3 years ago

How do you find holes of a function?

Mathematics

2 answers:

Sveta_85 [38]3 years ago

8 0

I think this helps, like a lot!

makvit [3.9K]3 years ago

7 0

That comment up above should help ^

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Which of the following relations describes a function? A. { (-4, 4), (-3, 3), (3, 3), (4, 4) } B. { (-3, 0), (0, 3), (3, 0), (0,

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The correct answer is A.

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4 years ago

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3 years ago

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3 years ago

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zlopas [31]

$\dfrac{5x-2}{(x-1)^2}=\dfrac a{x-1}+\dfrac b{(x-1)^2}$
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$\dfrac{x^3+x^2+x+2}{x^4+x^2}=\dfrac{x^3+x+x+2}{x^2(x^2+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac{cx+d}{x^2+1}$
$x^3+x^2+x+2=ax(x^2+1)+b(x^2+1)+(cx+d)x^2$
$x^3+x^2+x+2=(a+c)x^3+(b+d)x^2+ax+b$

When $x=0$ , you find that $b=2$ . It's also clear that $a=1$ . So the remaining constants must be $1+c=1\implies c=0$ and $2+d=1\implies d=-1$ , and so you get

$\dfrac{x^3+x^2+x+2}{x^4+x^2}=\dfrac1x+\dfrac2{x^2}-\dfrac1{x^2+1}$

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4 years ago

-2x-3=8<br> -2x+4=22<br><br> How to show work

Ray Of Light [21]

-2x-3=8
-2x=8+3
-2x=11
x=-11/2

-2x+4=22
-2x=22-4
-2x=18
x=-9

8 0

3 years ago