Question 235957
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Distance equals Rate times Time, or *[tex \Large d\ =\ rt]


Notice that, since rate is given in miles per hour, time must be in hours and distance in miles.


So for the day 1 trip, the time is *[tex \Large t] hours plus one minute (it took one minute longer than the *[tex \Large t] hours planned).  But one minute is *[tex \Large \frac{1}{60}] hour, so for the fixed distance *[tex \Large d], the day 1 trip can be described:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 40\left(t\ +\ \frac{1}{60}\right)]


Likewise, the day 2 trip can be described as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 45\left(t\ -\ \frac{1}{60}\right)]


Solve each of these equations for *[tex \Large t] in terms of *[tex \Large d]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{d}{40}\ -\ \frac{1}{60}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{d}{45}\ +\ \frac{1}{60}]


(Verification of the two re-arrangements is left as an exercise for the student)


Now you have two expressions for *[tex \Large t] in terms of *[tex \Large d], so set them equal:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{40}\ -\ \frac{1}{60}\ = \frac{d}{45}\ +\ \frac{1}{60}]


And finally, solve for *[tex \Large d]


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