a conical tank with vertex down is 10 feet across the top and 12 feet deep. If water is flowing out of the tank at a rate of 12 cubic feet per minute, find the rate at which the water level is dropping when radius of water level is 4ft.

The water level is dropping at a rate of 0.24 ft/s

Step-by-step explanation:

Here, we simply want to calculate the change in depth (height of the cone) , given the volume change

Mathematically, we have the volume of a cone as;

V = 1/3 * π * r^2 * h

we are given dv/dt as 12 ft^3/m

dv/dh = 1/3 * π * r^2

Substituting the value for the radius, we have

dv/dh = 1/3 * 22/7 * 4^2 = 50.29

dh/dt = dh/dv * dv/dt

dh/dv = 1/(dv/dh) = 1/50.29

Thus,

dh/dt = 1/50.29 * 12

dh/dt = 0.24 ft/s

i think there was a typo. you wrote pat brings 3 quarts of ice-cream to the party. each serving of ice-cream is 1 6 quart. how many total servings of ice-cream does pat bring to the party? i think the bold is supposed to be 1/6. so 1 quart of ice cream = 6 servings. 6 * 3 = 18.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15