It will just be -7/12 because anything times 1 stays the same
Answer:
Answer image is attached.
Step-by-step explanation:
Given rational expressions:
![1.\ \dfrac{x^2+x+4}{x-2}\\2.\ \dfrac{x^2-x+4}{x-2}\\3.\ \dfrac{x^2-4x+10}{x-2}\\4.\ \dfrac{x^2-5x+16}{x-2}](https://tex.z-dn.net/?f=1.%5C%20%5Cdfrac%7Bx%5E2%2Bx%2B4%7D%7Bx-2%7D%5C%5C2.%5C%20%5Cdfrac%7Bx%5E2-x%2B4%7D%7Bx-2%7D%5C%5C3.%5C%20%5Cdfrac%7Bx%5E2-4x%2B10%7D%7Bx-2%7D%5C%5C4.%5C%20%5Cdfrac%7Bx%5E2-5x%2B16%7D%7Bx-2%7D)
And the rewritten forms:
![(x-2)+\dfrac{6}{x-2}\\(x+3)+\dfrac{10}{x-2}\\(x+1)+\dfrac{6}{x-2}\\(x-3)+\dfrac{10}{x-2}](https://tex.z-dn.net/?f=%28x-2%29%2B%5Cdfrac%7B6%7D%7Bx-2%7D%5C%5C%28x%2B3%29%2B%5Cdfrac%7B10%7D%7Bx-2%7D%5C%5C%28x%2B1%29%2B%5Cdfrac%7B6%7D%7Bx-2%7D%5C%5C%28x-3%29%2B%5Cdfrac%7B10%7D%7Bx-2%7D)
We have to match the rewritten terms with the given expressions.
Let us consider the rewritten terms and let us solve them one by one by taking LCM.
![(x-2)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x-2)^{2}+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+4+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+10}{x-2}](https://tex.z-dn.net/?f=%28x-2%29%2B%5Cdfrac%7B6%7D%7Bx-2%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B%28x-2%29%5E%7B2%7D%2B6%20%7D%7Bx-2%7D%5C%5C%5CRightarrow%20%5Cdfrac%7Bx%5E2-4x%2B4%2B6%20%7D%7Bx-2%7D%5C%5C%5CRightarrow%20%5Cdfrac%7Bx%5E2-4x%2B10%7D%7Bx-2%7D)
So, correct option is 3.
![(x+3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x+3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{(x^2+3x-2x-6)+10}{x-2}\\\Rightarrow \dfrac{x^2+x+4}{x-2}](https://tex.z-dn.net/?f=%28x%2B3%29%2B%5Cdfrac%7B10%7D%7Bx-2%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B%28x%2B3%29%28x-2%29%2B10%7D%7Bx-2%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B%28x%5E2%2B3x-2x-6%29%2B10%7D%7Bx-2%7D%5C%5C%5CRightarrow%20%5Cdfrac%7Bx%5E2%2Bx%2B4%7D%7Bx-2%7D)
So, correct option is 1.
![(x+1)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x+1)(x-2)+6}{x-2}\\\Rightarrow \dfrac{x^{2} +x-2x-2+6}{x-2}\\\Rightarrow \dfrac{x^{2} -x+4}{x-2}](https://tex.z-dn.net/?f=%28x%2B1%29%2B%5Cdfrac%7B6%7D%7Bx-2%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B%28x%2B1%29%28x-2%29%2B6%7D%7Bx-2%7D%5C%5C%5CRightarrow%20%5Cdfrac%7Bx%5E%7B2%7D%20%2Bx-2x-2%2B6%7D%7Bx-2%7D%5C%5C%5CRightarrow%20%5Cdfrac%7Bx%5E%7B2%7D%20-x%2B4%7D%7Bx-2%7D)
So, correct option is 2.
![(x-3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x-3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{x^2-3x-2x+6+10}{x-2}\\\Rightarrow \dfrac{x^2-5x+16}{x-2}](https://tex.z-dn.net/?f=%28x-3%29%2B%5Cdfrac%7B10%7D%7Bx-2%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B%28x-3%29%28x-2%29%2B10%7D%7Bx-2%7D%5C%5C%5CRightarrow%20%5Cdfrac%7Bx%5E2-3x-2x%2B6%2B10%7D%7Bx-2%7D%5C%5C%5CRightarrow%20%5Cdfrac%7Bx%5E2-5x%2B16%7D%7Bx-2%7D)
So, correct option is 4.
The answer is also attached in the answer area.
Step-by-step explanation:
is it correct ?? think that d answer is helpful to u
Nadia Is correct I believe, Because if a pizza is cut into 8 slices each slice is going to be smaller. but if it were to be cut into fourths it would be bigger. They are indeed equivalent but still never the same in size.