QUESTION SIX (13 marks) a) The number of milligrams of a drug in a person's system after thours is given by the function D=20e^-0.4t i. Find the amount of the drug after 5 hours. (3mks) i. When will the amount of the drug be 0.1 milligrams? (3mks)

**a) i. Amount of drug after 5 hours:**<br /><br />To find the amount of the drug after 5 hours, we substitute *t* = 5 into the given function:<br /><br />D = 20e^(-0.4t)<br />D = 20e^(-0.4 * 5)<br />D = 20e^(-2)<br /><br />Using a calculator:<br /><br />e^(-2) ≈ 0.1353<br /><br />Therefore:<br /><br />D ≈ 20 * 0.1353 <br />D ≈ 2.706 milligrams<br /><br />**Answer:** After 5 hours, there will be approximately 2.706 milligrams of the drug in the person's system.<br /><br /><br />**a) ii. Time when the amount of drug is 0.1 milligrams:**<br /><br />To find the time when the amount of the drug is 0.1 milligrams, we set D = 0.1 and solve for *t*:<br /><br />0.1 = 20e^(-0.4t)<br /><br />Divide both sides by 20:<br /><br />0.005 = e^(-0.4t)<br /><br />Take the natural logarithm (ln) of both sides:<br /><br />ln(0.005) = ln(e^(-0.4t))<br /><br />Using the property of logarithms that ln(e^x) = x:<br /><br />ln(0.005) = -0.4t<br /><br />Solve for t:<br /><br />t = ln(0.005) / -0.4<br /><br />Using a calculator:<br /><br />ln(0.005) ≈ -5.298<br /><br />Therefore:<br /><br />t ≈ -5.298 / -0.4<br />t ≈ 13.25 hours<br /><br />**Answer:** The amount of the drug will be 0.1 milligrams after approximately 13.25 hours.<br />