80 = -1(80), -2(40), -4(20), -5(4),
Total answers include: -81, 79, -42, 16, -9, and 1.
None of them add up to 6!
Are you doing polynomial equations?
If so, you can solve it the hard way.
x^2 + 6x -80 = 0
(x^2 + 6x + 9) - 80 = 9
(x+3)^2 = 89
(x+3) = 9.43
x = 6.43
If you're not doing those then I'm afraid your question has no answer.
Hope that helps!
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.
The power of products property states that for number
enclosed in a bracket or parenthesis, if it is raised to a power, it must be
multiplied to the power of the enclosed number no matter how different the base
is. You cannot add it because it is not raised. You can only add it if they
have the same base. But in this problem, you will just multiply it. The breakdown
of the solution to this problem is shown below. So,
<span><span>• (2x⁵y²)³=(21x3x5*3y2*3)
= 6x15y6</span><span>
</span></span>