Chapter 39, Problem 22A

To determine

(a)

The value of $m+m{p}^2-r^2$.

Answer to Problem 22A

The value of $m+m{p}^2-r^2$ is $76$.

Explanation of Solution

Given:

The value of $m$ is $5$, $p$ is $4$ and $r$ is $3$.

Calculation:

The algebraic expression is given below:

  $m+m{p}^2-r^2$  ... (1)

Substitute $5$ for $m$, $4$ for $p$ and $3$ for $r$ in equation (1) as follows:

  $\begin{array}{c}m+m{p}^2-r^2=5+\left(5\times 4^2\right)-3^2\\ =5+80-9\\ =85-9\\ =76\end{array}$

Thus, the value of $m+m{p}^2-r^2$ is $76$.

Conclusion:

The value of $m+m{p}^2-r^2$ is $76$.

To determine

(b)

The value of ${\left(p+2\right)}^2{\left(m-r\right)}^2$.

Answer to Problem 22A

The value of ${\left(p+2\right)}^2{\left(m-r\right)}^2$ is $144$.

Explanation of Solution

Given:

The value of $m$ is $5$, $p$ is $4$ and $r$ is $3$.

Calculation:

The algebraic expression is given below:

  ${\left(p+2\right)}^2{\left(m-r\right)}^2$  ... (1)

Substitute $5$ for $m$, $4$ for $p$ and $3$ for $r$ in equation (1) as follows:

  $\begin{array}{c}{\left(p+2\right)}^2{\left(m-r\right)}^2={\left(4+2\right)}^2\times {\left(5-3\right)}^2\\ ={\left(6\right)}^2\times {\left(2\right)}^2\\ =36\times 4\\ =144\end{array}$

Thus, the value of ${\left(p+2\right)}^2{\left(m-r\right)}^2$ is $144$.

Conclusion:

The value of ${\left(p+2\right)}^2{\left(m-r\right)}^2$ is $144$.

To determine

(c)

The value of $\frac{{\left(pr\right)}^2}{2}-pr+m^3$.

Answer to Problem 22A

The value of $\frac{{\left(pr\right)}^2}{2}-pr+m^3$ is $185$.

Explanation of Solution

Given:

The value of $m$ is $5$, $p$ is $4$ and $r$ is $3$.

Calculation:

The algebraic expression is given below:

  $\frac{{\left(pr\right)}^2}{2}-pr+m^3$  ... (1)

Substitute $5$ for $m$, $4$ for $p$ and $3$ for $r$ in equation (1) as follows:

  $\begin{array}{c}\frac{{\left(pr\right)}^2}{2}-pr+m^3=\frac{{\left(4\times 3\right)}^2}{2}-\left(4\times 3\right)+5^3\\ =\frac{144}{2}-12+125\\ =72+113\\ =185\end{array}$

Thus, the value of $\frac{{\left(pr\right)}^2}{2}-pr+m^3$ is $185$.

Conclusion:

The value of $\frac{{\left(pr\right)}^2}{2}-pr+m^3$ is $185$.

To determine

(d)

The value of $\frac{p^3+3p-12}{m^2+15}$.

Answer to Problem 22A

The value of $\frac{p^3+3p-12}{m^2+15}$ is $1.5$.

Explanation of Solution

Given:

The value of $m$ is $5$, $p$ is $4$ and $r$ is $3$.

Calculation:

The algebraic expression is given below:

  $\frac{p^3+3p-12}{m^2+15}$  ... (1)

Substitute $5$ for $m$ and $4$ for $p$ in equation (1) as follows:

  $\begin{array}{c}\frac{p^3+3p-12}{m^2+15}=\frac{4^3+3\times 4-12}{5^2+15}\\ =\frac{64+12-12}{25+15}\\ =\frac{64}{40}\\ =1.6\end{array}$

Thus, the value of $\frac{p^3+3p-12}{m^2+15}$ is $1.5$.

Conclusion:

The value of $\frac{p^3+3p-12}{m^2+15}$ is $1.5$.

To determine

(e)

The value of $\frac{r^3}{3p-9}+m^2{\left(mp-6r\right)}^2$.

Answer to Problem 22A

The value of $\frac{r^3}{3p-9}+m^2{\left(mp-6r\right)}^2$ is $109$.

Explanation of Solution

Given:

The value of $m$ is $5$, $p$ is $4$ and $r$ is $3$.

Calculation:

The algebraic expression is given below:

  $\frac{r^3}{3p-9}+m^2{\left(mp-6r\right)}^2$  ... (1)

Substitute $5$ for $m$, $4$ for $p$ and $3$ for $r$ in equation (1) as follows:

  $\begin{array}{c}\frac{r^3}{3p-9}+m^2{\left(mp-6r\right)}^2=\frac{3^3}{3\times 4-9}+5^2{\left(5\times 4-6\times 3\right)}^2\\ =\frac{27}{3}+25{\left(2\right)}^2\\ =9+100\\ =109\end{array}$

Thus, the value of $\frac{r^3}{3p-9}+m^2{\left(mp-6r\right)}^2$ is $109$.

Conclusion:

The value of $\frac{r^3}{3p-9}+m^2{\left(mp-6r\right)}^2$ is $109$.

To determine

(f)

The value of $\frac{m^3-p^2+5}{r^2}+\frac{p+2m}{m+p+r}$.

Answer to Problem 22A

The value of $\frac{m^3-p^2+5}{r^2}+\frac{p+2m}{m+p+r}$ is $13.83$.

Explanation of Solution

Given:

The value of $m$ is $5$, $p$ is $4$ and $r$ is $3$.

Calculation:

The algebraic expression is given below:

  $\frac{m^3-p^2+5}{r^2}+\frac{p+2m}{m+p+r}$  ... (1)

Substitute $5$ for $m$, $4$ for $p$ and $3$ for $r$ in equation (1) as follows:

  $\begin{array}{c}\frac{m^3-p^2+5}{r^2}+\frac{p+2m}{m+p+r}=\frac{5^3-4^2+5}{3^2}+\frac{4+2\times 5}{5+4+3}\\ =\frac{125-16+5}{9}+\frac{14}{12}\\ =\frac{144}{9}+\frac{14}{12}\\ =\frac{456+42}{36}\end{array}$

Simplify the above equation as follows:

  $\begin{array}{c}\frac{m^3-p^2+5}{r^2}+\frac{p+2m}{m+p+r}=\frac{456+42}{36}\\ =\frac{498}{36}\\ =13.83\end{array}$

Thus, the value of $\frac{m^3-p^2+5}{r^2}+\frac{p+2m}{m+p+r}$ is $13.83$.

Conclusion:

The value of $\frac{m^3-p^2+5}{r^2}+\frac{p+2m}{m+p+r}$ is $13.83$.