Respuesta :
[tex]Hmm...\ and\ use\ \pi\approx\dfrac{22}{7}\ \ \ ;)\\\\the\ diameter\ (d):d=2r\\\\therefore:\ d=28\ in\to2r=28\ in\ \ \ |divide\ both\ sides\ by\ 2\\\\r=14\ in\\--------------\\The\ formula:C_O=2\pi r\ where\ r=14\ in\ and\ \pi\approx\dfrac{22}{7}\\\\subtitute:\\\\C_O\approx2\cdot\dfrac{22}{7}\cdot14=2\cdot22\cdot2=88\ (in)\\\\\huge\boxed{C_O=88\ in.}\leftarrow\boxed{c.}[/tex]