Question 388013
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Start with the basic formula, Distance equals Rate times Time.


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


But express it as time as a function of distance and rate:


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


So let's talk about the downstream trip first.  Going downstream the boat goes 15 mph PLUS the rate of the current, i.e. *[tex \Large 15\ +\ r], and the 80 mile trip is accomplished in *[tex \Large t] hours, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{80}{15\ +\ r}]


Now let's look at the upstream trip.  This time the boat speed relative to the shore is 15 mph MINUS the rate of the current, *[tex \Large 15\ -\ r], and the 80 mile trip is accomplished in *[tex \Large t\ +\ 4] hours, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ +\ 4\ =\ \frac{80}{15\ -\ r}]


A little algebra and fraction addition:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{80}{15\ -\ r}\ -\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{20\ +\ 4r}{15\ -\ r}]


Ah ha!  Now we have two things both equal to *[tex \Large t], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{80}{15\ +\ r}\ =\ \frac{20\ +\ 4r}{15\ -\ r}]


Cross-multiply, collect like terms, and put the quadratic into Standard Form.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r^2\ +\ 40r\ -\ 225\ =\ 0]


Factor and solve.  Discard the negative root -- unless you can find a river that flows upstream.



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