Mike's Movers charges a flat rate of 550 plus $25 per hour. If one of their customers was charged a total of $225, for how many

Mathematics

1 answer:

seropon [69]2 years ago

5 0

not 9 the answer should be 3

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Estimate the answer to the following problem to the nearest 10 13+46+87= a. 130 b.155 c.160 d.150

Lunna [17]

150

13+46= 59
59+87= 146

Round 46 to the nearest ten, which in this case would be 50 because the last number is larger than a 5.

Then add the 1 back in to make it 150.

3 0

3 years ago

Read 2 more answers

Write a response to the following using Claim, Evidence, Reasoning.

Bess [88]

Answer:

The answer is A 21

Step-by-step explanation:

5 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.