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Rashid [163]
3 years ago
7

Help?? (Ignore the answer choice I selected did that on accident)

Mathematics
1 answer:
beks73 [17]3 years ago
7 0
You have written the answer is it correct or not
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Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min, and its coarseness is such that it forms a pile in the shap
11Alexandr11 [23.1K]

Answer:

0.25 feet per minute

Step-by-step explanation:

Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min. Since we are told that the shape formed is a cone, the rate of change of the volume of the cone.

\dfrac{dV}{dt}=20$ ft^3/min

\text{Volume of a cone}=\dfrac{1}{3}\pi r^2 h

Since base diameter = Height of the Cone

Radius of the Cone = h/2

Therefore,

\text{Volume of the cone}=\dfrac{\pi h}{3} (\dfrac{h}{2}) ^2 \\V=\dfrac{\pi h^3}{12}

\text{Rate of Change of the Volume}, \dfrac{dV}{dt}=\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}

Therefore: \dfrac{3\pi h^2}{12}\dfrac{dh}{dt}=20

We want to determine how fast is the height of the pile is increasing when the pile is 10 feet high.

h=10$ feet$\\\\\dfrac{3\pi *10^2}{12}\dfrac{dh}{dt}=20\\25\pi \dfrac{dh}{dt}=20\\ \dfrac{dh}{dt}= \dfrac{20}{25\pi}\\ \dfrac{dh}{dt}=0.25$ feet per minute (to two decimal places.)

When the pile is 10 feet high, the height of the pile is increasing at a rate of 0.25 feet per minute

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