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Question
Two friends bought phones that cost different amounts but had identical batteries. After using the phones for a while, both friends had to turn them off to let the batteries cool. This table shows how the temperature of each battery changed over 5 minutes.

Answer 1

I don't see the table of temperatures — can you paste the table or a picture of it? Once I have the values I can compute rates and answer whatever the question asks.

Meanwhile, here’s how to analyze this kind of problem so you know what I’ll do once you share the data:

1. Identify the two temperature series: T1(t) and T2(t), where t is minutes after they were turned off (t = 0, 1, 2, 3, 4, 5).

2. Check if the temperature change looks linear:
- Compute successive differences (T(t+1) − T(t)). If the differences are roughly constant, a linear model is appropriate.

3. Find the cooling rate (slope) for each phone:
- Average rate = (T(final) − T(initial)) / (final time − initial time) = (T(5) − T(0)) / 5.
- Units: degrees per minute (°/min). A more negative slope = faster cooling.

4. Write linear equations if appropriate:
- T1(t) = T1(0) + m1·t
- T2(t) = T2(0) + m2·t
where m1 and m2 are the slopes found in step 3.

5. Answer typical questions:
- Which battery cooled faster? Compare slopes (more negative is faster).
- How much did each battery cool in 5 minutes? Compute T(5) − T(0).
- When are temperatures equal? Solve T1(t) = T2(t) for t (check that solution is in [0,5]).
- What is the temperature at a specific time? Plug t into the linear equation.

Example (illustrative):
- Suppose phone A: T at t=0 is 45°C and at t=5 is 30°C → slope = (30−45)/5 = −3 °/min, so T_A(t)=45−3t.
- Suppose phone B: T at t=0 is 42°C and at t=5 is 33°C → slope = (33−42)/5 = −1.8 °/min, so T_B(t)=42−1.8t.
- Which cools faster? Phone A (−3 < −1.8).
- When equal? 45−3t = 42−1.8t → 3−1.2t = 0 → t = 2.5 min.

Paste the actual table and I’ll do the real calculations and answer the specific question.