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3 years ago

2 ways to decompose the number 749

Mathematics

1 answer:

kolezko [41]3 years ago

6 0

Okay. You can decompose 749 with these ways:
700 + 40 + 9
or
500 + 200 + 40 + 9

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ruslelena [56]

I need more information to answer this question.

8 0

3 years ago

How do u solve this?

malfutka [58]

you can work it out or put in random numbers until the two sides are equal

3 0

3 years ago

A spinner has red, green, blue and orange sections. The probability of landing on red is 2/3,on green 1/12, on blue is 1/12. Wha

lesya [120]

Answer: P(landing on orange)=1 - [2/3 + 1/12 + 1/12]

so =1/6

7 0

3 years ago

Carl finishes 5kms.Run in 1 hour and 25 minutes.How many minutes did Carl ran?

alexdok [17]

Answer:

Carl ran for 85 minutes

Step-by-step explanation:

Since we know that he ran for 1 hour and 25 minutes and we know that 1 hour equals 60 minutes we can say that he ran for 60 + 25 minutes. Therefore he ran for 85 minutes.

6 0

3 years ago

Let f(x) = [infinity] xn n2 n = 1. find the intervals of convergence for f. (enter your answers using interval notation. ) find

inna [77]

Best guess for the function is

$\displaystyle f(x) = \sum_{n=1}^\infty \frac{x^n}{n^2}$

By the ratio test, the series converges for

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

When $x=1$ , $f(x)$ is a convergent $p$ -series.

When $x=-1$ , $f(x)$ is a convergent alternating series.

So, the interval of convergence for $f(x)$ is the closed interval $\boxed{-1 \le x \le 1}$ .

The derivative of $f$ is the series

$\displaystyle f'(x) = \sum_{n=1}^\infty \frac{nx^{n-1}}{n^2} = \frac1x \sum_{n=1}^\infty \frac{x^n}n$

which also converges for $|x|$ by the ratio test:

$\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{n+1} \cdot \frac n{x^n}\right| = |x| \lim_{n\to\infty} \frac{n}{n+1} = |x| < 1$

When $x=1$ , $f'(x)$ becomes the divergent harmonic series.

When $x=-1$ , $f'(x)$ is a convergent alternating series.

The interval of convergence for $f'(x)$ is then the closed-open interval $\boxed{-1 \le x < 1}$ .

Differentiating $f$ once more gives the series

$\displaystyle f''(x) = \sum_{n=1}^\infty \frac{n(n-1)x^{n-2}}{n^2} = \frac1{x^2} \sum_{n=1}^\infty \frac{(n-1)x^n}{n} = \frac1{x^2} \left(\sum_{n=1}^\infty x^n - \sum_{n=1}^\infty \frac{x^n}n\right)$

The first series is geometric and converges for $|x|$ , endpoints not included.

The second series is $f'(x)$ , which we know converges for $-1\le x$ .

Putting these intervals together, we see that $f''(x)$ converges only on the open interval $\boxed{-1 < x < 1}$ .

6 0

2 years ago