Find \(D(5)\), \(D(20)\), \(D(50)\), and \(D(65)\).
Substitute \(r=5\) into the equation
\(D(5) = \frac{33(5) + 16}{32} = \frac{165 + 16}{32} = \frac{181}{32} = 5.65625 \approx 5.7\)
Substitute \(r=20\) into the equation
\(D(20) = \frac{33(20) + 16}{32} = \frac{660 + 16}{32} = \frac{676}{32} = 21.125 \approx 21.1\)
Substitute \(r=50\) into the equation
\(D(50) = \frac{33(50) + 16}{32} = \frac{1650 + 16}{32} = \frac{1666}{32} = 52.0625 \approx 52.1\)
Substitute \(r=65\) into the equation
\(D(65) = \frac{33(65) + 16}{32} = \frac{2145 + 16}{32} = \frac{2161}{32} = 67.53125 \approx 67.5\)
\(D(5) \approx 5.7\), \(D(20) \approx 21.1\), \(D(50) \approx 52.1\), \(D(65) \approx 67.5\)
Choose the correct graph of \(D(r)\).
Analyze the equation
The equation \(D(r) = \frac{33r+16}{32}\) is a linear equation with a positive slope and a positive y-intercept.
Check the graphs
Graph C is the correct graph since it's a linear equation with a positive slope and a positive y-intercept.
Graph C.
Interpret the meaning of the slope in the context of this problem.
Find the slope.
The equation is \(D(r) = \frac{33r}{32} + \frac{16}{32}\), so the slope is \(\frac{33}{32}\), which is approximately 1.03.
Interpret the slope.
The slope represents the change in braking distance (\(D\)) for every 1 mph increase in speed (\(r\)). In this case, for every 1 mph increase in speed, the braking distance increases by approximately \(\frac{33}{32}\) feet.
For every 1-mph increase in speed, the braking distance increases by \(\frac{33}{32}\) feet.
\(D(5) \approx 5.7\), \(D(20) \approx 21.1\), \(D(50) \approx 52.1\), \(D(65) \approx 67.5\)
Graph C.
For every 1-mph increase in speed, the braking distance increases by \(\frac{33}{32}\) feet.