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Ulleksa [173]
3 years ago
8

Can someone help me out! it would be really appreciate :)

Mathematics
1 answer:
nirvana33 [79]3 years ago
5 0

Answer:

6x4+10x3−4x2−15x+5

Step-by-step explanation:

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Intentionally writing a check on an account with insufficient funds should not be done. True Fals
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The above statement is false because for intentionally writing a check on an account with sufficient funds, it is only considered illegal when it is intended for check kiting, it is written above $100 and has been done more than once.

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RideAnS [48]
So 1 would be 5r+20 and then the second one would be
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2 years ago
What is the value of a?
bearhunter [10]

Answer:

the second option : 5 1/3

Step-by-step explanation:

the height (h) in such a right-angled triangle is based on the 2 segments of the baseline : 3 and a

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with a and b being the baseline segments.

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8 0
2 years ago
Find the radius of convergence, r, of the series. ? n2xn 7 · 14 · 21 · ? · (7n) n = 1
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.
3 0
3 years ago
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