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bekas [8.4K]
3 years ago
5

I need help with my math

Mathematics
1 answer:
kherson [118]3 years ago
8 0

Answer:

Ok what is the problem

Step-by-step explanation:

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The sum of an integer and 6 is twice as big as the number itself. What is the integer?Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!I will bra
laila [671]

Answer: The integer is +6

Step-by-step explanation:

This is because it say TWICE so double that and 6 added twice is 12.

SO the integer is 6 not -6 because they would cancel out and you would end up with a 0

Hope that helps

7 0
3 years ago
Read 2 more answers
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)\}

5 0
3 years ago
Please help on this one!! I need it in ASAP :(
lisov135 [29]

Answer:

Hey there! here's your answer

Step-by-step explanation:

Speed = distance/Time

Speed = 3/5 ÷ 1/2

Speed = 3/5 * 2

Speed = 6/5 mph or 1.2 mph.

5 0
3 years ago
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G(x) = 1-3x when g(10)=
Tomtit [17]

Answer:

g(x)=1-3x

g(10)=1-3(10)=1-30=-29.

5 0
3 years ago
PRECALCULUS HELP A.0 b.3 C.4 D.7
Oliga [24]

All the numbers in each equation are identical, just the letter changed, which doesn't affect the equation.

Because they are both the same, the answers would be the same, so the difference would be 0.


8 0
3 years ago
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