Answer:
C)
Step-by-step explanation:
Have a great day
The correct option is (B) yes because all the elements of set R are in set A.
<h3>
What is an element?</h3>
- In mathematics, an element (or member) of a set is any of the distinct things that belong to that set.
Given sets:
- U = {x | x is a real number}
- A = {x | x is an odd integer}
- R = {x | x = 3, 7, 11, 27}
So,
- A = 1, 3, 5, 7, 9, 11... are the elements of set A.
- R ⊂ A can be understood as R being a subset of A, i.e. all of R's elements can be found in A.
- Because all of the elements of R are odd integers and can be found in A, R ⊂ A is TRUE.
Therefore, the correct option is (B) yes because all the elements of set R are in set A.
Know more about sets here:
brainly.com/question/2166579
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The complete question is given below:
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?
(A) yes, because all the elements of set A are in set R
(B) yes, because all the elements of set R are in set A
(C) no because each element in set A is not represented in set R
(D) no, because each element in set R is not represented in set A
Answer:
<h2>46•18</h2>
Step-by-step explanation:
<h3> I=PRT/100</h3><h2> 6740×<u>1</u><u>0</u>×7</h2><h2> 100</h2><h2> 674×1×7</h2><h2> 4618</h2><h2>the nearest hundredth is a 46•18 this is my answer</h2>
Answer:
81x^4+216x^3+216x^2+96x+16
Step-by-step explanation:
Answer:
Step-by-step explanation:
<u>We know that:</u>
- Area of shaded region = Area of square - Area of circles
- Radius of circle = 3 in
- Area of circle = πr²
- Area of square = s²
<u>Solution:</u>
- Area of shaded region = Area of square - Area of circles
- => Area of shaded region = (12²) - 4(22/7 x 3 x 3)
- => Area of shaded region = (144) - 4(22/7 x 9)
- => Area of shaded region = (144) - 4(198/7)
- => Area of shaded region = 144 - 792/7
- => Area of shaded region = 144 x 7/7 - 792/7
- => Area of shaded region = 1008/7 - 792/7
- => Area of shaded region = 1008/7 - 792/7
- => Area of shaded region = 216/7 in²
