Question

if a=(x:x is a factor of 24) &B= (x:x is a factor of 120) represent A&B in a venn diagram​

Answer 1

The Venn diagram for the given question is attached below.

A Venn diagram is a graphical representation of the relationships between sets. It consists of two or more overlapping circles or other shapes, each representing a set, with the overlap representing the intersection of the sets.

To represent A&B which is also written as $A\cap B$ in a Venn diagram, we need to first list the factors of both 24 and 120.

The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.

The factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, and 120.

To find $A\cap B$ , we need to identify the factors that are common to both sets. These are 1, 2, 3, 4, 6, 8, 12, and 24.

To create the Venn diagram, we draw two overlapping circles to represent sets A and B, and include the common factors in the overlapping region.

Learn more about Venn diagram at:

https://brainly.in/question/42145699

https://brainly.in/question/54106608

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Answer 2

Answer:

$\boxed{\sf \: \: A \: \cap \: B = \{1,2,3,4,6,8,12,24 \} \: }$

Step-by-step explanation:

Given set in set builder form as

$\sf \: A = \{x : x \: is \: a \: factor \: of \: 24 \}$

It means, Set A consists of those natural numbers which divides 24.

So, Set A in roster form is

$\sf \: A = \{1,2,3,4,6,8,12,24 \}$

Now, further given that

$\sf \: B = \{x : x \: is \: a \: factor \: of \: 120\}$

It means, Set B consists of those natural numbers which divides 120.

So, Set B in roster form is

$\sf \: B = \{1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120 \}$

So, we have

$\sf \: A = \{1,2,3,4,6,8,12,24 \}$

and

$\sf \: B = \{1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120 \}$

$\implies\sf \: A \: \sub \: B$

$\implies\sf \: A \: \cap\: B \: = \: A$

Hence,

$\implies\sf \: A \: \cap \: B = \{1,2,3,4,6,8,12,24 \}$

$\rule{190pt}{2pt}$

Additional Information

$\sf \: U' \: = \: \phi$

$\sf \: \phi ' \: = \: U$

$\sf \: A \: \cup \: A' \: = \: U$

$\sf \: A \: \cap \: A' \: = \: \phi$

$\sf \: (A\cup B)' \: = \: A' \: \cap \: B'$

$\sf \: (A\cap B)' \: = \: A' \: \cup \: B'$

$\sf \: n(A\cup B) = n(A - B) + n(A\cap B) + n(B - A)$