Unless unusually unusual unification under
1000/20 is the same as 100/2(you just removed a 0 from both of them)
10/2=5 add a 0
50
A ↔ B ↔ C ↔ D ↔ E ↔ F
8 7
???
AB + BC + CD = AD <em>segment addition postulate</em>
+ 8 + 7 = AD
+ 15 = AD
AD + 60 = 4AD
60 = 3AD
20 = AD
AB =
=
= 5
DE =
=
= 4
CD + DE + EF = CF <em>segment addition postulate</em>
7 + 4 + EF = CF
11 + EF = CF
Answer: 11 + EF
Note: You did not provide any info about EF. If you have additional information that you did not type in, calculate EF and add it to 11 to find the length of CF.
Answer:
Step-by-step explanation:
If m = 4, z = 9 and r = 1/6
1. 3 + m = 3 + 4 = 7
2. z - m = 9 - 4 = 5
3. 12r = 12 × 1/6 = 2
4. 60r - 4 = 60(1/6) - 4 = 10-4 = 6
5. 4m - 2 = 4(4) - 2 = 16-2 = 14
Answer:
y=8-x or y=9-x or y=243-x (The spot where the 8 is can be any number as long as the number infront of the 0 is still -1)
Step-by-step explanation: