Answer:
7.5 ft³/min
Step-by-step explanation:
Let x be the depth below the surface of the water. The height, h of the water is thus h = 10 - x.
Now, the volume of water V = Ah where A = area of isosceles base of trough = 1/2bh' where b = base of triangle = 4 ft and h' = height of triangle = 1 ft. So, A = 1/2 × 4 ft × 1 ft = 2 ft²
So, V = Ah = 2h = 2(10 - x)
The rate of change of volume is thus
dV/dt = d[2(10 - x)]/dt = -2dx/dt
Since dV/dt = 15 ft³/min,
dx/dt = -(dV/dt)/2 = -15 ft³/min ÷ 2 = -7.5 ft³/min
Since the height of the water is h = 10 - x, the rate at which the water level is rising is dh/dt = d[10 - x]/dt
= -dx/dt
= -(-7.5 ft³/min)
= 7.5 ft³/min
And the height at this point when x = 8 inches = 8 in × 1 ft/12 in = 0.67 ft is h = (10 - 0.67) ft = 9.33 ft
It's A. If U=16, 6+10=16
and if you plug in the numbers for the second equation it's 36=2(10)+16
