Its is true that C ⊆ D means Every element of C is present in D
According to he question,
Let C = {n ∈ Z | n = 6r – 5 for some integer r}
D = {m ∈ Z | m = 3s + 1 for some integer s}
We have to prove : C ⊆ D
Proof : Let n ∈ C
Then there exists an integer r such that:
n = 6r - 5
Since -5 = -6 + 1
=> n = 6r - 6 + 1
Using distributive property,
=> n = 3(2r - 2) +1
Since , 2 and r are the integers , their product 2r is also an integer and the difference 2r - 2 is also an integer then
Let s = 2r - 2
Then, m = 3r + 1 with r some integer and thus m ∈ D
Since , every element of C is also an element of D
Hence , C ⊆ D proved !
Similarly, you have to prove D ⊆ C
To know more about integers here
brainly.com/question/15276410
#SPJ4
Step-by-step explanation:
The answer is option D.
Answer:
B polynomial; binomial
Step-by-step explanation:
Yes, it is a polynomial, as it has two terms therefore it is a binomial.
I think it is they both have a 1 and a 0 in them.
Plug in the variables.
3(9)-2(3)/2-1
27-6/1
21/1
21 <==== the answer
I hope this helps!
~kaikers
