
switch sides

solve factoring

using the zero factor principle

solve

solve


Hope this helps
Answer:
1. a) (1-1) (1-2) (1-3) (2-1) (2-2) (3-1)
2. a) (5-1) (5-2) (5-3) (5-4) (5-5) (5-6)
3. d) (1-3) (2-3) (3-1) (3-2) (3-4) (3-5) (3-6) (4-3) (5-3) (6-3)
Step-by-step explanation:
1. added the numbers in the parenthesis and A had all of the sums less than 5
2. A had all the 5s in the first spot
3. D had the 3s in each parenthesis
The answer is 100 % because it’s the same but with two zeros
Y=2x+6
(This is me filling space)