Question 1122161
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I think this is right, but don't go making any large wagers.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ k(x^2\ +\ 15)^{1/2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y'(t)\ =\ \frac{2xk}{2\sqrt{x^2\,+\,15}}x'(t)]


But *[tex \LARGE \frac{y'(t)}{x'(t)}\ =\ \frac{1}{3}] when *[tex \LARGE x\ =\ 1], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2(1)k}{2\sqrt{(1)^2\,+\,15}}\ =\ \frac{1}{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{k}{4}\ =\ \frac{1}{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ \frac{4}{3}]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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