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3 years ago

I need help anyone ?? please

Mathematics

1 answer:

Ulleksa [173]3 years ago

5 0

Answer:

you need to show more for help

Step-by-step explanation:

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Please help me with this I domt understand

zzz [600]

Answer:

its A

Step-by-step explanation:

it is linear because this is happening every day, and it is negative because 1/2 is being removed every day

3 0

3 years ago

If you wanted to ask your friend to go to a movie today, you would sign "today"

Sphinxa [80]

Answer: ? Wha?

Step-by-step explanation:

7 0

3 years ago

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At a construction job for a house there are 16 painters. Of these painters, 4 of them are painting the

Nikolay [14]

For this case, we have the total of painters is 16, so to determine what percentage is painting the interior of the house, we make a rule of three:

16 ---------------> 100%

4 -----------------> x

Where "x" represents the% of painters who paint the interior of the house.

$x = \frac {4 * 100} {16}\\x = 25$

Thus, 25% of painters are painting the interior of the house.

Answer:

25%

8 0

3 years ago

Your reading material traces an example that shows how a giant like WalMart can make it difficult for small sellers. In the case

Sedbober [7]

Answer:

I'm pretty sure it is $3.04.. Hope this helps..

Step-by-step explanation:

3 0

3 years ago

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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