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Sever21 [200]
3 years ago
8

URGENT: Can someone help me solve these problems?

Mathematics
1 answer:
katrin2010 [14]3 years ago
3 0

Answer:

12

Step-by-step explanation:

21=31

31=7+2-3+0=12

This is correct because linear equations mean add subtract divide and multiply instead of finding roots and squares

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Distributive Property Using GCF<br><br> 25 + 30
postnew [5]

Hey there! I'm happy to help!

Let's look at the factors of each of these numbers.

25

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30

1,30

2,15

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The greatest common factor between these is 5. Therefore, we divide 25+30 by 5 and put the 5 on the outside.

5(5+6).

This has the same value but it is just written differently.

Have a wonderful day! :D

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defon
I'm guessing the series is supposed to be

\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}

By the ratio test, the series converges if the following limit is less than 1.

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7\cdot14\cdot21\cdot\cdots\cdot(7n)\cdot(7(n+1))}}{\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}}\right|

The first n terms in the numerator's denominator cancel with the denominator's denominator:

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7(n+1)}}{n^2x^n}\right|

|x^n| also cancels out and the remaining factor of |x| can be pulled out of the limit (as it doesn't depend on n).

\displaystyle|x|\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2}{7(n+1)}}{n^2}\right|=|x|\lim_{n\to\infty}\frac{|n+1|}{7n^2}=0

which means the series converges everywhere (independently of x), and so the radius of convergence is infinite.
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