Awater tank has the shape of an inverted right circular cone with base radius 3 meters and height 6 meters. water is being pumped into the tank at the rate of 12 meters3/sec. find the rate, in meters/sec, at which the water level is rising when the water is 2 meters deep. give 2 decimal places for your answer. type your answer in the space below. if your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35).

the rate of change is 3.8m/s

Step-by-step explanation:

The volume of the right circular cone is

The rate of change of volume is

In order to proceed further we have to define r in terms of h so tht we can compute the derivative above.

The ratio between h and r is

Therefore .

We plug that into the derivative above and get:

Thus

Now for the numerical part.

The rate of change of volume is , so when the water is 2 meters deep , therefore:

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

it is x equals 3   plese mark me really need it

step-by-step explanation: