Answer:
90 games
Step-by-step explanation:
Given: Total number of team (n) = 10
Each team will play twice with each other.
We know that every team will play two game with each other except with itself, therefore each team will play 
⇒ Each team will play = 
∴ Each team will have 18 games with each other in the league.
Now, as there are total 10 teams and each team have 18 games each, however, we have to make sure that no team have more than two games with other team.
Example: 
Both the above game are same and it is a mistake to take two different game.
∴ Total number of games played by all team= 
⇒ Total games = 
∴ Rana have to schedule 90 games in the Volleyball league.
∠4 = 90 (verticallyopposite)
∠2 = 68 (verticallyopposite)
∠6 = ∠2 + ∠4 (exterior angles = sum of opposite angles)
∠6 = 68 + 90 = 158°
Answer: 158°
Answer:
1/9
Step-by-step explanation:
Subtract them
5/9-4/9=1/9 Answer
Answer : 4123
Explanation:
Answer:
1. x = ±9
2.
3. 12 and -12.
4. Antoine is incorrect. There exists two solutions x=5 and x= -5.
Step-by-step explanation:
According to the questions,
Problem 1.
i.e.
i.e. x = ±9.
Problem 2.
i.e.
i.e.
i.e.
Problem 3. [tex]f(x)=x^{2}-144[tex]
To find the roots, we take, [tex]x^{2}-144=0[tex] i.e. [tex]x^{2}=144[tex] i.e. x = ±12.
Thus, the options are 12 and -12.
Problem 4. We have [tex]f(x)=x^{2}+25[tex]
For the roots, we take, [tex]x^{2}+25=0[tex] i.e. [tex]x^{2}=25[tex] i.e. x = ±5.
Thus, Antoine is not correct and two solutions namely x=5 and x= -5 exists.