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
2 hours and a half, or 2 hours and 30 minutes. :-)
-3+11p
first you add 1+2p that gets you -3 then you do 9p-4 and you get 11p from there you can either solve it, or leave it as is, depending on what your teacher wants.
Answer:
Using graph for units
D=1/2(6)(9)=27 sq units
E=1/2(7)(6)=21 sq units
F=1/2(6)(8)=24 sq units
Step-by-step explanation:
x-coordinates for the maximum points in any function f(x) by f'(x) =0 would be x = π/2 and x= 3π/2.
<h3>How to obtain the maximum value of a function?</h3>
To find the maximum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.
we want to find x-coordinates for the maximum points in any function f(x) by f'(x) =0
Given f(x)= 4cos(2x -π)
In general 
from x = 0 to x = 2π :
when k =0 then x = π/2
when k =1 then x= π
when k =2 then x= 3π/2
when k =3 then x=2π
Thus, X-coordinates of maximum points are x = π/2 and x= 3π/2
Learn more about maximum of a function here:
brainly.com/question/13333267
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