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}
Answer:
I think it is 9
Step-by-step explanation:
3×3 is 9 + 15= 24÷4=3×3=9
Let x = sue
and then let 3x-84 = marcia
therefore
(x)+(3x-84)=132
4x-84=132
4x=216
x=54
however just solving for x doesnt completely solve. it only shows sues money because x = sue
for marcia
3x-84
3(54)-84
162-84
78
therefore Marcia has 78$.
Answer:
x^87
Step-by-step explanation:
Multiplying x^22 and x^7 results in x^29.
Then we have:
(x^29)^3 = x^87
Recall that (a^b)^c = a^(bc) and that a^b*a^c = a^(b + c)
The answer is <span>0.43 that is what i got</span>
