Question 711800
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Let *[tex \LARGE x] represent the number of hours it takes for the larger pipe to fill the pool.  That means that the larger pipe can fill *[tex \LARGE \frac{1}{x}]th of the pool in one hour.  Likewise, the smaller pipe, that takes *[tex \LARGE 2x] hours to fill the pool, can fill *[tex \LARGE \frac{1}{2x}] of the pool in one hour.  Working together, they fill the pool in *[tex \LARGE 2\,\frac{1}{3}] hours, or *[tex \LARGE \frac{7}{3}] hours.  That means that the two pipes working together can fill *[tex \LARGE \frac{3}{7}] of the pool in one hour.  Since what one of the pipes can do in one hour plus what the other pipe can do in one hour is equal to what they can do together in one hour, we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{x}\ +\ \frac{1}{2x}\ =\ \frac{3}{7}]


Solve for *[tex \LARGE x]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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