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September 22, 2014

September 22, 2014

Posted by **becky** on Friday, August 11, 2006 at 12:00pm.

An enviromental study of a certain communtiy suggests that t years from now, the average level of carbon monoxide in the air will be Q(t)=0.05t^2+0.01t+3.4 parts per million. (Read as .05t raised to the second power). By approximately how much will the carbon momoxide level change during the next 6 months?

Q'(t) = 0.1t +0.1

change in t=.5

Q(.5)=.3+0.005+3.4 =3.705

Q'(.5) =0.05+0.1=.15

Q(.5)=.5(.15)+3.705=3.78

Am I right? If not, where did I go wrong?

I would not do the deraviative.

change in six months = Q(.5) - Q(0)

= 3.705-3.4

=.305

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