Question 343669
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Let *[tex \Large d] represent the distance in miles traveled by the first truck.


Let *[tex \Large t] represent the travel time of the first truck in minutes.


Since we are expressing time in minutes, convert 30 miles per hour to 1/2 mile per minute and 60 miles per hour to 1 mile per minute.


The first truck:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 0.5t]


If the first truck traveled *[tex \Large d] miles, then the second truck traveled *[tex \Large 12 - d] miles.  If the first truck traveled for *[tex \Large t] minutes, then the second truck, having been called 3 minutes later, traveled *[tex \Large t\ -\ 3] minutes.


The second truck:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12\ -\ d\ =\ t\ -\ 3]


(Remember, the second truck's rate was 1 mile per minute)


Solve for *[tex \Large d]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 15\ -\ t]


Now we have two expressions each of which is equal to *[tex \Large d], so set them equal to each other:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.5t\ =\ 15\ -\ t]


Solve for *[tex \Large t] to get the time for the first truck.  Substitute into *[tex \Large d\ =\ 0.5t] to find the distance for the second truck.


Subtract *[tex \Large t\ -\ 3] to get the time for the second truck.


Subtract *[tex \Large 12\ -\ d] to get the distance for the second truck.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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