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n200080 [17]
2 years ago
5

You have an ant farm with 27 ants. The population of ants in your farm will double every 3 months. The table shows the populatio

n growth of the ants over nine months. Decide whether the table represents a linear function or a nonlinear function. After one​ year, how many ants will there be in the ant​ farm?

Mathematics
1 answer:
disa [49]2 years ago
8 0

Answer:

It represents a linear function and there will be 216 ants after 1 year

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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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2 years ago
If anyone knows how to do this pls help I could really use it
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The perimeter of the triangle, namely the sum of all its 3 sides, is 110.

now, the three sides are in a 6:7:9 ratio, namely, if we divide 110 by (6+7+9), it will give us hmmm 110/22, a quotient of 22 equal pieces, namely how many times 22 would go into 110.

now, the small side, takes 6 of those pieces, the medium side takes 7 of them and the largest takes 9 of those pieces.

110/22 = 5

the small side is then 6(5).

the medium side is 7(5).

and the largest side is 9(5).

add them up, and you'd get the perimeter, 110 inches.
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How do the areas of the parallelogram compare?
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The first choice is right, area of 1 is 20, area of 2 is 16
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the perimeter of a rectangular picture frame is 32 in the length of the frame is 2 inches longer than the width what are the dim
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Answer:

the length is 9 in. and the width is 7 in.

Step-by-step explanation:

this is hard to explain but i promise that this is the correct answer.

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Jacob is planting flowers for his grandmother. This morning, he spent an hour planting
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Answer:

Ignore my answer

Step-by-step explanation:

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