The swimming pool is open when the high temperature is higher than 20^\circ\text{C}20∘C20, degrees, start text, C, end text. Lainey tried to swim on Monday and Thursday (which was 333 days later). The pool was open on Monday, but it was closed on Thursday. The high temperature was 30^\circ\text{C}30∘C30, degrees, start text, C, end text on Monday, but decreased at a constant rate in the next 333 days.

Inequality: 30 - 3d ≤ 20

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

Step-by-step 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.

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step-by-step explanation: