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Illusion [34]
3 years ago
11

Picture below 15 points​

Mathematics
2 answers:
Musya8 [376]3 years ago
8 0

Answer:

B) y=-1/4x+2

Step-by-step explanation:

Hope this helps

Law Incorporation [45]3 years ago
5 0

Answer:

b

Step-by-step explanation:

HOPE THIS HELPS

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Solve the inequality:<br><br> 3/10 is greater than or equal to k - 3/5
Ostrovityanka [42]

To solve the inequality 3/10 is greater than or equal to k - 3/5, we must first set it up. Since 3/10 is greater than or equal to something, we will have a greater than or equal to symbol, with the 'mouth' pointing towards 3/10. Setting this up we get:

3/10 ≥ k - 3/5

Now we want to get k by itself on one side. We can do this by adding 3/5 to each side.

3/10 + 3/5 ≥ k

Simplify

9/10 ≥ k

3 0
3 years ago
Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

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3 years ago
At a middle school, 30% of the
Llana [10]

Answer:

I don't understand percentages

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4 years ago
The probability distribution histogram shows the number of days a book is checked out of a library.
iren [92.7K]

Answer:

D. 0.65

Step-by-step explanation:

0.20 + 0.15 + 0.30

= 0.65.

7 0
2 years ago
What are some good editing apps i use alight motion and capcut
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:))))))

Step-by-step explanation:

videochamp, picsart

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