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.
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
i got it from khan academy
look in attachments
a = (-2, 0)
b= (-4, 0)
c = (-5,2)
d = (-1,2)