1. 2.5 Distributive Review Prize Show
1. Solve without a calculator. Be sure to show your work.
5 + 6(5 • 2 – 14  2)2
5 + 6(10 – 14  2)2
5 + 6(10 – 7)2
5 + 6(3)2
5 + 6(9)
= 59

2. 2.5 Distributive Review Prize Show1 2
1 + 2 = 180
10x – x + 10 = 180
15x = 180
x = 12
m1 = 10x – 10
= 10(12) – 10
= 110 2.
a) What do you call 1 & 2?
a linear pair
b) if 1 = 10x – 10, 2 = 5x + 10,
Find m 1.

3. 2.5 Distributive Review Prize Show1 2 4 3 A B C D E F 1
BCF
FCB
FCA
NOT A!
3. Give another name for:
a) CA CB
b) AB
AC, or AE
c) DC CD
d) DC
CD, DF, CF, FD, FC
e) ACF

4. 2.5 Distributive Review Prize Show1 2 4 3 A B C D E F
2.5 Distributive Review Prize Show
4. Name:
a) a pair of vertical angles
1 &  2 & 4
b) a linear pair
1 & 4 1 & 2
3 & 4 3 & 2
c) 3 collinear points
D,C,F A,B,C A,B,E A,C,E
B,C,E

5. 2.5 Distributive Review Prize Show5.
1 is acute. Find the restrictions on x. 1
6x - 12
0 < 6x – 12 < 90
12 < 6x < 102
2 < x < 17

6. 2.5 Distributive Review Prize Show1 2 4 3
1 = 3
7x + 7 = 10x – 2
9 = 3x
3 = x
m 1 = 7x + 7
= 7(3) + 7
= 28
1 = 7x + 7,
3 = 10x – 2 ,
Find m 1.

7. 2.5 Distributive Review Prize Show1 2 4 3
7. If m2 = 98,
a) is 1 right, acute, or obtuse?
(1 & 2 are a linear pair, so 1 = 82), acute
b) is 3 right, acute or obtuse?
(2 & 3 are a linear pair, so 3 = 82), acute
c) is 4 right, acute or obtuse?
(2 & 4 are vertical angles, so 4 = 98), obtuse

8. 2.5 Distributive Review Prize ShowM R K
8. K is the midpoint of RM
RK = 3x + 1,
KM = 5x – 15 ,
Find RM.
RK = KM
3x + 1 = 5x – 15
16 = 2x
8 = x
RM = 3x x – 15
= 3(8) (8) – 15
= 50