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.
Answer:
0.0049382716
PS: you could just look it up ;)
Step-by-step explanation:
From the given functions, it can be seen that the graph of g(x) represents a shift of 6 units to the left of the graph of f(x).
Given a function, f(x), a shift of k units to the left of f(x) is given by g(x + k).
Therefore, if g(x) = f(x + k), then k = 6.
12.
You must do 9 divided by 3/4
3/4 = .75
9 divided by .75 is 12
