3consider the set x = { 1 , 2 , 3 , 4 , 5 } and suppose you have two holes. also suppose that you have 10 pigeons: the 2 -element subsets of x . can you put these 10 pigeons into the two holes in a way that there is no 3 -element subset s = { a , b , c } ⊂ x for which all pigeons from s go in the same hole? then answer the same question if x = { 1 , 2 , 3 , 4 , 5 , 6 } with 15 = c ( 6 , 2 ) pigeons.

answer: 38.47 is your answer

step-by-step explanation: hope i you and who ever sees this answer have a great rest of your day : )

60%

22, 28, 34, 40, 46

b)   f(n) = 6n + 16

n = what is the number of the row ( 22 would be row 1) so if we replaced n with 1, we would get 22

c)   the 14th row will have 100 seats

(1)   (2)   (3)   (4)   (5)   (6)   (7)   (8)   (9)   (10) (11) (12) (13) (14)

22 | 28 | 34 | 40 | 46 | 52 | 58 | 64 | 70 | 76 | 82 | 88 | 94 | 100