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3 years ago

Pretty please help me with this‼️

1 answer:

3 0

For this case we have to, the distance of both numbers will be given by the absolute value of their subtraction, that is:

$d = | 4.7 - (- 11.2) | = | 4.7 + 11.2 | = | 15.9 | = 15.9$

This is equivalent to:

$d = | -11.2-4.7 | = | -11.2-4.7 | = | -15.9 | = 15.9$

ANswer:

$| 4.7 - (- 11.2) | = | -11.2-4.7 |$

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What is the slope and y-intercept of (0,8) (1,10) (2,12) (3,14) (4,16) (5,18)

Anna71 [15]

To get slope use

y2 - y1 / x2 - x1

10 - 8 / 1 - 0 = 2/1 = 2

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The y-intercept is the coordinate point that has 0 as the x value.

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3 years ago

What is the solution to -x2 + 5x - 10 = 0?

Korolek [52]

Answer:

No solution

Explanation

−x^2+5x−10=0

Step 1: Use quadratic formula with a=-1, b=5, c=-10.

you plug it in as shown below

and simplify

when you simplify

you get

as shown below

This is no solution....

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3 years ago

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Marat540 [252]

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3 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

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

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

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3 years ago

9. Tell whether the table is a linear equation and explain why

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3 years ago