Question 1160527
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Elapsed time, *[tex \Large t], for a trip of distance *[tex \Large d] at an average rate of *[tex \Large r] distance units per time units is given by *[tex \Large t\ =\ \frac{d}{r}]


She traveled 90km at the rate requested by the question so the outbound trip took *[tex \Large t_1\ =\ \frac{90}{r}].


The return trip was made at a rate 20 km/hr slower, hence the return trip time was: *[tex \Large t_2\ =\ \frac{90}{(r\,-\,20)}].


Since there weren't any side trips or any other anomalies, the total trip time was the sum of the time for the outbound trip and the time for the return trip, to wit:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t_1\ +\ t_2\ =\ 1.85\ \ ]hours.


which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{90}{r}\ +\ \frac{90}{(r\,-\,20)}\ =\ 1.85\ \ ]


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