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.
The degree of the polynomial will tell you the max amount of zeros you will have. 11 in this case.
However you can also have 9,7,5,3,1 (keep subtracting by 2)
So the answer is 1.
192 don't forget to square root it
Answer:
11 bucks
Step-by-step explanation:
