Answer: z=83/c - 7 + t
Step-by-step explanation:
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: x = 8
Step-by-step explanation:
8(3-2x) + 4(3x - 2) = -16
expanding the equation , we have
24 - 16x +12x - 8 = -16
16 - 4x = -16
collecting the like terms
4x = 32
x = 32/4
x = 8
Answer:
obtuse
Step-by-step explanation:
