Question

The temperature of a pot of water was 180.3 degrees, and it cooled at a rate of minus 2.5 degrees per minute. What was the temperature after 20 minutes? How many minutes did it take to cool from 180.3 degrees to 100.3 degrees?

Answer 1

Answer:

We want to know the temperature after 20 minutes if we replace x =20 we got:

$y = 180.3 -2.5*20 = 130.3$

Now we need to find how many minutes did it take to cool from 180.3 degrees to 100.3 degrees, so we can set up the following equation:

$100.3 = 180.3 - 2.5x$

And solving for x we got:

$x = \frac{100.3-180.3}{-2.5}= 32 minutes$

Step-by-step explanation:

For this case we can use a lineal model given by:

$y= mx+b$

Where mis the slope and b the initial value so then based on the info provided the model would be:

$y = 180.3 -2.5 x$

Where x represent the time and y the temeperature.

We want to know the temperature after 20 minutes if we replace x =20 we got:

$y = 180.3 -2.5*20 = 130.3$

Now we need to find how many minutes did it take to cool from 180.3 degrees to 100.3 degrees, so we can set up the following equation:

$100.3 = 180.3 - 2.5x$

And solving for x we got:

$x = \frac{100.3-180.3}{-2.5}= 32 minutes$