Question 322988
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Let *[tex \Large t] represent the time in hours it takes the faster guy to go 10 miles.  Then it takes *[tex \Large t + 3] hours for the slower guy to make the same trip.  Let *[tex \Large r] represent the slower speed.  Then the faster speed is *[tex \Large 2r]


Using *[tex \LARGE d\ =\ rt]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10\ =\ 2rt]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10\ =\ r(t\ +\ 3)]


Since 10 equals 10,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2rt\ =\ r(t\ +\ 3)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ rt\ =\ 3r]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ 3]


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r(3\ +\ 3)\ =\ 10]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ \frac{10}{6}\ =\ \frac{5}{3}] mph.


for the slow guy -- twice that for the fast guy.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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