Julia writes 2 fractions with the same denominator that have numerators 5 and 7 what could the denominator be if the sum is less
1 answer:
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Answer:
(1)Equivalence Relation
(2)Not Transitive, (0,3) is missing
(3)Equivalence Relation
(4)Not symmetric and Not Transitive, (2,1) is not in the set
Step-by-step explanation:
A set is said to be an equivalence relation if it satisfies the following conditions:
- Reflexivity: If
- Symmetry:
- Transitivity:
(1) {(0,0), (1,1), (2,2), (3,3)}
(3) {(0,0), (1,1), (1,2), (2,1), (2,2), (3,3)}
The relations in 1 and 3 are Reflexive, Symmetric and Transitive. Therefore (1) and (3) are equivalence relation.
(2) {(0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3)}
In (2), (0,2) and (2,3) are in the set but (0,3) is not in the set.
Therefore, It is not transitive.
As a result, the set (2) is not an equivalence relation.
(4) {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3)}
(1,2) is in the set but (2,1) is not in the set, therefore it is not symmetric
Also, (2,0) and (0,1) is in the set, but (2,1) is not, rendering the condition for transitivity invalid.
Hey there!
If z = 5
we replace "z" with "5"
So, let's test out the problem to see if it is true or false
Problem becomes:
Let's work this out
Problem becomes:
Problem becomes:
So, therefore it looks like this statement/ sentence is
!
~
If z = 5
we replace "z" with "5"
So, let's test out the problem to see if it is true or false
Problem becomes:
Let's work this out
Problem becomes:
Problem becomes:
So, therefore it looks like this statement/ sentence is
~