Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 1 - Section 1.7 - Addition and Subtraction of Fractions - Exercise - Page 40: 50


$\frac{1}{32}$ in

Work Step by Step

Diameter of 6011 welding rods = $\frac{1}{8}$ in. Diameter of super strength 100 rods = $\frac{3}{32}$ in. Difference of the diameters = $\frac{1}{8}$ in. - $\frac{3}{32}$ in =$\frac{1}{8}$x$\frac{4}{4}$ - $\frac{3}{32}$ = $\frac{4}{32}$ - $\frac{3}{32}$ = $\frac{1}{32}$ in
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.