All categories

3 years ago

What is the 100 nearest 513

Mathematics

1 answer:

ololo11 [35]3 years ago

6 0

Answer: 500

Step-by-step explanation:

it is closest and is under 550. if it were 551 through 599 it would be 600

You might be interested in

Plz help fast will mark the brainiest!!!

tiny-mole [99]

3x10^-6 is the answer

3 0

2 years ago

Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based

Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for $\mathrm{sinc}\,x$ about $x=0$ :

$\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the Taylor series for

$f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt$

is obtained by integrating the series above:

$f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

We have $f(0)=0$ , so $C=0$ and so

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

which converges by the ratio test if the following limit is less than 1:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}$

Like in the linked problem, the limit is 0 so the series for $f(x)$ converges everywhere.

7 0

3 years ago

The TV weather forecaster announces that one location has a 75% chance of sleet tomorrow and a 5% chance of snow. What is the pr

alexandr1967 [171]

Answer:

80%

Step-by-step explanation:

4 0

2 years ago

I need the answers for this question.

nadya68 [22]

Answer:

Step-by-step explanation:

7 0

3 years ago

Find the number of all 20 digit integers in which no two consecutive digits are the same

Arturiano [62]

For the first digit of the required 20-digit number, we can have 9 choices for the number, those are the digits 1 to 9. In the second digit of the number, we will only have 8 choices. That is because, we don't want to write a number in the second digit that is similar to the first digit and the third digit.

Similarly, in the third digit's place, we will only have 8 choices because we do not want the digit to be similar to the second digit and the fourth digit. This will go on up until the 19th digit. In the 20th digit, we will also have 9 choices (from 0-9) but we do not want the digit to be similar to the 19th digit.

From the explanation given above, we have the number of ways,

n = 9 x 8^(19 - 1) x 9 = 1.459 x 10^18

<em>ANSWER: 1.459 x 10^18</em>

5 0

3 years ago