Awnser: (15,15)
The image of (3,3) after a dilation by a scale factor of 5 centered at it's origin would be (15,15), I believe. From what I remember, if it's asking for a dilation by a scale factor, you're basically just multiplying that point (the 3,3) by whatever the scale factor is (5 in this case). So for example, here you would just multiply the (3,3) by 5, since 5 is our scale factor. And you would get (15,15). I hope this helps, I'm quite new to this.
Answer:
21/5 or 4.2
Step-by-step explanation:
2x+11=32-3x
5x+11=32
5x=21
x=21/5
The correct answer is: [A]: " 4x³ + x² + x + 3 " .
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Explanation:
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(3x³ + 2x² + 2x + 1) + (x³ − x² −<span> x + 2) =
</span>↔ (x³ − x² − x + 2) + (3x³ + 2x² + 2x + 1) =
<span>
</span>(x³ − x² − x + 2) + (3x³ + 2x² + 2x + 1) =
x³ − x² − x + 2 + 3x³ + 2x² + 2x + 1 ;
<span>
Combine the "like terms" ;
x</span>³ + 3x³ = + 4x³ ;
−x² + 2x² = + x² ;
− x + 2x = + x ;
+ 2 + 1 = + 3 ;
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And rewrite:
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→ " 4x³ + x² + x + 3 " ;
→ which is: "Answer choice: [A]: " 4x³ + x² + x + 3 " .
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a. Factorize the denominator:
![\dfrac{x+14}{x^2-2x-8}=\dfrac{x+14}{(x-4)(x+2)}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%2B14%7D%7Bx%5E2-2x-8%7D%3D%5Cdfrac%7Bx%2B14%7D%7B%28x-4%29%28x%2B2%29%7D)
Then we're looking for
such that
![\dfrac{x+14}{x^2-2x-8}=\dfrac a{x-4}+\dfrac b{x+2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%2B14%7D%7Bx%5E2-2x-8%7D%3D%5Cdfrac%20a%7Bx-4%7D%2B%5Cdfrac%20b%7Bx%2B2%7D)
![\implies x+14=a(x+2)+b(x-4)](https://tex.z-dn.net/?f=%5Cimplies%20x%2B14%3Da%28x%2B2%29%2Bb%28x-4%29)
If
, then
; if
, then
. So we have
![\dfrac{x+14}{x^2-2x-8}=\dfrac3{x-4}-\dfrac2{x+2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%2B14%7D%7Bx%5E2-2x-8%7D%3D%5Cdfrac3%7Bx-4%7D-%5Cdfrac2%7Bx%2B2%7D)
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)}](https://tex.z-dn.net/?f=%5Cdfrac%7B-3x%5E2%2B5x%2B6%7D%7Bx%5E3%2Bx%5E2%7D%3D%5Cdfrac%7B-3x%5E2%2B5x%2B6%7D%7Bx%5E2%28x%2B1%29%7D)
We want to find
such that
![\dfrac{-3x^2+5x+6}{x^2(x+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac c{x+1}](https://tex.z-dn.net/?f=%5Cdfrac%7B-3x%5E2%2B5x%2B6%7D%7Bx%5E2%28x%2B1%29%7D%3D%5Cdfrac%20ax%2B%5Cdfrac%20b%7Bx%5E2%7D%2B%5Cdfrac%20c%7Bx%2B1%7D)
Quick aside: for the second term, since the denominator has degree 2, we should be looking for another constant
such that the numerator of the second term is
. 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
anyway.
![\implies-3x^2+5x+6=ax(x+1)+b(x+1)+cx^2](https://tex.z-dn.net/?f=%5Cimplies-3x%5E2%2B5x%2B6%3Dax%28x%2B1%29%2Bb%28x%2B1%29%2Bcx%5E2)
If
, then
; if
, then
. 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](https://tex.z-dn.net/?f=-3x%5E2%2B5x%2B6%3Dax%5E2%2Bax%2Bbx%2Bb%2Bcx%5E2%3D%28a-2%29x%5E2%2B%28a%2B6%29x%2B6)
which tells us
and
; in both cases, we get
. Then
![\dfrac{-3x^2+5x+6}{x^2(x+1)}=-\dfrac1x+\dfrac6{x^2}-\dfrac2{x+1}](https://tex.z-dn.net/?f=%5Cdfrac%7B-3x%5E2%2B5x%2B6%7D%7Bx%5E2%28x%2B1%29%7D%3D-%5Cdfrac1x%2B%5Cdfrac6%7Bx%5E2%7D-%5Cdfrac2%7Bx%2B1%7D)
as required.