## Elementary Technical Mathematics

Let the specialist needs two pieces of duct 3$\frac{3}{4}$ ft and 2$\frac{1}{4}$ ft in length. the excess that will be left after the 3$\frac{3}{4}$ ft length iscut off from the 4 ft long piece = 4 - 3$\frac{3}{4}$ ft = 4 - $\frac{15}{4}$ ft = 4 $\times$ $\frac{4}{4}$ - $\frac{15}{4}$ ft =$\frac{16}{4}$ - $\frac{15}{4}$ ft = $\frac{1}{4}$ ft the excess that will be left after the 2$\frac{1}{4}$ ft length iscut off from the 4 ft long piece = 4 - 2$\frac{1}{4}$ ft = 4 -$\frac{9}{4}$ft = 4 $\times$ $\frac{4}{4}$ - $\frac{9}{4}$ft ft =$\frac{16}{4}$ - $\frac{9}{4}$ft ft = $\frac{7}{4}$ ft Total excess left = $\frac{1}{4}$ + $\frac{7}{4}$ = $\frac{8}{4}$ 2 ft