pollution is removed from a lake at a rate of $f(t)$ kg/day on day $t$. (a) explain the meaning of the…

pollution is removed from a lake at a rate of $f(t)$ kg/day on day $t$. (a) explain the meaning of the statement $f(12)=500$. on day choose one pollution is removed from the lake at a rate of choose one kg/day. (b) if $int_{5}^{15}f(t)dt = 4000$, give the units of the 5, the 15, and the 4000. 5 choose one 15 choose one 4000 choose one (c) give the meaning of $int_{5}^{15}f(t)dt = 4000$. from day 5 to day 15, pollution is removed at a rate of choose one and the amount of pollution removed is choose one

pollution is removed from a lake at a rate of $f(t)$ kg/day on day $t$. (a) explain the meaning of the statement $f(12)=500$. on day choose one pollution is removed from the lake at a rate of choose one kg/day. (b) if $int_{5}^{15}f(t)dt = 4000$, give the units of the 5, the 15, and the 4000. 5 choose one 15 choose one 4000 choose one (c) give the meaning of $int_{5}^{15}f(t)dt = 4000$. from day 5 to day 15, pollution is removed at a rate of choose one and the amount of pollution removed is choose one

Answer

Explanation:

Step1: Interpret function value

The function $f(t)$ gives the rate of pollution - removal in kg/day on day $t$. When $t = 12$ and $f(12)=500$, it means on day 12, pollution is removed from the lake at a rate of 500 kg/day.

Step2: Analyze integral units

The integral $\int_{a}^{b}f(t)dt$ represents the net - change in the amount of pollution removed. The variable $t$ is in days, and $f(t)$ is in kg/day. For $\int_{5}^{15}f(t)dt = 4000$, the limits 5 and 15 are in days, and since $f(t)$ has units of kg/day and we are integrating with respect to $t$ (in days), the result 4000 has units of kg.

Step3: Interpret the integral

The integral $\int_{5}^{15}f(t)dt$ represents the total amount of pollution removed from day 5 to day 15. Since $\int_{5}^{15}f(t)dt = 4000$, it means from day 5 to day 15, the total amount of pollution removed is 4000 kg.

Answer:

(a) On day 12, pollution is removed from the lake at a rate of 500 kg/day. (b) 5: days; 15: days; 4000: kg (c) From day 5 to day 15, the total amount of pollution removed is 4000 kg.