Chapter 7.2, Problem 15CYU

To determine

To Simplify: The Given expression

$\frac{f^{-3}{v}^0{t}^5}{2r{t}^3}$

Answer to Problem 15CYU

The simplified expression is $\frac{t^2}{2r{f}^3}$ .

Explanation of Solution

Given information:

$\frac{f^{-3}{v}^0{t}^5}{2r{t}^3}$

Concept Used:

Quotient of powers: It states that to divide two powers with same base, subtract the exponent.

For example, let a be any non-zero number and m and n be any integer.

  $\frac{a^m}{a^n}=a^{m-n}$

Zero Exponent Property:Any non-zero number raised to the power zero is 1.

Example:let a be any non-zero number

  $a^0=1$

Negative Exponent Property: $a^{-n}$ is the reciprocal of $a^n$ , where a is a non-zero number and n is an integer.

  $a^{-n}=\frac{1}{a^n}$

Calculation:

  $\frac{f^{-3}{v}^0{t}^5}{2r{t}^3}$

(Group the powers with the same base)

  $=\left(\frac{f^{-3}{v}^0}{2r}\right)\left(\frac{t^5}{t^3}\right)$

Using Quotient of Powers

  $=\left(\frac{f^{-3}{v}^0}{2r}\right)t^{5-3}$

Simplify.

  $=\left(\frac{f^{-3}{v}^0}{2r}\right)t^2$

Applying Zero Exponent Property.

  $\begin{array}{l}=\left(\frac{f^{-3}\cdot 1}{2r}\right)t^2\\ =\left(\frac{f^{-3}}{2r}\right)t^2\end{array}$

Applying Negative Exponent Property

  $\begin{array}{l}=\left(\frac{1}{2r{f}^3}\right)t^2\\ =\frac{t^2}{2r{f}^3}\end{array}$