1. Problem 38: Compare the rate of change of profits of companies A and B. - The rate of change of profit is the slope of the profit vs. year graph. - Vector B is steeper than vector A, indicating a greater rate of change. Answer: (b) the profit of B is greater. 2. Problem 39: Determine during which interval the average speed was greatest. - Average speed = change in distance / change in time. - Calculate for each interval: - First to second hour: $\frac{110 - 40}{2 - 1} = \frac{70}{1} = 70$ miles/hour - Second to fourth hour: $\frac{180 - 110}{4 - 2} = \frac{70}{2} = 35$ miles/hour - Sixth to eighth hour: $\frac{350 - 230}{8 - 6} = \frac{120}{2} = 60$ miles/hour - Eighth to tenth hour: $\frac{390 - 350}{10 - 8} = \frac{40}{2} = 20$ miles/hour - The greatest average speed is 70 miles/hour during the first to second hour. Answer: (a) the first hour to the second hour. 3. Problem 40: Find the equation of a line perpendicular to $2x - y = 7$. - Rewrite the given line in slope-intercept form: $$2x - y = 7 \implies y = 2x - 7$$ - The slope of this line is $m = 2$. - The slope of a line perpendicular to this is the negative reciprocal: $$m_{\perp} = -\frac{1}{2}$$ - Among the options, the line with slope $-\frac{1}{2}$ is: $$y = -\frac{1}{2}x + 6$$ Answer: (a) $y = -\frac{1}{2} x + 6$. Final answers: 38: (b) 39: (a) 40: (a)