Answer:
Step-by-step explanation:
Given
← factor numerator and denominator
= 
← cancel common factor (x + 1) on numerator/ denominator
=
Answer:
The relation is not a function
The domain is {1, 2, 3}
The range is {3, 4, 5}
Step-by-step explanation:
A relation of a set of ordered pairs x and y is a function if
- Every x has only one value of y
- x appears once in ordered pairs
<u><em>Examples:</em></u>
- The relation {(1, 2), (-2, 3), (4, 5)} is a function because every x has only one value of y (x = 1 has y = 2, x = -2 has y = 3, x = 4 has y = 5)
- The relation {(1, 2), (-2, 3), (1, 5)} is not a function because one x has two values of y (x = 1 has values of y = 2 and 5)
- The domain is the set of values of x
- The range is the set of values of y
Let us solve the question
∵ The relation = {(1, 3), (2, 3), (3, 4), (2, 5)}
∵ x = 1 has y = 3
∵ x = 2 has y = 3
∵ x = 3 has y = 4
∵ x = 2 has y = 5
→ One x appears twice in the ordered pairs
∵ x = 2 has y = 3 and 5
∴ The relation is not a function because one x has two values of y
∵ The domain is the set of values of x
∴ The domain = {1, 2, 3}
∵ The range is the set of values of y
∴ The range = {3, 4, 5}
1) 10 3/10
3) 2 1/2
4) 10 2/3
6) 6 1/2
7) 4 3/4
9) 1 8/9
10) 1 5/6
I believe it’s D. Y=5
If it’s not right , I apologize
Good luck
