The graph below represents which system of inequalities? (2 points)
graph of two intersecting lines. one line is solid, and goes through the points 0, negative 2 and 1, 0 and is shaded in below the line. the other line is dashed, and goes through the points 0, 3, 3, 0 and is shaded in above the line.
group of answer choices
y > 2x − 3
y > −x − 3
y < 2x − 2
y < −x + 3
y ≤ 2x − 2
y > −x + 3
none of the above
Option C, y ≤ 2x − 2 and y > −x + 3
Step-by-step explanation:
If a line passes through two points, then the equation of line is
[tex]y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]
One line is solid, and goes through the points (0, -2) and (1, 0) and is shaded in below the line. The equation of related line is
[tex]y-(-2)=\dfrac{0-(-2)}{1-0}(x-0)[/tex]
[tex]y+2=2x[/tex]
[tex]y=2x-2[/tex]
Everything is to the below of the line is shaded and the related line is solid. So, the sign of inequality must be ≤.
[tex]y\le 2x-2[/tex]
The other line is dashed, and goes through the points (0, 3), (3, 0) and is shaded in above the line. The equation of related line is
[tex]y-3=\dfrac{0-3}{3-0}(x-0)[/tex]
[tex]y-3=-x[/tex]
[tex]y=-x+3[/tex]
Everything is to the above of the line is shaded and the related line is dashed. So, the sign of inequality must be >.
[tex]y-x+3[/tex]
The system of inequity is
y ≤ 2x − 2
y > −x + 3
Therefore, correct option is C.
14
step-by-step explanation:
14 x 3 = 42
Solid...(-3,0)(-4,-1)...shaded below the line
slope = (-1-0) / (-4-(-3) = -1/(-4 + 3) = -1/-1 = 1
y = mx + b
-1 = 1(-4) + b
-1 = -4 + b
-1 + 4 = b
3 = b
this line is : y < = x + 3 <== (thats less then or equal)
solid...(1,1)(2,-1)...shaded below the line
slope = (-1-1) / (2-1) = -2/1 = -2
y = mx + b
1 = -2(1) + b
1 = -2 + b
1 + 2 = b
3 = b
this line is : y < = -2x + 3 <== (thats less then or equal)
Step-by-step explanation:
2 and 1.
step-by-step explanation:
plug in each answer. 0 doesn't work, 1 does. so does 2. 4 doesn't:
0 = 1 is not true.
2 = 2 is true.
4 = 4 is true.
8 = 16 is not true.
hope this !
[tex]15 points! k12 homies where you at? ?[/tex]