First, we need tot find a general expression for the amount of caffeine remaining in the body after certain time. As the problem states that every hour x percent of caffeine leaves the body, we must substract that percentage from the initial quantity of caffeine, by each hour passing. That expression would be:

$A= C_{0}(1-x)^t\\t: time \ in \ hours\\x: percentage \ of \ caffeine\ metabolized\\$

Then, to find the amount of caffeine metabolized per hour, we need to differentiate the previous equation. Following the differentiation rules we get:

$\frac{dA}{dt} =C_{0}(1-x)^t \ln (1-x)\\\frac{dA}{dt} =100*0.88\ln(0.88)\\\frac{dA}{dt} =-11.25 \frac{mg}{h}$

The rate is negative as it represents the amount of caffeine leaving the body at certain time.

3 0

3 years ago

Sawyer’s mom paid $1.50 for a 3/4 pound box of cereal. How much is that per pound? Question 4 options: $1.50 $1.75 $2.00 $2.50

Lynna [10]

The answer is $2.00, cause it's 50¢ a quarter.

8 0

3 years ago

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