POINT F IS ON LINE SEGMENT EG GIVEN FG EQUAL 8 ,EF EQUAL 2X PLUS 4 AND EG EQUAL 4X-10 DETERMINE THE NUMERICAL LENGTH OF EF

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

your answer is the frist one

hop i

1. sometimes

2. never true

3. sometimes

step-by-step explanation:

these are pretty self explanatory if you know what supp right and com angles are