Question

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The swimming pool is open when the high temperature is higher than 20∘C. Lainey tried to swim on Monday and Thursday (which was 3 days later). The pool was open on Monday, but it was closed on Thursday. The high temperature was 30∘C on Monday, but decreased at a constant rate in the next 3 days. Write an inequality to determine the rate of temperature decrease in degrees Celsius per day, d, from Monday to Thursday.

Answer 1

Answer:

• Inequality: 30 - 3d ≤ 20

• Rate of decrease from Monday to Thursday: greater than or equal to 10/3 degrees Celsius per day

Explanation:

Translate theconditions into mathematical terms:

• HIgh temperature on Monday = T₀ = 30°C.

• The temperature decrease at a constant rate in the next 3 days. Call d such rate, so T₃ = T₀ - 3d = 30 - 3d

That the pool is open when the high temperature is higher than 20°C means that, if the pool is closed, then the temperature is less than or equal to (≤) 20°C:

• 30 - 3d ≤ 20

And that is the requested inequality.

From that, you can solve for the rate of temperature decrease, d:

• Add 3d to both sides: 30 ≤ 20 + 3d

• Subtract 20 from both sides: 10 ≤ 3d

• Divide both sides by 3: 10/3 ≤ d

• Apply symmetry property: d ≥ 10/3

Hence, the rate of temperature decrease is greater than or equal to 10/3 degrees Celsius per day, from Monday to Thursday.

Answer 2

Answer:

The inequality is 30-3d≤20

The solution set is d≥10/3

Step-by-step explanation:

i got it from khan academy