Which pair shows equivalent expressions?
A.2(2/5x + 2)=2 2/5x + 1
B.2(2/5x + 2)=4/5x + 4
C.2(2/5x + 4)=4/5x + 2
D.2(2/5x + 4)=2 2/5x + 8
Solution:
![2(\frac{2}{5}x + 2)](https://tex.z-dn.net/?f=%202%28%5Cfrac%7B2%7D%7B5%7Dx%20%2B%202%29%20)
Let us distribute 2 inside the parenthesis.
That is, we use distributive property:
a(b+c)=ab+ac
![2(\frac{2}{5}x + 2) =\frac{2*2}{5}x+2*2](https://tex.z-dn.net/?f=%202%28%5Cfrac%7B2%7D%7B5%7Dx%20%2B%202%29%20%3D%5Cfrac%7B2%2A2%7D%7B5%7Dx%2B2%2A2%20%20)
So, ![2(\frac{2}{5}x + 2) =\frac{4}{5}x+4](https://tex.z-dn.net/?f=%202%28%5Cfrac%7B2%7D%7B5%7Dx%20%2B%202%29%20%3D%5Cfrac%7B4%7D%7B5%7Dx%2B4%20%20)
Answer:Option (b)
![2(\frac{2}{5}x+4)](https://tex.z-dn.net/?f=%202%28%5Cfrac%7B2%7D%7B5%7Dx%2B4%29%20%20)
Applying distributive property, a(b+c)=ab+ac
![2(\frac{2}{5}x+4) =2*\frac{2}{5} x+2*4](https://tex.z-dn.net/?f=%202%28%5Cfrac%7B2%7D%7B5%7Dx%2B4%29%20%3D2%2A%5Cfrac%7B2%7D%7B5%7D%20x%2B2%2A4%20)
![2(\frac{2}{5}x+4) =\frac{2*2}{5} x+2*4](https://tex.z-dn.net/?f=%202%28%5Cfrac%7B2%7D%7B5%7Dx%2B4%29%20%3D%5Cfrac%7B2%2A2%7D%7B5%7D%20x%2B2%2A4%20)
![2(\frac{2}{5}x+4) =\frac{4}{5} x+8](https://tex.z-dn.net/?f=%202%28%5Cfrac%7B2%7D%7B5%7Dx%2B4%29%20%3D%5Cfrac%7B4%7D%7B5%7D%20x%2B8%20)
So, Option (B) is correct.
Answer:
Step-by-step explanation:
x is at (0, 4)
Answer: 5
Step-by-step explanation:
If the sides of A is (8,4) and the coordinates of D is (8-1), the same value in both coordinates are 8 and the different values are 4 (A) and -1 (D). We will take the absolute values (AV) of both when added: |4 + -1|.
The answer will be<u><em> 5 grids</em></u> (or whatever measurements the question uses).
Answer:
Step-by-step explanation:
Let the solution to
2x^2 + x -1 =0
x^2+ (1/2)x -(1/2)
are a and b
Hence a + b = -(1/2) ( minus the coefficient of x )
ab = -1/2 (the constant)
A. We want to have an equation where the roots are a +5 and b+5.
Therefore the sum of the roots is (a+5) + (b+5) = a+ b +10 =(-1/2) + 10 =19/2.
The product is (a+5)(b+5) =ab + 5(a+b) + 25 = (-1/2) + 5(-1/2) + 25 = 22.
So the equation is
x^2-(19/2)x + 22 =0
2x^2-19x + 44 =0
B. We want the roots to be 3a and 3b.
Hence (3a) + (3b) = 3(a+b) = 3(-1/2) =-3/2 and
(3a)(3b) = 9(ab) =9(-1/2)=-9/2.
So the equation is
x^2 +(3/2) x -9/2 = 0
2x^2 + 3x -9 =0.