Question 487771
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 2l\ +\ 2w]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w\ =\ P\ -\ 2l]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ \frac{P\ -\ 2l}{2}\ =\ \frac{P}{2}\ -\ l]



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