Chapter 8.2, Problem 40PPS

a.

To determine

To find: the simplified expression of $R\left(x\right)$ .

Answer to Problem 40PPS

The simplified expression is $\frac{260x^2+400}{x\left(3x^2+20\right)}$ .

Explanation of Solution

Given information:

The expression is $R\left(x\right)=\frac{20}{x}+\frac{200x}{3x^2+20}$ .

Calculation: to find the simplified expression, first take LCM of denominator and then simplify,

  $\begin{array}{l}R\left(x\right)=\frac{20}{x}+\frac{200x}{3x^2+20}\\ R\left(x\right)=\frac{20\left(3x^2+20\right)+200x\left(x\right)}{x\left(3x^2+20\right)}\text{ [LCM of denominator]}\\ \text{R}\left(x\right)=\frac{260x^2+400}{x\left(3x^2+20\right)}\text{ [simplify]}\end{array}$

Thus, the simplified expression is $\frac{260x^2+400}{x\left(3x^2+20\right)}$ .

b.

To determine

To find: the rate when oil will pumping from the well in $50$ years.

Answer to Problem 40PPS

The rate is $1.72$ when oil will pumping from the well in $50$ years.

Explanation of Solution

Given information:

The expression is $R\left(x\right)=\frac{20}{x}+\frac{200x}{3x^2+20}$ .

Calculation: to find the rate, substitute $x=50$ in the expression,

  $\begin{array}{l}R\left(x\right)=\frac{260x^2+400}{x\left(3x^2+20\right)}\\ R\left(50\right)=\frac{260{\left(50\right)}^2+400}{50\left(3{\left(50\right)}^2+20\right)}\text{ [substitute]}\\ R\left(50\right)=\frac{650000+400}{50\left(7500+20\right)}\text{ [square and multiply]}\\ R\left(50\right)=\frac{650400}{376000}\text{ [simplify]}\\ R\left(50\right)=1.72\text{ [divide]}\end{array}$

Thus, the rate is $1.72$ .