Answer:

So it would takes approximately 6.9 hours to reach 32 F.
Step-by-step explanation:
For this case we have the following differential equationÑ

We can reorder the expression like this:

We can use the substitution
and
so then we have:

IF we integrate both sides we got:

If we apply exponential in both sides we got:

And if we replace w = u-T we got:

We can also express the solution in the following terms:

For this case we know that
since w ehave a cooloing,
, we have this model:
And if we want that the temperature would be 32F we can solve for t like this:



If we apply natural logs on both sides we got:


So it would takes approximately 6.9 hours to reach 32 F.
13 factored out is 13x - 39 add 39 to both sides it equals 13x=78 and 78 divided by 13 is 6 the anwser is x=6
For x y chart 19.36.2.27.13

and so on, and then, you graph those x,y coordinates
A translation is a rigid transformation, which means angle measures and distances from two points do not change.
Polygon ABCDEF is congruent to polygon A'B'C'D'E'F'
DE = D'E' = 7