Answer:
5
Step-by-step explanation:
The nth term of a geometric series is:
a_n = a₁ (r)^(n-1)
where a₁ is the first term and r is the common ratio.
Here, we have:
40 = a₁ (r)^(4-1)
160 = a₁ (r)^(6-1)
40 = a₁ (r)^3
160 = a₁ (r)^5
If we divide the two equations:
4 = r^2
r = 2
Now substitute into either equation to find a₁:
40 = a₁ (2)^3
40 = 8 a₁
a₁ = 5
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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91.000 091.
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slightly over 88, so you will need 89 drums.
The answer is B because every time Nelson mows a lawn, he will earn $15 each.
