Answer:choice b
Step-by-step explanation:
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Answer:
3p² + 32p + 13
Step-by-step explanation:
Okay, so lets first solve for 2b. 2b = 2(p + 5), which is equal to 2p + 10. Now, let's solve for 3a. 3a = 3(p² + 10p +1), simplifying to 3p² + 30p +3. After adding 2b and 3a, we are able to get 2p + 10 + 3p² + 30p + 3 = 3p² + 32p + 13
Answer:
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Step-by-step explanation:
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
#SPJ4
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
