Answer:
3.4625
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