Question

The swimming pool is open when the high temperature is higher than 20∘C. 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 30 ∘C 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:

$30-3d\leq 20$

Step by step explanation:

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 told that the pool was open on Monday. The high temperature was $30^oC$ on Monday. So the decrease in temperature from Monday to Thursday will be $30-3d$.

We have been given that the swimming pool is open when the high temperature is higher than $20^oC$.

As the pool was closed on Thursday. This means that temperature on Thursday was less than or equal to 20 degree Celsius. We can represent this information in an inequality as:

$30-3d\leq 20$

Therefore, the inequality $30-3d\leq 20$ can be used to determine the rate of temperature decrease in degrees Celsius per day, d, from Monday to Thursday.

Answer 2

Answer:

30 - 3d ≤ 20

Step-by-step explanation:

The temperature on Monday was 30°.  This is where our inequality begins.

The temperature has a steady decrease; this is why we subtract.

The decrease occurs over the next 3 days; letting d represent days, this gives us 3d.

The temperature on Thursday was low enough for the pool to be closed; this means it was less than or equal to 20.  

Altogether this gives us

30 - 3d ≤ 20

To solve this, we will subtract 30 from each side:

30 - 3d - 30 ≤ 20 - 30

-3d ≤ -10

Divide both sides by -3:

-3d/-3 ≤ -10/-3  (remember that we flip the inequality symbol when we divide by a negative)

d ≥ 3.33