, d_2$$, and $$\gcd(d_1, d_2) = 1$$, what is the degree of $$x_1 + x_2$$?),+0 # maths; confused 0 73 1 a. The degree of the polynomial \(p(x)\) is \(11\), and the degree of the polynomial \(q(x)\) is \(7\). Find all possible degrees of the polynomial \(p(x) + q(x)\). b. Suppose \(f\) is a polynomial such that \(f(0) = 47\), \(f(1) = 32\), \(f(2) = -13\), and \(f(3)=16\). What is the sum of the coefficients of \(f\)? Mar 16, 2019 #1 +5172 +2 \(deg(p(x)) > deg(q(x)) \Rightarrow deg((p+q)(x)) = deg(p(x))\) \(c_0 + c_1 + c_2 + \dots + c_n = f(1) =32\) . Mar 17, 2019 edited by Rom Mar 17, 2019 edited by Rom Mar 17, 2019,# Math Help - Polynomials 1. ## Polynomials What is the degree of the simplest polynomial with the zeroes:√2, 2√2 + i, and 2 + 2i ?? 2. My guess would be 6 since the simplest polynomial that I can come up with that has all of these as roots is (x² - 2)(x² - 4√2 x + 9)(x² - 4x + 8). 3. This means that $x - \sqrt{2}$, $x - (2 \sqrt{2} + \pi)$, $x - (2 + 2 \pi)$ are all factors of this polynomial. Therefore the simplest polynomial that has all these roots is : $(x - \sqrt{2})(x - (2 \sqrt{2} + \pi))(x - (2 + 2 \pi)) = 0$ Or, equivalently : $(x - \sqrt{2})(x - 2 \sqrt{2} - \pi)(x - 2 - 2 \pi) = 0$ And the degree of this polynomial is, trivially, 3 (can be checked by expanding). 4. Originally Posted by VMM What is the degree of the simplest polynomial with the zeroes:√2, 2√2 + i, and 2 + 2i ?? Roots or zeros of polynomials of degree greater than 2 - Topics in precalculus,the original equation. While this may look like a $120$ degree polynomial it is actually a $24$ degree polynomial in $X^5$. And here the approach stops. A similar approach will reduce a quadratic to a linear equation, a cubic to a quadratic and a quartic to a sextic which can still be solved due to some of its special properties. But Lagrange wasn't able to make this approach work for the quintic. -,fmpz_poly. Warning: the degree is $$2^n$$. >>> fmpz_poly.swinnerton_dyer(0) x >>> fmpz_poly.swinnerton_dyer(1) x^2 + (-2) >>> fmpz_poly.swinnerton_dyer(2) x^4 + (-10)*x^2 + 1 >>> fmpz_poly.swinnerton_dyer(3) x^8 + (-40)*x^6 + 352*x^4 + (-960)*x^2 + 576,of $\\mathbb{Q}$ of degree 4 that contains $\\mathbb{Q}(\\sqrt{5})$.,## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) We know that the degree of the product will be equal to the sum of the highest degree of each polynomial that is multiplied. Thus, we find: a) $=5+4=9$ b) $=7(4) =28$,# How do you find the degree and leading coefficient of the polynomial 5t^4+t^3-7? Aug 31, 2017 $\text{degree "4," leading coefficient } = 5$ #### Explanation: $\text{the degree of a polynomial is the value of the largest}$ $\text{power (exponent ) in the polynomial}$ $\text{the leading coefficient is the coefficient of the term of}$ $\text{largest power with the polynomial in standard form}$ $5 {t}^{4} + {t}^{3} - 7 \text{ is in standard form}$ $\Rightarrow \text{degree "=4" and leading coefficient } = 5$,# How do you find the degree and leading coefficient of the polynomial 3x^7-4x^5+x^3? Degree: $7$ Leading coefficient: $3$ The degree of the polynomial is simply the number of the highest exponent, i.e. color(red)(7. The leading coefficient is the coefficient in front of the quantity with the highest exponent, i.e. color(blue)(3.,degree d, on a k'-dimensional variety defined by polynomials of degree d_0, is bounded by (sd)^k' d_0^{k−k'} O(1)^k. Our most recent work takes this refinement of the dependence on the degrees even further, obtaining what could be considered a real analogue to the classical Bezout inequality over algebraically closed fields.,# Polynomials • May 15th 2010, 05:00 PM VMM Polynomials What is the degree of the simplest polynomial with the zeroes:√2, 2√2 + i, and 2 + 2i ?? • May 15th 2010, 07:24 PM NowIsForever My guess would be 6 since the simplest polynomial that I can come up with that has all of these as roots is (x² - 2)(x² - 4√2 x + 9)(x² - 4x + 8). • May 15th 2010, 07:29 PM Bacterius This means that $x - \sqrt{2}$, $x - (2 \sqrt{2} + \pi)$, $x - (2 + 2 \pi)$ are all factors of this polynomial. Therefore the simplest polynomial that has all these roots is : $(x - \sqrt{2})(x - (2 \sqrt{2} + \pi))(x - (2 + 2 \pi)) = 0$ Or, equivalently : $(x - \sqrt{2})(x - 2 \sqrt{2} - \pi)(x - 2 - 2 \pi) = 0$ And the degree of this polynomial is, trivially, 3 (can be checked by expanding). • May 16th 2010, 06:52 AM skeeter Quote: Originally Posted by VMM What is the degree of the simplest polynomial with the zeroes:√2, 2√2 + i, and 2 + 2i ?? Roots or zeros of polynomials of degree greater than 2 - Topics in precalculus,# What is the degree of 6p^3q^2? ##### 1 Answer May 18, 2018 $3$ #### Explanation: The degree of a polynomial is usually just taken to be the highest power that is present.,in general). In your case, $\pi$ has degree $2$.,$\sqrt[n]{11}$ over $\mathbb{Q}(\sqrt{-5})$ must have degree $n$, and so is $x^n-11$. In particular, that polynomial is irreducible. Dear @ZachL.: $[\mathbb{Q}(11^{1/n}):\mathbb{Q}] = n$. – Rankeya Dec 3 '12 at 8:20 $[\mathbb{Q}(\sqrt{-5},\sqrt[n]{11}):\mathbb{Q}(\sqrt{-5})] = n$ Why is this true? – user44322 Dec 3 '12 at 8:32,It has degree 1, just like the special case $x \mapsto 1-2x$. Oct 3 '20 at 14:22,How do you find the degree and leading coefficient of the polynomial 5t^4+t^3-7? Aug 31, 2017 $\text{degree "4," leading coefficient } = 5$ Explanation: $\text{the degree of a polynomial is the value of the largest}$ $\text{power (exponent ) in the polynomial}$ $\text{the leading coefficient is the coefficient of the term of}$ $\text{largest power with the polynomial in standard form}$ $5 {t}^{4} + {t}^{3} - 7 \text{ is in standard form}$ $\Rightarrow \text{degree "=4" and leading coefficient } = 5$,+0 # Pls help!!!!! 0 97 2 Suppose that $f(x)$ and $g(x)$ are polynomials of degree $4$ and $5$ respectively. What is the degree of $f(x^3) \cdot g(x^2)$? Feb 5, 2022 #1 +17 +2 "Suppose that $$f(x)$$ and $$g(x)$$ are polynomials of degree 4 and 5 respectively. What is the degree of $$f(x^3) \cdot g(x^2)$$?" Feb 5, 2022 #2 +117175 +1 f(x^3) will have a degree of 12 and g(x^2) will have a degree of 10 You can finish it Feb 6, 2022,total degree of $f$ because $\deg g'(x,y)y \leq \deg g(x,y) < \deg f(x,y). \ \Box$,$E(\mathbb{F}_{47}):y^2=x^3+x+38$ has order $61$ and $61|47^3-1$ so the embedding degree of $E$ is $3$ and therefore the MOV attack, presumably using some sort of distortion map and a ...,the authors also ask to find the interpolation of degree at most 2, and then compare the precision of the approximation of $$\cos(0.45)$$ by both polynomials. I guess the objective of the exercise is to let the reader see how more points approximate better, although in some cases that's totally false.(...TRUNCATED)

# Chapter 5 - Exponents, Polynomials and Functions - 5-1 Polynomial Functions - Practice and Problem-Solving Exercises: 9 $-3x+5$ The polynomial has a degree of one and has two terms. #### Work Step by Step RECALL: A polynomial is written in standard form if its terms are written in descending degree. Thus, the standard form of the given polynomial is: $-3x+5$. This polynomial has degree 1 and has two terms. It is a linear binomial. 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.,variables, coefficients, and operations of addition, subtraction, multiplication, and non - negative integer. The standard of a polynomial of degree nis anxn………+ a1x+a0 Examples: x + 2, 3y2 - 2y + 5, -6, 1/2y2 - 2/3 y + 3/4 ### What Does Standard Form of a Polynomial Means The standard form of a polynomial is a method of writing polynomials with the exponents in decreasing order. Polynomials are expressed in standard form to make the complex calculation easier. A polynomial is considered to be written in standard form, if it is expressed in such a way that the term with the highest degree is written first, followed by the term which has the next highest degree, and so on. Example : 14y4 - 5y3 - 11y2 - 11y + 8 You can find that in the above - given standard form of a polynomial, the exponents are placed in decreasing order. ### How to Write Polynomials In Standard Form? The like terms in the standard form of a polynomial are grouped, added, subtracted, and rearranged with the exponents of terms in decreasing order. Following are the steps to write a polynomial in standard form: 1. Write the terms. 2. Arrange all the like terms. 3. Find the exponent. 4. Write the term with the highest exponent first. 5.,# How do you write a polynomial in standard form, then classify it by degree and number of terms -2x^2 - 16x + -4x^3 +2x^4 +5? Jun 6, 2018 See below. #### Explanation: Standard form means that the terms are written in order form highest power to lowest power. $2 {x}^{4} - 4 {x}^{3} - 2 {x}^{2} - 16 x + 5$ The degree of the polynomial is $4$, as this is the highest power present in the polynomial. The polynomial has $5$ total terms.,# How do you write a polynomial in standard form, then classify it by degree and number of terms 3x^2 + 5x - 4 + 5x^3? Feb 3, 2018 Refer to the explanation. #### Explanation: A polynomial is written in descending order of exponents (powers), remembering that a number without an exponent is actually ${n}^{1}$. The given polynomial in standard form is: $5 {x}^{3} + 3 {x}^{2} + 5 x - 4$ The polynomial is 3rd degree because the highest exponent is 3. The terms in a polynomial are separated by $+$ or $-$ signs. There are four terms in the given polynomial.,# How do you write a polynomial in standard form, then classify it by degree and number of terms c^2-2+4c? Aug 9, 2017 See a solution process below: #### Explanation: First, rewrite this polynomial from the term with the highest exponent to the lowest exponent to put it in standard terms: ${x}^{2} + 4 c - 2$ The degree of the polynomial is found by looking at the term with the highest exponent on its variable(s). This polynomial has a $2$ as its largest exponent and consists of three terms. Therefore, this is a $\textcolor{red}{2 \text{nd}}$ degree $\textcolor{red}{\text{tri}}$nomial.,# How do you write 11t+2t^2-3+t^5 in standard form and what is the leading coefficients? ##### 1 Answer Apr 16, 2017 See the solution explanation below: #### Explanation: To write this expression in standard form we put the terms in order of the highest exponent to lowest: $11 t + 2 {t}^{2} - 3 + {t}^{5}$ becomes: ${t}^{5} + 2 {t}^{2} + 11 t - 3$ The leading coefficient for the ${t}^{5}$ term is $1$: $1 {t}^{5} + 2 {t}^{2} + 11 t - 3$,# How do you write a polynomial in standard form, then classify it by degree and number of terms 4x^2 - 3x^3 + 2? Jul 10, 2018 See below: #### Explanation: To write our polynomial in standard form, we want the term with the highest degree first. The ${x}^{3}$ term will come first. We have $- 3 {x}^{3} + 4 {x}^{2} + 2$ The degree is the highest exponent we see, and we see that this is $3$, so we can say this is a third-degree polynomial. We see that our addition signs divide this expression into three sections- our three terms. Hope this helps!,# Write in Standard Form (-2x^4+7x^2y-7)+(-9x^3+7xy+7) Write in Standard Form (-2x^4+7x^2y-7)+(-9x^3+7xy+7) To write a polynomial in standard form, simplify and then arrange the terms in descending order. Remove parentheses. Combine the opposite terms in .,# Write the standard form of a cubic polynomial with real coefficients. Question: Write the standard form of a cubic polynomial with real coefficients. Solution: The most general form of a cubic polynomial with coefficients as real numbers is of the form $f(x)=a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are real number and $a \neq 0$,# How do you write a polynomial in standard form, then classify it by degree and number of terms x^7y^2 + 4x^3y + 10x^3? Feb 22, 2018 Currently is in standard form, degree$= 9$; There are $3$ terms #### Explanation: Our polynomial is in standard form when the exponents of $x$ are descending, and the exponents of $y$ are descending. Right now, it is in standard form because from left to right, the exponents on $x$ go from $7$ to $3$ to $3$, and on the $y$ terms, the exponents go from $2$ to $1$ to $0$. NOTE: ${y}^{0}$ can go at the end of the last term since it is equal to $1$ and will not change the meaning of the expression. Degree: This would be a $9 t h$ degree polynomial, because the highest exponent is on the term ${x}^{7} {y}^{2}$, and if we add the exponents, we get $9$ as the degree. Terms: The terms are separated by the addition signs, thus we have $3$ terms. Hope this helps!,# How do you write a polynomial in standard form, then classify it by degree and number of terms 8b^4 - 9b^3 - 2a^2? Jul 24, 2018 Standard form: $8 {b}^{4} - 9 {b}^{3} - 2 {a}^{2}$ Degree: $4$ Terms: $3$ #### Explanation: In standard form, these terms will be expressed from highest to lowest exponent, which it is now: $8 {b}^{4} - 9 {b}^{3} - 2 {a}^{2}$ Classifying by degree means to find the highest exponent of the terms, so the degree is $4$. A term is split by the addition and subtraction operations, so there are $3$ terms. Hope this helps!,# How do you write 3x^4+2x^3-5-x^4 in standard form? $2 {x}^{4} + 2 {x}^{3} - 5$ The standard form of a polynomial (which is what we have here) is written from highest exponent to lowest exponent. In addition, there should only be 1 of each exponent/variable pair (like x^4 or x^3). For example, here we have $3 {x}^{4}$ and $- {x}^{4}$; but we can only have 1 term with an ${x}^{4}$ in it. To fix this, we need to combine these terms by adding $3 {x}^{4}$ to $- {x}^{4}$; the result is $2 {x}^{4}$ (remember that $3 {x}^{4} + \left(- {x}^{4}\right)$ is the same thing as $3 {x}^{4} - {x}^{4}$). From here, we just need to organize a bit - from highest exponent to lowest. Our highest is, of course, $4$. Then we have $3$ - and since we don't have ${x}^{2}$ or $x$, we ignore them. Now, what about the $- 5$? Well, since ${x}^{0} = 1$, we can write $- 5$ as $- 5 {x}^{0}$, which is the same thing as $- 5 \cdot 1$ - which equals $- 5$. That makes 0 our lowest exponent. Cool, huh? Finally, we simply write it. We start with the term with the highest exponent ($2 {x}^{4}$), and then go down from there. So, our polynomial in standard form,How do you write a polynomial in standard form, then classify it by degree and number of terms 3x^4 − 9x^3 − 3x^2 + 6? Dec 23, 2017 Refer tot he explanation. Explanation: Given: $3 {x}^{4} - 9 {x}^{3} - 3 {x}^{2} + 6$ You have already written the polynomial in standard form, from highest to lowest exponent and then the constant. This is a 4th degree polynomial because the highest exponent is $4$ in ${x}^{4}$. It has four terms, which are separated by addition and subtraction symbols.,# How do you write a polynomial in standard form, then classify it by degree and number of terms 8-6w-12w-8w^2-7-3w^3? Jul 6, 2017 See explanation. #### Explanation: $8 - 6 w - 12 w - 8 {w}^{2} - 7 - 3 {w}^{3}$ First you can reduce the like terms: $8 - 18 w - 8 {w}^{2} - 3 {w}^{3}$ Now we can order the terms by decreasing powers of $w$: $- 3 {w}^{3} - 8 {w}^{2} - 18 w + 8$ This polynomial is in the standard form. From this form you see, that the polynomial: • has $4$ terms, • is a polynomial of third degree. (the highest exponent to which the unknown is raised is $3$),# How do you write the polynomial so that the exponents decrease from left to right, identify the degree, and leading coefficient of the polynomial 5n^3+2n-7? ##### 1 Answer Mar 19, 2017 It is already in standard form. the degree is 3 and the leading coefficient is 5. #### Explanation: 1. When the exponents decrease from left to right, the polynomial is in standard form. It is already in standard form the way you wrote it. $5 {n}^{3} + 2 n - 7$ 2. The degree of a polynomial is the highest exponent, which should be the first one if the polynomial is in standard form. the degree of this polynomial is 3. 3. The leading coefficient is just that: the first coefficient when the polynomial is in standard form. In other words, the coefficient of the term with the highest exponent. the leading coefficient of this polynomial is 5.,# Write in Standard Form 3x^3+x^2-8x^5+2x^3-4x To write a polynomial in standard form, simplify and then arrange the terms in descending order.,# How do you write a polynomial in standard form, then classify it by degree and number of terms x^6 + 3x^3? ${x}^{6} + 3 {x}^{3}$ The polynomial is already written in standard form. It is a 6th degree polynomial because the greatest exponent is $6$. There are two terms separated by a plus sign.,# How do you write a polynomial in standard form, then classify it by degree and number of terms 8g-3g^3 +4g^2 -1? ##### 1 Answer Jul 24, 2018 Standard form: $- 3 {g}^{3} + 4 {g}^{2} + 8 g - 1$ Degree: $3$ Terms: $4$ #### Explanation: In standard form, the terms are expressed from the largest to smallest exponent: $- 3 {g}^{3} + 4 {g}^{2} + 8 g - 1$ Classifying by degree means to find the highest exponent of the terms, so the degree is $3$. A term is a value split by the addition and subtraction operations, so there are 4# terms. Hope this helps!,# What is the standard form of a polynomial 8x(3x+4-x^2)? Jun 6, 2016 The answer is $- 8 {x}^{3} + 24 {x}^{2} + 32 x$ #### Explanation: All you basically need to do is remember what standard form is -- standard for is always in the form $a {x}^{2} + b x + c$ (the first exponent does not necessarily always have to be 2, but they must be in descending order). The math to get this for your particular problem is not difficult -- all you have to do is multiply the $8 x$ to all 3 terms. The only trick here is that you have to list the terms in descending order of power. This is also known as descending degree. If you have that, you should honestly have no problem with these kinds of problems. Here's an article that explains this all quite nicely. Want to learn about standard form in polynomials? Watch my video on it. Hope that was helpful :),# How do you write a polynomial in standard form, then classify it by degree and number of terms 8x + 5x^3 – 5? Jul 24, 2018 Standard form: $5 {x}^{3} + 8 x - 5$ Degree: $3$ Terms: $3$ #### Explanation: In standard form, these terms will be expressed from highest to lowest exponent: $5 {x}^{3} + 8 x - 5$ Classifying by degree means to find the highest exponent of the terms, so the degree is $3$. A term is a value split by the addition and subtraction operations, so there are $3$ terms. Hope this helps!(...TRUNCATED)

# Chapter 5 - Exponents, Polynomials and Functions - 5-1 Polynomial Functions - Practice and Problem-Solving Exercises: 9 $-3x+5$ The polynomial has a degree of one and has two terms. #### Work Step by Step RECALL: A polynomial is written in standard form if its terms are written in descending degree. Thus, the standard form of the given polynomial is: $-3x+5$. This polynomial has degree 1 and has two terms. It is a linear binomial. 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.,# What is the standard form of a polynomial 8x(3x+4-x^2)? Jun 6, 2016 The answer is $- 8 {x}^{3} + 24 {x}^{2} + 32 x$ #### Explanation: All you basically need to do is remember what standard form is -- standard for is always in the form $a {x}^{2} + b x + c$ (the first exponent does not necessarily always have to be 2, but they must be in descending order). The math to get this for your particular problem is not difficult -- all you have to do is multiply the $8 x$ to all 3 terms. The only trick here is that you have to list the terms in descending order of power. This is also known as descending degree. If you have that, you should honestly have no problem with these kinds of problems. Here's an article that explains this all quite nicely. Want to learn about standard form in polynomials? Watch my video on it. Hope that was helpful :),# How do you write a polynomial in standard form, then classify it by degree and number of terms -4x^4+x^2-10+12x^4-7x^2? Mar 1, 2017 $8 {x}^{4} - 6 {x}^{2} - 10$ #### Explanation: Standard form$= 8 {x}^{4} - 6 {x}^{2} - 10$ Degree ( Highest power ) = 4 No. of terms = 3 #Aru R. sir , you think right. It is just ordering them based on degree.,# How do you write a polynomial in standard form, then classify it by degree and number of terms -2x^2 - 16x + -4x^3 +2x^4 +5? Jun 6, 2018 See below. #### Explanation: Standard form means that the terms are written in order form highest power to lowest power. $2 {x}^{4} - 4 {x}^{3} - 2 {x}^{2} - 16 x + 5$ The degree of the polynomial is $4$, as this is the highest power present in the polynomial. The polynomial has $5$ total terms.,## Precalculus (10th Edition) $x^{2}+8x+16$ The standard form of a polynomial is $a_{n}x^{n}+a_{n}x^{n-1}+\cdots+a_{1}x+a_{0}$ (written in order, highest degree first. If a term is missing, it means that the coefficient is 0) --- Use the perfect square formula $(A+B)^{2}=A^{2}+2AB+B^{2}$ $(x+4)^{2}=x^{2}+2\cdot x\cdot 4+4^{2}$ = $x^{2}+8x+16$,## Precalculus: Mathematics for Calculus, 7th Edition Type: Polynomial Terms: x, -$x^2$, $x^3$, -$x^4$ Degree: 4 The type of the monomial is a monomial if it has one term, a binomial if it has two terms, and a trinomial if it has three terms. This polynomial has 4 terms, therefore it is not referred to by any of these terms and is simply called a four term polynomial. The terms are simply the values separated by plus and minus signs: x, -$x^2$, $x^3$, -$x^4$. The degree is just the highest exponent of the variable in any of the terms, and if no x is present, the degree is 0. Therefore, the degree of this polynomial is 4.,+0 # maths; confused 0 73 1 a. The degree of the polynomial \(p(x)\) is \(11\), and the degree of the polynomial \(q(x)\) is \(7\). Find all possible degrees of the polynomial \(p(x) + q(x)\). b. Suppose \(f\) is a polynomial such that \(f(0) = 47\), \(f(1) = 32\), \(f(2) = -13\), and \(f(3)=16\). What is the sum of the coefficients of \(f\)? Mar 16, 2019 #1 +5172 +2 \(deg(p(x)) > deg(q(x)) \Rightarrow deg((p+q)(x)) = deg(p(x))\) \(c_0 + c_1 + c_2 + \dots + c_n = f(1) =32\) . Mar 17, 2019 edited by Rom Mar 17, 2019 edited by Rom Mar 17, 2019,# How do you write the polynomial 4x + x + 2 in standard form and how many terms and degree is it? May 27, 2015 $4 x + x + 2$ $= 5 x + 2$ (this is standard form) It has 2 terms: $5 x$ and $2$ The degree of a polynomial is the degree of the term with the highest degree. The degree of a term is the sum of its factor exponents $5 x = {5}^{1} {x}^{1}$ has degree 2. $2 = {2}^{1}$ has degree 1 $5 x + 2$ has degree 2.,# How do you write the polynomial 1 - 2s + 5s^4 in standard form and how many terms and degree is it? $5 {s}^{4} - 2 s + 1$. Its degree is 4 Standard form is with the highest power terms preceding the other terms. The given polynomial in standard form would be $5 {s}^{4} - 2 s + 1$.,## Algebra 1 $-2q+7$ $linear$ $binomial$ In order to solve this problem, we need to put the equation in standard form, and name the polynomial. To put the equation in standard form, we must order the monomials within from highest to lowest. Since $-2q$ has a higher degree than $7$, it will go first. So the polynomial is already in standard form ($-2q+7$). Now we need to name the polynomial. It is linear because its highest degree is 1, and it is a binomial because it has two terms. Therefore, it is a linear binomial.,$$\mathbb{Q}$$ are $$d_1, d_2$$, and $$\gcd(d_1, d_2) = 1$$, what is the degree of $$x_1 + x_2$$?),# How do you write a polynomial in standard form, then classify it by degree and number of terms x^4y^2 + 4x^3y^5 + 10x? May 29, 2018 color(orange)(5x^3y^5 + x^4y^2 + 10x is the standard form. Degree of polynomial: $8$ Number of terms: $3$ #### Explanation: Polynomials in two variables are algebraic expressions consisting of terms in the form . The degree of each term in a polynomial in two variables is the sum of the exponents in each term and the degree of the polynomial is the largest such sum. ${x}^{4} {y}^{2} + 5 {x}^{3} {y}^{5} + 10 x$ $5 {x}^{3} {y}^{5} + {x}^{4} {y}^{2} + 10 x$ is the standard form. Degree of polynomial: $8$ Number of terms: $3$,# How do you write the polynomial so that the exponents decrease from left to right, identify the degree, and leading coefficient of the polynomial 5n^3+2n-7? ##### 1 Answer Mar 19, 2017 It is already in standard form. the degree is 3 and the leading coefficient is 5. #### Explanation: 1. When the exponents decrease from left to right, the polynomial is in standard form. It is already in standard form the way you wrote it. $5 {n}^{3} + 2 n - 7$ 2. The degree of a polynomial is the highest exponent, which should be the first one if the polynomial is in standard form. the degree of this polynomial is 3. 3. The leading coefficient is just that: the first coefficient when the polynomial is in standard form. In other words, the coefficient of the term with the highest exponent. the leading coefficient of this polynomial is 5.,What is the degree of the following monomial 4xyz? The degree of a monomial is the sum of the degrees of each letter. So the degree is $3$.,a polynomial that you'll eventually want to factor, don't expand the polynomial. Keep it together. What I'm saying here is do not try to expand $(14x^2 +7x-21)^2$ into a 5 term 4th degree polynomial in x. Keep it factored. One final hint: You can factor $2x^2+x-3$,fmpz_poly. Warning: the degree is $$2^n$$. >>> fmpz_poly.swinnerton_dyer(0) x >>> fmpz_poly.swinnerton_dyer(1) x^2 + (-2) >>> fmpz_poly.swinnerton_dyer(2) x^4 + (-10)*x^2 + 1 >>> fmpz_poly.swinnerton_dyer(3) x^8 + (-40)*x^6 + 352*x^4 + (-960)*x^2 + 576,# How do you write a polynomial in standard form, then classify it by degree and number of terms 1 - 2s + 5s^4? Jan 15, 2018 #### Explanation: When you want to write a polynomial in standard form, we need to put the variable with the highest degree first, and the number last. Our highest degree here is the $5 {s}^{4}$, and our number is $1$. Our standard form polynomial is $5 {s}^{4} - 2 s + 1$. To classify it by a degree, we also look at the variable with the highest degree, $5 {s}^{4}$. So our degree of the polynomial is 4th degree . There are three terms in this polynomial: $1$, $- 2 s$, and $5 {s}^{4}$, meaning that this polynomial is a trinomial. Hope this helps!,# What is the standard form of a polynomial (x+3)(x-2) ? Apr 15, 2016 ${x}^{2} + x - 6$ ${x}^{2} \textcolor{w h i t e}{\text{XXxXX}}$: degree 2 $x \left(= {x}^{1}\right) \textcolor{w h i t e}{\text{x}}$: degree 1 $6 \left(= 6 {x}^{0}\right)$: degree 0,It has degree 1, just like the special case $x \mapsto 1-2x$. Oct 3 '20 at 14:22,Chapter 4 - Section 4.2 - Adding and Subtracting Polynomials - 4.2 Exercises: 23 binomial; 1 Work Step by Step The polynomial $7m-22$ can be rewritten as $7m^{1}-22m^{0}$. Therefore, this polynomial is a binomial, because it consists of 2 terms. The degree of a term is the power to which the variable in the term is raised. The degree of a polynomial is equal to the largest degree among all of the terms in the polynomial. Therefore, the degree of this polynomial is 1. 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.(...TRUNCATED)

# How do you evaluate (4(12-3))/(6-3)? $12$ $\frac{4 \left(12 - 3\right)}{6 - 3} = \frac{4 \times 9}{3}$ =$\frac{36}{3}$ =$12$,Evaluate $\left ( \frac{4}{9} \right ) \left ( \frac{36}{40} \right )$ 15. Evaluate $\left ( \frac{12}{14} \right ) \left ( \frac{7}{8} \right )$,- \left\lceil nx+1\right\rceil \\ =&\left\lceil x+\frac{1}{n}\right\rceil+\left\lceil x+\frac2n\right\rceil+\dots +\left\lceil x\right\rceil - \left\lceil nx\right\rceil \\ =&\lceil x\rceil+\left\lceil x+\frac1n\right\rceil+\dots +\left\lceil x+1-\frac{1}{n}\right\rceil - \left\lceil nx\right\rceil\\ =&f(x) \end{align} So $f$ is $\frac1n$-periodic. And we only need to evaluate it over $\left(0,\frac1n\right]$, in which case $$f(x) = n-1$$ This is because if $x\in(0,\frac{1}{n}]$, then $x,x+\frac{1}{n},\dots,x+\frac{n-1}{n}$ and $nx\in(0,1]$ and $$\left\lceil x\right\rceil = \left\lceil x+\frac{1}{n}\right\rceil=\dots=\left\lceil x+\frac{n-1}{n}\right\rceil=\left\lceil nx\right\rceil=1$$ • Nice approach! But shouldn't you have $\lceil nx+1\rceil$ when you evaluate $f(x+1/n)$? I think that would change the picture here... – bartgol Aug 17 '15 at 16:38 • I think the lest term of $f\left(x+\frac1n\right)$ has a typo. – user142971 Aug 17 '15 at 16:38 • @bartgol: You came first by ~3 sec!! – user142971 Aug 17 '15 at 16:39 • @user36790 Basically, if a function has a period $T$, then studying it over $[0,T)$ (or $(0,T]$, $[a,a+T)$, etc.) is enough to calculate it over all of $\Bbb R$. In this case, we have $T=1/n$ and we know that if $x\in (0,T]$, then $f(x)$ is equal to $n-1$. So if $x\in (T,2T]$, then $f(x) = f(x-T) = n-1$. If $x\in (2T,3T]$, then $f(x)=f(x-2T)=n-1$ and so on. So we know that $f(x) = n-1$ for all $x\in(0,\infty)$. The same thing can be done for $(\infty,0]$ – Kitegi Aug 17 '15 at 16:56 • @user36790 Well, since $x\in(0,\frac1n]$, then $x,x+\frac1n,\dots,x+1-\frac1n$,Evaluate $\left|65\right|$|65| ##### Question 14 What is the value of $\left|-155\right|$|155|?,\frac{3 \left \lceil \frac{415}{47} \right \rceil}{7} \right \rceil = \left \lceil \frac{3 \cdot 9}{7} \right \rceil = \left \lceil \frac{27}{7} \right \rceil = 4$. So, being $i = 2$ and $n = 4$, we can say that the twelfth dash is $\mathrm{t}(12) = (2, 4)$. Now, as we have computed the twelfth dash, we can compute its value by means of the dashed line function given by (1): $$\mathrm{t\_value}(x) = T(\mathrm{t}(x)) = T(2, 4) = 4 \cdot 4 = 16$$ That is the answer to the second question: the twelfth person who will go jogging, will do it in 16 days. A note for the most curious people: if you want you can try formulas (2), (3) and (4) with different values of $x$, even very big ones, for verifying their correctness. However formula (5) holds only for $i = 2$, while in general $n$ is given by: $$n = \begin{cases} \left \lceil \frac{4 \left \lceil \frac{35x - 5}{47} \right \rceil + 1}{7} \right \rceil & \text{if i = 1}\\\left \lceil \frac{3 \left \lceil \frac{35x - 5}{47} \right \rceil}{7} \right \rceil & \text{if i = 2}\\\left \lceil \frac{5 \left \lceil \frac{32x - 7}{47} \right \rceil}{8} \right \rceil & \text{if i = 3}\end{cases}$$ Moreover, this is not the only possible exact formula, but we’ll see the others in specific,# $x\le\lceil\sqrt{x}\rceil\left\lceil\frac{x}{\lceil\sqrt{x}\rceil}\right\rceil$? $x\le\lceil\sqrt{x}\rceil\left\lceil\frac{x}{\lceil\sqrt{x}\rceil}\right\rceil$ How do I show this? I made a plot, and it looks true: - Hint: $\frac{x}{\lceil\sqrt{x}\rceil} \leq \lceil\frac{x}{\lceil\sqrt{x}\rceil}\rceil$. - I of course assume that you're only considering $x \gt 0$. – t.b. Apr 16 '11 at 22:32 Divide both sides by $\lceil \sqrt x\rceil$. -,# How do you evaluate (-6)^-5? $- \frac{1}{7776}$ ${\left(- 6\right)}^{- 5} = \frac{1}{- 6} ^ 5 = \frac{1}{\left(- 6\right) \left(- 6\right) \left(- 6\right) \left(- 6\right) \left(- 6\right)}$ = $- \frac{1}{7776}$,\left(r-\frac 12 \sigma ^ 2\right) t, \sigma ^2 t\right)$$,- Here is the attempt to evaluate $\text{Li}_1\left(\frac12\right)$ and $\text{Li}_2\left(\frac12\right)$. +1 – Tunk-Fey Aug 24 '14 at 17:20 @Tunk-Fey Thanks for your link. However, those values are well known. Any way, it's always nice to see that derivation. – Felix Marin Aug 24 '14 at 18:03,correspond to $d=1$ and $m = \lceil 53/12 \rceil = 5$.,last equation by trial and error. I just kept diddling with it until I got something that worked. That is to say $$\left \lceil \dfrac{k^2 + k + 3}{6} \right \rceil - \left \lceil \dfrac{k^2 - k + 2}{6} \right \rceil = \left \lfloor \dfrac{k+1}{3} \right \rfloor$$ • can you just explain, just for one row of the first table, how you are finding third column from second column @steven – Sushil Verma Dec 14 '16 at 5:31 HINT: If $k=6a,k^2+k+3=(6a)(6a+1)+3,k^2-k+2=6a(6a-1)+2$ $$\left\lceil\frac{k^2+k+3}6\right\rceil - \left\lceil\frac{k^2-k+2}6\right\rceil=6a+1+1-(6a-1+1)=2$$ If $k=6a+1,k^2+k+3=(6a+1)(6a+2)+3=6a(6a+3)+5$ $,k^2-k+2=(6a+1)(6a)+2$ $$\left\lceil\frac{k^2+k+3}6\right\rceil - \left\lceil\frac{k^2-k+2}6\right\rceil=6a+3+1-(6a+1+1)=2$$ • There is something strange here as I would expect the values to be close to $\dfrac{k}{3}$ or $2a$ – Henry Dec 13 '16 at 13:59 Make a table showing $k$, $k^2+k+3$, and $k^2-k+2$ modulo $6$: $$\begin{array}{c|c|c} k&k^2+k+3&k^2-k+2\\ \hline 0&3&2\\ 1&5&2\\ 2&3&4\\ 3&3&2\\ 4&5&2\\ 5&3&4 \end{array}$$ Thus, if $k\equiv0\pmod3$, then $$\left\lceil\frac{k^2+k+3}6\right\rceil-\left\lceil\frac{k^2-k+2}6\right\rceil=\frac{k^2+k+6}6-\frac{k^2-k+6}6=\frac{k}3\;.$$ If $k\equiv1\pmod3$, then $$\left\lceil\frac{k^2+k+3}6\right\rceil-\left\lceil\frac{k^2-k+2}6\right\rceil=\frac{k^2+k+4}6-\frac{k^2-k+6}6=\frac{k-1}3\;.$$ And if $k\equiv2\pmod3$, then $$\left\lceil\frac{k^2+k+3}6\right\rceil-\left\lceil\frac{k^2-k+2}6\right\rceil=\frac{k^2+k+6}6-\frac{k^2-k+4}6=\frac{k+1}3\;.$$ If you wish to combine these into a single formula, one possibility is $$\left\lceil\frac{k^2+k+3}6\right\rceil-\left\lceil\frac{k^2-k+2}6\right\rceil=\left\lfloor\frac{k+1}3\right\rfloor\;.$$ By following the hint I gave you yesterday, you can write $$R=\lceil\frac{k^2+k+3}{6}\rceil - \lceil\frac{k^2-k+2}{6}\rceil =\frac{k^2+k+3}{6} +u- \frac{k^2-k+2}{6} - v =\frac{2 k+1}{6} + (u-v)$$ where $0 \le u, v < 1$, which imply that $-1 < u-v < 1$. It means that the result $R$ is an integer which distance from $\frac{2k+1}{6}$ is smaller than,Evaluate $3x^4 + 2x^2y^2 – y^4 – y^2 + 4$ if $x = \sin \left(\frac{71\pi}{7919}\right)$ and $y = \cos \left(\frac{71\pi}{7919} \right)$.,$$m\cdot r$$ trials. Hence, the answer is $$1-\left( \frac47\right)^{m\cdot r}$$,# How do you evaluate 122 - [45 - (32 + 8) - 23]? $122 - \left[45 - 40 - 23\right]$ $122 - \left[- 18\right]$ $122 + 18$ $140$,1} \left(x\right) = \frac{7 x - 25}{x - 2}$ To evaluate ${h}^{- 1} \left(9\right)$, simply substitute $x$ with $9$ to get ${h}^{- 1} \left(9\right) = \frac{7 \cdot 9 - 25}{9 - 2} = \frac{38}{7}$,formula gives you $$z$$ rounded up to the next multiple of 5: $$5\left\lceil \frac{z}{5} \right\rceil$$ For example, for $$z=24$$, $$5\lceil \frac{24}{5} \rceil=5\lceil 4.8 \rceil=5\cdot 5=25$$ and for $$z=30$$, $$5\lceil \frac{30}{5} \rceil=5\lceil 6\rceil=5\cdot 6=30$$. Messenger,# Some Quickies Part III of my (never-ending) series “A Means to an End” should be up soon, but I wanted to record this little proof now before I forget it. Incidentally, these observations will be relevant to the to-be-had discussion in my upcoming posts, but for now I’ll treat it as stand-alone material. Recall that a pair of whole numbers $(m, n)$ is said to be an isoperimetric pair if $\displaystyle m + n = \left\lceil\,\frac{mn}{\left\lceil\,\sqrt{mn}\,\right\rceil}\,\right\rceil + \left\lceil\,\sqrt{mn}\,\right\rceil.$ This definition arises from the work of my REU students. Specifically, a rectangular arrangement of $mn$ squares will require the smallest perimeter possible if and only if $(m, n)$ is an isoperimetric pair. Also, if you take the expression on the right-hand side and replace $+$ with $\cdot$, we stumble across our previously celebrated “next pronic or square” function $SP(mn)$. Something else to point out is that the formula $SP(x) = \left\lceil\,\frac{x}{\left\lceil\,\sqrt{x}\right\rceil}\right\rceil\,\cdot\,\left\lceil\,\sqrt{x}\,\right\rceil$ presents the output as a product of its “square” or “pronic” factors. In particular $\left\lceil\,\sqrt{x}\,\right\rceil =\left\lceil\,\frac{x}{\left\lceil\,\sqrt{x}\right\rceil}\right\rceil \text{ or }\left\lceil\,\frac{x}{\left\lceil\,\sqrt{x}\right\rceil}\right\rceil + 1$ These observations are helpful in proving the following neat, little facts. Proposition 1: All pairs of the form $(n, n+2)$ are isoperimetric. Moreover, whenever $n > 1$, all pairs of the form $(n, n+3)$ are also isoperimetric. Proof (sketch): For the first claim, one uses the,comparison test with $$\sum \left(\frac1{n^{2/3}}\right)^\alpha$$ for $$\frac23\alpha >1$$ that is $$\alpha>\frac32$$.,evaluated at $a$ in the limit $h\rightarrow 0.$,n ≥ 0, where b = 1, 2, 3, 4, or 5, then \begin{align} k &= \left\lceil{\frac{90^\circ}{\theta}}\right\rceil = \left\lceil{\frac{90(6n+b)}{540}}\right\rceil = \left\lceil{{n+\frac{b}{6}}}\right\rceil = n+1 \\ \\ m &= \left\lceil\frac{450^\circ}{\theta}\right\rceil = \left\lceil\frac{450(6n+b)}{540}\right\rceil = \left\lceil{5n+\frac{5b}{6}}\right\rceil = 5n+b \end{align} Hence $\cos\left(\frac{m+k-1}{2}\theta\right) = \cos\left(\frac{6n+b}{2}\frac{540^\circ}{6n+b}\right) = \cos(270^\circ) = 0 \Rightarrow N(\theta) = 0.$ Now suppose $$\theta = \dfrac{360^\circ}{2p+1}$$. First consider if p = 2n is even. Then \begin{align} k &= \left\lceil{\frac{90^\circ}{\theta}}\right\rceil = \left\lceil{\frac{90(4n+1)}{360}}\right\rceil = \left\lceil{{n+\frac{1}{4}}}\right\rceil = n+1 \\ \\ m &= \left\lceil\frac{450^\circ}{\theta}\right\rceil = \left\lceil\frac{450(4n+1)}{360}\right\rceil = \left\lceil{5n+\frac{5}{4}}\right\rceil = 5n+2 \end{align} Hence $\sin\left(\frac{m-k}{2}\theta\right) = \sin\left(\frac{4n+1}{2}\frac{360^\circ}{4n+1}\right) = \sin(180^\circ) = 0 \Rightarrow N(\theta) = 0.$ Now if p = 2n+1 is odd, then \begin{align} k &= \left\lceil{\frac{90^\circ}{\theta}}\right\rceil = \left\lceil{\frac{90(4n+3)}{360}}\right\rceil = \left\lceil{{n+\frac{3}{4}}}\right\rceil = n+1 \\ \\ m &= \left\lceil\frac{450^\circ}{\theta}\right\rceil = \left\lceil\frac{450(4n+3)}{360}\right\rceil = \left\lceil{5n+\frac{15}{4}}\right\rceil = 5n+4 \end{align} Hence $\sin\left(\frac{m-k}{2}\theta\right) = \sin\left(\frac{4n+3}{2}\frac{360^\circ}{4n+3}\right) = \sin(180^\circ) = 0 \Rightarrow N(\theta) = 0.$ Here is a graph of the zeros for 40° ≤ θ ≤ 180°. Notice that between each pair of red zeros are three black zeros.(...TRUNCATED)

last equation by trial and error. I just kept diddling with it until I got something that worked. That is to say $$\left \lceil \dfrac{k^2 + k + 3}{6} \right \rceil - \left \lceil \dfrac{k^2 - k + 2}{6} \right \rceil = \left \lfloor \dfrac{k+1}{3} \right \rfloor$$ • can you just explain, just for one row of the first table, how you are finding third column from second column @steven – Sushil Verma Dec 14 '16 at 5:31 HINT: If $k=6a,k^2+k+3=(6a)(6a+1)+3,k^2-k+2=6a(6a-1)+2$ $$\left\lceil\frac{k^2+k+3}6\right\rceil - \left\lceil\frac{k^2-k+2}6\right\rceil=6a+1+1-(6a-1+1)=2$$ If $k=6a+1,k^2+k+3=(6a+1)(6a+2)+3=6a(6a+3)+5$ $,k^2-k+2=(6a+1)(6a)+2$ $$\left\lceil\frac{k^2+k+3}6\right\rceil - \left\lceil\frac{k^2-k+2}6\right\rceil=6a+3+1-(6a+1+1)=2$$ • There is something strange here as I would expect the values to be close to $\dfrac{k}{3}$ or $2a$ – Henry Dec 13 '16 at 13:59 Make a table showing $k$, $k^2+k+3$, and $k^2-k+2$ modulo $6$: $$\begin{array}{c|c|c} k&k^2+k+3&k^2-k+2\\ \hline 0&3&2\\ 1&5&2\\ 2&3&4\\ 3&3&2\\ 4&5&2\\ 5&3&4 \end{array}$$ Thus, if $k\equiv0\pmod3$, then $$\left\lceil\frac{k^2+k+3}6\right\rceil-\left\lceil\frac{k^2-k+2}6\right\rceil=\frac{k^2+k+6}6-\frac{k^2-k+6}6=\frac{k}3\;.$$ If $k\equiv1\pmod3$, then $$\left\lceil\frac{k^2+k+3}6\right\rceil-\left\lceil\frac{k^2-k+2}6\right\rceil=\frac{k^2+k+4}6-\frac{k^2-k+6}6=\frac{k-1}3\;.$$ And if $k\equiv2\pmod3$, then $$\left\lceil\frac{k^2+k+3}6\right\rceil-\left\lceil\frac{k^2-k+2}6\right\rceil=\frac{k^2+k+6}6-\frac{k^2-k+4}6=\frac{k+1}3\;.$$ If you wish to combine these into a single formula, one possibility is $$\left\lceil\frac{k^2+k+3}6\right\rceil-\left\lceil\frac{k^2-k+2}6\right\rceil=\left\lfloor\frac{k+1}3\right\rfloor\;.$$ By following the hint I gave you yesterday, you can write $$R=\lceil\frac{k^2+k+3}{6}\rceil - \lceil\frac{k^2-k+2}{6}\rceil =\frac{k^2+k+3}{6} +u- \frac{k^2-k+2}{6} - v =\frac{2 k+1}{6} + (u-v)$$ where $0 \le u, v < 1$, which imply that $-1 < u-v < 1$. It means that the result $R$ is an integer which distance from $\frac{2k+1}{6}$ is smaller than,$s_n \le \boxed{2\left\lceil \frac n2 \right \rceil} = 2k$, which we showed is achievable above.,\frac{3 \left \lceil \frac{415}{47} \right \rceil}{7} \right \rceil = \left \lceil \frac{3 \cdot 9}{7} \right \rceil = \left \lceil \frac{27}{7} \right \rceil = 4$. So, being $i = 2$ and $n = 4$, we can say that the twelfth dash is $\mathrm{t}(12) = (2, 4)$. Now, as we have computed the twelfth dash, we can compute its value by means of the dashed line function given by (1): $$\mathrm{t\_value}(x) = T(\mathrm{t}(x)) = T(2, 4) = 4 \cdot 4 = 16$$ That is the answer to the second question: the twelfth person who will go jogging, will do it in 16 days. A note for the most curious people: if you want you can try formulas (2), (3) and (4) with different values of $x$, even very big ones, for verifying their correctness. However formula (5) holds only for $i = 2$, while in general $n$ is given by: $$n = \begin{cases} \left \lceil \frac{4 \left \lceil \frac{35x - 5}{47} \right \rceil + 1}{7} \right \rceil & \text{if i = 1}\\\left \lceil \frac{3 \left \lceil \frac{35x - 5}{47} \right \rceil}{7} \right \rceil & \text{if i = 2}\\\left \lceil \frac{5 \left \lceil \frac{32x - 7}{47} \right \rceil}{8} \right \rceil & \text{if i = 3}\end{cases}$$ Moreover, this is not the only possible exact formula, but we’ll see the others in specific,then $$16^{3/4}=(16^{\frac{1}{4}})^{3}=(2)^3=\boxed{8}$$ - Have you read the other answers? – Servaes May 18 '14 at 16:09,$f(x) = 11 + \left\lceil {\frac{x}{3}} \right\rceil$ involves the ceiling function(the least integer greater or equal to the number). For some examples: $f(1)=12,~f(15)=16~\&~f(32)=22$. I would guess that the text simply wants you to draw the graph. Thank you for your post! Okay I think I understand that. But how do you get f(1)=12? I think I'll be able to draw the graph once I understand how you get the series. 4. Here are more examples. $\left\lceil {\frac{1} {3}} \right\rceil = 1,~\left\lceil {\frac{3} {3}} \right\rceil = 1,~\left\lceil {\frac{4} {3}} \right\rceil = 2,~\left\lceil {\frac{6} {3}} \right\rceil = 2,~\left\lceil {\frac{8} {3}} \right\rceil = 3$ The next integer. 5. Originally Posted by Plato Here are more examples. $\left\lceil {\frac{1} {3}} \right\rceil = 1,~\left\lceil {\frac{3} {3}} \right\rceil = 1,~\left\lceil {\frac{4} {3}} \right\rceil = 2,~\left\lceil {\frac{6} {3}} \right\rceil = 2,~\left\lceil {\frac{8} {3}} \right\rceil = 3$ The next integer. Ohh! Now it makes sense! Thank you so much!,$\binom 3 3 \left(\frac12\right)^3\left(1-\frac12\right)^{3-3}=\frac18$,. . . . . . . . . . $\ln x \:=\:-\ln12 \:=\:\ln(12)^{-1}$ x . . . . . . . . . . . . $\ln x \:=\:\ln\left(\frac{1}{12}\right)$ . . . . . . . . . . . . . . . $x \:=\:\frac{1}{12}$,to find the next number that will increase the difference. $62$ doesn't because both both Cozy's and Dash's number of jumps increases, but $63$ does, and $64$. $65$ actually gives a difference of $20$ jumps, but $66$ goes back down to $19$ (because Dash had to take another jump when Cozy didn't). We don't need to go any further because the difference will stay above $19$ onward. Therefore, the possible numbers of steps in the staircase are $63$, $64$, and $66$, giving a sum of $193$. The sum of those digits is $13$, so the answer is $\boxed{D}$ ## Solution 3 We're looking for natural numbers $x$ such that $\left \lceil{\frac{x}{5}}\right \rceil + 19 = \left \lceil{\frac{x}{2}}\right \rceil$. Let's call $x = 10a + b$. We now have $2a + \left \lceil{\frac{b}{5}}\right \rceil + 19 = 5a + \left \lceil{\frac{b}{2}}\right \rceil$, or $19 - 3a = \left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil$. Obviously, since $b \le 10$, this will not work for any value of $a$ under $6$. In addition, since obviously $\frac{b}{2} \ge \frac{b}{5}$, this will not work for any value over six, so we have $a = 6$ and $\left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil = 1.$ This can be achieved when $\left \lceil{\frac{b}{5}}\right \rceil = 1$ and $\left \lceil{\frac{b}{2}}\right \rceil = 2$, or,formula gives you $$z$$ rounded up to the next multiple of 5: $$5\left\lceil \frac{z}{5} \right\rceil$$ For example, for $$z=24$$, $$5\lceil \frac{24}{5} \rceil=5\lceil 4.8 \rceil=5\cdot 5=25$$ and for $$z=30$$, $$5\lceil \frac{30}{5} \rceil=5\lceil 6\rceil=5\cdot 6=30$$. Messenger,248 views How many of the integers $1,2, \dots, 120,$ are divisible by none of $2,5$ and $7 ?$ 1. $40$ 2. $42$ 3. $43$ 4. $41$ We know that, $n( A \cup B \cup C) = n(A) + n(B) + n(C) – n(A \cap B) – n(B \cap C) – n(A \cap C) + n ( A \cap B \cap C)$ We have $n(U) = 120$ Now, • $n(2) = \left \lceil \frac{120}{2} \right \rceil = 60$ • $n(5) = \left \lceil \frac{120}{5} \right \rceil = 24$ • $n(7) = \left \lceil \frac{120}{7} \right \rceil = 17$ • $n(2 \cap 5) = \left \lceil \frac{120}{\text{LCM}(2,5)} \right \rceil = \left \lceil \frac{120}{10} \right \rceil= 12$ • $n(5 \cap 7) = \left \lceil \frac{120}{\text{LCM}(5,7)} \right \rceil = \left \lceil \frac{120}{35} \right \rceil = 3$ • $n(2 \cap 7) = \left \lceil \frac{120}{\text{LCM}(2,7)} \right \rceil = \left \lceil \frac{120}{14} \right \rceil= 8$ • $n(2 \cap 5 \cap 7) = \left \lceil \frac{120}{\text{LCM}(2,5,7)} \right \rceil= \left \lceil \frac{120}{70} \right \rceil = 1$ So, $n(2 \cup 5 \cup 7) = 60 + 24 + 17 – 12 – 3 – 8 + 1 = 79$ Therefore, number not divisible by $2,5,$ and $7 = n \overline{(2 \cup 5 \cup 7)} = n(U) – n(2 \cup 5 \cup 7) = 120 – 79,such that $1-\left(1-\frac{1}{d}\right)\left(1-\frac{2}{d}\right)\cdots\left(1-\frac{n-1}{d}\right)\geq \frac{1}{2}.$ The classical birthday problem thus corresponds to determining n(365). The first 99 values of n(d) are given here: d n(d) 1–2 3–5 6–9 10–16 17–23 24–32 33–42 43–54 55–68 69–82 83–99 2 3 4 5 6 7 8 9 10 11 12 A number of bounds and formulas for n(d) have been published.[8] For any d≥1, the number n(d) satisfies[9] $\frac{3-2\ln2}{6} These bounds are optimal in the sense that the sequence $n(d)-\sqrt{2d\ln2}$ gets arbitrarily close to $(3-2\ln2)/6\approx 0.27$, while it has $9-\sqrt{86\ln2}\approx 1.28$ as its maximum, taken for d=43. The bounds are sufficiently tight to give the exact value of n(d) in 99% of all cases, for example n(365)=23. In general, it follows from these bounds that n(d) always equals either $\left\lceil\sqrt{2d\ln2}\right\rceil$ or $\left\lceil\sqrt{2d\ln2}\right\rceil+1$ where $\lceil x \rceil$ denotes the ceiling function. The formula $n(d) = \left\lceil\sqrt{2d\ln2}\right\rceil$ holds for 73% of all integers d.[10] The formula $n(d) = \left\lceil\sqrt{2d\ln2}+\frac{3-2\ln2}{6}\right\rceil$ holds for almost all d, i.e., for a set of integers d with asymptotic density 1.[10] The formula $n(d)=\left\lceil \sqrt{2d\ln2}+\frac{3-2\ln2}{6}+\frac{9-4(\ln2)^2}{72\sqrt{2d\ln2}} \right\rceil$ holds for all d up to 1018, but it is conjectured that there are infinitely many counter-examples to this formula.[11] The formula $n(d)=\left\lceil \sqrt{2d\ln2}+\frac{3-2\ln2}{6}+\frac{9-4(\ln2)^2}{72\sqrt{2d\ln2}} -\frac{2(\ln2)^2}{135d}\right\rceil$ holds too for all d up to 1018, and it is conjectured that this formula holds for all d.[11],39 views How many of the integers $1,2, \dots, 120,$ are divisible by none of $2,5$ and $7 ?$ 1. $40$ 2. $42$ 3. $43$ 4. $41$ 1 Answer We know that, $n( A \cup B \cup C) = n(A) + n(B) + n(C) – n(A \cap B) – n(B \cap C) – n(A \cap C) + n ( A \cap B \cap C)$ We have $n(U) = 120$ Now, • $n(2) = \left \lceil \frac{120}{2} \right \rceil = 60$ • $n(5) = \left \lceil \frac{120}{5} \right \rceil = 24$ • $n(7) = \left \lceil \frac{120}{7} \right \rceil = 17$ • $n(2 \cap 5) = \left \lceil \frac{120}{\text{LCM}(2,5)} \right \rceil = \left \lceil \frac{120}{10} \right \rceil= 12$ • $n(5 \cap 7) = \left \lceil \frac{120}{\text{LCM}(5,7)} \right \rceil = \left \lceil \frac{120}{35} \right \rceil = 3$ • $n(2 \cap 7) = \left \lceil \frac{120}{\text{LCM}(2,7)} \right \rceil = \left \lceil \frac{120}{14} \right \rceil= 8$ • $n(2 \cap 5 \cap 7) = \left \lceil \frac{120}{\text{LCM}(2,5,7)} \right \rceil= \left \lceil \frac{120}{70} \right \rceil = 1$ So, $n(2 \cup 5 \cup 7) = 60 + 24 + 17 – 12 – 3 – 8 + 1 = 79$ Therefore, number not divisible by $2,5,$ and $7 = n \overline{(2 \cup 5 \cup 7)} = n(U) – n(2 \cup 5 \cup 7) = 120,\frac{3}{2 x - 3}$ $- 6 \left(2 x - 3\right) = - 3$ $- 12 x + 18 = - 3$ $- 12 x = - 21$ $x = \frac{21}{12}$ $x = \frac{7}{4}$ Hopefully this helps!,$\frac{1}{3}(18 a+54 b)$ e. $12\left(\frac{1}{2} a+\frac{3}{4} b\right)$ Check back soon! 1,≈ r^{-2N} \left(2\pi r N\right)^{-1}$, ${{2N}\choose{2n}} ≈ r^{-2N} \left(4\pi r N\right)^{-\frac12}$. The ratio of these is $\left(\pi r N\right)^{\frac12}$ 3. Feb 24, 2013 Thank you!,that our result is $\frac 32 \left( 10^{-1} + 10^{-2} + \cdots \right) = \frac 32 \cdot \frac 19 = \boxed{\frac 16}$.,We can see that if we halve $x$ we end up getting $\left\lceil\frac{-1\pm\sqrt{1+4x}}{2}\right\rceil$. This is approximately the number divided by $\sqrt{2}$. $\frac{200}{\sqrt{2}} = 141.4$ and since $142$ looks like the only number close to it, it is answer $\boxed{(C) 142}$ ~Lopkiloinm We can look at answer choice $C$, which is $142$ first. That means that the number of numbers from $1$ to $142$ is roughly the number of numbers from $143$ to $200$. The number of numbers from $1$ to $142$ is $\frac{142(142+1)}{2}$ which is approximately $10000.$ The number of numbers from $143$ to $200$ is $\frac{200(200+1)}{2}-\frac{142(142+1)}{2}$ which is approximately $10000$ as well. Therefore, we can be relatively sure the answer choice is $\boxed{(C) \text{ } 142}.$ ~IceMatrix,at $\left(16 , \frac{1}{36}\right)$,Mar 1 at 20:18 $$\sqrt{18-\left(\frac{a}{\sqrt2}\right)^2} = \sqrt{18-\frac{a^2}{2}} = \sqrt{\frac{36-a^2}{2}};$$ as pointed out by @John Wayland Bales in a comment to the question, $$\sqrt2$$ on the right side of the equation in the question should be $$2$$. It is $$18-\left(\frac{a}{\sqrt{2}}\right)^2=18-\frac{a^2}{2}=\frac{36-a^2}{2}$$ • I think you meant $36 \mathbf - a^2$ – J. W. Tanner Mar 1 at 20:13 • Yes I meant this, thank you for your hint! – Dr. Sonnhard Graubner Mar 1 at 20:15,17 \sin \frac{5 \pi}{12} = {17}^{2} \sin \left(\frac{\pi}{6}\right)$ Now $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$ So: $A r e a = {17}^{2} / 2 = \frac{289}{2} = 144.5$(...TRUNCATED)

\frac{3 \left \lceil \frac{415}{47} \right \rceil}{7} \right \rceil = \left \lceil \frac{3 \cdot 9}{7} \right \rceil = \left \lceil \frac{27}{7} \right \rceil = 4$. So, being $i = 2$ and $n = 4$, we can say that the twelfth dash is $\mathrm{t}(12) = (2, 4)$. Now, as we have computed the twelfth dash, we can compute its value by means of the dashed line function given by (1): $$\mathrm{t\_value}(x) = T(\mathrm{t}(x)) = T(2, 4) = 4 \cdot 4 = 16$$ That is the answer to the second question: the twelfth person who will go jogging, will do it in 16 days. A note for the most curious people: if you want you can try formulas (2), (3) and (4) with different values of $x$, even very big ones, for verifying their correctness. However formula (5) holds only for $i = 2$, while in general $n$ is given by: $$n = \begin{cases} \left \lceil \frac{4 \left \lceil \frac{35x - 5}{47} \right \rceil + 1}{7} \right \rceil & \text{if i = 1}\\\left \lceil \frac{3 \left \lceil \frac{35x - 5}{47} \right \rceil}{7} \right \rceil & \text{if i = 2}\\\left \lceil \frac{5 \left \lceil \frac{32x - 7}{47} \right \rceil}{8} \right \rceil & \text{if i = 3}\end{cases}$$ Moreover, this is not the only possible exact formula, but we’ll see the others in specific,last equation by trial and error. I just kept diddling with it until I got something that worked. That is to say $$\left \lceil \dfrac{k^2 + k + 3}{6} \right \rceil - \left \lceil \dfrac{k^2 - k + 2}{6} \right \rceil = \left \lfloor \dfrac{k+1}{3} \right \rfloor$$ • can you just explain, just for one row of the first table, how you are finding third column from second column @steven – Sushil Verma Dec 14 '16 at 5:31 HINT: If $k=6a,k^2+k+3=(6a)(6a+1)+3,k^2-k+2=6a(6a-1)+2$ $$\left\lceil\frac{k^2+k+3}6\right\rceil - \left\lceil\frac{k^2-k+2}6\right\rceil=6a+1+1-(6a-1+1)=2$$ If $k=6a+1,k^2+k+3=(6a+1)(6a+2)+3=6a(6a+3)+5$ $,k^2-k+2=(6a+1)(6a)+2$ $$\left\lceil\frac{k^2+k+3}6\right\rceil - \left\lceil\frac{k^2-k+2}6\right\rceil=6a+3+1-(6a+1+1)=2$$ • There is something strange here as I would expect the values to be close to $\dfrac{k}{3}$ or $2a$ – Henry Dec 13 '16 at 13:59 Make a table showing $k$, $k^2+k+3$, and $k^2-k+2$ modulo $6$: $$\begin{array}{c|c|c} k&k^2+k+3&k^2-k+2\\ \hline 0&3&2\\ 1&5&2\\ 2&3&4\\ 3&3&2\\ 4&5&2\\ 5&3&4 \end{array}$$ Thus, if $k\equiv0\pmod3$, then $$\left\lceil\frac{k^2+k+3}6\right\rceil-\left\lceil\frac{k^2-k+2}6\right\rceil=\frac{k^2+k+6}6-\frac{k^2-k+6}6=\frac{k}3\;.$$ If $k\equiv1\pmod3$, then $$\left\lceil\frac{k^2+k+3}6\right\rceil-\left\lceil\frac{k^2-k+2}6\right\rceil=\frac{k^2+k+4}6-\frac{k^2-k+6}6=\frac{k-1}3\;.$$ And if $k\equiv2\pmod3$, then $$\left\lceil\frac{k^2+k+3}6\right\rceil-\left\lceil\frac{k^2-k+2}6\right\rceil=\frac{k^2+k+6}6-\frac{k^2-k+4}6=\frac{k+1}3\;.$$ If you wish to combine these into a single formula, one possibility is $$\left\lceil\frac{k^2+k+3}6\right\rceil-\left\lceil\frac{k^2-k+2}6\right\rceil=\left\lfloor\frac{k+1}3\right\rfloor\;.$$ By following the hint I gave you yesterday, you can write $$R=\lceil\frac{k^2+k+3}{6}\rceil - \lceil\frac{k^2-k+2}{6}\rceil =\frac{k^2+k+3}{6} +u- \frac{k^2-k+2}{6} - v =\frac{2 k+1}{6} + (u-v)$$ where $0 \le u, v < 1$, which imply that $-1 < u-v < 1$. It means that the result $R$ is an integer which distance from $\frac{2k+1}{6}$ is smaller than,$f(x) = 11 + \left\lceil {\frac{x}{3}} \right\rceil$ involves the ceiling function(the least integer greater or equal to the number). For some examples: $f(1)=12,~f(15)=16~\&~f(32)=22$. I would guess that the text simply wants you to draw the graph. Thank you for your post! Okay I think I understand that. But how do you get f(1)=12? I think I'll be able to draw the graph once I understand how you get the series. 4. Here are more examples. $\left\lceil {\frac{1} {3}} \right\rceil = 1,~\left\lceil {\frac{3} {3}} \right\rceil = 1,~\left\lceil {\frac{4} {3}} \right\rceil = 2,~\left\lceil {\frac{6} {3}} \right\rceil = 2,~\left\lceil {\frac{8} {3}} \right\rceil = 3$ The next integer. 5. Originally Posted by Plato Here are more examples. $\left\lceil {\frac{1} {3}} \right\rceil = 1,~\left\lceil {\frac{3} {3}} \right\rceil = 1,~\left\lceil {\frac{4} {3}} \right\rceil = 2,~\left\lceil {\frac{6} {3}} \right\rceil = 2,~\left\lceil {\frac{8} {3}} \right\rceil = 3$ The next integer. Ohh! Now it makes sense! Thank you so much!,248 views How many of the integers $1,2, \dots, 120,$ are divisible by none of $2,5$ and $7 ?$ 1. $40$ 2. $42$ 3. $43$ 4. $41$ We know that, $n( A \cup B \cup C) = n(A) + n(B) + n(C) – n(A \cap B) – n(B \cap C) – n(A \cap C) + n ( A \cap B \cap C)$ We have $n(U) = 120$ Now, • $n(2) = \left \lceil \frac{120}{2} \right \rceil = 60$ • $n(5) = \left \lceil \frac{120}{5} \right \rceil = 24$ • $n(7) = \left \lceil \frac{120}{7} \right \rceil = 17$ • $n(2 \cap 5) = \left \lceil \frac{120}{\text{LCM}(2,5)} \right \rceil = \left \lceil \frac{120}{10} \right \rceil= 12$ • $n(5 \cap 7) = \left \lceil \frac{120}{\text{LCM}(5,7)} \right \rceil = \left \lceil \frac{120}{35} \right \rceil = 3$ • $n(2 \cap 7) = \left \lceil \frac{120}{\text{LCM}(2,7)} \right \rceil = \left \lceil \frac{120}{14} \right \rceil= 8$ • $n(2 \cap 5 \cap 7) = \left \lceil \frac{120}{\text{LCM}(2,5,7)} \right \rceil= \left \lceil \frac{120}{70} \right \rceil = 1$ So, $n(2 \cup 5 \cup 7) = 60 + 24 + 17 – 12 – 3 – 8 + 1 = 79$ Therefore, number not divisible by $2,5,$ and $7 = n \overline{(2 \cup 5 \cup 7)} = n(U) – n(2 \cup 5 \cup 7) = 120 – 79,39 views How many of the integers $1,2, \dots, 120,$ are divisible by none of $2,5$ and $7 ?$ 1. $40$ 2. $42$ 3. $43$ 4. $41$ 1 Answer We know that, $n( A \cup B \cup C) = n(A) + n(B) + n(C) – n(A \cap B) – n(B \cap C) – n(A \cap C) + n ( A \cap B \cap C)$ We have $n(U) = 120$ Now, • $n(2) = \left \lceil \frac{120}{2} \right \rceil = 60$ • $n(5) = \left \lceil \frac{120}{5} \right \rceil = 24$ • $n(7) = \left \lceil \frac{120}{7} \right \rceil = 17$ • $n(2 \cap 5) = \left \lceil \frac{120}{\text{LCM}(2,5)} \right \rceil = \left \lceil \frac{120}{10} \right \rceil= 12$ • $n(5 \cap 7) = \left \lceil \frac{120}{\text{LCM}(5,7)} \right \rceil = \left \lceil \frac{120}{35} \right \rceil = 3$ • $n(2 \cap 7) = \left \lceil \frac{120}{\text{LCM}(2,7)} \right \rceil = \left \lceil \frac{120}{14} \right \rceil= 8$ • $n(2 \cap 5 \cap 7) = \left \lceil \frac{120}{\text{LCM}(2,5,7)} \right \rceil= \left \lceil \frac{120}{70} \right \rceil = 1$ So, $n(2 \cup 5 \cup 7) = 60 + 24 + 17 – 12 – 3 – 8 + 1 = 79$ Therefore, number not divisible by $2,5,$ and $7 = n \overline{(2 \cup 5 \cup 7)} = n(U) – n(2 \cup 5 \cup 7) = 120,formula gives you $$z$$ rounded up to the next multiple of 5: $$5\left\lceil \frac{z}{5} \right\rceil$$ For example, for $$z=24$$, $$5\lceil \frac{24}{5} \rceil=5\lceil 4.8 \rceil=5\cdot 5=25$$ and for $$z=30$$, $$5\lceil \frac{30}{5} \rceil=5\lceil 6\rceil=5\cdot 6=30$$. Messenger,such that $1-\left(1-\frac{1}{d}\right)\left(1-\frac{2}{d}\right)\cdots\left(1-\frac{n-1}{d}\right)\geq \frac{1}{2}.$ The classical birthday problem thus corresponds to determining n(365). The first 99 values of n(d) are given here: d n(d) 1–2 3–5 6–9 10–16 17–23 24–32 33–42 43–54 55–68 69–82 83–99 2 3 4 5 6 7 8 9 10 11 12 A number of bounds and formulas for n(d) have been published.[8] For any d≥1, the number n(d) satisfies[9] $\frac{3-2\ln2}{6} These bounds are optimal in the sense that the sequence $n(d)-\sqrt{2d\ln2}$ gets arbitrarily close to $(3-2\ln2)/6\approx 0.27$, while it has $9-\sqrt{86\ln2}\approx 1.28$ as its maximum, taken for d=43. The bounds are sufficiently tight to give the exact value of n(d) in 99% of all cases, for example n(365)=23. In general, it follows from these bounds that n(d) always equals either $\left\lceil\sqrt{2d\ln2}\right\rceil$ or $\left\lceil\sqrt{2d\ln2}\right\rceil+1$ where $\lceil x \rceil$ denotes the ceiling function. The formula $n(d) = \left\lceil\sqrt{2d\ln2}\right\rceil$ holds for 73% of all integers d.[10] The formula $n(d) = \left\lceil\sqrt{2d\ln2}+\frac{3-2\ln2}{6}\right\rceil$ holds for almost all d, i.e., for a set of integers d with asymptotic density 1.[10] The formula $n(d)=\left\lceil \sqrt{2d\ln2}+\frac{3-2\ln2}{6}+\frac{9-4(\ln2)^2}{72\sqrt{2d\ln2}} \right\rceil$ holds for all d up to 1018, but it is conjectured that there are infinitely many counter-examples to this formula.[11] The formula $n(d)=\left\lceil \sqrt{2d\ln2}+\frac{3-2\ln2}{6}+\frac{9-4(\ln2)^2}{72\sqrt{2d\ln2}} -\frac{2(\ln2)^2}{135d}\right\rceil$ holds too for all d up to 1018, and it is conjectured that this formula holds for all d.[11],to find the next number that will increase the difference. $62$ doesn't because both both Cozy's and Dash's number of jumps increases, but $63$ does, and $64$. $65$ actually gives a difference of $20$ jumps, but $66$ goes back down to $19$ (because Dash had to take another jump when Cozy didn't). We don't need to go any further because the difference will stay above $19$ onward. Therefore, the possible numbers of steps in the staircase are $63$, $64$, and $66$, giving a sum of $193$. The sum of those digits is $13$, so the answer is $\boxed{D}$ ## Solution 3 We're looking for natural numbers $x$ such that $\left \lceil{\frac{x}{5}}\right \rceil + 19 = \left \lceil{\frac{x}{2}}\right \rceil$. Let's call $x = 10a + b$. We now have $2a + \left \lceil{\frac{b}{5}}\right \rceil + 19 = 5a + \left \lceil{\frac{b}{2}}\right \rceil$, or $19 - 3a = \left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil$. Obviously, since $b \le 10$, this will not work for any value of $a$ under $6$. In addition, since obviously $\frac{b}{2} \ge \frac{b}{5}$, this will not work for any value over six, so we have $a = 6$ and $\left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil = 1.$ This can be achieved when $\left \lceil{\frac{b}{5}}\right \rceil = 1$ and $\left \lceil{\frac{b}{2}}\right \rceil = 2$, or,$s_n \le \boxed{2\left\lceil \frac n2 \right \rceil} = 2k$, which we showed is achievable above.,apparently I failed) $$\left\lceil \frac{3N}{10} \right\rceil - \left\lceil \frac{2N}{10} \right\rceil = 12$$ Now, let $$N=5q+r$$, where $$0 \le r < 5$$. Then we get $$\left\lceil \frac{3(5q+r)}{10} \right\rceil - \left\lceil \frac{2(5q+r)}{10} \right\rceil = 12 \\ \left\lceil \frac{3}{2}q+\frac{3}{10}r \right\rceil - \left\lceil q+\frac{r}{5} \right\rceil = 12$$ Since we can factor out integral parts (namely $$q$$) from the ceilings, $$\left\lceil \frac{q}{2}+\frac{3}{10}r \right\rceil - \left\lceil \frac{r}{5} \right\rceil = 12$$ The maximal possible value of $$q$$ is $$q=25$$, because $$\frac{3}{10}r \ge \frac{r}{5}$$, so if $$\frac{q}{2} \ge 13$$ then the LHS would be always at least 13. If we plug it in, we get $$\left\lceil \frac{1}{2}+\frac{3}{10}r \right\rceil - \left\lceil \frac{r}{5} \right\rceil = 0$$ Then the first term is at least 1 (because of $$\frac{1}{2}$$) and the second is at most 1 (because $$0 \le r < 5$$), so the only possible value for both terms is 1. Finally we solve $$\frac{1}{2} + \frac{3}{10}r \le 1 \Rightarrow r \le \frac{5}{3} \\ \frac{r}{5} > 0 \Rightarrow r > 0$$ which gives that the only possible integral value of $$r$$ is $$r=1$$, and therefore $$N=126$$. The expected participants were $$3N = 378$$. • @Downvoter, can you elaborate? – Bubbler Feb 10 at 5:49 The original number can be written as something like $$30k+3r$$ ($$r<10$$). Then the remaining number of people must be $$20k+2r$$. So, there are $$3k+\left\lceil3r/10\right\rceil$$ sanitizers,# $x\le\lceil\sqrt{x}\rceil\left\lceil\frac{x}{\lceil\sqrt{x}\rceil}\right\rceil$? $x\le\lceil\sqrt{x}\rceil\left\lceil\frac{x}{\lceil\sqrt{x}\rceil}\right\rceil$ How do I show this? I made a plot, and it looks true: - Hint: $\frac{x}{\lceil\sqrt{x}\rceil} \leq \lceil\frac{x}{\lceil\sqrt{x}\rceil}\rceil$. - I of course assume that you're only considering $x \gt 0$. – t.b. Apr 16 '11 at 22:32 Divide both sides by $\lceil \sqrt x\rceil$. -,. . . . . . . . . . $\ln x \:=\:-\ln12 \:=\:\ln(12)^{-1}$ x . . . . . . . . . . . . $\ln x \:=\:\ln\left(\frac{1}{12}\right)$ . . . . . . . . . . . . . . . $x \:=\:\frac{1}{12}$,I: Suppose that $p \equiv 1 \mod{8}$. This tells us that $p = 8k+1$ for some integer k. In this case we get (19) \begin{align} \frac{p-1}{2}-\left\lceil \frac{p}{4}\right\rceil+1 = \frac{8k+1-1}{2}-\left\lceil \frac{8k+1}{4}\right\rceil+1 = 4k - (2k+1) + 1 = 2k. \end{align} Since this number is even, that means we have $\left(\frac{2}{p}\right) = (-1)^{2k} = 1$. Case II: Suppose that $p \equiv 3 \mod{8}$. This tells us that $p = 8k+3$ for some integer k. In this case we get (20) \begin{align} \frac{p-1}{2}-\left\lceil \frac{p}{4}\right\rceil+1 = \frac{8k+3-1}{2}-\left\lceil \frac{8k+3}{4}\right\rceil+1 = 4k+1 - (2k+1) + 1 = 2k+1. \end{align} Since this number is even, that means we have $\left(\frac{2}{p}\right) = (-1)^{2k+1} = -1$. Case III: Suppose that $p \equiv 5 \mod{8}$. This tells us that $p = 8k+5$ for some integer k. In this case we get (21) \begin{align} \frac{p-1}{2}-\left\lceil \frac{p}{4}\right\rceil+1 = \frac{8k+5-1}{2}-\left\lceil \frac{8k+5}{4}\right\rceil+1 = 4k+2 - (2k+2) + 1 = 2k+1. \end{align} Since this number is even, that means we have $\left(\frac{2}{p}\right) = (-1)^{2k+1} = -1$. Case IV: Suppose that $p \equiv 7 \mod{8}$. This tells us that $p = 8k+7$ for some integer k. In this case we get (22) \begin{align} \frac{p-1}{2}-\left\lceil \frac{p}{4}\right\rceil+1 = \frac{8k+7-1}{2}-\left\lceil \frac{8k+7}{4}\right\rceil+1 = 4k+3 - (2k+2) + 1 = 2k+2. \end{align} Since this number is even, that means we have $\left(\frac{2}{p}\right) = (-1)^{2k+2} = -1$. $\square$ Now that,for $i = 2$: $$d_2(x) := \left \lceil \frac{n_1 x}{n_1 + n_2} \right \rceil \tag{5}$$ Let’s verify the previous theorem with reference to the first example of the post The downcast problem. That example was about the downcast of $\mathrm{t}$ from $(3,4)$ to $(3)$, that for convenience we rewrite below: In this case $(n_1, n_2) = (3, 4)$ and $i = 1$, because the dashed subline we are considering is $(3) = (n_1)$. By placing these values into (4), the formula becomes: $$d_1(x) = \left \lceil \frac{n_2 x + 1}{n_1 + n_2} \right \rceil = \left \lceil \frac{4 x + 1}{7} \right \rceil$$ Let’s apply the formula for all the first row ordinals that are visible in the figure: 1, 3, 5, 6, 8, 10. So we’ll obtain: • $d_1(1) = \left \lceil \frac{4 \cdot 1 + 1}{7} \right \rceil = \left \lceil \frac{5}{7} \right \rceil = 1$ • $d_1(3) = \left \lceil \frac{4 \cdot 3 + 1}{7} \right \rceil = \left \lceil \frac{13}{7} \right \rceil = 2$ • $d_1(5) = \left \lceil \frac{4 \cdot 5 + 1}{7} \right \rceil = \left \lceil \frac{21}{7} \right \rceil = 3$ • $d_1(6) = \left \lceil \frac{4 \cdot 6 + 1}{7} \right \rceil = \left \lceil \frac{25}{7} \right \rceil = 4$ • $d_1(8) = \left \lceil \frac{4 \cdot 8 + 1}{7},will be at position $7+p_1$ in the full stack. The third time you lay the cards on the table, my card will appear at position $p_2=\lceil\frac{7+p_1}{3}\rceil$ in its column. Finally, when you pick up the cards for the third time, my card will be at position $7+p_2$ in the full stack. Putting this all together, my card will be at position $$7+p_2=7+\left\lceil\frac{7+\lceil\frac{7+x}{3}\rceil}{3}\right\rceil$$ in the full stack. The trick is that this is equal to $11$ for all $x$ in the range $1\leq x\leq 7$. For a proof of this last statement, as jpmc26 mentions, one can apply the identities $\lceil\frac{m+\lceil x\rceil}{n}\rceil=\lceil\frac{m+x}{n}\rceil$ and $\lceil n+x\rceil = n+\lceil x\rceil$ (for real $x$, integer $m$, and positive integer $n$) to show that $$7+\left\lceil\frac{7+\lceil\frac{7+x}{3}\rceil}{3}\right\rceil = 7+\left\lceil\frac{7+\frac{7+x}{3}}{3}\right\rceil = 7 + \left\lceil 3 + \frac{x+1}{9}\right\rceil = 10 + \left\lceil\frac{x+1}{9}\right\rceil \enspace,$$ which is clearly equal to $11$ for $1\leq x\leq 7$. • You can use the fact that $\left \lceil \frac{7+\lceil y \rceil}{3} \right \rceil = \left \lceil \frac{7+ y }{3} \right \rceil$ (see here) to show that the last expression is equal to $7 + \left \lceil 3 + \frac{x + 1}{9} \right \rceil = 10 + \left \lceil \frac{x + 1}{9} \right \rceil$, which makes the fact it's always 11 much more obvious. Dec 12 '16 at 23:14 • Good point. I just,\frac 1 \frac 1 3 \right\rceil + 1 = 3 code(A) is 001 For B \bar F(B) = p(A) + \frac 12 p(B) = \frac 13 + \frac 12 \cdot \frac 14 = 0.4583333... In binary, Z(B) = 0.01110101010101... L(B) = \left\lceil \log_2 \frac 1 \frac 1 4 \right\rceil + 1 = 3 code(B) is 011 For C \bar F(C) = p(A) + p(B) + \frac 12 p(C) = \frac 13 + \frac 14 + \frac 12 \cdot \frac 16 = 0.66666... In binary, Z(C) = 0.101010101010... L(C) = \left\lceil \log_2 \frac 1 \frac 1 6 \right\rceil + 1 = 4 code(C) is 1010 For D \bar F(D) = p(A) + p(B) + p(C) + \frac 12 p(D) = \frac 13 + \frac 14 + \frac 16 + \frac 12 \cdot \frac 14 = 0.875 In binary, Z(D) = 0.111 L(D) = \left\lceil \log_2 \frac 1 \frac 1 4 \right\rceil + 1 = 3 code(D) is 111 ## Algorithm analysis ### Prefix code Shannon–Fano–Elias coding produces a binary prefix code, allowing for direct decoding. Let bcode(x) be the rational number formed by adding a decimal point before a binary code. For example, if code(C)=1010 then bcode(C) = 0.1010. For all x, if no y exists such that bcode(x) \le bcode(y) < bcode(x) + 2^{-L(x)} then all the codes,$\binom 3 3 \left(\frac12\right)^3\left(1-\frac12\right)^{3-3}=\frac18$,if its i/3 instead of i/2 can you do it for i/3 ??? No. Try a few values of $i$, and work out $\left\lceil\frac{i}{3}\right\rceil$: $i = 0:\left\lceil\frac{i}{3}\right\rceil=0$ $i = 1:\left\lceil\frac{i}{3}\right\rceil=1$ $i = 2:\left\lceil\frac{i}{3}\right\rceil=1$ $i = 3:\left\lceil\frac{i}{3}\right\rceil=1$ $i = 4:\left\lceil\frac{i}{3}\right\rceil=2$ ... and so on. So instead of taking pairs of values of $n$, even and odd, you'll have to take groups of 3, depending on the remainder 0, 1 or 2, that $n$ leaves when divided by 3. I haven't worked it out, but I suspect that you'll need to link $f(n)$ and $f(n-3)$ now, rather than $f(n)$ and $f(n-2)$. Try it and see.,= (-1)^{2k} = 1$. Case II: Suppose that $p \equiv 3 \mod{8}$. This tells us that $p = 8k+3$ for some integer k. In this case we get (19) \begin{align} \frac{p-1}{2}-\left\lceil \frac{p}{4}\right\rceil+1 = \frac{8k+3-1}{2}-\left\lceil \frac{8k+3}{4}\right\rceil+1 = 4k+1 - (2k+1) + 1 = 2k+1. \end{align} Since this number is even, that means we have $\left(\frac{2}{p}\right) = (-1)^{2k+1} = -1$. Case III: Suppose that $p \equiv 5 \mod{8}$. This tells us that $p = 8k+5$ for some integer k. In this case we get (20) \begin{align} \frac{p-1}{2}-\left\lceil \frac{p}{4}\right\rceil+1 = \frac{8k+5-1}{2}-\left\lceil \frac{8k+5}{4}\right\rceil+1 = 4k+2 - (2k+2) + 1 = 2k+1. \end{align} Since this number is even, that means we have $\left(\frac{2}{p}\right) = (-1)^{2k+1} = -1$. Case IV: Suppose that $p \equiv 7 \mod{8}$. This tells us that $p = 8k+7$ for some integer k. In this case we get (21) \begin{align} \frac{p-1}{2}-\left\lceil \frac{p}{4}\right\rceil+1 = \frac{8k+7-1}{2}-\left\lceil \frac{8k+7}{4}\right\rceil+1 = 4k+3 - (2k+2) + 1 = 2k+2. \end{align} Since this number is even, that means we have $\left(\frac{2}{p}\right) = (-1)^{2k+2} = -1$. $\square$ Recall that for an arbitrary integer $n = \pm 2^e p_1^{e_1}\cdots p_k^{e_k}$ and an odd prime p not dividing n, we have (22) \begin{align} \left(\frac{n}{p}\right) = \left(\frac{\pm 1}{p}\right)\left(\frac{2}{p}\right)^e\left(\frac{p_1}{p}^{e_1}\right)\cdots\left(\frac{p_k}{p}\right)^{e_k}. \end{align} Now these first two terms we know how to compute, based on the residue class of p modulo 4 and 8 (respectively).,\right \rceil = \left \lceil \frac{33}{7} \right \rceil = 5$ • $d_1(10) = \left \lceil \frac{4 \cdot 10 + 1}{7} \right \rceil = \left \lceil \frac{41}{7} \right \rceil = 6$ As we expected, the behaviour the function $d_1$ is exactly like the one of the arrow shown in Figure 2. Instead for $i = 2$, i.e. for the downcast from $(3, 4)$ to $(4)$, formula (5) becomes: $$d_2(x) = \left \lceil \frac{n_1 x}{n_1 + n_2} \right \rceil = \left \lceil \frac{3 x}{7} \right \rceil$$ Applying this formula to the ordinals of the second row, that are 2, 4, 7 and 9, we again obtain a sequence of consecutive natural numbers (1, 2, 3 and 4), that are the corresponding ordinals in the dashed subline $(4)$. For example $$d_2(4) = \left \lceil \frac{3 \cdot 4}{7} \right \rceil = \left \lceil \frac{12}{7} \right \rceil = 2$$ in fact the fourth dash of the dashed line $(3,4)$ (the one with value 8) is the second within the dashed subline $(3,4)[2] = (4)$. Now we have all the tools required for computing the value of the $x$-th dash of a second order linear dashed line $T$, provided that it belongs to row $i$. In fact we can state the following Theorem: Formula for computing the second order linear $\mathrm{t\_value}$ function Let $T$(...TRUNCATED)

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