Question 991207
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ S\ =\ L\ -\ rL]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ S\ =\ L(1\ -\ r)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ r\ =\ \frac{S}{L}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -r\ =\ \frac{S}{L}\ -\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ 1\ -\ \frac{S}{L}]


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

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