answer:1. We are given that Egor participated in 13 games, and Nikita participated in 27 games. We need to find out how many games Innokentiy played. 2. Initially, one player looks on while the other two play. After each game, the loser steps down and is replaced by the observer. Since there are no draws, each game results in one player losing and stepping aside. 3. Given that Nikita played 27 games, we will start by examining the dynamics of play: - Since only Nikita can play every game without skipping two in a row, the upper bound on games played is at most 27 because every game must have one loser who steps down. 4. Next, consider the total number of games: - Let's denote the total number of games played as (N). - Since Egor participated in 13 games and Nikita in 27, the remaining number of games where Egor was not a participant must be (N - 13). - Since there are no draws and each game's loser steps down for the observer, every game will have only one player changing. 5. Now, analyze game participation: - For Nikita to have played in every game, (N = 27), because he was a part of every single game. - In (N) games, Nikita played 27, Egor played 13, so the rest of the games must involve Innokentiy and Nikita. 6. Calculate Innokentiy's games: - Innokentiy’s number of played games is the complement of Nikita's total games (where Egor is not a participant). - Hence, the total number of games played by Innokentiy is (N - text{games played by Egor}) which equals (27 - 13). - Thus, Innokentiy played (27 - 13 = 14) games. Therefore, the number of games Innokentiy played is: [ boxed{14} ]

answer:For the function to be well-defined, we need to satisfy the following conditions: begin{cases} 2cos x- sqrt {2}geqslant 0 & (1) 2sin x-1neq 0 & (2) end{cases} From (1), we get cos xgeqslant dfrac { sqrt {2}}{2}, which implies 2kpi- dfrac {π}{4}leqslant xleqslant 2kpi+ dfrac {π}{4}, where k in mathbb{Z}. From (2), we get sin xneq dfrac {1}{2}, which implies xneq 2kpi+ dfrac {π}{6} or xneq 2kpi+ dfrac {5π}{6}, where k in mathbb{Z}. Taking the intersection of the above sets, we obtain the domain of x: 2kpi- dfrac {π}{4}leqslant xleqslant 2kpi+ dfrac {π}{4}, text{ and } xneq 2kpi+ dfrac {π}{6}, text{ where } k in mathbb{Z}. Therefore, the domain of the function is boxed{{x;|;2kpi- dfrac {π}{4}leqslant xleqslant 2kpi+ dfrac {π}{4}, text{ and } xneq 2kpi+ dfrac {π}{6}, text{ where } k in mathbb{Z}}}.

answer:Let's denote the number of children who got on the Ferris wheel as ( x ). The cost for the Ferris wheel for ( x ) children is ( 5x ) dollars. Since all 5 children went on the merry-go-round, the cost for the merry-go-round is ( 3 times 5 = 15 ) dollars. Each child bought 2 cones of ice cream, so each child spent ( 8 times 2 = 16 ) dollars on ice cream. For all 5 children, the total cost for ice cream is ( 16 times 5 = 80 ) dollars. The total cost of all activities is the sum of the costs for the Ferris wheel, the merry-go-round, and the ice cream, which is ( 110 ) dollars. So we have the equation: [ 5x + 15 + 80 = 110 ] Solving for ( x ): [ 5x + 95 = 110 ] [ 5x = 110 - 95 ] [ 5x = 15 ] [ x = frac{15}{5} ] [ x = 3 ] Therefore, boxed{3} children got on the Ferris wheel.

answer:On day n, Barry increases the length of the object by a factor of frac{n+3}{n+2} for n geq 1. The overall increase through day n is given by: [ left( frac{4}{3} right) left( frac{4}{3} right) left( frac{5}{4} right) cdots left( frac{n+2}{n+1} right) left( frac{n+3}{n+2} right) ] This sequence simplifies by cancelling intermediate terms and yields: [ frac{4}{3} cdot frac{4}{3} cdot frac{(n+3)/3}{2} ] Setting this equal to 50, and solving for n, we have: [ frac{4 cdot 4 cdot (n+3)}{3 cdot 3 cdot 2} = 50 implies frac{16(n+3)}{18} = 50 implies frac{8(n+3)}{9} = 50 implies 8(n+3) = 450 implies n+3 = 56.25 implies n = 53.25 ] Since n must be an integer, and knowing the cumulative factor must equal exactly 50, the nearest integer approximation satisfying all requirements is n = boxed{53}.