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3 years ago

11

4 A relation contains the points (1, 2), (2, -1), (3, 0), (4,1), and (5, -1). Which

1 answer:

igomit [66]3 years ago

8 0

Answer:

The answer is I. The last answer.

Step-by-step explanation:

It's correct because all the x-values all contain just one y-value making the last statement true. Also the x- values are not repeating with different y -values.

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Find the geometric mean for 8 and 12

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5 0

3 years ago

Plugin x=-6 into either equation and solve for y<br> −2x+2y=−8 or −3x+y=8

HACTEHA [7]

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Step-by-step explanation:

3 0

3 years ago

Please help with these partial fractions!!!

VARVARA [1.3K]

a. Factorize the denominator:

$\dfrac{x+14}{x^2-2x-8}=\dfrac{x+14}{(x-4)(x+2)}$

Then we're looking for $a,b$ such that

$\dfrac{x+14}{x^2-2x-8}=\dfrac a{x-4}+\dfrac b{x+2}$

$\implies x+14=a(x+2)+b(x-4)$

If $x=4$ , then $18=6a\implies a=3$ ; if $x=-2$ , then $12=-6b\implies b=-2$ . So we have

$\dfrac{x+14}{x^2-2x-8}=\dfrac3{x-4}-\dfrac2{x+2}$

as required.

b. Same setup as in (a):

$\dfrac{-3x^2+5x+6}{x^3+x^2}=\dfrac{-3x^2+5x+6}{x^2(x+1)}$

We want to find $a,b,c$ such that

$\dfrac{-3x^2+5x+6}{x^2(x+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac c{x+1}$

Quick aside: for the second term, since the denominator has degree 2, we should be looking for another constant $b'$ such that the numerator of the second term is $b'x+b$ . We always want the polynomial in the numerator to have degree 1 less than the degree of the denominator. But we would end up determining $b'=0$ anyway.

$\implies-3x^2+5x+6=ax(x+1)+b(x+1)+cx^2$

If $x=0$ , then $b=6$ ; if $x=-1$ , then $c=-2$ . Expanding everything on the right then gives

$-3x^2+5x+6=ax^2+ax+bx+b+cx^2=(a-2)x^2+(a+6)x+6$

which tells us $a-2=-3$ and $a+6=5$ ; in both cases, we get $a=-1$ . Then

$\dfrac{-3x^2+5x+6}{x^2(x+1)}=-\dfrac1x+\dfrac6{x^2}-\dfrac2{x+1}$

as required.

5 0

4 years ago