Module 1
mixed review
assessment readiness
1. look at each equation and possible solution. tell whether each solution is correct.
a. –31= -30,x= -10
b. 25 – 2y = 13; y=6
c. { +2=6; 2= 16

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⭐

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer