Basic Math Operations - Mathematics and Statistics for the Environment - page 55

What would have happened if we had multiplied the dimensions instead of dividing?

gal
ft

×

3 3 3

ft × ( gal/ft ) = ft

3

Multiply the numerator and denominator to obtain

3

ft × gal

3

ft

and cancel common terms to obtain

3

ft × gal

3

ft

Obviously, by multiplying the two dimensions (ft 3 and gal/ft 3 ), the answer will be in gallons,
which is what we want. Thus, because the math setup is correct, we would then multiply the num-
bers to obtain the number of gallons:

(140 0 ft 3 ) × (7.48 gal l /f ft3) 3 ) = 10,472 gal

Now, let's try another problem with exponents. We wish to obtain an answer in square feet. If we
are given the two terms—70 ft 3 /s and 4.5 ft/s—is the following math setup correct?

(70 ft 3 /s) × (4.5 ft /s)

First, only the dimensions are used to determine if the math setup is correct. By multiplying the two
dimensions, we get

3

ft
s

ft
s

2

(ft /s) ft/s × () = ×

Multiply the terms in the numerators and denominators of the fraction:

3 4

2

ft ft
ss

×
×

ft
s

=

Obviously, the math setup is incorrect because the dimensions of the answer are not square feet;
therefore, if we multiply the numbers as shown above, the answer will be wrong.
Let's try division of the two dimensions instead:

3

ft
s
ft
s

3

(ft /s) =

Invert the denominator and multiply to get

3

( ) × ( ) × ft

s

s
ft

ft ×× ft ft s ft ×× ft ft s

= ft 2

=×= =

s t × s t ×