X-3=0
x=3
4x+2y-1=0
4(3)+2y-1=0
12+2y-1=0
11+2y=0
2y=-11
y=-5.5
To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.
jk=ac and j+k=b
The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...
2k^2-5k-18=0
2k^2+4k-9k-18=0
2k(k+2)-9(k+2)=0
(2k-9)(k+2)=0
so k=-2 and 9/2
k=(-2, 4.5)
Answer:
(6,3)
Step-by-step explanation:
y=2/3 x - 1
y=-1/2 x + 6
Since both equations are equal to y, we can set them equal
2/3 x - 1 =-1/2 x + 6
We have fractions, so I will multiply by 6 to clear the fractions
6(2/3 x - 1) =(-1/2 x + 6)6
Distribute
4x -6 = -3x +36
Add 3x to each side
4x+3x -6 = -3x+3x +36
7x -6 = 36
Add 6 to each side
7x-6+6 = 36+6
7x = 42
Divide each side by 7
7x/7 = 42/7
x =6
Now we need to find y
y =2/3x -1
y = 2/3(6) -1
y = 4-1
y=3
(6,3)
A, dilation because the shape got larger. The other three options just move it around
Answer: 4a⁴b⁵
Step-by-step explanation:
