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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Find the area of the triangle faces (b*h/2) and multiply that by the depth.
a)
[(6*7)/2]*8
(42/2)*8
21*8
⭐️168cm^3⭐️
b) 396ft^3
c) 63m^3
Answer:
5% increase
Step-by-step explanation:
First, find how much it increased:
6300 - 6000
= 300
Then, to find the percent increase, divide 300 by 6000
300/6000
= 0.05
So, there was a 5% increase in the population
Answer:
ldk
Step-by-step explanation:
sorry