Answer: [tex]\begin{gathered} a)\text{ }The\text{ copper concentration at the beginning of 2033 is 3.2439 milligrams per liter} \\ \\ b)\text{ 3.4815 years after 2000} \end{gathered}[/tex]
Explanation:
Given:
[tex]\begin{gathered} C(t)\text{ = }\frac{4t\text{ + 1}}{8\text{ + t}} \\ where\text{ C\lparen t\rparen = concentration over time t} \\ t\text{ = time after year 2000} \end{gathered}[/tex]
To find:
a) the copper concentration at the beginning of the year 2033
b) the time the maximum contaminate level of 1.3 m/L will be reached
a) year 2033 = 2000 + 33
since t is the time after year 2000
t = 33
To get the copper concentration, we will substitute for t in the formula:
[tex]\begin{gathered} C(t)\text{ = }\frac{4(33)\text{ + 1}}{8\text{ + 33}} \\ C(t)\text{ = }\frac{133}{41} \\ C(t)\text{ = 3.2439} \\ \\ The\text{ copper concentration at the beginning of 2033 os 3.2439 milligrams per liter} \end{gathered}[/tex]
b) To get the time it takes to get to the maximum contaminate level of 1.3 mg/l, we will substitute 1.3 for C(t) in the function:
[tex]\begin{gathered} 1.3\text{ = }\frac{4t\text{ + 1}}{8\text{ + t}} \\ 1.3(8\text{ + t\rparen = 4t + 1} \\ 1.3(8)\text{ + 1.3\lparen t\rparen = 4t + 1} \\ 10.4\text{ + 1.3t = 4t + 1} \\ 10.4\text{ - 1 = 4t - 1.3t} \\ 9.4\text{ = 2.7t} \end{gathered}[/tex]
divide both sides by 2.7:
[tex]\begin{gathered} \frac{9.4}{2.7}\text{ = t} \\ t\text{ = 3.4815} \end{gathered}[/tex]
