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January 28, 2015

January 28, 2015

Posted by **Ann** on Friday, August 6, 2010 at 10:18pm.

The cost, in millions of dollars, to remove x % of pollution in a lake modeled by the equation C=6,000/200-2x^2

What is the cost to remove 20% of the pollutant?

Please help! Thanks!

- Axia Math 117 -
**Reiny**, Friday, August 6, 2010 at 11:26pmFirst of all, I don't think you meant to type the equation as you did, or else why not reduce it to

C = 3000 - 2x^2

So I will assume you meant

C = 6000/(200-2x^2)

Secondly, the way you defined x % , we would have to replace x with 20, and then your result would be

C = 6000/(200 - 800) which will result in a negative cost, thus making no sense.

If we replace 20% with 0.2, we would get

C = 6000/(200 - .08) = 30.012

but the cost of removing 90% would then be

6000/(200 - 1.62) = 30.24

which is only slightly higher than removing 20%.

This does not make sense to me, so please check the typing of the equation, and the definition of x

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