Question 989499
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 2L\ +\ 2W]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2W\ =\ P\ -\ 2L]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ W\ =\ \frac{P}{2}\ -\ L]


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

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