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.