In Section 5.1, the rate of change in the concentration of a drug was modeled as 1.701(0.841*) r(x) = (0.11872x - 3.5858

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answerhappygod
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In Section 5.1, the rate of change in the concentration of a drug was modeled as 1.701(0.841*) r(x) = (0.11872x - 3.5858

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In Section 5 1 The Rate Of Change In The Concentration Of A Drug Was Modeled As 1 701 0 841 R X 0 11872x 3 5858 1
In Section 5 1 The Rate Of Change In The Concentration Of A Drug Was Modeled As 1 701 0 841 R X 0 11872x 3 5858 1 (63.28 KiB) Viewed 69 times
In Section 5.1, the rate of change in the concentration of a drug was modeled as 1.701(0.841*) r(x) = (0.11872x - 3.5858 when 0 ≤ x ≤ 20 when 20 < x≤ 29 where r was measured in µg/mL per day and x is the number of days after the drug was administered. Evaluate the following definite integrals and interpret the answers. (Round your answers to three decimal places.) (a) √²0 r(x r(x) dx µg/mL Interpret your answer. The concentration of the drug ---Select--- by µg/mL during the first 20 days after the drug was administered. [2³ mo r(x) dx μg/mL Interpret your answer. The concentration of the drug ---Select--- by μg/mL between the end of the 20th day and the end of the 29th day after the drug was administered. 1.²⁹ r(x) dx µg/mL Interpret your answer. At the end of the 29th day after the drug was first administered, the concentration of the drug was μg/mL. Read It (b) (c) Need Help?
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