Let a represent the number of ounces of perfume a. let b represent the number of ounces of perfume b. perfume a costs $11 per ounce, and perfume b costs $19 per ounce. the perfume maker wants to make a bottle that costs $71. write an equation in standard form that models the cost of each perfume in the 5-ounce bottle.

The system of equations in standard form is equal to

The number of ounces of perfume A is

The number of ounces of perfume B is

Step-by-step explanation:

Let

a-----> the number of ounces of perfume A

b----> the number of ounces of perfume B

we know that

-----> equation A

-----> equation B

substitute equation B in equation A and solve for b

Find the value of a

----->

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.