Question 1038816
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Let *[tex \Large x] represent the length of the prism.  Then *[tex \Large x\ -\ 5] must represent the height and width.  So the volume as a function of *[tex \Large x] is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(x)\ =\ x(x\ -\ 5)^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(x)\ =\ x^3\ -\ 10x^2\ +\ 25x]


So if the volume is to be 250 cubic inches:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3\ -\ 10x^2\ +\ 25x\ =\ 250]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3\ +\ 25x\ -\ (10x^2\ +\ 250)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(x^2\ +\ 25)\ -\ 10(x^2\ +\ 25)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 10)(x^2\ +\ 25)\ =\ 0]


Since *[tex \LARGE x^2\ +\ 25] has no real zeros, the only real solution is *[tex \LARGE x\ =\ 10]


Hence, the dimensions of the prism are 10 by 5 by 5.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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