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Mathematics

2 answers:

Pie2 years ago

8 0

Answer:

1. Positive, 1+2=3

2. Negative, -1-2=-3

Step-by-step explanation:

If you look at both in a graphing perspective, the point (1,2) is in Quadrant I. likewise, adding 2 to the x-coordinate will also result in the point (3,2), also in Quadrant I, where the x coordinate is positive. The point (-1,2) is in Quadrant II, and adding -2 to the x coordinate keeps it in Quadrant II, where the x-coordinate is negative.

aksik [14]2 years ago

4 0

Answer:1positive. 2 m

Negative

Step-by-step explanation:

Rule of signs.

+ +=+

- - =+

+ - =-

If you have 2 negative tiles and add 3 negative tiles you have 5 negative tiles

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I think this is what you meant.

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.