HELP I WILL GIVE BRAINLIEST I PROMISE

Mathematics

2 answers:

Irina-Kira [14]3 years ago

8 0

Answer:

m ≤ $14.25

Step-by-step explanation:

Aleks04 [339]3 years ago

8 0

Answer:

m ≤ $80.75

Step-by-step explanation:

hope it helps

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4 9/10 divided by 2 3/5

lora16 [44]

<span>4 9/10 divided by 2 3/5

</span>4 9/10 = 49/10
2 3/5 = 13/5

so
(49/10) / (13/5)
=49/10 * 5/13
= 49/26
or
= 1 23/26

8 0

3 years ago

PLEASE HELP ME WITH THIS

Harrizon [31]

Answer:

The last option.

Step-by-step explanation:

Hope this helps!

Wait hold on

6 0

3 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.