A trough of water is 8 meters deep and its ends are in the shape of isosceles triangles whose width is 5 meters and height is 2 meters. If water is being pumped in at a constant rate of 6 m3Isec. At what rate is the height of the water changing when the water has a height of 120 cm?

0.3 m/s

Step-by-step explanation:

The first thing is to attach the allusive graphic to the question. Now yes, let's move on to the solution that would be:

If the through is completely filtered the its volume will:

V = l * [1/2 w * h] = 1/2 l * w * h

Now we derive with respect to time and we are left with:

dV / dt = 1/2 * l * w * dh / dt

We solve by dh / dt and we have:

dh / dt = (2 / (l * w)) * (dV / dt)

We know that l = 8 and w = 5, in addition to dV / dt = 6, we replace:

dh / dt = (2 / (8 * 5)) * (6)

dh / dt = 0.3

Therefore the rate at which the height of the water changes is 0.3 m / s

if x decreases by 4 then y increases by 10

step-by-step explanation:

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)