Question 908662
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Let *[tex \Large t] represent the time of each of the downstream and upstream trips.  Let *[tex \Large s_c] represent the speed of the current.  Distance equals rate times time, hence time equals distance divided by rate.  The current goes against the speed of the boat upstream and adds to the speed of the boat downstream.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{2}{20\ -\ s_c}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{4}{20\ +\ s_c}]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{2}{20\ -\ s_c}\ =\ \frac{4}{20\ +\ s_c}]


Simply solve for *[tex \Large s_c]


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