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Novay_Z [31]
3 years ago
6

In a group of nine people, one person knows exactly two others, two people know exactly 5 others, and another 2 people know exac

tly 6 others. Show that there are three people who all know each other
Mathematics
1 answer:
Rus_ich [418]3 years ago
8 0

Answer:

1, 2 and 5 all know each other.

Step-by-step explanation:

I will denote the people by numbers: 1 2 3 4 5 6 7 8 9

The following information is given in the question.

1 knows two others

2 knows five others

3 knows five others

4 knows six others

5 knows six others

Let's assume the worst case: They all know different persons, so it is difficult to find three people who all know each other.

1 knows 2 and 5

2 knows 1 3 4 5 6

3 knows 9 8 7 6 2

By now, 2, 5 and 6 are known to two people, and others are known to just one person.

4 knows 9 8 7 6 5 3

5 knows 1 2 4 7 8 9

As a result 1, 2 and 5 all know each other.

1 knows 2 and 5

2 knows 1 and 5

5 knows 1 and 2

The reason for this is that there are three groups:

1, (2,3) and (4,5)

As much as we try to separate the people that they know there are nine different persons and at least three of them will be spread to these groups in common.

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kakasveta [241]

Answer:

92

Step-by-step explanation:

w=7

x=4

y=8

z=1

8^2 + 2(4) + 3 (7) -1

64 + 2 * 4 + 3  * 7 -1

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6 0
3 years ago
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The quotient of 2 and t
tino4ka555 [31]

Answer: 2t/12

Step-by-step explanation:The quotient of is the result of dividing two terms

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3 years ago
Need help on this please
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2 years ago
These two trapezoids are similar What is the correct way to complete the similarity statement?
pentagon [3]

Option A:

\mathrm{ABCD} \sim \mathrm{GFHE}

Solution:

ABCD and EGFH are two trapezoids.

To determine the correct way to tell the two trapezoids are similar.

Option A: \mathrm{ABCD} \sim \mathrm{GFHE}

AB = GF (side)

BC = FH (side)

CD = HE (side)

DA = EG (side)

So, \mathrm{ABCD} \sim \mathrm{GFHE} is the correct way to complete the statement.

Option B: \mathrm{ABCD} \sim \mathrm{EGFH}

In the given image length of AB ≠ EG.

So, \mathrm{ABCD} \sim \mathrm{EGFH} is the not the correct way to complete the statement.

Option C: \mathrm{ABCD} \sim \mathrm{FHEG}

In the given image length of AB ≠ FH.

So, \mathrm{ABCD} \sim \mathrm{FHEG} is the not the correct way to complete the statement.

Option D: \mathrm{ABCD} \sim \mathrm{HEGF}

In the given image length of AB ≠ HE.

So, \mathrm{ABCD} \sim \mathrm{HEGF} is the not the correct way to complete the statement.

Hence, \mathrm{ABCD} \sim \mathrm{GFHE} is the correct way to complete the statement.

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2 years ago
How many lines of symmetry?
Alex73 [517]

Answer:

i think 4

sorryif its wrong

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2 years ago
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