An environmental study of a certain community suggests that t years from now, the average level of carbon monoxide in the air will be Q(t)=0.05t^2 + 0.1t + 3.4 parts per million. By approximately how much will the carbon monoxide lvel change during the coming 6 months?

3.4625

Step-by-step explanation:

Given the equation:

Q(t)=0.05t^2 + 0.1t + 3.4 parts per million

Where t = time in years

By approximately how much will the carbon monoxide lvel change during the coming 6 months?

t = 6 months = 6/12 = 0.5 year

Q(0.5) = 0.05(0.5)^2 + 0.1(0.5) + 3.4 parts per million

Q(0.5) = 0.05(0.25) + 0.05 + 3.4

Q(0.5) = 0.0125 + 0.05 + 3.4

Q = 3.4625

Carbon monoxide level will change by approximately 3.4625 parts per million in 6 months

relating to or denoting a system of numbers and arithmetic based on the number ten, tenth parts, and powers of ten.

step-by-step explanation:

so what is the problem what are they asking you to solve

step-by-step explanation:

so what is the problem what are they asking you to solve