Respuesta :
[tex]\bf \begin{array}{ccll}
\stackrel{\stackrel{x}{hours}}{time}&\stackrel{\stackrel{y}{miles}}{distance}\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
4&212\\
\boxed{6}&\boxed{318}\\
8&424\\
\boxed{10}&\boxed{530}
\end{array}\\\\
-------------------------------[/tex]
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 6}}\quad ,&{{ 318}})\quad % (c,d) &({{ 10}}\quad ,&{{ 530}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{530-318}{10-6}\implies \cfrac{212}{4} \\\\\\ \stackrel{\textit{average rate of change}}{\cfrac{53}{1}}[/tex]
recall the top is distance, and the bottom is hours, so 53 miles for every 1 hour. So the object or vehicle is moving at 53 mph on average.
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 6}}\quad ,&{{ 318}})\quad % (c,d) &({{ 10}}\quad ,&{{ 530}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{530-318}{10-6}\implies \cfrac{212}{4} \\\\\\ \stackrel{\textit{average rate of change}}{\cfrac{53}{1}}[/tex]
recall the top is distance, and the bottom is hours, so 53 miles for every 1 hour. So the object or vehicle is moving at 53 mph on average.
Answer:
The car travels 53 miles per hour.
Step-by-step explanation:
time (hours) 4 6 8 10
distance (miles) 212 318 424 530
The rate of change can be given as:
[tex]\frac{318-212}{6-4}[/tex] = [tex]\frac{106}{2}[/tex] = 53 mph
[tex]\frac{424-318}{8-6}[/tex] = [tex]\frac{106}{2}[/tex] = 53 mph
Hence, the rate of change is 53 mph or we can say the car travels 53 miles per hour.