You can count from 34-28 and the numbers you counted (the numbers in between 34-28) is your answer.
The question is asking how many ways can ten team be randomly assigned to 5 games. They can be randomly assigned in nPrC
r = n!/( n-r)! ways.
So we have that 10 teams can be randomly assigned to five games in 10C5 ways.
10C5 = 10!/ (10-5)! = 10!/ 5! = 30240 ways. So they can be arranged in 30240 ways.
(-3,6) is were the lines intersect
y = x² - 12x + 8
First step is to move the 8 to the other side by subtract 8 from both sides. This gives you:
y - 8 = x² - 12x
Now we complete the square. Divide the b term (in this case 12) by 2, then square the result. You will then add that square to both sides. This gives you:
y - 8 + 36 = x² - 12x + 36
Combine like terms and factor:
y + 28 = (x - 6)²
y = (x - 6)² - 28. Option A should be your answer.
5/6=0.83~
Meanwhile 3/4=0.75.
5/6 is bigger than 3/4.
Good luck!