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PilotLPTM [1.2K]
3 years ago
11

How do I find the length and the width?

Mathematics
2 answers:
stiv31 [10]3 years ago
8 0
Online calculator to calculate the dimensions (length<span> and </span>width<span>) of a rectangle given the area A and perimeter P of the rectangle. Then these equations are solved for L and W which are the </span>length<span> and </span>width<span> of the rectangle. Enter the perimeter P and area A as positive real numbers and press "enter".</span>
Gnesinka [82]3 years ago
4 0
A = L * W
A = 91
L = W + 6

91 = W(W + 6)
91 = W^2 + 6W
W^2 + 6W - 91 = 0
(W - 7)(W + 13) = 0

W - 7 = 0
W = 7

W + 13 = 0
W = -13.......extraneous solution...this one does not work

so the width is 7 inches <====
L = W + 6
L = 7 + 6
L = 13...and the length is 13 inches <====

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I guess the series is

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We have

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Recall that

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In our limit, we have

\dfrac n{n+1}=\dfrac{n+1-1}{n+1}=1-\dfrac1{n+1}

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\implies\displaystyle2\lim_{n\to\infty}\left(\frac n{n+1}\right)^n=2\frac{\lim\limits_{n\to\infty}\left(1-\frac1{n+1}\right)^{n+1}}{\lim\limits_{n\to\infty}\left(1-\frac1{n+1}\right)}=\frac{2e}1=2e

which is greater than 1, which means the series is divergent by the ratio test.

On the chance that you meant to write

\displaystyle\sum_{n=1}^\infty\frac{2^n}{n!n^n}

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\displaystyle\lim_{n\to\infty}\left|\frac{\frac{2^{n+1}}{(n+1)!(n+1)^{n+1}}}{\frac{2^n}{n!n^n}}\right|=2\lim_{n\to\infty}\frac1{(n+1)^2}\left(\frac n{n+1}\right)^2

=\displaystyle2\left(\lim_{n\to\infty}\frac1{(n+1)^2}\right)\left(\lim_{n\to\infty}\left(\frac n{n+1}\right)^n\right)=2\cdot0\cdot e=0

which is less than 1, so this series is absolutely convergent.

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