Answer:
4032 different tickets are possible.
Step-by-step explanation:
Given : At a race track you have the opportunity to buy a ticket that requires you to pick the first and second place horses in the first two races. If the first race runs 9 horses and the second runs 8.
To find : How many different tickets are possible ?
Solution :
In the first race there are 9 ways to pick the winner for first and second place.
Number of ways for first place - 
Number of ways for second place - 
In the second race there are 8 ways to pick the winner for first and second place.
Number of ways for first place - 
Number of ways for second place - 
Total number of different tickets are possible is


Therefore, 4032 different tickets are possible.
As we can tell, the first number is 3, so we'll have a y-intercept in our equation and overall. Secondly, we can tell it's increasing by 7 each time, so we can tell that's the rate. But, it won't just be 3+7x, if we plug 1 in, it would be 10 and not 3. We need to put in the parenthesis 3+7(x-1). If we put one in, we would receive three. Let's make sure this works.
3+7(5-1)
3+7(4)
3+28
31
31 is the fifth terms.
So the expression would be 3+7(x-1).
Let's find the 100th term.
3+7(100-1)
3+7(99)
3+693
696
So the 100th term would be 696.
Answer:
32º
Step-by-step explanation:
Fº-32 ×5/9
5/9Fº-160/9
5/9Fº =160/9
(5Fº)9=9(160)
45Fº=1440
divide both sides by 45
Fº=32
A 2 - - 7 = 9
b - 3 - - 4 = 1
c - 5 - 6 = - 11
d - 3 - - 5 = 2
H =8
6 times 6 is 36 and times 8 is 288.