## Algebra 1

Published by Prentice Hall

# Chapter 7 - Exponents and Exponential Functions - 7-1 Zero and Negative Exponents - Practice and Problem-Solving Exercises - Page 417: 36

#### Answer

$\frac{14}{m^{2}t^{5}}$

#### Work Step by Step

1. First, get rid of the zero exponent. Remember that anything to the power of zero become $1$. So here, $s^{0}$ will become $1$. 2a. Now, we want to make the negative exponents positive. To do that, we're going to start by moving the "$t^{-5}$" from the top, to the bottom of our fraction. Now we should have "$\frac{7}{2^{-1}m^{2}t^{5}}$" 2b. After that, we're going to move the "$2^{-1}$" to the top. This would make our fraction look like this: "$\frac{7(2^{1})}{m^{2}t^{5}}$" 3. Finally, we just have to simplify the top of our fraction. Remember that a value to the power of one will always be equal to that same value. So in our case, $2^{1}$ will equal $2$. With that, our fraction now looks like "$\frac{7(2)}{m^{2}t^{5}}$" Here, all we need to do is multiply $7$ by $2$ and we have our answer of $\frac{14}{m^{2}t^{5}}$.

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.