There are 4 teams in total and each team has 7 members. One of the team will be the host team.
Tournament committee will be made from 3 members from the host team and 2 members from each of the three remaining teams. Selecting the members for tournament committee is a combinations problem. We have to select 3 members out 7 for host team and 2 members out of 7 from each of the remaining 3 teams.
So total number of possible 9 member tournament committees will be equal to:

This is the case when a host team is fixed. Since any team can be the host team, there are 4 possible ways to select a host team. So the total number of possible 9 member tournament committee will be:

Therefore, there are 2917215 possible 9 member tournament committees
Answer:
62.8
Step-by-step explanation:
10*2=20
20*3.14= 62.8
HOPE IT HELPS
1 ) cot x * sin x = cos x
(cos x / sin x) * sin x = cos x
cos x = cos x
Answer: B ) cot x = cos x / sin x
2 ) ( sin² x + cos² x ) / cos x = sec x
1/cos x = sec x
sec x = sec x
Answer: C ) cos² x + sin² x = 1
Answer:
x = 4
Step-by-step explanation:
∆ABC ~ ∆DEF
--> AB/DE = BC/EF
--> x / 3 = 8 / 6
--> x / 3 = 4 / 3
--> x = 4
Answer:
Step-by-step explanation:
Use synthetic division:
The coefficients of 2x^3 - 1 are {2, 0, 0, -1}. Setting up synthetic division, we get:
1 / 2 0 0 -1
2 2 2
---------------------------
2 2 2 1
Note that the remainder is 1. The same result would be obtained through long division.
The quotient is 2x^2 + 2x + 2 with a remainder of 1.