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Evgen [1.6K]
3 years ago
10

Suppose that 40 deer are introduced in a protected wilderness area. The population of the herd

Mathematics
1 answer:
Scorpion4ik [409]3 years ago
8 0

The horizontal asymptote of the function is the minimum number of deer in the area.

  • The equation of horizontal asymptote is: \mathbf{y = 40}
  • The horizontal asymptote means that, the number of deer will never be less than 40

The equation is given as:

\mathbf{f(x) = \frac{40 + 2x}{1 + 0.05x}}

Expand the numerator

\mathbf{f(x) = \frac{40(1 + 0.05x)}{1 + 0.05x}}

Cancel out the common factor

\mathbf{f(x) = 40}

Hence, the equation of horizontal asymptote is:

\mathbf{y = 40}

The horizontal asymptote means that, the number of deer will never be less than 40

Read more about horizontal asymptotes at:

brainly.com/question/4084552

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Answer:

2

Step-by-step explanation:

-7 - (-8) = 1

1 + (-3) = -2

-2 + 6 = 4

4 - 2 = 2

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What is the solution set? m-18<36
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Answer:

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Add 18 to both sides.

  m -18 +18 ≤ 36 +18

  m ≤ 54 . . . . . simplify

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What integer multiplied by negative 4 gives negative 32
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8 multiplied by -4 gives -32
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45% of the pizzas in the freezer in the store are cheese pizzas. There are 120 pizzas in the
Zepler [3.9K]

Answer:

54 cheese pizzas

Step-by-step explanation:

We know that there are 120 pizzas in the freezer, and 45% of them are cheese pizzas.

We want to find out how many cheese pizzas are in the freezer, so we should multiply 45% and 120.

45% * 120

Convert 45 % to a decimal. Divide 45 by 100 or move the decimal place 2 spots to the left.

45/100= 0.45      or       45.0 --> 4.5 ---> 0.45

0.45 * 120

Multiply 0.45 and 120

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4 0
3 years ago
"(b) use part (a) to find a power series for f(x) = 1 (4 + x)3 ."
Firdavs [7]
Assuming

f(x)=\dfrac1{(4+x)^3}

Recall that for |x|,

\displaystyle\sum_{n\ge0}x^n=\frac1{1-x}

Denote the above by s(x). Then

s'(x)=\displaystyle\sum_{n\ge1}nx^{n-1}=\frac1{(1-x)^2}
s''(x)=\displaystyle\sum_{n\ge2}n(n-1)x^{n-2}=\frac2{(1-x)^3}

Now,

\dfrac1{(4+x)^3}=\dfrac1{4^3}\dfrac1{\left(1-\left(-\frac x4\right)\right)^3}=\dfrac1{2^7}\dfrac2{\left(1-\left(-\frac x4\right)\right)^3}

which means we have

f(x)=\dfrac1{2^7}s''(x)=\dfrac1{128}\displaystyle\sum_{n\ge2}n(n-1)\left(-\frac x4\right)^{n-2}

which is valid for \left|-\dfrac x4\right|, or |x|.
4 0
3 years ago
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