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Andre45 [30]
3 years ago
12

Predict the next three numbers in the pattern. 4 comma 2 comma 1 comma one half comma one fourth ​,

Mathematics
1 answer:
kolbaska11 [484]3 years ago
4 0
I think that the next numbers are 1/3, 1/5 and 1/6
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Simply 4/6 to its equivalent fraction of 2/3
Is 1/3 of a cup enough for the 2/3 a cup needed? No. So he doesn't have enough
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3 years ago
Which of these relations on the set {0, 1, 2, 3} are equivalence relations? If not, please give reasons why. (In other words, if
Delicious77 [7]

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 \forall x \in A, x \rightarrow x
  • Symmetry: \forall x,y \in A, $if x \rightarrow y,$ then y \rightarrow x
  • Transitivity: \forall x,y,z \in A, $if x \rightarrow y,$ and y \rightarrow z, $ then x \rightarrow z

(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.

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3 years ago
What is the LCM of 13 and 26
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The lest common multiple is 26. If you do multiples of each number

13:13,26
26: 26

Both of their lest common multiples would be 26
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castortr0y [4]

Answer:

2 5 7

Step-by-step explanation:

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