Question 749184
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20 hours?  Really?  She couldn't find a closer park?


We know that distance equals rate times time, so one way to describe the distance to the park is 7 miles per hour times *[tex \LARGE t] hours, presuming we allow *[tex \LARGE t] to represent the number of hours that she ran.  With that presumption, and the fact that the entire trip took 20 hours, we can say that she must have walked for *[tex \LARGE 20\ -\ t] hours.  Hence, another way to describe the distance (from the park back home this time) is *[tex \LARGE 3(20\ -\ t)].  Since it is a pretty safe bet that nobody picked up her house and moved it while she was gone, the distance from home to the park has to be equal to the distance from the park to home.  Therefore we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 7t]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 3(20\ -\ t)]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7t\ =\ 3(20\ -\ t)]


Solve for *[tex \LARGE t] and then plug the value of *[tex \LARGE t] back into *[tex \LARGE d\ =\ 7t] to find *[tex \LARGE d]


Must be some spectacular park if she is going to run a marathon and a half just to get there.


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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