Question 596772
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Let *[tex \Large r] represent the speed of the plane in still air.  Let *[tex \Large r_w] represent the wind speed.


We are given that *[tex \Large r\ -\ r_w\ =\ 336] mph, but *[tex \Large r\ -\ r_w] is also the ground speed against the wind, at which speed the aircraft required 8 hours to fly the distance.  Hence the distance flown is *[tex \Large 8\,\cdot\,336\ =\ 2688] miles.


From that we calculate the time for the downwind trip: *[tex \Large \frac{2688}{7}\ =\ 384] mph.  But the ground speed downwind is given by *[tex \Large r\ +\ r_w], hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left{r\ +\ r_w\ =\ 384\cr r\ -\ r_w\ =\ 336]


Solve the system for *[tex \Large r] and *[tex \Large r_w]


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