Elementary Technical Mathematics

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

Chapter 2 - Section 2.7 - Engineering Notation - Exercises - Page 130: 31

Answer

$$15.6 \times 10^{-18}. $$

Work Step by Step

We simplify the numerator and denominator by adding the exponents and multiplying the like terms and then by subtracting the exponents and dividing the like terms, as follows: $$\frac{ (5.15\times 10^{9} )(65.3\times 10^{-6} )}{(27\times 10^{6} )(800\times 10^{12} )}=\frac{ (5.15 )(65.3 )\times 10^{9-6}}{(27 )(800)\times 10^{6+12} } \\=\frac{336.295\times 10^{3}}{21600\times 10^{18} }=\frac{336.295}{21600 }\times 10^{3-18}\\ =0.0156\times 10^{-15}=15.6 \times 10^{-18}. $$
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.