4*2 + 3[37 – (2x5)]= Exponents are this *
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.
Answer:
C
Step-by-step explanation:
Given
2x² + x - 1 = 2 ( subtract 2 from both sides )
2x² + x - 3 = 0
Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.
product = 2 × - 3 = - 6 and sum = + 1
The factors are - 2 and + 3
Use these factors to split the x- term
2x² - 2x + 3x - 3 = 0 ( factor the first/second and third/fourth terms )
2x(x - 1) + 3(x - 1) = 0 ← factor out (x - 1) from each term
(x - 1)(2x + 3) = 0
Equate each factor to zero and solve for x
x - 1 = 0 ⇒ x = 1
2x + 3 = 0 ⇒ 2x = - 3 ⇒ x = -