The cost revenue graphs will be as shown in the figure
From the graph the break-even point is at n=15 and C=300.
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:
The answer is 13
Step-by-step explanation:
91 / 7 = 13
Answer:
<h2>f(-3) = 22</h2>
Step-by-step explanation:
f(x)=x² - 2x + 7
To find f(-3) substitute the value of x that's - 3 into f(x). That is for every x in f (x) replace it with - 3
We have
f(-3) = (-3)² - 2(-3) + 7
= 9 + 6 + 7
We have the final answer as
f(-3) = 22
Hope this helps you
Answer:
Domain: all real x values, Range: -4_<x_<infinity
Step-by-step explanation:
The Domain is the X values, and the range is the Y values