Question 993720
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The length is 8 more than the width, so *[tex \Large w\ =\ l\ -\ 8], then since the perimeter is *[tex \Large P\ =\ 2l\ +\ 2w],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  2l\ +\ 2(l\ -\ 8)\ =\ 52]


Solve for *[tex \Large l]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \