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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6x = 15
I apologize if this is wrong!
whats is y? I can't solve the problem without it
Answer:
and I have
Step-by-step explanation:
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Make each a ratio, and set them equal to each other.
Larger rectangle = 3/5
smaller rectangle = x/3
3/5 = x/3
cross multiply
3/5(5)(3) = x/3(3)(5)
3(3) = x(5)
9 = 5x
Isolate the x, divide 5 from both sides
9/5 = 5x/5
x = 9/5
x = 1.8
1.8 in. is your answer
hope this helps
