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.

Inequality: 30 - 3d ≤ 20Rate 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 + 3dSubtract 20 from both sides: 10 ≤ 3dDivide both sides by 3: 10/3 ≤ dApply 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.

The inequality is 30-3d≤20

The solution set is d≥10/3

Step-by-step explanation:

i got it from khan academy

He secknd step was the answer to them

look in attachments

step-by-step explanation:

a = (-2, 0)

b= (-4, 0)

c = (-5,2)

d = (-1,2)