Question 234574
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The time is the same for both the upstream and downstream trips ("in the same amount of time"), so just call it *[tex \Large t].


Since

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


we can also say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{d}{r}]


So for the upstream trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{9}{r\ -\ 3}]


And for the downstream trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{11}{r\ +\ 3}]


But *[tex \Large t\ =\ t], so we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{9}{r\ -\ 3}\ =\ \frac{11}{r\ +\ 3}]


Just cross-multiply and solve for *[tex \Large r]


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