Question

use for questions 7 - 9. mischa recorded the battery life of her laptop each hour after unplugging it.

1. find the rate of change after the first two hours of use after unplugging her laptop.
2. find the rate of change from two hours of use to two hours and 30 minutes of use.
3. find the rate of change from three hours and 30 minutes of use to five hours of use.

Explanation:

Question 7

Step1: Identify points

At \( t = 0 \), battery life \( y_1 = 100 \) (assuming the top point is 100). At \( t = 2 \), \( y_2 = 70 \).

Step2: Calculate rate of change

Rate of change formula: \( \frac{y_2 - y_1}{t_2 - t_1} \)
\( \frac{70 - 100}{2 - 0} = \frac{-30}{2} = -15 \)

Step1: Identify points

At \( t = 2 \) (2 hours), \( y_1 = 70 \). At \( t = 2.5 \) (2.5 hours), \( y_2 = 40 \) (from the graph at 3 hours? Wait, maybe correction: At \( t = 2 \), \( y = 70 \); at \( t = 3 \), \( y = 40 \)? Wait, 2.5 hours is between 2 and 3. Wait, the graph: at \( t = 2 \), \( y = 70 \); at \( t = 3 \), \( y = 40 \). So time difference \( 3 - 2 = 1 \) hour? No, 2 hours to 2.5 hours: \( t_1 = 2 \), \( t_2 = 2.5 \); \( y_1 = 70 \), \( y_2 = 40 \)? Wait, maybe the point at 2.5 hours (2 hours 30 minutes) is 50? Wait, the graph: at \( t = 3 \), the point is around 40? Wait, maybe I misread. Let's re-express: At \( t = 2 \) (2 hours), battery life is 70. At \( t = 2.5 \) (2.5 hours), let's see the graph: between 2 and 3, the point at \( t = 3 \) is 40? Wait, no, the graph has at \( t = 2 \), \( y = 70 \); at \( t = 3 \), \( y = 40 \). So time from 2 to 3 is 1 hour, but 2 to 2.5 is 0.5 hours. Wait, maybe the correct points: At \( t = 2 \), \( y = 70 \); at \( t = 3 \), \( y = 40 \). So for 2 hours to 2.5 hours (0.5 hours), the change: \( \frac{40 - 70}{3 - 2} = -30 \) per hour? Wait, no, 2 to 2.5: time difference 0.5 hours. If at \( t = 2 \), \( y = 70 \); at \( t = 2.5 \), \( y = 55 \)? No, maybe the graph: at \( t = 2 \), \( y = 70 \); at \( t = 3 \), \( y = 40 \). So the rate from 2 to 3 hours is \( \frac{40 - 70}{3 - 2} = -30 \). But 2 to 2.5 hours: since it's a straight line? Wait, maybe the problem is from 2 hours (t=2, y=70) to 3 hours (t=3, y=40), but the question is 2 hours to 2.5 hours. Wait, maybe the graph has at t=2, y=70; t=3, y=40. So the slope between 2 and 3 is \( \frac{40 - 70}{3 - 2} = -30 \). So for 0.5 hours (2 to 2.5), the rate would be the same? Wait, maybe the problem is considering from t=2 (70) to t=2.5 (let's say y=55? No, maybe the graph at t=2.5 is 50? Wait, maybe I made a mistake. Let's use the formula: rate of change = (change in y)/(change in t). At t=2 (2 hours), y=70. At t=2.5 (2.5 hours), y=40? No, that can't be. Wait, the graph: at t=0, y=100; t=2, y=70; t=3, y=40; t=4, y=30; t=5, y=0. So from t=2 (70) to t=3 (40): time 1 hour, change in y -30, so rate -30 per hour. But from t=2 to t=2.5 (0.5 hours), the change in y would be -15 (since -30 per hour, so -15 per 0.5 hours). Wait, maybe the correct calculation: \( t_1 = 2 \), \( y_1 = 70 \); \( t_2 = 2.5 \), \( y_2 = 55 \)? No, the graph at t=3 is 40, so linear between 2 and 3: slope \( \frac{40 - 70}{3 - 2} = -30 \). So for \( t_2 - t_1 = 0.5 \) hours, \( \frac{y_2 - 70}{0.5} = -30 \) → \( y_2 - 70 = -15 \) → \( y_2 = 55 \). But maybe the problem expects using t=2 (70) and t=3 (40), but the time is 1 hour. Wait, the question is from 2 hours to 2 hours 30 minutes (0.5 hours). So maybe the points are t=2 (70) and t=2.5 (let's say the point at 2.5 is 50? No, the graph at t=3 is 40. Wait, maybe the correct points: t=2 (70), t=3 (40). So time difference 1 hour, but the question is 0.5 hours. Wait, maybe the problem has a typo, but assuming t=2 (70) and t=3 (40), and time 1 hour, but the question is 0.5 hours. Alternatively, maybe the graph at t=2 is 70, t=2.5 is 50. Then rate of change: \( \frac{50 - 70}{2.5 - 2} = \frac{-20}{0.5} = -40 \). Wait, this is confusing. Maybe the correct approach: look at the graph, at t=2 (2 hours), y=70; at t=3 (3 hours), y=40. So from 2 to 3 hours (1 hour), rate is -30. From 2 to 2.5 hours (0.5 hours), the rate would be the same as the slope, which is -30 per hour. Wait, maybe the problem expects using t=2 (70) and t=3 (40), but the time is 1 hour, but the question is 0.5 hours. I think there's a misinterpretati…

Step1: Identify points

At \( t = 3.5 \) (3.5 hours), \( y_1 = 30 \) (wait, at t=4, y=30; at t=3, y=40; so at t=3.5, y=35? No, the graph at t=4 is 30, t=5 is 0. Wait, t=3.5 (3 hours 30 minutes) and t=5 (5 hours). So \( t_1 = 3.5 \), \( y_1 = 30 \) (at t=4, y=30; so at t=3.5, y=30? Wait, no: at t=4, y=30; at t=5, y=0. So \( t_1 = 3.5 \), \( y_1 = 30 \); \( t_2 = 5 \), \( y_2 = 0 \). Time difference \( 5 - 3.5 = 1.5 \) hours. Change in y \( 0 - 30 = -30 \).

Step2: Calculate rate

\( \frac{-30}{1.5} = -20 \)

Answer:

-15

Question 8