Question 608581
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If A can do a job in <i>x</i> time periods, then A can do *[tex \Large \frac{1}{x}] of the job in 1 time period.  Likewise, if B can do the same job in <i>y</i> time periods, then B can do *[tex \Large \frac{1}{y}] of the job in 1 time period.  For this problem you will need C doing *[tex \Large \frac{1}{z}] of the job in 1 time period


So, working together, they can do


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{1}{x}\ +\ \frac{1}{y}\ + \frac{1}{z}\ =\ \frac{xy\ +\ yz\ +\ xz}{xyz} ]


of the job in 1 time period.


Therefore, they can do the whole job in:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{1}{\frac{xy + yz\ +\ xz}{xyz}}\ =\ \frac{xyz}{xy + yz\ +\ xz}]


time periods. 


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