Given:
A(3,0)
B(1,-2)
C(3,-5)
D(7,-1)
1) reflect across x=-4
essentially calculate the difference between the x=-4 line and Px and "add" it in the other direction to x=-4
A(-4-(3-(-4)),0)=A(-11,0)
B(-4-(1-(-4)),-2)=B(-9,-2)
C(-4-(3-(-4),-5))=C(11,-5)
D(-4-(7-(-4)),-1)=D(-15,-1)
2) translate (x,y)->(x-6,y+8)
A(-3,8)
B(-5,6)
C(-3,3)
D(1,7)
3) clockwise 90° rotation around (0,0), flip the x&y coordinates and then decide the signs they should have based on the quadrant they should be in
A(0,-3)
B(-2,-1)
C(-5,-3)
D(-1,-7)
D) Dilation at (0,0) with scale 2/3, essentially multiply all coordinates with the scale, the simple case of dilation, because the center point is at the origin (0,0)
A((2/3)*3,(2/3)*0)=A(2,0)
B((2/3)*1,(2/3)*-2)=B(2/3,-4/3)
C((2/3)*3,(2/3)*-5)=C(2,-10/3)
D((2/3)*7,(2/3)*-1)=D(14/3,-2/3)
Answer:
CLASS FREQUENCIES RELATIVE FREQUENCIES
A 60 0.5
B 12 0.1
C 48 0.4
TOTAL 120 1
Step-by-step explanation:
Given that;
the frequencies of there alternatives are;
Frequency A = 60
Frequency B = 12
Frequency C = 48
Total = 60 + 12 + 48 = 120
Now to determine our relative frequency, we divide each frequency by the total sum of the given frequencies;
Relative Frequency A = Frequency A / total = 60 / 120 = 0.5
Relative Frequency B = Frequency B / total = 12 / 120 = 0.1
Relative Frequency C = Frequency C / total = 48 / 120 = 0.4
therefore;
CLASS FREQUENCIES RELATIVE FREQUENCIES
A 60 0.5
B 12 0.1
C 48 0.4
TOTAL 120 1
Answer:
mc013-1.j pg
Step-by-step explanation:
The answer is A
Answer:
SAS Postulate
Step-by-step explanation:
You can use the SAS (side, angle, side) postulate that says "if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent"
Side AB is proportionate to DE and
Side AC is proportionate to DF.
Angle A and Angle D are the same; and is between the two sides
I hope this helps.
