Question 199485
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Slight mis-step there. You wrote: *[tex \Large y = \frac{k}{x}].  That means *[tex \Large y] varies <b><i>inversely</i></b> as *[tex \Large x].  If *[tex \Large y] varies <b><i>as</i></b> *[tex \Large x] (meaning directly as), then you want:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = kx]


So if *[tex \Large x = 2] when *[tex \Large y = 8], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8 = 2k \ \ \Rightarrow\ \ k = 4]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = 4x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 17 = 4x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{17}{4}]


On the other hand, if you simply left out the word 'inversely' when you posted the problem (and I think that may be the case because I have seen this particular problem before) then you did it correctly.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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