Solve the problem.
-The water is drained from a tank by opening a valve at the bottom of the tank. The depth of the water in the tank thours after the valve is opened is $d(t)=6\left(1-\frac{t}{12}\right)^{2}$ meters. If the valve is opened at $1 \mathrm{pm}$, what is the depth of water at $10 \mathrm{pm}$ ?
A) 2.62 meters
B) 1.50 meters
C) 0.38 meters
D) 0.17 meters

Question

Quizplus · Accepted Answer

The valve is opened at 1 pm, so at 10 pm, 9 hours have passed. Substitute $t = 9$ into the equation $d(t) = 6\left(1-\frac{t}{12}\right)^{2}$ to find the depth: $d(9) = 6\left(1-\frac{9}{12}\right)^{2} = 6\left(\frac{1}{4}\right)^{2} = 6 \times \frac{1}{16} = 0.375$ meters, which rounds to 0.38 meters.

Solve the Problem d(t)=6(1−t12)2d(t)=6\left(1-\frac{t}{12}\right)^{2}d(t)=6(1−12t​)2 Meters If the Valve Is Opened At from a Tank

Solve the Problem $d(t)=6\left(1-\frac{t}{12}\right)^{2}$ Meters If the Valve Is Opened At from a Tank