Question CLEAR work showing how to solve all 3 parts of this problem would be very appreciated and result in a thumbs up! An inverted cylindrical cone, 24 ft deep and 12 ft across at the top, is being filled with water at a rate of 16 ft/min. At what rate is the water rising in the tank when the depth of the water is: 1 foot? Answer= 10 feet? Answer= 23 feet? Answer=

givi-- An Imented colindical conet 248 deep and 12 \mathrm{ft} cocors the torp.filled wth woth at a rate of 16 \mathrm{fp}^{3} / \mathrm{mes}\begin{array}{l}h=2: 4 y \\\gamma=12 y \\\left.\frac{d v}{d t} \Rightarrow 16 \mathrm{jp}^{3}\right|_{\mathrm{mn}}\end{array}wher V is volume\begin{array}{l}v \Rightarrow \frac{1}{3} \pi r^{2} h \\{\left[r \Rightarrow \frac{h}{2}\right)=\frac{16}{h}=\frac{12}{2 t}=\frac{1}{2}} \\v \Rightarrow \frac{1}{3} \pi\left(\frac{h}{2}\right)^{3} h \\v \Rightarrow \frac{1}{3}+\frac{1}{4} \pi \frac{h^{3}}{4} \\\frac{d v}{d t}=\frac{\pi}{12} \times 3 h^{2} \frac{d h}{d t} \\\frac{d v}{d t}=\frac{\pi}{4} h^{2} \times \frac{d h}{d t} \\\frac{d v}{d t} \Rightarrow \frac{\pi h^{2}}{4} \times \frac{d h}{d t}\end{array}16=\frac{\pi}{2} \times h^{2} \times \frac{d h}{d t}at h=1 ft\begin{array}{l}\frac{d h}{d t}=\frac{3 z^{8}}{\pi} \times \frac{1}{12^{2}} \\{\left[\frac{d h}{d t} \Rightarrow \frac{8}{32} \Rightarrow 0.848\right]}\end{array}at h=10 \mathrm{ft}\begin{aligned}\frac{d h}{d t} & =\frac{32}{\pi} \times \frac{1}{10^{2}} \\\frac{d h}{d t} & =0.101\end{aligned}at h=23 \mathrm{fs}\begin{array}{l}\frac{d h}{d t}=\frac{32}{\pi} \times \frac{1}{23^{2}} \\{\left[\frac{d h}{d t} \Rightarrow 0000.019 \quad 11\right]}\end{array} ...