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Nadya [2.5K]
3 years ago
8

there were 27,376 animals at the shelter. Last week.8,476 animals were adopted. How many animals were left at the animal shelter

Mathematics
2 answers:
astraxan [27]3 years ago
8 0
18,900 what grade are you in?
katrin [286]3 years ago
8 0
You need to subtract be some animal was adopted so that means they were taking from the shelter and brought to a new family. so 27376-8476=28900. there is 28900 animals left in the shelter
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gulaghasi [49]

Answer:

24π or about 75.4 mm

Step-by-step explanation:

The circumference of a circle is denoted by C = 2πr, where r is the radius.

Here, we see that r = 12, so plugging this in:

C = 2πr

C = 2π * 12 = 24π ≈ 75.4 mm

The answer is thus 24π or about 75.4 mm.

<em>~ an aesthetics lover</em>

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There are 1,570 souvenir paperweights that need to be packed in boxes. Each box will hold 17 paperweights. How many boxes will b
arlik [135]

Answer:

93 boxes will be needed.

Step-by-step explanation:

Divide 1570 ÷ 17

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2 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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ki77a [65]

The greatest common factor is 2r^{4} s^{2}


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Then we use the smallest number of each variable. There are 4 r's in both equations. So, that is the number that we take. There are 2 s's in the first term, so we take that number.

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Tresset [83]

Answer:

Step-by-step explanation:

1a = yes

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