Respuesta :
ANSWER FOR A)
we know that the linear equation in slope intercept form is equal to
[tex]y=mx+b[/tex]where m is the slope of unit rate of the linear equation, b is the y-intercept or initial value of the linear equation.
Let x be the number of years since 2011 and y the car value. In this problem the year 2011 represent x = 0 so the year 2015, represent x = 4 years (2015-2011) we have the ordered pairs (x1, y1) = (0, 35,000) and (x2, y2) = (4, 15,000). With this points we are going to find the slope (m), the formula to calculate the slope between two points is equal to
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1}=\frac{15,000-35,000}{4-0} \\ m=\frac{-20,000}{4}=-5,000 \end{gathered}[/tex]this is per year and is negative because is a decreasing function. We have when x = 0 that
[tex]\begin{gathered} 35,000=-5,000(0)+b \\ b=35,000 \end{gathered}[/tex]substitute the given values equation is
[tex]y=-5,000x+35,000[/tex]ANSWER FOR B)
If we want to know the current value of Joe's car we need to take in count the current year that is 2021 that represents x = 10 (2021 - 2011), and substitute in the equation of the point A we have that
[tex]\begin{gathered} y=-5,000(10)+35,000 \\ y=-50,000+35,000=-15,000 \end{gathered}[/tex]As the function is a decreasing function we have that in this year actually Joe owe money for the car. But may be this would not be the case, we need more information to give a better model that takes in count another factors, but for this model the value of Joe's car is - 15,000.
ANSWER FOR C)
To know this we need to find when y = 20,000, so we are going to substitute in the equation of the point A,
[tex]\begin{gathered} 20,000=-5,000x+35,000 \\ 20,000-35,000=-5,000x \\ \frac{20,000-35,000}{-5,000}=x \\ x=\frac{-15,000}{-5,000}=3 \end{gathered}[/tex]So when x=3, in the year 2014 the Joe's car is going to have a value of 20,000 dollars.
