Question

one spring day, jordan noted the time of day and the temperature, in degrees fahrenheit. during that time, the temperature rose, stayed steady, and fell, at different rates. on the set of axes below, jordan created a graph of temperature over time.
which of the following statements is true of the temperature at 8 p.m.?
answer
it was dropping at a rate of 1°f every 2 hours.
it was rising at a rate of 1°f every 2 hours.
it was dropping at a rate of 2°f every 1 hour.
it was rising at a rate of 2°f every 1 hour.

Explanation:

Step1: Analyze the graph's time and temp

At 6 pm (18:00), temp is 64°F. At midnight (24:00), temp is 57°F? Wait, no, let's check the points. Wait, the falling segment: from 6 pm (let's say 6 pm is time \( t_1 \), temp \( T_1 = 64 \)) to, say, 8 pm ( \( t_2 = 20 \) ), what's the temp? Wait, the graph: after 6 pm, the temp starts falling. Let's take two points: at 6 pm (x=6 pm, y=64) and at midnight (x=12 am, y=57? Wait, no, the points: one at 6 pm (64), then a point at, say, 8 pm? Wait, the options are about rate. Let's check the slope. The falling part: from 6 pm (64°F) to, say, 8 pm (let's see the grid). Wait, the time axis: 6 am, Noon (12 pm), 6 pm, Midnight (12 am). So between 6 pm and midnight (6 hours), the temp goes from 64 to, let's see the last point: around 57? Wait, no, the options: let's check the rate. The first option: dropping at 1°F every 2 hours. Let's calculate slope (rate) for the falling segment. Let's take two points: at 6 pm (time \( t = 6 \) pm, temp \( T = 64 \)) and at 8 pm ( \( t = 8 \) pm, temp \( T = 63 \))? Wait, no, maybe better: from 6 pm (64) to midnight (12 am, 6 hours later), temp goes from 64 to 57? No, the last point is around 57? Wait, no, the graph: the falling line starts at 6 pm (64) and goes to a point at, say, 8 pm (63), 10 pm (62), midnight (57)? Wait, no, the options: let's check the rate. The slope (rate of change) is \( \frac{\Delta T}{\Delta t} \). For the falling segment, \( \Delta T \) is negative (dropping). Let's take two points: at 6 pm (64°F) and at 8 pm (let's see the grid: each square is 1 hour? Wait, the x-axis: 6 am to Noon is 6 hours, so each grid square is 1 hour. So from 6 pm (x=6 pm) to, say, 8 pm (x=8 pm), that's 2 hours. The temp at 6 pm is 64, at 8 pm: let's see the y-axis, each square is 1°F. So at 8 pm, temp is 63? Wait, no, the point after 6 pm: the first falling point is at, say, 8 pm? Wait, the graph: after 6 pm, the temp decreases. Let's take two points: (6 pm, 64) and (8 pm, 63). Then \( \Delta T = 63 - 64 = -1 \), \( \Delta t = 8 - 6 = 2 \) hours. So rate is \( \frac{-1}{2} \) °F per hour, which is dropping at 1°F every 2 hours. Let's check other options:

• "rising at 1°F every 2 hours": no, it's falling.
• "dropping at 2°F every 1 hour": rate would be -2, but our calculation is -0.5 (1°F per 2 hours), so no.
• "rising at 2°F every 1 hour": no, falling.

So the first option: "It was dropping at a rate of \( 1^\circ \)F every 2 hours" is correct.

Step2: Verify each option

• Option 1: Dropping 1°F every 2 hours. As calculated, slope \( \frac{-1}{2} \) (°F per hour), so every 2 hours, \( -1 \) °F (dropping 1°F). Correct.
• Option 2: Rising? No, it's falling. Eliminate.
• Option 3: Dropping 2°F per hour? Rate would be -2, but our slope is -0.5. Eliminate.
• Option 4: Rising? No, falling. Eliminate.

Answer:

It was dropping at a rate of \( 1^\circ \)F every 2 hours.