Find the distance between the two points. Simplify radical if possible and give approximation to one decimal.

2sqrt13
7.2
If DF bisects angle EFG, find angle EFG.
124º
Triangle ABC is translated based on the vector shown. Write the coordinates of A'.

A'(-1, 1)
Write a vector that maps the given pre-image to the image.

< 8 , 4 >
EFGH = NPLM, Find x

x = 11/2 or 5.5 m
Find the midpoint between (-5, 6) & (9, 0)
M(2, 3)
Solve for x

x = 13
Complete the motion rule that would describe a reflection over the y-axis.
(x,y) →
(x,y) → (-x,y)
List the points for the image reflected across the line y=x. A(4,-8) B(6,5) C(7,-2) D(-4,-9)
A'(-8,4) B'(5,6) C'(-2,7) D'(-9,-4)
EFGH = NPLM, Find y

y = 9
Find the distance between the two points. Simplify radical if possible and give approximation to one decimal.
A(-5,4) B(4,-9)
5sqrt10
15.8
Use the Angle Addition Postulate to find the value of x. (A straight line is 180º)

2x + 15 +13x + 30 + 90 = 180
x = 3
Triangle ABC has coordinate points A(-2, -2), B(1, 4) and C(-2, 5). Graph the pre-image and the image of the triangle after translating along ⟨-2, 3⟩. Label both figures and list the coordinates of the image.
A'(-4, 1), B'(-1, 7), C'(-4, 8)

Rotate the following figure 180º. List the points of the image.

D'(3,-1)
E'(1,1)
F'(1,4)
G'(3,3)
If the line segment AB has an endpoint A(-5, 2) and a midpoint of M(3,0). Find the other endpoint, B.
B(11, -2)

BD = 12
State the motion rule for the following transformation:

(x,y) → (-y,x)
_______ is the result of A translated down and left, reflected over a vertical line, and rotated 180º.

B
Find A', B', C', A", B", C" given the motion rule
A(3,4) B(-7,2) C(-3,-4)
(x,y) → (-x,y) → (2y, x)
A'(-3,4)
B'(7,2)
C'(3,-4)
A"(8,-3)
B"(4,7)
C"(-8,3)
Line segment LS has endpoint L(-4,5) and S(6,-2). The segment is first reflected over the y-axis, then translated with the vector <-4,9>. Lastly, the segment is rotated 180º. What is the final location of S"'.
S"'(10,-7)