Question 901599
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The time for the outbound trip is *[tex \LARGE t_1\ =\ \frac{100}{45}]


The time for the return trip is *[tex \LARGE t_2\ =\ \frac{100}{80}]


The average speed for the entire trip is the total distance divided by the total time:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ S_{avg}\ =\ \frac{200}{t_1\ +\ t_2}]


Just do the arithmetic.


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