Consider the sets below. U = {x | x is a real number} A = {x | x is an odd integer} R = {x | x = 3, 7, 11, 27} Is R ⊂ A? yes, because all the elements of set A are in set R yes, because all the elements of set R are in set A no, because each element in set A is not represented in set R no, because each element in set R is not represented in set A

The correct option is "Yes because all the elements of set R are in set A"

Given the following sets;

U = {x | x is a real number}

A = {x | x is an odd integer}

R = {x | x = 3, 7, 11, 27}

The elements of the set A are A = {1, 3, 5, 7, 9, 11...}

R ⊂ A can be interpreted as R is a subset of A i.e all the elements of R can be found in A.

Since all the elements of R are all odd integers and can be fi=ound in A, hence R ⊂ A is TRUE.

The correct option is "Yes because all the elements of set R are in set A"

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yes, becoz all the elements of set R is in set A

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you'll have to attache the image of the bar graph in order for us to answer : )

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