Question 621133
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Let *[tex \LARGE r] mph represent the speed of the boat in still water.  Then the speed downstream in a 5 mph current is *[tex \LARGE r\ +\ 5] mph and the speed upstream in the same 5 mph current is *[tex \LARGE r\ -\ 5] mph.


The time to go downstream is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{240}{r\ +\ 5}]


The time to go upstream


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{160}{r\ -\ 5}]


And these two times are equal, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{240}{r\ +\ 5}\ =\ \frac{160}{r\ -\ 5}]


Cross multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 240(r\ -\ 5)\ =\ 160(r\ +\ 5)]


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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