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Yuki888 [10]
4 years ago
11

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Mathematics
2 answers:
Ostrovityanka [42]4 years ago
6 0

Answer:

D

Step-by-step explanation: i got that ? and i got it right actually not kidding!

Lilit [14]4 years ago
4 0

Answer:  A & B are the same answer --> 96, max

<u>Step-by-step explanation:</u>

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:

y=\dfrac{60}{0.625}\quad \rightarrow \quad y=96

This is the maximum number of moose that the forest can sustain at one time.

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Which expression is equivalent to m^2-13m-30?
sergiy2304 [10]

Answer: First option

Step-by-step explanation:

You have the quadratic equation given in the problem:

m^2-13m-30

To find an equivalent expression you cacn factorize. Find two numbers whose sum is -13 and whose product is -30.

These numbers would be -15 and 2.

Therfore, you obtain the following equivalent expression:

(m-15)(m+2)

If you don't want to apply the method above, you can use the quadratic formula:

x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}

Where:

a=1\\b=-13\\c=-30

When you susbstitute values you obtain that:

x=15\\x=-2

Then you can rewrite the equation as (m-15)(m+2)

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