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3 years ago

25 is 10% of what number?

Mathematics

2 answers:

ZanzabumX [31]3 years ago

6 0

Answer: 40

Step-by-step explanation:

This is saying one-fourth, or 25% , of a number x , is equal to 10 . Therefore, 25% of 40 is 10

larisa86 [58]3 years ago

4 0

Answer:

250

Step-by-step explanation:

10%x10=100%

25x10=250

probably not the best explanation but I hope it helps!

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The largest major arc is?

Alika [10]

Answer:

$\widehat {JMG}$

Step-by-step explanation:

$\widehat {JMG}$ is the largest major arc.

7 0

3 years ago

Solve. 3x+4y=36y=-1/2x+8

quester [9]

Answer:

7.5

Step-by-step explanation:

3x+4y=36y=-1/2+8=

5 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

Of 10 • Write in standard form 9000000

Klio2033 [76]

Answer: 9 * 10^6

Step-by-step explanation:

4 0

3 years ago

Help please i dont get thiss

melomori [17]

Answer:

4/6

Step-by-step explanation:

just keep in multiplying the fraction upp and down

5 0

3 years ago

Read 2 more answers