An inverted conical tank starts the day with 250 ft^3 of crayon wax in it. As the factory commences work, the tank is filled with an additional 40 ft^3 of wax per minute. The height of the wax is modeled by H(V)=3 piV/25. A. Write a function , V(t) to model the volume of wax in the tank after t minutes. B. Find an expression for the composition (HoV)(t) C. The composition in B (above) can be described as the of the wax in terms of

A. V(t) = 40t + 250 B. (HoV)(t) = 24πt/5 + 30π C. The composition in B (above) can be described as the height of the wax in terms of time.

Step-by-step explanation:

A. Let the rate of change of volume V with respect to time be dV/dt = 40 ft³/min

Solving this, V = 40t + C. At the start of the day, that is t = 0, V = 250 ft³

Substituting these values, we have

250 ft³ = 40(0) + C

C = 250 ft³

So, V(t) = 40t + 250

B. Since H(V) = 3πV/25

(HoV)(t) = 3π(40t + 250)/25

= 24πt/5 + 30π

C. The composition in B (above) can be described as the height of the wax in terms of time.