Question 1046368
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48 seconds is 48/60ths of a minute which is to say 8/10ths of a minute.  Hence the "with the wind" leg took 1.8 minutes.  Since the wind speed is given in km/hr, we convert the elapsed time to hours.  1.8 divided by 60 is 0.03 hours and 3 divided by 60 is 0.05 hours.  Distance equals rate times time.  Let *[tex \Large r] km/hr be the speed of the plane.  The total rate with the wind is the sum of the plane's speed and the wind speed.  The rate against the wind is the difference, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 0.03(r\ +\ 5)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 0.05(r\ -\ 5)]


But since *[tex \Large d\ =\ d]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.05(r\ -\ 5)\ =\ 0.03(r\ +\ 5)]


Solve for *[tex \Large r], then calculate *[tex \Large 0.03(r\ +\ 5)].  Then check your work by calculating *[tex \Large 0.05(r\ -\ 5)] and make sure the distances are the same.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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