A. Let x represent the amount of air in the lungs at time t in seconds. Then, the function that models the amount of air in the lungs is:
x(t) = 0.08 + (0.82 - 0.08) * (1 - cos(πt/4))/2
B. Substitute t = 5.5 into the function to determine the amount of air in the lungs at 5.5 seconds:
x(5.5) = 0.08 + (0.82 - 0.08) * (1 - cos(π * 5.5 / 4))/2
x(5.5) = 0.08 + 0.74 * (1 - cos(5.5π/4))/2
x(5.5) = 0.08 + 0.37 * (1 - cos(11π/8))
x(5.5) = 0.08 + 0.37 * (1 - (-sqrt(2)/2))
x(5.5) = 0.08 + 0.37 * (1 + sqrt(2)/2)
x(5.5) = 0.08 + 0.37 + 0.228
x(5.5) = 0.678 liters
Therefore, the amount of air in the lungs at 5.5 seconds is 0.678 liters.
An average seated adult breathes in and out every 4 seconds. The average minimum amount of air in the lungs is 0.08 liter, and the average maximum amount of air in the lungs is 0.82 liter. Suppose the lungs have a minimum amount of air at t = 0, where t is time in seconds.
A. Write a function that models the amount of air in the lungs.
B. Determine the amount of air in the lungs at 5.5 seconds.
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