select the correct answer from each drop - down menu.\nthe finance department at a regional toy company has…

select the correct answer from each drop - down menu.\nthe finance department at a regional toy company has been tracking the income and costs of a new line of dolls. they have determined that the income and costs can be modeled by the equations below, where x is the number of dolls sold, in hundreds, and y is the total dollar amount, in thousands.\nincome: $y=-0.4x^{2}+3x + 45$\ncosts: $y = 1.5x+20$\nconsider the system of equations that can be used to determine the number of dolls for which the company will break - even.\nhow many total possible solutions of the form (x,y) are there for this situation?\nof any possible solutions of the form (x,y), how many are viable for this situation?

select the correct answer from each drop - down menu.\nthe finance department at a regional toy company has been tracking the income and costs of a new line of dolls. they have determined that the income and costs can be modeled by the equations below, where x is the number of dolls sold, in hundreds, and y is the total dollar amount, in thousands.\nincome: $y=-0.4x^{2}+3x + 45$\ncosts: $y = 1.5x+20$\nconsider the system of equations that can be used to determine the number of dolls for which the company will break - even.\nhow many total possible solutions of the form (x,y) are there for this situation?\nof any possible solutions of the form (x,y), how many are viable for this situation?

Answer

Answer:

  1. two possible solutions
  2. one viable solution

Explanation:

Step1: Set income equal to costs

$-0.4x^{2}+3x + 45=1.5x + 20$

Step2: Rearrange to quadratic form

$-0.4x^{2}+3x-1.5x + 45 - 20=0$ $-0.4x^{2}+1.5x + 25 = 0$

Step3: Use discriminant formula

For $ax^{2}+bx + c = 0$ ($a=-0.4$, $b = 1.5$, $c = 25$), the discriminant $\Delta=b^{2}-4ac$. $\Delta=(1.5)^{2}-4\times(-0.4)\times25=2.25 + 40=42.25>0$, so there are two possible solutions.

Step4: Consider viability

Since $x$ represents the number of dolls sold in hundreds and must be non - negative. Solving the quadratic equation $-0.4x^{2}+1.5x + 25 = 0$ using the quadratic formula $x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-1.5\pm\sqrt{42.25}}{2\times(-0.4)}$. One root is negative and one root is positive. So there is one viable solution (the non - negative root).