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3 years ago

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

Mathematics

2 answers:

vodka [1.7K]3 years ago

8 0

Answer: A) y = 96,

The maximum number of moose the forest can sustain at one time.

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

The horizontal asymptote (H.A.) is the value of y that the graph cannot cross.

Since the degree of the numerator equals the degree of the denominator, divide their coefficients to find the H.A.

$y=\dfrac{60}{0.625}\\\\\\\large\boxed{y=96}$

Oliga [24]3 years ago

7 0

The answer is A: y=96 The maximum number of moose that the forest can sustain at one time.

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Please Help Me! I will give all my point to who ever answers my question!!!

love history [14]

Step-by-step explanation:

The equation that is given is only for the specific place of that object. To find the velocity, you need to take the derivative of the equation. This will give you:

$V=2+\frac{1}{2} t$

Now, to find the average velocity of this object, plug in the values given to you. It's between the time interval [1, 2] so these are the two numbers you'll plug into the velocity equation. Finding this average is like finding any other average.

$V(1)=2+\frac{1}{2} (1)\\V(1)=2+\frac{1}{2}=\frac{5}{2}\\V(2)=2+\frac{1}{2}(2)\\V(2)=2+1=3$

$V_{avg}=\frac{3-\frac{5}{2} }{2-1} \\V_{avg}=\frac{\frac{6}{2}-\frac{5}{2} }{1} \\V_{avg}=\frac{1}{2}$

Average velocity is 0.5 sec

To find instantaneous velocity just find the velocity at time one. Think about the name "instantaneous velocity," it's the velocity in that <u>instant</u>.

We already found this, so I don't need more work (it's displayed above).

The instantaneous velocity when $t=1$ is 2.5 sec.

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3 years ago

Find the perimeter of this figure. Please show work.

Elena-2011 [213]

Answer:

40.56 ft

Step-by-step explanation:

The perimeter is the sum of the lengths of the "sides" of this figure. Starting from the left side and working clockwise, the sum is ...

P = left side (8 ft) + top side (10 ft) + semicircle (1/2×8 ft×π) + bottom side (10 ft)

= 28 ft + 4π ft

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3 years ago

What is the reciprocal of 3 1/5

Talja [164]

Answer:

5/16

Step-by-step explanation:

We need to change the mixed number to an improper fraction to find the reciprocal

3 1/5 = (5*3+1)/5 = 16/5

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5/16

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3 years ago

Read 2 more answers

Jayce looks around at an assembly. He notices his younger sister to his right and his older brother 48 feet ahead of him. If Jay

olganol [36]

Answer:

I think it's 50

Step-by-step explanation:

50 8s inbetween the two

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csf%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%

xxTIMURxx [149]

To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is

$\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }$

Let a = 1 and b the cosine product, and write them as

$\dfrac{a - b}{x^2}$

with

$b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}$

Now we use the identity

$a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)$

to rationalize the numerator. This gives

$\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}$

As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to

$\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}$

For the remaining limit, use the Taylor expansion for cos(x) :

$\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)$

where $\mathcal{O}(x^4)$ essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.

Then

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2$

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 6x^2 + 12x^4 - 8x^6\right) \left(1 - 9x^2 + \frac{81x^4}4\right)$

$\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)$

so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with

$\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52$

Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.

Edit: some scratch work suggests the limit is 10 for n = 6.

6 0

2 years ago