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.
33.967741935483870967741935483871 just divide the years into the wins and you will have wins per year. This answer has to be rounded but I like to give the entire answer as if you did this yourself
Hope this helps some
The <u>correct answer</u> is:
As x→-∞, y→-3.
As x→∞, y→∞.
Explanation:
As our values of x get further into the negative numbers, the value of 2ˣ will approach 0. This is because raising a number to a negative exponent "flips" the number below the denominator and raises it to a power; we end up with smaller and smaller fractions, eventually so small that they nearly equal 0.
This will make the value of the function 0-3=-3.
As x gets larger and larger (towards ∞), the value of y, 2ˣ, continues to grow as well. Since it continues to grow exponentially, we say the value approaches ∞.
Answer:
35
Step-by-step explanation:
1/5x - 3 = 4
first move 3 to the other side with addition
1/5x = 7
then you can divide 7 by 1/5
7/1 • 5/1 (when dividing fractions the second one is always the opposite reciprocal)
=35/1 = 35
