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

Just need someone to double check this problem. I just don't feel comfortable with my answer.

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

## Answer This Question

## Related Questions

- micro - hi, i need some help with my econ homework. i know what the law is, ...
- Math check - Average cost = q^2¡V15q + 48 Calculate the output level (q), which ...
- math - The average cost of manufacturing a quantity q of a good, is defined to ...
- Pre-Calc Help please? - The more you study for a certain exam, the better your ...
- social studies - I just needed some general help.................can you give me...
- Math - "Find the slope of the line passing through J(0,5) and K(-1,2)" I did the...
- Math - Hi, would someone mind just checking the answer I got for this formula ...
- DIFFICULT MATH - Okay I've been working on this for 10 hours straight and cannot...
- physics - This is a sample question: A man jogs at a speed of 1.2 m/s. His dog ...
- 9th grade algebra/Ms Sue - No, I did not check Reiny's answer that night. I did ...

More Related Questions