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kari74 [83]
3 years ago
11

C= 44+(p-1)(32)/2 what is p?

Mathematics
1 answer:
Vilka [71]3 years ago
6 0

Answer:   \bold{p=\dfrac{c-28}{16}}

<u>Step-by-step explanation:</u>

c=44+\dfrac{(p-1)(32)}{2}\\\\c=44+(p-1)(16)\quad \rightarrow\quad \text{simplified}\ (32\div2=16)\\\\c=44+16p-16\quad \rightarrow\quad \text{distributed 16 into p - 1}\\\\c=28+16p\quad \rightarrow\quad \text{added like terms (44 - 16)}\\\\c-28=16p\quad \rightarrow\quad \text{subtracted 28 from both sides}\\\\\dfrac{c-28}{16}=p\quad \rightarrow\quad \text{divided 16 from both sides}

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MissTica

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No, it is not correct.</span>

<span> </span>To compare this answer, do this<span>.</span>

<span>\frac{45}{100} is Jayne's 0.45 in fraction form.</span>

<span /><span>If you simplify it you get \frac{7}{20}</span>

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4 years ago
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I guess the series is

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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

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\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

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

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~Hope I helped!~
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