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3 years ago

Please help please please

Mathematics

1 answer:

kap26 [50]3 years ago

8 0

A. equation: p=$10.75h
variables=p(profit) h(hours)
B. use this equation and plug in 37 for h then solve.
hope this helps

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What is the quasten. you did not say eny thing pasifik

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Express the series in summation notation.<br> 2 + 4 + 6 + 8 + 10 + 12

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Answer:

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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

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

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

In our limit, we have

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

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

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

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On the chance that you meant to write

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

we have

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

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

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Step-by-step explanation:

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8. $\frac{3}{x - 3} = \frac{6}{8}$

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$3*8 = 6(x - 3)$

$24 = 6x - 18$

$24 + 18 = 6x$ (addition property of equality)

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