Question 984082
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ rt]


So if *[tex \Large d\ =\ 330], and the speed TO Grandma's was *[tex \Large r] and the time it took was *[tex \Large t] then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{330}{r}]


But we also know that the return trip speed was *[tex \Large r\ +\ 11] and the return trip time was *[tex \Large t\ -\ 1], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ -\ 1\ =\ \frac{330}{r\ +\ 11}]


which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{330}{r\ +\ 11}\ +\ 1]


Since *[tex \Large t\ =\ t],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{330}{r}\ =\ \frac{330}{r\ +\ 11}\ +\ 1]


Solve for *[tex \Large r] then calculate *[tex \Large r\ +\ 11]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \