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:
There is a 4% decrease
Step-by-Step:
if the previous total was 50, then each animal represents 2%. 50*2=100. if you are increasing the dogs by 20% of 20, then you are increasing them by 4 equaling out to 24 dogs. If your decreasing the cats by 20% of 30, then you are decreasing them by 6 equaling out to 24 cats. The new total of animals is now 48. Seeing how each animal still represent 2%, the new percentage is 96%. Subtracting the previous percentage from the new percentage gives a 4% difference.
50 * 2 = 100%
20 * 1.2 = 24
20 * .8 = 24
24 + 24 = 48
48 * 2 = 96%
100% - 96% = 4%
For this case, we perform the conversions:
First roll:
We make a rule of three to determine the number of "c" boxes that can be packed with 300 meters of adhesive tape.
1 -----------> 4.2
c -----------> 300
You can pack 71 boxes.
Second roll:
We make a rule of three to determine the number of "c" boxes that can be packed with 70 meters of adhesive tape.
1 -----------> 4.2
c -----------> 70
You can pack 16 boxes.
Third roll:
1 -----------> 4.2
c -----------> 50
You can pack 11 boxes.
Thus, in total you can pack
Answer:
98 boxes
