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
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Answer:
-56
Step-by-step explanation:
Just to help I'm going to bolden the changes
So I'm assuming the question asking to sub in -2 in for x
= 8 - (3 x (-2) -2) ^2
= 8- (-6-2)^2
= 8- (-8)^2
= 8-64
= -56
Answer:
Step-by-step explanation:
There are several ways to solve this quartic equation. But since the coefficients, they repeat a=1,b=2,c=1,d=2, but e=0, and they are multiple of each other, then it is more convenient to work with factoring as the method of solving it.
As if it was a quadratic one.
Step-by-step explanation:
maybe check khan if the app dosent help khan academy
Answer:
$76
Step-by-step explanation:
10 guests -----> Costs $9.50
80 guests -----> Costs $9.50 x 8 = $76.00 (because 80 is 8 times 10)