Question 447163
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Remember <i>Distance = Rate times Time</i>


Let *[tex \Large r] represent the speed of the plane in still air.  Let *[tex \Large r_w] represent the speed of the wind.  The rate of the plane against the wind is *[tex \Large r\ -\ r_w] and the rate of the plane with the wind is *[tex \Large r\ +\ r_w].


Using *[tex \Large d\ =\ rt]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4(r\ -\ r_w)\ =\ 1680]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3.5(r\ +\ r_w)\ =\ 1680]


Divide the first equation by 4 and the second by 3.5:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ -\ r_w\ =\ 420]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ +\ r_w\ =\ 480]


Add the two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ +\ r\ +\ r_w\ -\ r_w\ =\ 420\ +\ 480]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2r\ =\ 900]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ 450]


and then substitute 450 back into either of the orginal equations to compute *[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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