Question

The swimming pool is open when the high temperature is higher than 20 degrees celcius. Lainey tried to swim on Monday and Thursday (which was 33 days later). The pool was open on Monday, but it was closed on Thursday. The high temperature was 30C on Monday, but decreased at a constant rate in the next 33 days. Write an inequality to determine the rate of temperature decrease in degrees Celsius per day, d, from Monday to Thursday.

Answer 1

Answer:

30 - 33·d ≤ 20 --> d ≥ 10/33 = 0.3030 C°/day

Answer 2

Answer:

$30-3d\leq 20$

Step-by-step explanation:

Given :

The swimming pool is open when the high temperature is higher than 20 degrees Celsius.

Lainey tried to swim on Monday and Thursday (which was 3 days later).

To Find: . Write an inequality to determine the rate of temperature decrease in degrees Celsius per day, d, from Monday to Thursday.

Solution :

The pool was open on Monday, but it was closed on Thursday.

The high temperature was 30°C on Monday.

Let d represent the rate of temperature decrease in degrees Celsius per day from Monday to Thursday.  

As temperature decreased at a constant rate in the next 3 days.

So the rate of temperature will decrease 3d degree Celsius in 3 days.  

We are also given that  The swimming pool is open when the high temperature is higher than 20 degrees Celsius.

Since it was closed on Thursday . So, temperature on Thursday was less than 20°C.

So, the inequality becomes:

$30-3d\leq 20$

Hence an inequality to determine the rate of temperature decrease in degrees Celsius per day, d, from Monday to Thursday:

$30-3d\leq 20$