Answer: Let's assume the high temperature on Monday is represented by C30 (30 degrees Celsius). Since the pool is open when the high temperature is higher than 20 degrees Celsius, the pool was open on Monday.
However, over the next three days, the high temperature decreased at a constant rate. Let's denote the rate of decrease as "r" (in degrees Celsius per day).
Since the high temperature on Monday was C30, we can calculate the high temperature on Thursday by subtracting the decrease in temperature over three days:
High temperature on Thursday = C30 - 3r
We know that the pool was closed on Thursday, so the high temperature on Thursday must have been lower than or equal to 20 degrees Celsius.
Therefore, we can set up the inequality:
C30 - 3r ≤ 20
Now, we can solve this inequality to find the range of values for the rate of decrease (r) that would satisfy the condition:
C30 - 3r ≤ 20
Substituting C30 = 30, we have:
30 - 3r ≤ 20
Subtracting 30 from both sides:
-3r ≤ -10
Dividing by -3 (and reversing the inequality since we are dividing by a negative number):
r ≥ 10/3
Therefore, for the pool to be closed on Thursday, the rate of decrease in temperature (r) must be greater than or equal to 10/3 degrees Celsius per day.