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2 years ago

Please answer 4 and 5. Show your work.

Mathematics

1 answer:

Amanda [17]2 years ago

8 0

Answer:

4: 32a^5 5: 2a / bc^2

Step-by-step explanation:

4. (2a)^3 / (2a)^-2

= (2a)^3 * (2a)^2

= 8a^3 * 4a^2

= 32a^5

5. 10a^7 b^9 c^6 / 5a^6 b^10 c^8

= 2a^7 b^9 c^6 / a^6 b^10 c^8

= 2a / bc^2

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2 years ago

Write a polynomial with the zeros equaling to -1,1,8

Andreyy89

Answer:x³-8x²-x+8

Step-by-step explanation:

x are equal to -1,1,8 respectively

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(x-(-))(x-1)(x-8)

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3 years ago

For the function g(x)=3-8(1/4)^2-x

Reika [66]

Using function concepts, it is found that:

a) The y-intercept is y = 2.5.

b) The horizontal asymptote is x = 3.

c) The function is decreasing.

d) The domain is $(-\infty,\infty)$ and the range is $(-\infty,3)$ .
e) The graph is given at the end of the answer.

------------------------------------

The given function is:

$g(x) = 3 - 8\left(\frac{1}{4}\right)^{2-x}$

------------------------------------

Question a:

The y-intercept is g(0), thus:

$g(0) = 3 - 8\left(\frac{1}{4}\right)^{2-0} = 3 - 8\left(\frac{1}{4}\right)^{2} = 3 - \frac{8}{16} = 3 - 0.5 = 2.5$

The y-intercept is y = 2.5.

------------------------------------

Question b:

The horizontal asymptote is the limit of the function when x goes to infinity, if it exists.

$\lim_{x \rightarrow -\infty} g(x) = \lim_{x \rightarrow -\infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2+\infty} = 3 - 8\left(\frac{1}{4}\right)^{\infty} = 3 - 8\frac{1^{\infty}}{4^{\infty}} = 3 -0 = 3$

--------------------------------------------------

$\lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2-\infty} = 3 - 8\left(\frac{1}{4}\right)^{-\infty} = 3 - 8\times 4^{\infty} = 3 - \infty = -\infty$

Thus, the horizontal asymptote is x = 3.

--------------------------------------------------

Question c:

The limit of x going to infinity of the function is negative infinity, which means that the function is decreasing.

--------------------------------------------------

Question d:

Exponential function has no restrictions in the domain, so it is all real values, that is $(-\infty,\infty)$ .
From the limits in item c, the range is: $(-\infty,3)$

--------------------------------------------------

The sketching of the graph is given appended at the end of this answer.

A similar problem is given at brainly.com/question/16533631

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2 years ago