A box is partially filled with liquid. The length of the box is 11 inches. The width of the box is 17 inches. If the volume of the liquid is increasing at a rate of 562 cubic inches per minute, what is the rate, in inches per minute, at which the height of the liquid is changing when the height of the liquid is 4 inches?

dh/dt = 3.01 inches/minute

The height of the liquid is changing at the rate of 3.01 inches/minute

Step-by-step explanation:

Volume of liquid in the box is ;

V = length × width × height

V = lbh 1

If the volume of the liquid is increasing at a rate of 562 cubic inches per minute.

dV/dt = 562 cubic inches per minute

Length l = 11 inches

Width b = 17 inches

From equation 1; differentiating the equation we have;

dV/dt = bh(dl/dt) + lh(db/dt) + lb(dh/dt) 2

Since the length and width of water is not changing then;

dl/dt = db/dt =0

Equation 2 becomes;

dV/dt = lb(dh/dt)

dh/dt = (dV/dt)/lb

Substituting the values;

dh/dt = 562/(11×17) inch/minute

dh/dt = 3.01 inches/minute

The height of the liquid is changing at the rate of 3.01 inches/minute

add a pic of the graph, then ill edit this "answer" and solve it for you : 3

step-by-step explanation:

x² + 11x + 18 < 0  

[-11 ± sqrt(121 - 72)]/2 = (-11 ± 7)/2  

solution: (-9, -2)

step-by-step explanation: