Question 1162960
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*[tex \Large d\ =\ rt], so *[tex \Large t\ =\ \frac{d}{r}]


Let *[tex \Large r] represent the speed without the current.  Then the upstream speed is *[tex \Large r\ -\ 3] and the downstream speed is *[tex \Large r\ +\ 3].


For the upstream trip, the distance is 200, the rate is *[tex \Large r\ -\ 3], and the time is *[tex \Large t], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{200}{r\,-\,3}]


For the downstream trip, the distance is 260, the rate is *[tex \Large r\ +\ 3], and the time is *[tex \Large t], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{260}{r\,+\,3}]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{260}{r\,+\,3}\ =\ \frac{200}{r\,-\,3}]


Solve for *[tex \Large r]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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