<PRE><B>Question 749002</B>
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The absent guy can do the work in 14 hours, so he can do *[tex \LARGE \frac{1}{14}] of the work in 1 hour.  The guy doing the work can do the work in *[tex \LARGE x] hours, so he can do *[tex \LARGE \frac{1}{x}] of the work in 1 hour.  Together, they could do *[tex \LARGE \frac{1}{14}\ +\ \frac{1}{x}] of the work in 1 hour, but we know that they can do *[tex \LARGE \frac{1}{8}] of the work in 1 hour working together, hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{14}\ +\ \frac{1}{x}\ =\ \frac{1}{8}]


Solve for *[tex \LARGE x]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
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