first off let's notice that the height is 11 meters and the volume of the cone is 103.62 cubic centimeters, so let's first convert the height to the corresponding unit for the volume, well 1 meters is 100 cm, so 11 m is 1100 cm.
![\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ V=\stackrel{cm^3}{103.62}\\ h=\stackrel{cm}{1100} \end{cases}\implies 103.62=\cfrac{\pi r^2 (1100)}{3} \\\\\\ 3(103.62)=1100\pi r^2\implies \cfrac{3(103.62)}{1100\pi }=r^2 \\\\\\ \sqrt{\cfrac{3(103.62)}{1100\pi }}=r\implies \stackrel{cm}{0.00510199305952} \approx r](https://tex.z-dn.net/?f=%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20V%3D%5Cstackrel%7Bcm%5E3%7D%7B103.62%7D%5C%5C%20h%3D%5Cstackrel%7Bcm%7D%7B1100%7D%20%5Cend%7Bcases%7D%5Cimplies%20103.62%3D%5Ccfrac%7B%5Cpi%20r%5E2%20%281100%29%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%203%28103.62%29%3D1100%5Cpi%20r%5E2%5Cimplies%20%5Ccfrac%7B3%28103.62%29%7D%7B1100%5Cpi%20%7D%3Dr%5E2%20%5C%5C%5C%5C%5C%5C%20%5Csqrt%7B%5Ccfrac%7B3%28103.62%29%7D%7B1100%5Cpi%20%7D%7D%3Dr%5Cimplies%20%5Cstackrel%7Bcm%7D%7B0.00510199305952%7D%20%5Capprox%20r)
Answer:
-288
Step-by-step explanation:
n=1/2 divided by -144. You then just plug that number which is .00347 into n in the equation and use a scientific calculator to find the answer.
Answer:
graph C
Step-by-step explanation:
because the vertex is h,k. h is -3 k is -25. C is the only graph with a vertex at that location.
Answer:
Rearrange the equation so "y" is on the left and everything else on the right.
Plot the "y=" line (make it a solid line for y≤ or y≥, and a dashed line for y< or y>)
Shade above the line for a "greater than" (y> or y≥) or below the line for a "less than" (y< or y≤).
Step-by-step explanation:
sorry if wrong
