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Ksenya-84 [330]
3 years ago
13

A natural number greater than 1 that has only itself and 1 as factors is called alan

Mathematics
2 answers:
lina2011 [118]3 years ago
8 0
I think the answer should be divisible because you can’t really divide 1 without getting the same # your dividing by.
Leno4ka [110]3 years ago
6 0

Answer:

A natural number greater than 1 that has only itself and 1 as factors is called a  <u>prime</u> number

Step-by-step explanation:

  • A prime number is a natural number greater than 1, that has only two factors;- the natural number itself and 1.
  • Example:
  • The factors of 13 are {1,13}, therefore 13 is a prime number.
  • The factors of 2 are  {1.2}, therefore 2 is also a prime number.
  • Composite numbers have two or more factors.
  • An exponential number is a power of natural numbers. An example is 9=3^2
  • The first choice is correct
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U= {1,2,3,4,5,6,7,8,}, A={1,4,5,7}, B= {2,5,6,7,}, and C= {3,4,6,7}
choli [55]

Answer:

  a. {1, 2, 3, 4, 5, 6, 7}

  b. {1, 2, 3, 4, 5, 6, 8}

  c. {4, 5, 7}

  d. {1, 4, 5}

Step-by-step explanation:

The union of two sets is the list of elements in either set. The intersection of two sets is the list of elements in both sets. The complement of a set is the list of elements in the universal set that are not in the set being complemented. The complement of a set can be indicated with an apostrophe: A' is the complement of set A, for example.

<h3>a. AU(BUC)</h3>

B∪C = {2, 5, 6, 7} ∪ {3, 4, 6, 7} = {2, 3, 4, 5, 6, 7}

A∪(B∪C) = {1, 4, 5, 7} ∪ {2, 3, 4, 5, 6, 7} = {1, 2, 3, 4, 5, 6, 7}

<h3>b. (AN(BNC))'</h3>

B∩C = {2, 5, 6, 7} ∩ {3, 4, 6, 7} = {6, 7}

A ∩ (B∩C) = {1, 4, 5, 7} ∩ {6, 7} = {7}

(A∩(B∩C))' = {7}' = {1, 2, 3, 4, 5, 6, 8}

<h3>c. (ANB)U(ANC)</h3>

A∩Β = {1, 4, 5, 7} ∩ {2, 5, 6, 7} = {5, 7}

Α∩C = {1, 4, 5, 7} ∩ {3, 4, 6, 7} = {4, 7}

(A∩B)∪(A∩C) = {5, 7} ∪ {4, 7} = {4, 5, 7}

<h3>d. (ANB')U(ANC')</h3>

(A∩B')∪(A∩C') = A∩(B'∪C') = A∩(B∩C)'

(B∩C)' = {6, 7}' = {1, 2, 3, 4, 5, 8}

A∩(B∩C)' = {1, 4, 5, 7} ∩ {1, 2, 3, 4, 5, 8} = {1, 4, 5}

__

<em>Additional comment</em>

In part (d) we made use of De Morgan's law for sets:

  B'∪C' = (B∩C)'

We also made use of the distributive property for sets:

  A∩(B∪C) = (A∩B)∪(A∩C)

6 0
2 years ago
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