Question 1117527
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ 8x^2\ +\ 40x]


Factor out *[tex \Large ax] where *[tex \Large a\ \in\ \mathbb{R},\ a\ >\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax\(\frac{8x}{a}\ +\ \frac{40}{a}\)]


Substitute *[tex \Large x\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3a\(\frac{24}{a}\ +\ \frac{40}{a}\)]


The dimensions of the playground are *[tex \Large 3a] by *[tex \Large \frac{24}{a}\ +\ \frac{40}{a}] where *[tex \Large a] is any positive real number.  In other words, there are an infinite number of solutions to this problem.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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