Question 797237
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Let *[tex \Large r] represent the rate in still water.


Let *[tex \Large r_c] represent the rate of the current


Let *[tex \Large d] represent the distance.


For the downstream trip:  *[tex \Large d\ =\ 4(r\ +\ r_c)]


For the upstream trip:  *[tex \Large d\ =\ 8(r\ -\ r_c)]


Since *[tex \Large d\ =\ d],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8(r\ -\ r_c)\ =\ 4(r\ +\ r_c}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_c\ =\ \frac{r}{3}]


Substitute:


For the downstream trip:  *[tex \Large d\ =\ 4\left(r\ +\ \frac{r}{3}\right)]


For the upstream trip:  *[tex \Large d\ =\ 8\left(r\ -\ \frac{r}{3}\right)]


Add the two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2d\ =\ 12r\ -\ \frac{4}{3}r]


Collect like terms:


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


Since *[tex \Large d\ =\ rt,\ \ ]*[tex \Large t\ =\ \frac{16}{3}] hours.



John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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