## Elementary Technical Mathematics

19$\frac{27}{32}$ in
Let the diameter of the first circle = 12$\frac{5}{8}$ in therefore radius of the first circle = $\frac{1}{2}$ x $\frac{101}{8}$ = $\frac{101}{16}$ in Let the diameter of the last circle = 15$\frac{9}{16}$ in therefore radius of the last circle = $\frac{1}{2}$ x $\frac{249}{16}$ = $\frac{249}{32}$ in Therefore the length of x = $\frac{101}{16}$ in + 5$\frac{3}{4}$ in + $\frac{249}{32}$ in = $\frac{101}{16}\times$ $\frac{2}{2}$ + $\frac{23}{4}\times$ $\frac{8}{8}$ +$\frac{249}{32}$ =$\frac{202}{32}$+$\frac{184}{32}$ + $\frac{249}{32}$ = $\frac{635}{32}$ =19$\frac{27}{32}$ in