Gravel is being dumped from a conveyor belt at a rate of 40 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 11 ft high? (Round your answer to two decimal places.)

0.42ft/mn

Step-by-step explanation:

we have the following information to answer this question

dv/dt = 40 feet

height = 11 ft

volume = 1/3πr²h

= 1/3π(h/2)²h

= 1/3πh³/4

= πh³/12

dv/dt = π3h²/12

= πh²/4

dh/dt = 4/πh²dv/dt

= 4(40)÷22/7(11)²

= 160/380.29

= 0.42 ft/min

The height of the pile is therefore increasing by 0.42ft/min at a height of 11 feets

idk

step-by-step explanation: i just don't

6,250

step-by-step explanation:

to get the area, multiply 360 by 160.(57,600)

to find out how many bags of fertilizer it will take, first figure out how many square feet one bag can hold. divide 1000 by 8(125) to get the unit rate(how many square feet one pound of fertilizer can hold)and multiply it by 50(6250). if i haven't made a mistake, that is the answer.