Answer:
19 and 23/55
Step-by-step explanation:
Divide using long division. The whole number portion will be the number of times the denominator of the original fraction divides evenly into the numerator of the original fraction, and the fraction portion of the mixed number will be the remainder of the original fraction division over the denominator of the original fraction.
Answer:
Step-by-step explanation:
It appears you're trying to simplify ...
The final factor cancels the denominator of the fraction, so we have ...
_____
The applicable rules of exponents are ...
(ab)^c = (a^c)(b^c)
(a^b)(a^c) = a^(b+c)
(a^b)^c = a^(bc)
1) 14
2) -10
3) 23
Tell me if you need details
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
brainly.com/question/15276410
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Answer:
1.3c - 4.69
Step-by-step explanation:
5.2 - 3.9 = 1.3
2.65 - 7.35 = -4.69
