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Polluted water flows into a pond. The concentration of pollutant in the pond at the t minutes is modelled by the equation c(t)= 9-90,000(1/10000+3t), where c is measured in kilograms per cubic metre.

a) When will the concentration of pollutant in the pond reach 6 kg/m^3?

b) What will happen to the concreatration of pollutant over time?

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4 answers
  1. I have a feeling that there are some brackets in there

    e.g. is it
    c(t)= 9-90,000(1/ (10000+3t) )

    check and come back

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  2. yea there is, i forgot to add it in there.

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  3. so , assuming my guess was correct .....

    6 = 9-90,000(1/ (10000+3t) )
    -3 = - 90,000(1/ (10000+3t) )
    3(10000 + 3t) = 90000
    30000 + 9t = 90000
    9t = 60000
    t = appr 6666.7 minutes or appr 111.1 days

    for the second part look at the term
    90,000(1/ (10000+3t) )
    as t gets larger, the denominator gets larger and as a result 1/ (10000+3t) gets smaller and smaller and eventually approaches zero.
    -90,000(1/ (10000+3t) ) will get smaller and smaller and you are left with
    c(t)= 9 - (very very small)

    so the concentration will settle in at 9 kg/m^3

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  4. Is it not 4.6 days?

    For: t = appr 6666.7 minutes or appr 111.1 days

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