At a competition with 6 runners, 6 medals are awarded for first place through sixth place. each medal is different. how many ways are there to award the medals? decide if the situation involves permutation or a combination, and then find the number of ways to award the medals

There are 720 ways to award the medals

Step-by-step explanation:

* Lets explain the difference between permutations and combinations

- Both permutations and combinations are collections of objects

- Permutations are for lists (order matters)  

- Combinations are for groups (order doesn't matter)  

- A permutation is an ordered combination.  

- Permutation is nPr, where n is the total number and r is the number  

 of choices

# Example: chose the first three students from the group of 10

  students, n = 10 and r = 3,then 10P3 is 720

- Combinations is nCr, where n is the total number and r is the number  

  of the choices

# Example: chose a group of three students from the group of 10

  students n = 10 and r = 3,then 10C3 is 120

* Lets solve the problem

- There are six runner

- There are 6 medals awarded for first place through sixth place

- Each medal is different

- The order is important because they arranged from 1st position to

  the 6th position

∴ We will use the permutation

∵ There are 6 medals for 6 runners

∵ 6P6 = 6 × 5 × 4 × 3 × 2 × 1 = 720

∴ There are 720 ways to award the medals

permutation; 720 ~apex

Step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.