Question 997857
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The small hose can fill the pool in 6 hours, so it can fill *[tex \Large \frac{1}{6}] of the pool in one hour.  The large hose can fill *[tex \Large \frac{1}{x}] of the pool in one hour.  Working together they can fill *[tex \Large \frac{1}{2}] of the pool in one hour, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{6}\ +\ \frac{1}{x}\ =\ \frac{1}{2}]


Solve for *[tex \Large x]


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

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