A(t) = 10(.08)^t A drug is eliminated from the body through urine. Suppose that

Question

A(t) = 10(.08)^t A drug is eliminated from the body through urine. Suppose that for an initial dose of 10 milligrams, the amount A(t)in the body t hours later is given by the equation above. A) Estimate the amount of the drug in the body 8 hoursafter the initial dose. B) What percentage of the drug still in the body iseliminated each hour? Please show all steps A(t) = 10(.08)^t A drug is eliminated from the body through urine. Suppose that for an initial dose of 10 milligrams, the amount A(t)in the body t hours later is given by the equation above. A) Estimate the amount of the drug in the body 8 hoursafter the initial dose. B) What percentage of the drug still in the body iseliminated each hour? Please show all steps

Explanation / Answer

So A(t) = 10(0.8)^8 = 10*0.16777 = 1.6777 mg left in thebody
B) After each hour, 80% of the drug is left. We can look at this a few ways: i)After one hour, the amount goes from 10 mg to 8 mg, 80% ofthe original
ii)In t hours, the equation is A(t) = 10*0.8*0.8*0.8....wherewe multiply by 0.8 t times. 0.8 = 80%, meaning that after each hourthe amount is mutliplied by 80%.

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