(1 point) Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 6 cubic feet per minute. If the pool has radius 3 feet and height 6 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 2 feet

dh/dt = 2/3*π  ft/min

Step-by-step explanation:

We have a right circular cylinder, and water is pouring at a constant rate, we must expect the height  of the water will rise a constant rate, therefore it does not matter the height

V(c) = π*r²*h     ⇒  V(c) = π*(3)²*h  ⇒ V(c) =9*π*h

DV(c)/dh  =  9*π

DV(c)/dh = DV(c)/dt  *  dt/ dh    ⇒   9*π  =   DV(c)/dt  *  dt/ dh  

dh/dt  =   DV(c)/dt /  9*π

DV(c)/dt  = 6 ft³/m     ( from problem statement )

Then

dh/dt = 6/9*π ft/min

dh/dt = 2/3*π  ft/min

the required function is .

step-by-step explanation:

from the given table it is noticed that the graph passing through (-3,7), (0,4), (3,1) and (8,-4).

it is a linear function because the value of y is decrease by one as the value of x increase by 1.

the slope of the function is

the slope of the function is -1 and the y-intercept is 4.

the slope intercept form is

where, m is slope and b is y-intercept.

therefore the required function is . the graph of this function is shown below.

answer: idk

step-by-step explanation: