Chapter 7.2, Problem 72PFA

a.

To determine

To express: The product $(v^{-2}{w}^3{x}^2{y}^{-6})(v{w}^{-2}{x}^{-3}{y}^4)$ as a fraction with all the current factors having positive exponents.

Answer to Problem 72PFA

The product $(v^{-2}{w}^3{x}^2{y}^{-6})(v{w}^{-2}{x}^{-3}{y}^4)$ can be expressed as $(\frac{w^3{x}^2}{v^2{y}^6})(\frac{v{y}^4}{w^2{x}^3})$

Explanation of Solution

Given information:The product $(v^{-2}{w}^3{x}^2{y}^{-6})(v{w}^{-2}{x}^{-3}{y}^4)$

In the first term $(v^{-2}{w}^3{x}^2{y}^{-6})$ ,

1. $v^{-2}$ can be expressed as a factor with positive exponent as $\frac{1}{v^2}$,
2. $w^3$ already has positive exponent 3,
3. $x^2$ already has positive exponent 2,
4. $y^{-6}$ can be expressed as a factor with positive exponent as $\frac{1}{y^6}$ .

Thus, $(v^{-2}{w}^3{x}^2{y}^{-6})$ can be expressed as $(\frac{w^3{x}^2}{v^2{y}^6})$

In the second term $(v{w}^{-2}{x}^{-3}{y}^4)$ ,

1. $v$ already has positive exponent 1,
2. $w^{-2}$ can be expressed as a factor with positive exponent as $\frac{1}{w^2}$ ,
3. $x^{-3}$ can be expressed as a factor with positive exponent as $\frac{1}{x^3}$ ,
4. $y^4$ already has positive exponent 4.

Thus, $(v{w}^{-2}{x}^{-3}{y}^4)$ can be expressed as $(\frac{v{y}^4}{w^2{x}^3})$

Conclusion:

The product $(v^{-2}{w}^3{x}^2{y}^{-6})(v{w}^{-2}{x}^{-3}{y}^4)$ can be expressed as $(\frac{w^3{x}^2}{v^2{y}^6})(\frac{v{y}^4}{w^2{x}^3})$

b.

To determine

To find: The simplified form of the expression $(\frac{w^3{x}^2}{v^2{y}^6})(\frac{v{y}^4}{w^2{x}^3})$

Answer to Problem 72PFA

D. $\frac{w}{vx{y}^2}$

Explanation of Solution

Given information:The product $(v^{-2}{w}^3{x}^2{y}^{-6})(v{w}^{-2}{x}^{-3}{y}^4)$

Calculation:

  $\begin{array}{l}(\frac{w^3{x}^2}{v^2{y}^6})(\frac{v{y}^4}{w^2{x}^3})\\ =\frac{w^3\times x^2\times v\times y^4}{v^2\times y^6\times w^2\times x^3}\\ =\frac{w^{3-2}}{v^{2-1}\times x^{3-2}\times y^{6-4}}\\ =\frac{w}{vx{y}^2}\end{array}$

Conclusion:

The simplified form of the expression $(\frac{w^3{x}^2}{v^2{y}^6})(\frac{v{y}^4}{w^2{x}^3})$ is $\frac{w}{vx{y}^2}$

c.

To determine

To find: The variables which cannot equal 0.

Answer to Problem 72PFA

A. $v$

C. $x$

D. $y$

Explanation of Solution

Given information:The product $(v^{-2}{w}^3{x}^2{y}^{-6})(v{w}^{-2}{x}^{-3}{y}^4)$

  $(\frac{w^3{x}^2}{v^2{y}^6})(\frac{v{y}^4}{w^2{x}^3})=\frac{w}{vx{y}^2}$

From the product, it is observed that the numerator contains the variable $w$ and the denominator contain the variables $v,x$ and $y$ .

Thus, $v,x$ and $y$ cannot be equal to 0 because if denominator is 0, the fraction becomes undefined.

Conclusion:

  $v,x$ and $y$ cannot be equal to 0.