Water is leaking out of an inverted conical tank at a rate of 11600.011600.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. the tank has height 13.013.0 meters and the diameter at the top is 5.05.0 meters. if the water level is rising at a rate of 22.022.0 centimeters per minute when the height of the water is 4.04.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute

Win=436584.2477 cm3/min

Step-by-step explanation:

Volume for a conical tank

V=π/3 r^2 h

We need a relation between r and h

r/h=252.5/1301.3=0.194 --> r=0.194h

sust. in Volume equation

V=π/3 (0.194r)^2 h=0.0394h^3

taking derivative with time respect in both sides

dV/dt=3(0.0394) h^2 dh/dt    and we know when h=404cm, dh/dt=22.022 cm/min

dV/dt=3(0.0394)(404)^2 (22.022)=424984.2477 cm^3/min

Also, we know

dV/dt=Win-Wout and Wout=Win-11600.01160

Finally

Win=424984.2477+11600.0116

Answer

Win=436584.2477 cm3/min

answer:  

step-by-step explanation:

here x represents the height of the pole,

and, the height of the shadow of the pole is 12 feet.

also, the angle of elevation from the top of the pole is 40°,

therefore,

by the low of sine,

which is the required equation for finding the value of x,

ιт ιѕ (9χ^5+10)(81χ^10-90χ^5+100)

step-by-step explanation:

нσρє ι ¢συℓ∂ нєℓρ уσυ συт тσ∂α