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3 years ago

Which of the following is a factor of x^3-4x^2-18x+9

2 answers:

7 0

Hello,

$x^3-4x^2-18x+9\\
=x^3+3x^2-7x^2-21x+3x+9\\
=x^2(x+3)-7x(x+3)+3(x+3)\\
=(x+3)(x^2-7x+3)\\
=(x+3)(x- \frac{7- \sqrt{37}}{2} )(x- \frac{7+ \sqrt{37}}{2} )$

monitta3 years ago

3 0

(x-2)(x^2-7x+3) most likely

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Redo !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! the problem is in the imaage

kvasek [131]

The answer is: 40
Step by step explanation:

8 0

3 years ago

Consider the series ∑n=1[infinity]2nn!nn. Evaluate the the following limit. If it is infinite, type "infinity" or "inf". If it d

Vikki [24]

I guess the series is

$\displaystyle\sum_{n=1}^\infty\frac{2^nn!}{n^n}$

We have

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}}{\frac{2^nn!}{n^n}}\right|=2\lim_{n\to\infty}\left(\frac n{n+1}\right)^n$

Recall that

$e=\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)^n$

In our limit, we have

$\dfrac n{n+1}=\dfrac{n+1-1}{n+1}=1-\dfrac1{n+1}$

$\left(\dfrac n{n+1}\right)^n=\dfrac{\left(1-\frac1{n+1}\right)^{n+1}}{1-\frac1{n+1}}$

$\implies\displaystyle2\lim_{n\to\infty}\left(\frac n{n+1}\right)^n=2\frac{\lim\limits_{n\to\infty}\left(1-\frac1{n+1}\right)^{n+1}}{\lim\limits_{n\to\infty}\left(1-\frac1{n+1}\right)}=\frac{2e}1=2e$

which is greater than 1, which means the series is divergent by the ratio test.

On the chance that you meant to write

$\displaystyle\sum_{n=1}^\infty\frac{2^n}{n!n^n}$

we have

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{2^{n+1}}{(n+1)!(n+1)^{n+1}}}{\frac{2^n}{n!n^n}}\right|=2\lim_{n\to\infty}\frac1{(n+1)^2}\left(\frac n{n+1}\right)^2$

$=\displaystyle2\left(\lim_{n\to\infty}\frac1{(n+1)^2}\right)\left(\lim_{n\to\infty}\left(\frac n{n+1}\right)^n\right)=2\cdot0\cdot e=0$

which is less than 1, so this series is absolutely convergent.

6 0

3 years ago

What is the equation of the translated function?

Musya8 [376]

For this case the main function is:
f (x) = x ^ 2
We are going to apply the following transformations:

Vertical translations
Suppose that k> 0:
To graph y = f (x) + k, move the graph of k units up.
We have then:
f (x) = x ^ 2 + 5

Horizontal translations
Suppose that h> 0
To graph y = f (x + h), move the graph of h units to the left.
We have then:
f (x) = (x + 1) ^ 2 + 5

Answer:
f (x) = (x + 1) ^ 2 + 5

3 0

3 years ago

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rusak2 [61]

The correct answer would be B.)

Multiplying -5x-7= 35

Multiplying two negative numbers gives you a positive

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3 years ago

Which of the following is a solution of y - x > -3?

Colt1911 [192]

X<<span>y+<span>3 its simple here is your answer</span></span>

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3 years ago