Points equidistant from DE EF are in the bisector of angle DEF
points equidistant from EF DF are in the bisector of angle EFD
the sought after point is the intersection of bisectricess of triangle
Answer:
11.8294Step-by-step explanation:
Answer:
it looks like 7
Step-by-step explanation:
Answer:
3j + 6
Step-by-step explanation:
5j - (2j - 6) Distribute the negative inside the parentheses
5j - 2j + 6 Combine like terms
3j + 6