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3 years ago

How would you graph 10<=20 on a number line graph

Mathematics

2 answers:

Mnenie [13.5K]3 years ago

8 0

Answer:

put a point on 10 then an arrow up to 20

Step-by-step explanation:

ValentinkaMS [17]3 years ago

7 0

Answer:

you would put a point on 10 then an arrow up to 20

Step-by-step explanation:

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a case of plastic spoons has 8.400 spoons. there are 8 boxes of spoons in the case. how many spoons are in each box

Veseljchak [2.6K]

It's either 1050 or 1.050 if that was suppose to be a decimal or a comma after the eight

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3 years ago

–10p = –11p − 15 What is the answer

tensa zangetsu [6.8K]

Answer: p = −15

Step-by-step explanation:

Add 11p to both sides.

−10p+11p=−11p−15+11p

p=−15

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3 years ago

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Help me What is 1/2 divided by 2/3

ycow [4]

Answer: 0.75 = 3/4

Step-by-step explanation: To solve, flip the second fraction and multiply the numerators and denominators together and simplify.

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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

4plss help its timed...

galina1969 [7]

Answer:

width = 5.29 ft

Step-by-step explanation:

a² + 6² = 8²

a² = 64 - 36

a² = 28

a = √28

a = 5.29 ft

3 0

2 years ago

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