how do we do this brainly keeps kicking me out seeming they dont want you guys to help me a pervious brainly tutor got it wromng two partsTo 3 decimal place

[tex]\begin{gathered} differentiating\text{ the volume with respect to t:} \\ \frac{dV}{dt}\text{ = \pi r^^b2 }\frac{dh}{dt} \\ \\ 6\text{ = \pi r^^b2}\times\frac{dh}{dt} \\ \\ diamter\text{ = 2\lparen radius\rparen} \\ radius\text{ = r = 6/2 = 3} \\ \\ 6\text{ = \pi\lparen3\rparen^^b2 }\times\frac{dh}{dt} \end{gathered}[/tex][tex]\begin{gathered} 6\text{ = 9\pi}\times\frac{dh}{dt} \\ \frac{6}{9π}\text{ = }\frac{dh}{dt} \\ \\ the\text{ rate at which the level is rising = }\frac{6}{9\pi\text{ }}\frac{in}{min} \\ \\ The\text{ rate at which the level is rising = 0.212 in/min} \end{gathered}[/tex]

b) Volume of the cone is given as:

[tex]V=\text{ }\frac{1}{3}πr²h[/tex]

To get the rate at which the level of the cone is falling, we will differentiate with respect to h:

First, we need to write r in terms of h before differentiating

from the diagram, d = h

d = 2r

As a result, h = 2r and r = h/2

[tex]\begin{gathered} Substitute\text{ for r:} \\ V\text{ = }\frac{1}{3}π(\frac{h}{2})²\times h \\ \\ V\text{ = }\frac{1}{3}π(\frac{h^2}{4})\times h\text{ = }\frac{1}{12}\pi h^3 \\ \\ differentiation\text{ with respect to h:} \\ \frac{dV}{dt}=\frac{1}{12}\pi(3h^2\frac{dh}{dt}) \\ \\ \frac{dV}{dt}=\frac{1}{4}\pi(h^2\frac{dh}{dt}) \end{gathered}[/tex][tex]\begin{gathered} \frac{dV}{dt}=\text{ }\frac{1}{4}(4^2)\pi\times\frac{dh}{dt};\text{ h = 4inches deep} \\ \\ \frac{dV}{dt}=\text{ }\frac{16\pi}{4}\times\frac{dh}{dt} \\ \\ \frac{dV}{dt}=\text{ 4}\times\pi\times\frac{dh}{dt} \\ \frac{dV}{dt}=\text{ 10 }\frac{in^3}{min} \\ But\text{ since the volume is decreasing with time, the sign will be negative} \\ \frac{dV}{dt}\text{ for the cone = -10} \\ \\ -10\text{ = 4}\times\pi\times\frac{dh}{dt} \\ \\ \frac{dh}{dt}=\text{ }\frac{-10}{4\pi}\frac{in}{min}\text{ = -0.796} \\ \end{gathered}[/tex]

The rate at which the level of the cone is falling is in/min