Polluted water from a factory flows into a pond. The concentration of pollutant,
c, in the pond at time t minutes is modelled by the equation
c(t)=8-8000(1/1000+2t)
where c is measured in kg/m3. When will the concentration of pollutant in the
pond reach 6 kg/m3?
To find the time when the concentration of pollutant in the pond reaches 6 kg/m^3, we need to solve the equation:
c(t) = 6
Substituting the given equation for c(t):
8 - 8000(1/1000 + 2t) = 6
Simplifying the equation:
8 - 8 - 16000t = 6
-16000t = -2
Dividing both sides by -16000:
t = -2 / -16000
t = 1/8000
Therefore, the concentration of the pollutant in the pond will reach 6 kg/m^3 after 1/8000 minutes.
To find when the concentration of pollutant in the pond will reach 6 kg/m3, we need to solve the equation for t.
The equation is given as:
c(t) = 8 - 8000(1/1000 + 2t)
We set c(t) to be equal to 6 kg/m3 and solve for t:
6 = 8 - 8000(1/1000 + 2t)
Now, let's simplify the equation:
6 = 8 - 8000/1000 - 16000t
6 = 8 - 8 - 16000t
6 = -16000t
To isolate t, we divide both sides of the equation by -16000:
t = 6 / -16000
Finally, we calculate the answer:
t ≈ -0.000375 minutes
Therefore, the concentration of pollutant in the pond will reach 6 kg/m3 approximately after -0.000375 minutes. It's important to note that this negative value indicates that the concentration reduction occurred before time 0.
you want t when c=6
8-8000(1/1000+2t) = 6
8000(1/1000+2t) = 2
(1/1000+2t) = 2/8000
1/1000 + 2t = 1/4000
2t = 1/4000 - 1/1000
2t = -3/4
Bzzt
You must have meant
8-8000(1/(1000+2t)) = 6
8000(1/(1000+2t)) = 2
1/(1000+2t) = 1/4000
1000+2t = 4000
2t = 3000
t = 1500
that's just over 1 day.