Polluted water flows into a pond. The concentration of pollutant, c, in the pond at time, t, minutes is modeled by the equation c(t) = 9 - 90 000 (1/ 10 000 + 3t)
a) When will the concentration of pollutant in the pound reach 6kg/m^3?
I suspect you meant:
c(t) = 9 - 90 000 (1/ (10 000 + 3t) )
I will also assume that c(t) is in kg/m^3
then
6 = 9 - 90 000 (1/ (10 000 + 3t))
90000(1/(10000+3t)) = 3
30000 + 9t = 90000
9t = 60000
t = 6666 2/3 minutes ?
check my arithmetic
To find the time at which the concentration of pollutant in the pond reaches 6 kg/m^3, we can plug this concentration value into the equation c(t) and solve for t.
Given the equation: c(t) = 9 - 90,000 (1/10,000 + 3t)
Replace c(t) with 6: 6 = 9 - 90,000 (1/10,000 + 3t)
Now we can solve for t:
6 = 9 - 90,000 (0.0001 + 3t)
6 = 9 - 9 + 3t
6 = 3t
t = 6/3
t = 2
Therefore, the concentration of pollutant in the pond will reach 6 kg/m^3 at 2 minutes.