Solution:
Given the table below:
To determine the function that models the total cost c, to rent a truck for any number of miles m, we use the line equation passing through two points.
The equation of a line that passes through two points is expressed as
[tex]\begin{gathered} y-y_1=(\frac{y_2-y_1}{x_2-x_1})(x-x_1)\text{ ---- equation 1} \\ where \\ (x_1,y_1)\text{ and \lparen x}_2,y_2)\text{ are the points through which the line passes.} \end{gathered}[/tex]
In this case,
[tex]\begin{gathered} y\Rightarrow c \\ x\Rightarrow m \\ this\text{ gives} \\ c-c_1=(\frac{c_2-c_1}{m_2-m_1})(m-m_1)\text{ ------ equation 2} \end{gathered}[/tex]
From the table of values,
[tex]\begin{gathered} c_1=200 \\ m_1=100 \\ c_2=350 \\ m_2=200 \end{gathered}[/tex]
Substituting these values into equation 2, we have
[tex]\begin{gathered} c-200=(\frac{350-200}{200-100})(m-100) \\ \Rightarrow c-200=\frac{150}{100}(m-100) \\ \end{gathered}[/tex]
Multiply through by 100,
[tex]\begin{gathered} 100(c-200)=100(\frac{150}{100}(m-100)) \\ \Rightarrow100(c-200)=150(m-100) \end{gathered}[/tex]
Open parentheses,
[tex]100c-20000=150m-15000[/tex]
Add 20000 to both sides of the equation,
[tex]\begin{gathered} 100c-20000+20000=150m-15000+20000 \\ \Rightarrow100c=150m+5000 \end{gathered}[/tex]
Divide both sides by the coefficient of c.
The coefficient of c is 100.
Thus,
[tex]\begin{gathered} \frac{100c}{100}=\frac{150m+5000}{100} \\ c=\frac{150}{100}m+\frac{5000}{100} \\ \Rightarrow c=1.5m+50 \end{gathered}[/tex]
Hence, the function that models the total cost, c, to rent a large truck for any number of miles m is expressed as
[tex]c=1.5m+50[/tex]
