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Rainbow [258]
3 years ago
10

PLEASE HELP ITS DUE TODAY!!!

Mathematics
2 answers:
Flura [38]3 years ago
4 0

6/8 = p/12= tan∅

so p= 9

brainliest

DanielleElmas [232]3 years ago
3 0
C.9 I attached a photo of my calculator with my work lmk if you need any further explanation or a step by step answer

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Assuming the series is

\displaystyle\sum_{n\ge1}\frac{x^n}{2n-1}

The series will converge if

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|

We have

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

So the series will certainly converge if -1, but we also need to check the endpoints of the interval.

If x=1, then the series is a scaled harmonic series, which we know diverges.

On the other hand, if x=-1, by the alternating series test we can show that the series converges, since

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and is strictly decreasing.

So, the interval of convergence for the series is -1\le x.
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Answer:

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Step-by-step explanation:

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Heres a something that might help:

Download pdf
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