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IceJOKER [234]
3 years ago
11

Show that in any group of 101 people there are at least two persons having the same number friends (It is assumed that if a pers

on x is a friend of y then y is also a friend of x and we are only considering the friends amongst the group of people). How would you modify your proof if the number of people were not 101?
Mathematics
1 answer:
vladimir1956 [14]3 years ago
7 0

Answer:

Step-by-step explanation:

Let us assume that if possible in the group of 101, each had a different number of friends.  Then the no of friends 101 persons have 0,1,2,....100 only since number of friends are integers and non negative.

Say P has 100 friends then P has all other persons as friends.  In this case, there cannot be any one who has 0 friend.  So a contradiction.  Hence proved

Part ii: EVen if instead of 101, say n people are there same proof follows as

if different number of friends then they would be 0,1,2...n-1

If one person has n-1 friends then there cannot be any one who does not have any friend.

Thus same proof follows.

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Help me Please..
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Just general definitons:

a TRInomial has 3 terms (tri means three)
a BInomial has 2 terms (bi means two)
a MONOmial has 1 term (mono means one)

the degree is the highest exponent found in the algebraic expression

so they should be pretty easy to solve with that information, but just in case:

1. trinomial, degree of 4
2. binomial, degree of 3
3. monomial, degree of 2

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2 years ago
Se sabe q la suma de tres numeros es de 850. El primer es un tercio del segundo y el tercer número es el doble del segundo.¿Cual
murzikaleks [220]

Answer:

Primer número = x = 85

Segundo número = y = 255

Tercer número = z = 510

Step-by-step explanation:

Se sabe que la suma de tres números es 850. El primero es un tercio del segundo y el tercer número es el doble del segundo.

Representamos:

Primer número = x

Segundo número = y

Tercer número = z

Se sabe que la suma de tres números es 850.

x + y + z = 850 ...... Ecuación 1

El primero es un tercio del segundo

x = 1/3 años

El tercer número es el doble del segundo.

z = 2 años

Sustituimos 1 / 3y por x y 2y por z en la Ecuación 1

1 / 3y + y + 2y = 850

y / 3 + 3y = 850

Multiplica ambos lados por 3

y + 9y = 850 × 3

10 años = 2550

y = 2550/10

y = 255

Resolviendo para x

x = 1/3 años

x = 1/3 × 255

x = 85

z = 2 años

z = 2 × 255

z = 510

Por eso,

Primer número = x = 85

Segundo número = y = 255

Tercer número = z = 510

5 0
3 years ago
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