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Sagot :
Given that the population can be represented by the equation;
[tex]P(t)=\frac{2t^2+75}{2t^2+150}[/tex]The current population (Initial population) is the population at time t=0;
Substituting;
[tex]t=0[/tex][tex]\begin{gathered} P(0)=\frac{2t^2+75}{2t^2+150}=\frac{2(0)^2+75}{2(0)^2+150}=\frac{75}{150} \\ P(0)=0.5\text{ million} \end{gathered}[/tex]Therefore, the current population of the habitat is;
[tex]0.5\text{ million}[/tex]The long term population would be the population as t tends to infinity;
[tex]\begin{gathered} \lim _{t\to\infty}P(t)=\frac{2t^2+75}{2t^2+150}=\frac{2(\infty)^2+75}{2(\infty)^2+150}=\frac{\infty}{\infty} \\ \lim _{t\to\infty}P(t)=\frac{4t}{4t}=1 \end{gathered}[/tex]Therefore, the long term population of the habitat is;
[tex]P(\infty)=1\text{ million}[/tex]
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