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Answer:
about 0.177 mg/mL
Step-by-step explanation:
The maximum is found where the derivative of C(t) is zero.
dC/dt = 1.35e^(-2.802t) -(1.35t)2.802e^(-2.802t) = 0
Solving for t gives ...
t = 1/2.802
So, the maximum C(t) is ...
C(1/2.802) = 1.35/2.802e^(-1) ≈ 0.177 . . . . . mg/mL
The maximum average BAC during the first 6 hours is about 0.177 mg/mL.
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The maximum occurs about 21 minutes 25 seconds after consumption.
