The region is in the first quadrant, and the axis are continuous lines, then x>=0 and y>=0
The region from x=0 to x=1 is below a dashed line that goes through the points:
P1=(0,2)=(x1,y1)→x1=0, y1=2
P2=(1,3)=(x2,y2)→x2=1, y2=3
We can find the equation of this line using the point-slope equation:
y-y1=m(x-x1)
m=(y2-y1)/(x2-x1)
m=(3-2)/(1-0)
m=1/1
m=1
y-2=1(x-0)
y-2=1(x)
y-2=x
y-2+2=x+2
y=x+2
The region is below this line, and the line is dashed, then the region from x=0 to x=1 is:
y<x+2 (Options A or B)
The region from x=2 to x=4 is below the line that goes through the points:
P2=(1,3)=(x2,y2)→x2=1, y2=3
P3=(4,0)=(x3,y3)→x3=4, y3=0
We can find the equation of this line using the point-slope equation:
y-y3=m(x-x3)
m=(y3-y2)/(x3-x2)
m=(0-3)/(4-1)
m=(-3)/3
m=-1
y-0=-1(x-4)
y=-x+4
The region is below this line, and the line is continuos, then the region from x=1 to x=4 is:
y<=-x+2 (Option B)
Answer: The system of inequalities would produce the region indicated on the graph is Option B
Maybe after the dog walked the street
If its 8z+3z=z
add like factors
so 8z=3z=11z
11z=z
Answer:
The equation of the line parallel is 6x - 5y = - 36
