Question 632407
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Let *[tex \LARGE r] represent RRH's rate, so *[tex \LARGE r\ -\ 6] must represent BBW's rate.  Let *[tex \LARGE t] represent the time it took RRH to make the trip and *[tex \LARGE t\ +\ 1] must then represent BBW's time.


Starting with Distance Equals Rate Times Time, we can easily derive that Distance divided by Rate equals Time, so RRH's trip is described by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{432}{r}\ =\ t]


And BBW's trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{432}{r\ -\ 6}\ =\ t\ +\ 1]


A little algebra music, Sammy (and I leave it as an exercise for the student to verify)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{438\ -\ r}{r\ -\ 6}\ =\ t]


Now that we have two different expressions in *[tex \LARGE r] that are both equal to *[tex \LARGE t], set them equal to each other:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{432}{r}\ =\ \frac{438\ -\ r}{r\ -\ 6}]


Cross-multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 432(r\ -\ 6)\ =\ 438r\ -\ r^2]


Simplify to standard quadratic form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r^2\ -\ 6r\ -\ 2590\ =\ 0]


Solve for the positive root, then calculate *[tex \LARGE r\ -\ 6]


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