Answer :
Approximately 19.9919 pounds of bananas are removed from the display table during the first 2 hours the store is open.
The function has a constant term 10, which represents the minimum rate of banana removal even if no customers are present.
The second function is g(t), which models the rate at which bananas are added to the display table.
Now, to answer the question, we need to find out how many pounds of bananas are removed from the display table during the first 2 hours. We can do this by integrating the function f(t) from 0 to 2, which gives us the total amount of bananas removed during this time interval. Using the integral notation, we have:
Total amount of bananas removed = ∫₀² f(t) dt
Using the formula for f(t), we can evaluate this integral as follows:
∫₀² [10 + (0.81)sin(100t)] dt = [10t - (0.81/100)cos(100t)] from 0 to 2
Plugging in the limits of integration, we get:
[10(2) - (0.81/100)cos(200)] - [10(0) - (0.81/100)cos(0)] = 20 - 0.0081 ≈ 19.9919 pounds
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Complete Question:
When a certain grocery store opens, it has 50 pounds of bananas on a display table. Customers remove bananas from the display table at a rate modeled by
f(t) = 10 + (0.81)sin 100 for 0
where f(t) is measured in pounds per hour and is the number of hours after the store opened. After the store has been open for three hours, store employees add bananas to the display table at a rate modeled by
8(t) = 3 + 2.4 In (12 + 2+) for 3
where g(t) is measured in pounds per hour and is the number of hours after the store opened.
(a) How many pounds of bananas are removed from the display table during the first 2 hours the store is open?
