Answer:
\(11\ ft\)
Step-by-step explanation:
we have
\(4\frac{1}{2}\ ft\)
\(4\ ft\)
\(2\frac{1}{2}\ ft\)
we know that
The length of all three tables when pushed end to end is the sum of the three lengths
Convert mixed number to an improper fraction
\(4\frac{1}{2}\ ft=\frac{4*2+1}{2}=\frac{9}{2}\ ft\)
\(2\frac{1}{2}\ ft=\frac{2*2+1}{2}=\frac{5}{2}\ ft\)
therefore
\(\frac{9}{2}\ ft+4\ ft+\frac{5}{2}\ ft\)
Rewrite \(4\ ft\) as \(\frac{8}{2}\ ft\)
so
\(\frac{9}{2}\ ft+\frac{8}{2}\ ft+\frac{5}{2}\ ft=\frac{22}{2}\ ft=11\ ft\)
