Five warm-ups for grades 9–12. Each takes three to five minutes. Answers and teacher notes are in the collapsed box under each prompt.
ELA
A rhetorical device is a pattern of language that gives a sentence extra force. Match each sentence with one device from this bank: anaphora, antithesis, asyndeton, chiasmus, rhetorical question. Then write one sentence of your own, on any topic, that uses two of the devices.
- We will walk until the road ends, we will walk until our legs give out, we will walk until we are home.
- The words were small; the silence they left was enormous.
- Don’t work for the weekend; make the weekend work for you.
- She packed, paid, left, never looked back.
- Who among us has never wanted one more day of summer?
Answer and teacher notes
1. Anaphora: repeats the same words at the start of each clause.
2. Antithesis: sets opposite ideas side by side in balanced form.
3. Chiasmus: reverses the order of key words in the second half (work, weekend / weekend, work).
4. Asyndeton: leaves out conjunctions between items.
5. Rhetorical question: asks for effect, not for an answer.
Your sentence: Answers will vary. Sample: “We train in the rain, we train in the cold, and what do we get? Stronger.”
Common mistake: confusing antithesis with chiasmus. Antithesis needs opposite ideas. Chiasmus needs a reversed word order, and the ideas don’t have to be opposites. Ask students what each device does to the reader: asyndeton speeds a sentence up, and anaphora builds a rhythm that makes an idea feel certain. These sample sentences were written for practice. Extension: find one of these devices in a speech, advertisement, or song title, then explain what it adds that a plain sentence would lose.
Math
A prize wheel at a community fair has 20 equal sections. One spin costs $3. One section pays $20, three sections pay $5, six sections pay $2, and ten sections pay nothing. (These numbers are made up for practice.) (a) Find the expected payout for one spin. (b) Find the expected net gain or loss for a player. (c) If the wheel is spun 400 times, about how much money should the fair expect to keep? (d) What is the probability that a spin pays at least the $3 cost? (e) The fair wants the expected payout to be exactly $2. If only the $20 prize changes, what should the new prize be?
Answer and teacher notes
(a) (1 × 20 + 3 × 5 + 6 × 2 + 10 × 0) ÷ 20 = 47 ÷ 20 = $2.35. (b) 2.35 − 3 = −$0.65, a loss of 65 cents per spin on average. (c) 400 × 0.65 = $260. (d) The $20 and $5 sections: 4 ÷ 20 = 0.20, or 20%. (e) (x + 15 + 12) ÷ 20 = 2, so x + 27 = 40, and x = $13.
Check on (e): (13 + 15 + 12) ÷ 20 = 40 ÷ 20 = 2. ✓ Common mistake on (a): averaging the four prizes (20, 5, 2, 0) without weighting them by how many sections each covers. Remind students that expected value is a long-run average. No single spin pays $2.35. On (c), $260 is an estimate because 400 spins won’t match the probabilities exactly. Extension: find the standard deviation of one spin’s payout. (E[X²] = (400 + 75 + 24) ÷ 20 = 24.95. Variance = 24.95 − 2.35² ≈ 19.43. Standard deviation ≈ $4.41.)
Science
Tire pressure warning lights often come on during the first cold mornings of fall. A tire holds a nearly fixed volume of air, so its pressure follows Gay-Lussac’s law: P1/T1 = P2/T2, where P is absolute pressure and T is absolute temperature in kelvins. A gauge shows pressure above the atmosphere, which is about 14.7 psi. A tire reads 35 psi on a gauge at 70°F, and the temperature falls overnight to 30°F. (a) Convert both temperatures to kelvins. (b) Find the absolute pressure at 70°F, the new absolute pressure at 30°F, and the new gauge reading. (c) About how much pressure was lost per 10°F? CBS Detroit reports a loss of roughly 1 to 2 psi for every 10-degree drop. How does your answer compare? (d) Suppose someone leaves out the 14.7 psi and uses 35 psi directly. What new reading would they get? (e) Explain the pressure drop in terms of air molecules.
Answer and teacher notes
(a) K = (°F − 32) × 5/9 + 273.15. 70°F ≈ 294.26 K, and 30°F ≈ 272.04 K. (b) Absolute pressure: 35 + 14.7 = 49.7 psi. New absolute pressure: 49.7 × 272.04 ÷ 294.26 ≈ 45.95 psi. New gauge reading: 45.95 − 14.7 ≈ 31.25 psi. (c) The loss is about 3.75 psi over 40°F, or about 0.9 psi per 10°F. That is close to the low end of the reported range. (d) 35 × 272.04 ÷ 294.26 ≈ 32.4 psi, a drop of only about 2.6 psi, which understates the real change. (e) Cooler molecules move more slowly, so they hit the tire wall less often and with less force. That means less pressure.
Common mistake: using Fahrenheit or Celsius in the ratio. The law only works with an absolute scale. Check on (b): 45.95 ÷ 272.04 ≈ 0.1689, and 49.7 ÷ 294.26 ≈ 0.1689. ✓ Real tires can flex, so the volume stays only nearly constant, and this is why a simple model and a reported rule of thumb may not match exactly. Extension: at what Fahrenheit temperature would this tire’s gauge read 30 psi? (44.7 ÷ 49.7 × 294.26 ≈ 264.66 K, or about 16.7°F.)
Sources: LibreTexts: Gay-Lussac’s Law (statement and equation); CBS Detroit: Cold Air and Tire Pressure (1 to 2 psi per 10 degrees)
Social Studies
UNESCO’s constitution came into force on November 4, 1946, after twenty of its signatories accepted it. Its preamble says that “wars begin in the minds of men.” In the 1960s, the organization faced a real test. Egypt was building the Aswan High Dam to boost farming and electric power, and its reservoir would have flooded ancient monuments in Nubia. UNESCO launched an international campaign in 1960. At Abu Simbel, workers cut the temples into more than 1,000 blocks of about 30 tons each and rebuilt them 64 meters higher and 180 meters inland. The whole campaign cost $80 million, half of it donated by some 50 countries. (a) How many years ago did UNESCO’s constitution come into force? (b) Estimate the total mass of the blocks. (c) List one benefit of the dam and one cost. (d) Why might countries pay to save monuments in another country? (e) How might saving shared heritage connect to the preamble’s idea about where peace begins?
Answer and teacher notes
(a) 2026 − 1946 = 80 years. (b) 1,000 × 30 = about 30,000 tons at minimum, since “more than 1,000” is a lower limit. (c) Benefit: more farmland and electricity. Cost: flooding of monuments and land. (d) Answers will vary: the monuments matter to everyone’s history, a shared project builds goodwill, and once lost they could not be replaced. (e) Answers will vary: working together and valuing other cultures can change how people think about each other.
Source check: UNESCO says more than 1,000 blocks of about 30 tons, while JSTOR Daily says over 7,000 blocks of 20 to 30 tons. Ask students how to handle the disagreement: they might check what each one is counting, or look for a primary record. UNESCO reports that the campaign ran from 1960 to 1980, saved 22 monuments and complexes, and gave rise to the World Heritage Convention. The temples reopened on September 22, 1968. The constitution was done at London on November 16, 1945, and Wikipedia reports that 41 countries signed it at that conference. Extension: who else lived in the area that would be flooded, and what happened to them? Encourage students to research this as part of weighing the project.
Sources: UNESCO World Heritage Centre: The Story of Abu Simbel (dam, campaign, blocks, heights, cost, World Heritage link); JSTOR Daily: Moving Abu Simbel out of the Nile (block count); Constitution of UNESCO (preamble, Article XV, adoption); Wikipedia: UNESCO (date in force, 41 signatories)
Current Events
The European Space Agency’s Hera spacecraft is closing in on the asteroid pair Didymos and Dimorphos. On October 18, from about 1,280 km away, it took its first clear view of them, and detailed images of Dimorphos are scheduled for release on November 25. Hera will study what happened after NASA’s DART spacecraft struck Dimorphos on September 26, 2022, at about 6.1 km/s. Dimorphos, about 151 meters across, orbits Didymos at an average distance of 1.15 km and a speed of about 18 cm/s. The impact shortened its orbit from 11 hours 55 minutes to 11 hours 23 minutes. (a) By what percent did the orbital period decrease? (b) Treat the orbit as a circle and estimate its period using T = 2πr/v. Compare your result with the measured times. (c) Hera slowed to about 20 m/s relative to the asteroids. How many times faster was DART’s impact speed? (d) Johns Hopkins APL says debris blasted off the asteroid boosted DART’s push. Use conservation of momentum to explain why. (e) Why might scientists choose to test on a small moon instead of a lone asteroid?
Answer and teacher notes
(a) 11 h 55 min = 715 min, and the change is 32 min. 32 ÷ 715 ≈ 4.5%. (b) T = 2π(1,150 m) ÷ (0.18 m/s) ≈ 40,143 s ≈ 11.2 hours. That is close to the measured 11 to 12 hours. The inputs are rounded averages, so an exact match isn’t expected. (c) 6,100 ÷ 20 = 305 times. (d) Debris thrown backward carries momentum in that direction. To conserve total momentum, the asteroid gains extra momentum in the other direction, so the push is larger than the spacecraft’s momentum alone. (e) Answers will vary: a small moon’s orbit is short and easy to track, so a small change shows up in telescope measurements from Earth.
Common mistake on (b): forgetting to convert units. Change km to meters and cm/s to m/s first. APL compares the debris effect to air rushing out of a balloon, and it reported a change of 32 minutes with an uncertainty of about 2 minutes. ScienceDaily rounds the change to roughly 33 minutes, so students should notice how uncertainty appears in reporting. The debris reached as far as 10,000 km into space. Extension: why would slowing Dimorphos make its orbit shorter? (A slower moon falls into a smaller orbit, which it completes in less time.)
Sources: ScienceDaily: NASA Smashed Into an Asteroid. Now Hera Is Closing In (Hera dates and speeds, asteroid sizes, orbit distance and speed, debris); Johns Hopkins APL: DART Impact Changed Asteroid’s Motion (orbital period before and after, debris, measurement)
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