Y+47x12.6 rewrite the equation below as a function of x

(m3 - 6m2 - 16m + 63) divided by (m - 7)

Arlecino [84]

Therefore( $m^3 -6m^2-16m +63$ )÷ (m-7) $= (m^2 +m-9)$

Step-by-step explanation:

$m^3 -6m^2-16m +63$

$=m^3 -7m^2$ $+m^2 -7m$ $-9m+63$

$= m^2(m-7)$ $+m(m-7)$ $-9(m-7)$

$= (m-7 ) (m^2 +m-9)$

Therefore( $m^3 -6m^2-16m +63$ )÷ (m-7) $= (m^2 +m-9)$

3 0

3 years ago

Thirty-nine randomly selected students in a college took the stat final. The sample mean was 68.9 with standard deviation 8.3 .

SCORPION-xisa [38]

Answer:

(65.5775 , 72.2225)

Step-by-step explanation:

Confidence Interval is the range around sample mean, likely to consist the population parameter.

Formula = x' + z (s / √n)

= 68.9 + 2.5 (8.3 / √39)

= 68.9 + 3.3225

(65.5775 , 72.2225)

7 0

3 years ago

Laura bought 8 pounds of flour for $4. How many dollars did she pay per pound of flour

LenKa [72]

Answer:

.50

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

he given table shows the growth of Anytown, NC by providing the estimated population from 1970 to 2000. Year Population 1970 4,8

Morgarella [4.7K]

The population is increasing by 1500 each year and starts at 1970 population ( = 4800) so the equation is
y=1500x+4800

4 0

4 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.

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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.

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

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

Thus, the horizontal asymptote is x = 3.

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Question c:

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

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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: $(-\infty,3)$

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

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

8 0

3 years ago