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Sagot :
To answer this question, we will use the following general form of an exponential model:
[tex]T=P(1\pm r)^t,[/tex]where P is the initial amount, r is the rate in decimal form, and t is the time.
In this case, since the herbicide degrades, the sign inside the parenthesis will be a minus sign, P=200, r=0.11, and t will be the time in weeks, therefore the number of gallons after t weeks can be modeled by the following equation:
[tex]T=200(1-0.11)^t.[/tex]Evaluating the above equation at t=2, we get:
(Answer part 1)
[tex]T=200(1-0.11)^2=158.42.[/tex]If we set T=50, and solve for t, we get:
(Answer part 2)
[tex]\begin{gathered} 50=200(1-0.11)^t, \\ \frac{50}{200}=(0.89)^t, \\ \frac{1}{4}=0.89^t, \\ \ln (\frac{1}{4})=t\ln 0.89, \\ t=\frac{\ln (\frac{1}{4})}{\ln 0.89}\approx12. \end{gathered}[/tex]Answer:
There will be 158 gallons after 2 weeks.
In about 12 weeks the landscaper has to put another dose.
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