How do you find the value of x and the length of the indicated segment

Brut [27]

If you would like to solve x^5 * x^4 * x^3, you can do this using the following steps:

x^5 * x^4 * x^3 = x^(5 + 4 + 3) = x^12

The correct result would be x^12.

8 0

2 years ago

Read 2 more answers

A and b share money in 3:2 A pays £125, how much does b pay?

spin [16.1K]

83.33 should be the right answer.
Hope it helps.

7 0

3 years ago

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A sporting goods store is offering a 10% discount on in-line skates that normally cost $110.99. How much will the in-line skates

IgorLugansk [536]

Answer:

$90.90

Step-by-step explanation:

The item normally costs $110.99. If the sporting goods store is offering a 10% discount, it means that the price including the discount will be:

$110.99 x 0.90 = $99.891.

the in-line skates cost with the discount $99.89, not including taxes.

5 0

3 years ago

Can someone help me asap

scoundrel [369]

M=4/5
you find a point that is exact like (20,60)

7 0

2 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

1 year ago