Answer: A & B are the same answer --> 96, max
Step-by-step explanation:
Consider m is the degree of the numerator (top) and n is the degree of the denominator (bottom). Then the horizontal asymptote (H.A.) is based on the relationship between m and n:
- If m > n, then there is no H.A.
- If m = n, then y = coefficient of numerator ÷ coefficient of denominator
- If m < n, then y = 0
In the given problem, m = 1 and n = 1 so the H.A. is:
[tex]y=\dfrac{60}{0.625}\quad \rightarrow \quad y=96[/tex]
This is the maximum number of moose that the forest can sustain at one time.
