Question

Question 10 (1 point)

Answer 1

Answer:

From the values obtained, we can see that after the initial 10mg/L values obtained in the first 1 and 2 minutes, the concentration has been dipping and it will continue to do so.

Step-by-step explanation:

The concentration monitored ar time, t > 0 is represented by :

C(t) = 30t / (t² + 2.)

At, t = 1

C(1) = 30(1) / (1 + 2) = 30/3 = 10

At t = 2

C(2) = 30(2) / (2² + 2) = 60/(4+2) = 60/6 = 10

At t = 3

C(3) = 30(3) / (3² + 2) = 90/ 11 = 8.18

At t = 4

C(4) = 30(4) / (4²+2) = 120/18 = 6.67

At t = 5

C(5) = 30(5) / (5²+2) = 150/ 27 = 5.55

At t = 10

C(10) = 30(10) / (10²+2) = 300/102 = 2.94

From the values obtained, we can see that after the initial 10mg/L values obtained in the first 1 and 2 minutes, the concentration has been dipping and it will continue to do so.