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Hatshy [7]
3 years ago
7

A number has 2,5 and 7 as its prime factors. what are the four smallest values it and take​

Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
3 0

Answer:

  70, 140, 280, 350

Step-by-step explanation:

Obviously, it must have the factors 2, 5, 7 as a minimum, so the smallest value is 2×5×7 = 70.

Any of these primes can be added to the product. In increasing order, the smallest additional factors will be 2, 4, 5, 7, 8, 10, ...

So, the four smallest numbers with prime factors of 2, 5, and 7 are ...

  70 = 2·5·7

  140 = 2²·5·7

  280 = 2³·5·7

  350 = 2·5²·7

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Which of these relations on the set {0, 1, 2, 3} are equivalence relations? If not, please give reasons why. (In other words, if
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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 \forall x \in A, x \rightarrow x
  • Symmetry: \forall x,y \in A, $if x \rightarrow y,$ then y \rightarrow x
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(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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Answer:

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Step-by-step explanation:

<u>Length of Chord QS</u>:

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