Question 331366
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You don't bother to mention the distance units that the plane travels.  Not a real problem except that it makes rendering the units a bit awkward.


The rate of the plane relative to the ground when flying against the wind is *[tex \Large r\ -\ r_w] where *[tex \Large r] is the rate in still air and *[tex \Large r_w] is the rate of the wind.  Flying with the wind, *[tex \Large r\ +\ r_w]


We know that distance equals rate times time, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5580\ =\ 9(r\ -\ r_w)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4200\ =\ 5(r\ +\ r_w)]


Putting the two equations into standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9r\ -\ 9r_w\ =\ 5580]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5r\ +\ 5r_w\ =\ 4200]


Solve the system of equations 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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