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elena-14-01-66 [18.8K]
2 years ago
9

Which graph shows the solution to the inequality Negative 3 x minus 7 less-than 20

Mathematics
2 answers:
Aleksandr-060686 [28]2 years ago
6 0

Answer:

A number line goes from negative 10 to positive 10. An open circle appears at negative 9. The number line is shaded from negative 9 through positive 10.

Step-by-step explanation:

we have

-3x-7

Solve for x

Adds 7 both sides

-3x

-3x

Divide by -3 both sides

Remember that

When you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

so

x> 27/(-3)\\x>-9

The solution is the interval (-9,∞)

therefore

A number line goes from negative 10 to positive 10. An open circle appears at negative 9. The number line is shaded from negative 9 through positive 10.

Lisa [10]2 years ago
5 0

Answer:

A number line goes from negative 10 to positive 10. An open circle appears at negative 9. The number line is shaded from negative 9 through positive 10.

Step-by-step explanation:

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Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
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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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1) Jodi liked to collect stamps. On 3 different days she bought 6 stomps. Then she
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Answer:

4

Step-by-step explanation:

6-4+2*5=12

12/3

4

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