Question

If A= 3.5.7.9.11 ). B = 17.9.11. 13). C = (11. 13. 15)and D = (15.17):
find
(a) A N (BUD) (b) (AnB)(BUC)
​

Answer 1

Answer:

Step-by-step explanation:

xGiven, A={3,5,7,9,11} , B={7,9,11,13} , C={11,13,15} and D={15,17}

(i)

The intersection between sets A and B is denoted as A∩B .

So, the intersection between sets A and B is,

A∩B={3,5,7,9,11}∩{7,9,11,13}={7,9,11}

Hence, A∩B={7,9,11} .

(ii)

The intersection of the sets B and C is denoted as B∩C .

Given, B={7,9,11,13} and C={11,13,15}

So, the intersection between sets B and C is,

B∩C={7,9,11,13}∩{11,13,15}={11,13}

Hence, B∩C={11,13} .

(iii)

The intersection of the sets A , C and D is denoted as A∩C∩D .

Given, A={3,5,7,9,11} , C={11,13,15} and D={15,17}

So, A∩C∩D can be calculated as

A∩C∩D={3,5,7,9,11}∩{11,13,15}∩{15,17}={11}∩{15}=ϕ

Hence, A∩C∩D=ϕ .

(iv)

The intersection of the sets A and C is denoted as A∩C .

Given, A={3,5,7,9,11} and C={11,13,15}

So, A∩C can be calculated as

A∩C={2,3,4,5,6}∩{9,11,13,15}=ϕ

Hence, A∩C=ϕ .

(v)

The intersection of the sets B and D is denoted as B∩D .

Given, B={7,9,11,13} and D={15,17}

So, the intersection between B and D is

B∩D={7,9,11,13}∩{15,17}=ϕ

Hence, B∩D=ϕ .

(vi)

Given, A={3,5,7,9,11} , B={7,9,11,13} and C={11,13,15}

So, A∩(B∪C) can be calculated as

A∩(B∪C)={3,5,7,9,11}∩({7,9,11,13}∪{11,13,15})={3,5,7,9,11}∩{7,9,11,13,15}={7,9,11}

Hence, A∩(B∪C)={7,9,11} .

(vii)

The intersection of the sets A and D is denoted as A∩D .

Given, A={3,5,7,9,11} and D={15,17}

So, A∩D can be calculated as

A∩D={2,3,4,5,6}∩{15,17}=ϕ

Hence, A∩D=ϕ .

(viii)

Given, A={3,5,7,9,11} , B={7,9,11,13} and D={15,17}

So, A∩(B∪D) can be calculated as

A∩(B∪D)={3,5,7,9,11}∩({7,9,11,13}∪{15,17})={3,5,7,9,11}∩{7,9,11,13,15,17}={7,9,11}

Hence, A∩(B∪D)={7,9,11} .

(ix)

Given, A={3,5,7,9,11} , B={7,9,11,13} and C={11,13,15}

So, (A∩B)∩(B∪C) can be calculated as

(A∩B)∩(B∪C)=({3,5,7,9,11}∩{7,9,11,15})∩({7,9,11,13}∪{11,13,15,})={7,9,11}∩{7,9,11,13,15}={7,9,11}

Hence, (A∩B)∩(B∪C)={7,9,11} .

(x)

Given, A={3,5,7,9,11} , B={7,9,11,13} , C={11,13,15} and D={15,17}

So, (A∪D)∩(B∪C) can be calculated as

(A∪D)∩(B∪C)=({3,5,7,9,11}∪{15,17})∩({7,9,11,13}∪{11,13,15,})={3,5,7,9,11,15,17}∩{7,9,11,13,15}={7,9,11,15}

Hence, (A∪D)∩(B∪C)={7,9,11,15} .