Question 397720
We have {{{P(t) = 50(1+.5t)/(2+.01t) = (50 + 25t)/(2 + .01t)}}}


A: Just replace t = 0 to obtain {{{P(0) = 25}}}.


B: Replace t with 60 (since 60 months equals 5 years)


{{{P(60) = (50 + 25(60))/(2 + .01(60)) = 596.1538}}} (round to 596)


C: A horizontal asymptote occurs when {{{lim(x->infinity, (50 + 25t)/(2 + .01t))}}} is a finite number. In this case, {{{lim(x->infinity, (50 + 25t)/(2 + .01t)) = 25/.01 = 2500}}}, so 2500 is the maximum attainable population.