Acup of coffee contains 130 mg of caffeine. if caffeine is eliminated from the body at a rate of 11% per hour, how much caffeine will remain in the body after 3 hours? write the exponential function and solve. a. 4072-06-01-04-00_files/i0260000.jpg; 184 mg b. 4072-06-01-04-00_files/i0260001.jpg; about 346 mg c. 4072-06-01-04-00_files/i0260002.jpg; less than 1 mg d. 4072-06-01-04-00_files/i0260003.jpg; about 92 mg

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y(t) = a*e^(kt)
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now:
130 mg
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one hour from intake:
130 * (1 - 0.11) = 115.7 mg
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y(1) = 115.7 = 130*e^(k(1))
115.7 = 130*e^k
115.7/130 = e^k
ln( 115.7/130 ) = ln( e^k )
ln( 115.7/130 ) = k
k = -0.1165338

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three hours from intake:
y(3) = 130*e^(3k)
y(3) = 91.64597
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three hours after drinking a cup of coffee, 91.6 mg of caffeine will remain in the body

Answer with explanation:

part a:

the x-intercept of the graph represents that there was no profit when the price of the eraser was $ 8 and also there will be no profit when no eraser is sold.

and the maximum value of the graph represents that there was a maximum profit of $ 2760 obtained by the company when the company sold erasers at a cost of $ 4 .

( since, the vertex of the graph is at (4,270) )

the interval where the function is increasing is: (-∞,4)

this represents that the profit is increasing when the erasers are sold at less than $ 4

and the function is decreasing in the interval (4,∞)

this represents that the profit is decreasing when the erasers are sold at a price greater than $ 4.

part b:

the average rate of change from x=1 to x=4 is calculated by:

 

it represents that the profit of the company is increasing at the rate of $ 50 per increase in the prize of the eraser.