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Diano4ka-milaya [45]
3 years ago
15

Find the slope between the points ( − 2 , 5 ) (−2,5) and ( 4 , − 7 ) (4,−7)

Mathematics
1 answer:
emmainna [20.7K]3 years ago
6 0

Answer:

1) undefined

2) 0

Step-by-step explanation:

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If x2 = 30, what is the value of x?<br> A. ±60 B. ±15 C. ±square root of 30 D.±square root of 15
anygoal [31]

Answer:

  C.  ±square root of 30

Step-by-step explanation:

Apply the square root function to both sides of the equation:

  \sqrt{x^2}=\sqrt{30}\\\\|x|=\sqrt{30}\\\\x=\pm\sqrt{30}

_____

The absolute value equation has two solutions. They match choice C.

6 0
3 years ago
Which decimals as the same value as 2 divided by 5
LenKa [72]

Answer:

anything that ends in 0 like .10

5 0
2 years ago
Read 2 more answers
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
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dusya [7]

Yo sup??

let the number of pencils and pens be 8x and 14x

x=2

therefore the number of pens and pencils are

16 pencils and 28 pens

Hope this helps

6 0
3 years ago
The square root of 68
Anna71 [15]

the  square root is 8.246211251

8 0
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