Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 7 - Exponents and Exponential Functions - 7-1 Zero and Negative Exponents - Practice and Problem-Solving Exercises: 36

Answer

The answer is $\frac{14}{t^{5}m^{2}}$

Work Step by Step

The answer is $\frac{14}{t^{5}m^{2}}$ for the expression $\frac{7s^{0}t^{-5}}{2^{-1}m^{2}}$. To simplify this expression, we have to change the exponents from negative to positive, and the only way to do it is take the reciprocal of the negative exponent. Also, any number to the power of 0 becomes 1, so $s^{0}$=1. $\frac{7s^{0}t^{-5}}{2^{-1}m^{2}}$. = $\frac{7\times(1)\times2^{1}}{t^{5}m^{2}}$ = $\frac{7\times(1)\times2}{t^{5}m^{2}}$ = $\frac{14}{t^{5}m^{2}}$
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