#### Explanation:

Did you know you can, and may, treat units of measurement in the same way you do the numbers. Very useful in applied maths, physics, engineering and so on.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Looking at just the units as a guide to how to go about solving this.

The objective is to end up with just $We are told that there is a decrease of$ per hour: written as " "$/h So to change $/h into just $ we multiply by $h$So her have: $/hxxh" "->" "($1.20)/(1 h)xx3.5h =>1.2xx3.5xx$/(cancel(h))xxcancel(h)

=> 3.00xx $ normally written as $3.00

As this is a decrease you could if you chose write this as

The change is -\$3.00