Question 374999
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We know that distance equals rate times time.  But that also means that time is equal to distance divided by rate.  Since the current adds to the rate of the boat going down stream and subtracts going upstream, we can write an expression that represents the time going either way.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{5}{16\ +\ r}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{3}{16\ -\ r} ]


Where *[tex \Large t] is the time for each trip and *[tex \Large r] is the rate of the current.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5}{16\ +\ r}\ =\ \frac{3}{16\ -\ r}]


Cross-multiply, collect terms, and 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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