Question 509517
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Use time equals distance divided by rate.  Let*[tex \Large r] represent the initial rate.  Then the reduced rate is *[tex \Large r\ -\ 20].  Let *[tex \Large t] represent the time it took to travel the first 50 miles.  Then *[tex \Large 1.5\ -\ t] is the time it took to travel the last 5 miles.


We know that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{50}{r}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1.5\ -\ t\ =\ \frac{5}{r\ -\ 20}]


Substituting from the first equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1.5\ -\ \frac{50}{r}\ =\ \frac{5}{r\ -\ 20}]


A little Algebra music, Sammy:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1.5\ =\ \frac{50}{r}\ +\ \frac{5}{r\ -\ 20}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1.5\ =\ \frac{50r\ -\ 1000\ +\ 5r}{r^2\ -\ 20r}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1.5r^2\ -\ 85r\ +\ 1000\ =\ 0]


Multiply by 2 and the quadratic factors rather tidily.  One of the roots doesn't work because it is less than 20, which means that the rate would be negative (driving in reverse to go backwards in time?) for the 5 mile second leg of the trip.


Solve for *[tex \Large r] and then calculate *[tex \Large r\ -\ 20]


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