Question 528679
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If the average speed for a 60 mile trip was 54 miles per hour, then the elapsed time for the trip had to be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{60}{54}\ =\ \frac{10}{9}] hour.


The first 30 miles was completed in 40 minutes which is *[tex \Large \frac{2}{3}] hour so the second half of the trip had to have been completed in:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{10}{9}\ -\ \frac{2}{3}\ =\ \frac{4}{9}] hour.


Hence the speed for the last 30 miles must be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\ 30\ }{\frac{4}{9}}\ =\ 30\ \cdot\ \frac{9}{4}\ =\ 67.5] mph.


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