Part B:
The solution in Part 1 is -39.1, so we can berak this result in tens, ones and theths so:

So if we sum the we have the same answer

Part C:
the operation we can rewrite is:

and using the distribution propertie we can write it like:

Part D:
and then we operate using the assosiative propertie:

Part E:
Finally we simplyfy the expression:

Part F:
now he meassure the temperature of the water and it dropes -0.6 FÂș so we have to rest this to the last temperature so:
