Question 397357
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Let *[tex \Large t] represent the time that it took him to come down the hill.


Using <i>distance equals rate times time</i> the distance he hiked down the hill must be *[tex \Large 2.4t].


Since the entire time hiking was 11 hours, the time it took to climb up the hill is 11 hours minus the time it took to climb down the hill.  Hence, the distance up the hill must be *[tex \Large 1.6(11\ -\ t)]


Making the wild assumption that the hill neither grew nor shrank any significant amount overnight, we can say that the distance down the hill is equal to the distance up the hill, therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 1.6(11\ -\ t)]


Which can be written:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{d}{1.6}\ -\ 11]


And further simplified:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{d\ -\ 17.6}{1.6}]


And we can also say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 2.4t]


Which can be written:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{d}{2.4}]


Since *[tex \Large t\ =\ t] we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{2.4}\ =\ \frac{d\ -\ 17.6}{1.6}]


Just solve for *[tex \Large d].  Hint:  Cross-multiply and then collect terms.


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