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

the correct answer is x/3+10≥32

hope this !

the solution is x=1

step-by-step explanation:

we are given our   equation as

let's assume

left side function is

right side function is

now, we can draw graph

and then we can get intersection point

the intersection point will be solution

so, the solution is

x=1