Water is leaking out of a container at a rate of -50(t-10) milliliters per hour (where t is the number of hours). How many milliliters of water leak out of the container between t=0 and t=2 ? Choose 1 answer: (A) 100 (B) 400 (C) 500 (D) 900

Understand the rate: Understand the rate of water leakage. The rate of water leakage is given by the function -50(t-10) milliliters per hour. This means that the rate of leakage depends on the time t, and the negative sign indicates that water is leaking out of the container.Calculate total leakage: Calculate the total leakage between t=0 and t=2. To find the total amount of water leaked, we need to integrate the rate of leakage from t=0 to t=2. The integral of the rate function will give us the total volume leaked over that time period.Set up integral: Set up the integral for the leakage from t=0 to t=2. The integral of the rate function -50(t-10) from t=0 to t=2 is: ∫ from 0 to 2 of -50(t-10) dtCalculate integral: Calculate the integral. ∫ from 0 to 2 of -50(t-10) dt = [-50 * (1/2 * t^2 - 10t)] from 0 to 2 = [-25t^2 + 500t] from 0 to 2Evaluate at bounds: Evaluate the integral at the upper and lower bounds. First, evaluate at t=2: -25(2^2) + 500(2) = -25(4) + 1000 = -100 + 1000 = 900 Then, evaluate at t=0: -25(0^2) + 500(0) = 0Find total leakage: Find the difference between the upper and lower evaluations to get the total leakage. Total leakage = 900 - 0 = 900 milliliters

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Water is leaking out of a container at a rate of
-50(t-10) milliliters per hour (where
t is the number of hours).
How many milliliters of water leak out of the container between
t=0 and
t=2 ?
Choose 1 answer:
(A)
100
(B) 400
(C) 500
(D) 900

Water is leaking out of a container at a rate of $-50(t-10)$ milliliters per hour (where $t$ is the number of hours). $\newline$ How many milliliters of water leak out of the container between $t=0$ and $t=2$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\mathbf{1 0 0}$ $\newline$ (B) $400$ $\newline$ (C) $500$ $\newline$ (D) $900$

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Q. Water is leaking out of a container at a rate of

-50(t-10)

milliliters per hour (where

t

is the number of hours).

\newline

How many milliliters of water leak out of the container between

t=0

and

t=2

\newline

Choose

1

answer:

\newline

(A)

\mathbf{1 0 0}

\newline

(B)

400

\newline

(C)

500

\newline

(D)

900

Understand the rate: Understand the rate of water leakage. $\newline$ The rate of water leakage is given by the function $-50(t-10)$ milliliters per hour. This means that the rate of leakage depends on the time $t$ , and the negative sign indicates that water is leaking out of the container.
Calculate total leakage: Calculate the total leakage between $t=0$ and $t=2$ . To find the total amount of water leaked, we need to integrate the rate of leakage from $t=0$ to $t=2$ . The integral of the rate function will give us the total volume leaked over that time period.
Set up integral: Set up the integral for the leakage from $t=0$ to $t=2$ . The integral of the rate function $-50(t-10)$ from $t=0$ to $t=2$ is: $\int_{0}^{2} -50(t-10) \, dt$
Calculate integral: Calculate the integral. $\newline$ $\int_{0}^{2} -50(t-10) \, dt = [-50 \times (\frac{1}{2} \times t^2 - 10t)]$ from $0$ to $2$ $\newline$ = [ $-25$ t^ $2$ + $500$ t] from $0$ to $2$
Evaluate at bounds: Evaluate the integral at the upper and lower bounds. $\newline$ First, evaluate at $t=2$ : $\newline$ $-25(2^2) + 500(2) = -25(4) + 1000 = -100 + 1000 = 900$ $\newline$ Then, evaluate at $t=0$ : $\newline$ $-25(0^2) + 500(0) = 0$
Find total leakage: Find the difference between the upper and lower evaluations to get the total leakage. $\newline$ Total leakage = $900 - 0 = 900$ milliliters