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2 years ago

Answers or correcting???

Mathematics

1 answer:

never [62]2 years ago

7 0

Answer:

Answers

Step-by-step explanation:

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If the two lines below are perpendicular and the slope of the red line is -7,

Tems11 [23]

Answer:

C. ⅐

Step-by-step explanation:

Recall: the slope of a line that is perpendicular to another is the negative reciprocal of the slope of the other line that it is perpendicular to.

Thus:

Slope of red line = -7

The green line that is perpendicular to the red line will have a slope that is the negative reciprocal of -7.

Negative reciprocal of -7 = ⅐

The slope of the green line is therefore ⅐

5 0

2 years ago

Gabriel saves 40% of his monthly paycheck for college. He earned $270 last month. How much money did Gabriel save

Sati [7]

The best thing to do is to find 10%, and you can do this by dividing $270 by 10, and therefore, $27 is equal to 10%. To find 40%, you've got to multiply $27 by 4, and this gives you $108. Therefore, gabriel saved $108 last month :)

4 0

3 years ago

Read 2 more answers

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

12x - 16 = 68<br> Solve for x

cricket20 [7]

Answer:

x = 7

Step-by-step explanation:

12x - 16 = 68

Add 16 to each side

12x - 16+16 = 68+16

12x =84

Divide by 12

12x/12 = 84/12

x = 7

6 0

3 years ago

Read 2 more answers

I am not good at algebra so I REALLY NEED HELP WITH THE PROBLEM 8n+4n^2-8n!

NeX [460]

Answer:

4n^2

Step-by-step explanation:

8n+4n^2-8n

The first 8n + the other 8n is 0 because you combine like terms

6 0

3 years ago