A water tank is drained. The following function gives the volume, in liters, of the water remaining in the tank t minutes after the drain is opened: V(t)=3000(1-0.05 t)^(2) What is the instantaneous rate of change of the volume after 10 minutes? Choose 1 answer: (A) - 150 liters per minute (B) -150 minutes per liter (C) 750 liters per minute D 750 minutes per liter

Find Derivative of V(t): We need to find the derivative of V(t) to get the rate of change. V(t) = 3000(1 - 0.05t)^2 Let's find V'(t), the derivative of V with respect to t.Apply Chain Rule: Using the chain rule, V'(t) = 2 * 3000(1 - 0.05t) * -0.05 V'(t) = -300 * (1 - 0.05t)Substitute t = 10: Now we substitute t = 10 into V'(t) to find the instantaneous rate of change at t = 10 minutes. V'(10) = -300 * (1 - 0.05 * 10)Simplify Expression: Simplify the expression. V'(10) = -300 * (1 - 0.5) V'(10) = -300 * 0.5Calculate Final Value: Calculate the final value. V'(10) = -150

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A water tank is drained. The following function gives the volume, in liters, of the water remaining in the tank
t minutes after the drain is opened:

V(t)=3000(1-0.05 t)^(2)
What is the instantaneous rate of change of the volume after 10 minutes?
Choose 1 answer:
(A) -
150 liters per minute
(B) -150 minutes per liter
(C) 750 liters per minute
D 750 minutes per liter

A water tank is drained. The following function gives the volume, in liters, of the water remaining in the tank $t$ minutes after the drain is opened: $\newline$ $V(t)=3000(1-0.05 t)^{2}$ $\newline$ What is the instantaneous rate of change of the volume after $10$ minutes? $\newline$ Choose $1$ answer: $\newline$ (A) $-150$ liters per minute $\newline$ (B) $-150$ minutes per liter $\newline$ (C) $750$ liters per minute $\newline$ (D) $750$ minutes per liter

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Q. A water tank is drained. The following function gives the volume, in liters, of the water remaining in the tank

t

minutes after the drain is opened:

\newline

V(t)=3000(1-0.05 t)^{2}

\newline

What is the instantaneous rate of change of the volume after

10

minutes?

\newline

Choose

1

answer:

\newline

(A)

-150

liters per minute

\newline

(B)

-150

minutes per liter

\newline

(C)

750

liters per minute

\newline

(D)

750

minutes per liter

Find Derivative of V(t): We need to find the derivative of $V(t)$ to get the rate of change. $V(t) = 3000(1 - 0.05t)^2$ Let's find $V'(t)$ , the derivative of $V$ with respect to $t$ .
Apply Chain Rule: Using the chain rule, $V'(t) = 2 \times 3000(1 - 0.05t) \times -0.05$ $\newline$ $V'(t) = -300 \times (1 - 0.05t)$
Substitute $t = 10$ : Now we substitute $t = 10$ into $V'(t)$ to find the instantaneous rate of change at $t = 10$ minutes. $\newline$ $V'(10) = -300 \times (1 - 0.05 \times 10)$
Simplify Expression: Simplify the expression. $\newline$ $V'(10) = -300 \times (1 - 0.5)$ $\newline$ $V'(10) = -300 \times 0.5$
Calculate Final Value: Calculate the final value. $V'(10) = -150$