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EleoNora [17]
4 years ago
9

Please Please I really need Help!!

Mathematics
1 answer:
Snowcat [4.5K]4 years ago
7 0
There is no question?
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Find the mean and the median.
SCORPION-xisa [38]

Answer:

The mean is large so

Yeah u add and later divide by how many numbers there are

Median

128 and 129

8 0
3 years ago
Which of the following correctly identifies the set of outputs? (4 points)
spayn [35]

Answer:

the answer is c

Step-by-step explanation:

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3 years ago
The graph shows which inequality? The equation of the boundary line is y = –x + 4. y –x + 4 y ≥ –x + 4
DaniilM [7]

Answer:

y≤-x+4

Step-by-step explanation:

x intercept =4

y intercept=4

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8 0
4 years ago
Let C = {n ∈ Z | n = 6r – 5 for some integer r} and D = {m ∈ Z | m = 3s + 1 for some integer s}.
noname [10]

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

#SPJ4

7 0
2 years ago
Identify the area of segment MNO to the nearest hundredth.
Tatiana [17]

Step-by-step explanation:

911

5 0
3 years ago
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