Question 316576
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Distance equals Rate times Time:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ rt]


Which can (more conveniently for this problem) be expressed as:


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


When the plane flies against the wind, the actual speed is the speed in still air MINUS the speed of the wind.  With the wind, speed in still air PLUS wind speed.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{259}{210\ +\ v}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{161}{210\ -\ v}]


Having two expressions equal to the elapsed time, *[tex \LARGE t], regardless of how long that might be, we can set them equal to each other, eliminating time of flight as a factor in the calculation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{259}{210\ +\ v}\ =\  \frac{161}{210\ -\ v}]


Now all that is left is to cross multiply and solve for *[tex \LARGE v].


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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