Question

If a factory continuously pumps pollutants into the air at the rate of the quotient of the square root of t and 15 tons per day, then the amount dumped after 3 days is?

Answer 1

Given

`D(t)=(10t)/(1+2t)`

To find how much waste will the pumping station deliver during the 8 hour period.

Step-by-step explanation:

It is given that,

`\begin{equation*} D(t)=(10t)/(1+2t) \end{equation*}`

Then for t=1,

`\begin{gathered} D(1)=(10(1))/(1+2(1)) \\ =(10)/(3) \\ =3.33 \end{gathered}`

For t=2,

`\begin{gathered} D(2)=(10(2))/(1+2(2)) \\ =(20)/(5) \\ =4 \end{gathered}`

For t=3,

`\begin{gathered} D(3)=(10(3))/(1+2(3)) \\ =(30)/(7) \\ =4.29 \end{gathered}`

For t=4,

`\begin{gathered} D(4)=(10(4))/(1+2(4)) \\ =(40)/(9) \\ =4.44 \end{gathered}`

For t=5,

`\begin{gathered} D(5)=(10*5)/(1+2(5)) \\ =(50)/(11) \\ =4.55 \end{gathered}`

For t=6,

`\begin{gathered} D(6)=(10(6))/(1+2(6)) \\ =(60)/(13) \\ =4.62 \end{gathered}`

For t=7,

`\begin{gathered} D(7)=(10(7))/(1+2(7)) \\ =(70)/(15) \\ =4.67 \end{gathered}`

For t=8,

`\begin{gathered} D(8)=(10(8))/(1+2(8)) \\ =(80)/(17) \\ =4.71 \end{gathered}`

Then,

`\begin{gathered} Total\text{ waste}=3.33+4+4.29+4.44+4.55+4.62+4.67+4.71 \\ =34.61 \end{gathered}`

Hence, the total waste delivered is 34.61.