First substitute .5x in for y to get a solvable equation. You should get 6x+8(.5x)=30. Then multiply 8 times .5 to get 6x+4x=30. Then combine like terms (6x+4x) to get 10x=30 and divide both sides by 10 to get x=3. Then plug this x value (3) into the other equation (y=.5x) to find the value of y. Y=.5(3) so y=1.5. Therefore the answer is B.
Answer:
= 20 ft
Step-by-step explanation:
The perimeter is found by
P = 2(l+w)
= 2(6+4)
=2(10)
= 20 ft
Answer:
x=1
Step-by-step explanation:
Find the negative reciprocal of the slope of the original line and use the point-slope formula y
−
y
1
=
m
(
x
−
x
1
) to find the line perpendicular to −
2
+
2
y
=
5
.
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
Solution :
Group Before After
Mean 693.75 743.75
Sd 155.37 143.92
SEM 54.93 50.88
n 8 8
Null hypothesis : The preparation course not effective.
Alternative hypothesis : The preparation course is effective in improving the exam scores.
(after - before)
