Answer: The correct line is
![\textup{Line 1 :}x^2+3x+2=(x+1)(x+2)=(x+1.5)^2-0.25.](https://tex.z-dn.net/?f=%5Ctextup%7BLine%201%20%3A%7Dx%5E2%2B3x%2B2%3D%28x%2B1%29%28x%2B2%29%3D%28x%2B1.5%29%5E2-0.25.)
Step-by-step explanation: We are given the following two sets of quadratic expressions in various forms:
![\textup{Line 1: }x^2+3x+2=(x+1)(x+2)=(x+1.5)^2-0.25,\\\\\textup{Line 2 :}x^2+5x+6=(x+2)(x+3)=(x+2.5)^2+6.25.](https://tex.z-dn.net/?f=%5Ctextup%7BLine%201%3A%20%7Dx%5E2%2B3x%2B2%3D%28x%2B1%29%28x%2B2%29%3D%28x%2B1.5%29%5E2-0.25%2C%5C%5C%5C%5C%5Ctextup%7BLine%202%20%3A%7Dx%5E2%2B5x%2B6%3D%28x%2B2%29%28x%2B3%29%3D%28x%2B2.5%29%5E2%2B6.25.)
We are to select one of the lines from above that represent three equivalent expressions.
We can see that there are three different forms of a quadratic expression in each of the lines:
First one is the simplified form, second is the factorised form and third one is the vertex form.
So, to check which line is correct, we need to calculate the factorised form and the vertex form from the simplified form.
We have
![\textup{Line 1: }\\\\x^2+3x+2\\\\=x^2+2x+x+2\\\\=x(x+2)+1(x+2)\\\\=(x+1)(x+2),](https://tex.z-dn.net/?f=%5Ctextup%7BLine%201%3A%20%7D%5C%5C%5C%5Cx%5E2%2B3x%2B2%5C%5C%5C%5C%3Dx%5E2%2B2x%2Bx%2B2%5C%5C%5C%5C%3Dx%28x%2B2%29%2B1%28x%2B2%29%5C%5C%5C%5C%3D%28x%2B1%29%28x%2B2%29%2C)
and
![x^2+3x+2\\\\=x^2+2\times x\times 1.5+(1.5)^2-(1.5)^2+2\\\\=(x+1.5)^2-2.25+2\\\\=(x+1.5)^2-0.25.](https://tex.z-dn.net/?f=x%5E2%2B3x%2B2%5C%5C%5C%5C%3Dx%5E2%2B2%5Ctimes%20x%5Ctimes%201.5%2B%281.5%29%5E2-%281.5%29%5E2%2B2%5C%5C%5C%5C%3D%28x%2B1.5%29%5E2-2.25%2B2%5C%5C%5C%5C%3D%28x%2B1.5%29%5E2-0.25.)
So,
![\textup{Line 1 :}x^2+3x+2=(x+1)(x+2)=(x+1.5)^2-0.25.](https://tex.z-dn.net/?f=%5Ctextup%7BLine%201%20%3A%7Dx%5E2%2B3x%2B2%3D%28x%2B1%29%28x%2B2%29%3D%28x%2B1.5%29%5E2-0.25.)
Thus, Line 1 contains three equivalent expressions.
Now,
![\textup{Line 2: }\\\\x^2+5x+6\\\\=x^2+3x+2x+6\\\\=x(x+3)+2(x+3)\\\\=(x+2)(x+3),](https://tex.z-dn.net/?f=%5Ctextup%7BLine%202%3A%20%7D%5C%5C%5C%5Cx%5E2%2B5x%2B6%5C%5C%5C%5C%3Dx%5E2%2B3x%2B2x%2B6%5C%5C%5C%5C%3Dx%28x%2B3%29%2B2%28x%2B3%29%5C%5C%5C%5C%3D%28x%2B2%29%28x%2B3%29%2C)
and
![x^2+5x+6\\\\=x^2+2\times x\times 2.5+(2.5)^2-(2.5)^2+6\\\\=(x+2.5)^2-6.25+6\\\\=(x+2.5)^2-0.25\neq (x+2.5)^2+6.25.](https://tex.z-dn.net/?f=x%5E2%2B5x%2B6%5C%5C%5C%5C%3Dx%5E2%2B2%5Ctimes%20x%5Ctimes%202.5%2B%282.5%29%5E2-%282.5%29%5E2%2B6%5C%5C%5C%5C%3D%28x%2B2.5%29%5E2-6.25%2B6%5C%5C%5C%5C%3D%28x%2B2.5%29%5E2-0.25%5Cneq%20%28x%2B2.5%29%5E2%2B6.25.)
So,
![\textup{Line 2 :}x^2+3x+2=(x+1)(x+2)=(x+1.5)^2+6.25.](https://tex.z-dn.net/?f=%5Ctextup%7BLine%202%20%3A%7Dx%5E2%2B3x%2B2%3D%28x%2B1%29%28x%2B2%29%3D%28x%2B1.5%29%5E2%2B6.25.)
Thus, Line 2 does not contain three equivalent expressions.
Hence, Line 1 is correct.
Step-by-step explanation:
2*5*9*x = 540
x = 540/90 = 54/9 = 6
option D
Selections 2, 3, 5, 6 are polynomials.
1 and 4 are not. The coefficients don't have to be integers, but the powers of the variables need to be positive integers. In 1, you have x^-1. in 4, you have x^(1/2).
Answer:
flipping coin
Step-by-step explanation:
Step-by-step explanation:
1/6 divided by 3=1/2.
I think this is the answer