A water trough is 9 m long and has a cross-section in the shape of an isosceles trapezoid that is 20 cm wide at the bottom, 70 cm wide at the top, and has height 50 cm. If the trough is being filled with water at the rate of 0.2 m3/min how fast is the water level rising when the water is 20 cm deep?

dv = surface area * dh

so

dv/dt = surface area * dh/dt

width at surface = 40 + (80-40)(30/40)

= 40 + 30 = 70 cm = 0.70 m

so

surface area = 9 * .7 = 6.3 m^2

so

.3 m^3/min = 6.3 m^2 * dh/dt

and

dh/dt = .047 meters/min or 4.7 cm/min

Step-by-step explanation:

answer: the equation matches the regression

step-by-step explanation:

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680