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natka813 [3]
3 years ago
14

A = {1, 3, 5, 7, 9}

Mathematics
2 answers:
lutik1710 [3]3 years ago
8 0

The upside down U means the numbers the sets have in common.

Do what is in parenthesis first:

(B ∩ C) = 6 is the only common number in set B and Set C.

A ∩ (B ∩ C) = Would be all the numbers in A plus the number 6 from (B ∩ C).

The answer would be A) {1, 3, 5, 6, 7, 9}

Zielflug [23.3K]3 years ago
5 0

The intersection of two sets is a new set that contains all of the elements that are in both sets

A = {1, 3, 5, 7, 9}

B = {2, 4, 6, 8, 10}

C = {1, 5, 6, 7, 9}

Set B and C has 6 in both sets

B ∩ C = {6}

A ∩ (B ∩ C) = {}

Answer: C) {}

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How many 4-digit multiples of 21 and 15 are there and 4-digit multiples of either 21 or 15 are there?
salantis [7]

Answer:

943

Step-by-step explanation:

21| 1000

   | ----------

       47 | 13

21-13=8

1000+8=1008

21|9999

  |---------

     476|3

9999-3=9996

1008,1029,...,9996

let n be number of terms

a=1008,d=21,l=9996

9996=1008+(n-1)21

9996-1008=(n-1)21

8988=(n-1)21

n-1=8988/21

n-1=428

n=428+1=429

again 15|1000

            |---------

                66|10

15-10=5

1000+5=1005

15|9999

  | ---------

      666|9

9999-9=9990

1005,1020,...,9990

a=1005,d=15,l=9990

let n be number of terms.

9990=1005+(n-1)15

9990-1005=(n-1)15

8985=(n-1)15

n-1=8985/15

n-1=599

n=599+1=600

21=3×7

15=3×5

L.C.M.=3×5×7=105

105)1000(9

      -945

     ____

         55

105-55=50

1000+50=1050

105)9999(

     -945

   ---------

         549

       - 525

        -------

            24

        -------

9999-24=9975

1050,1155,...,9975

let n be number of terms.

d=105

l=a+(n-1)d

9975=1050+(n-1)105

9975-1050=(n-1)105

8025=(n-1)105

n-1=8925/105

n-1=85

n=85+1=86

so there are 86 multiples of 21 and 15

(A∪B)=A+B-(A∩B)

multiples of 21 or 15=429+600-86=1029-86=943

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