Question 625738
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The volume of a cone as a function of its radius and height is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_1(r,h)\ =\ \frac{1}{3}\pi{r^2}h]


So if you double the radius, then the radius of the new, larger cone is *[tex \LARGE 2r], so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_2(r,h)\ =\ \frac{1}{3}\pi(2r)^2h]


A little algebra music, Sammy:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_2(r,h)\ =\ \frac{4}{3}\pi{r^2}h]


Divide the volume of the large cone by the volume of the smaller one:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{V_2}{V_1}\ =\ \frac{\frac{4}{3}\pi{r}^2h}{\frac{1}{3}\pi{r^2}h}\ =\ 4]


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