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:
its the top right answer choice
you'd simply just need to subtract 80 from 210 since its 10 per visit, and after 8 visits well its self explanatory
Step-by-step explanation:
its the top right answer choice
you'd simply just need to subtract 80 from 210 since its 10 per visit, and after 8 visits well its self explanatory
Answer:
Step-by-step explanation:
<u>Exponential Function</u>
When we need to express the exponential relation between two variables, we use the equation
where C, r are constants to be determined by using the given points from the table
For x=0, y=2, thus
We find that C=2
The equation is now
Now we use the point x=1, y=1
We find that
Thus, the equation is
Rearranging
The required equation is
We can easily verify the last two points are also obtained by using the equation
Answer:
Step-by-step explanation:
By properties of parallelograms we know that opposite are the same so we have
so for the angles we have
