Chapter 8.1, Problem 49PPE

(a)

To determine

To calculate: Equation of line p and q.

Answer to Problem 49PPE

Equation of p is $y=2x-1$ and q is $y=\frac{1}{2}x+3$

Explanation of Solution

Given information:

Points on line q:(2,4);(4,5)

Points on line p:(2,3);(3,5)

Formula used:

  $\begin{array}{l}y=mx+C\\ m=\frac{y_2-y_1}{x_2-x_1}\end{array}$

Calculation: 

Line of q:

  $\begin{array}{l}y=mx+C\\ m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{5-4}{4-2}\\ m=\frac{1}{2}\\ y=\frac{1}{2}x+C\\ 5=\frac{1}{2}(4)+C\\ C=3\\ y=\frac{1}{2}x+3\end{array}$

Line of p:

  $\begin{array}{l}y=mx+C\\ m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{5-3}{3-2}\\ m=2\\ y=2x+C\\ 5=2(3)+C\\ C=-1\\ y=2x-1\end{array}$

(b)

To determine

To calculate: Function of D(x).

Answer to Problem 49PPE

  $x=\frac{16}{3}$

Explanation of Solution

Given information: 

  $\begin{array}{l}y=\frac{1}{2}x+3\\ y=2x-1\end{array}$

Calculation: 

Common points of equation is:

  $\begin{array}{l}y=\frac{1}{2}x+3\\ -\frac{y=2x-1}{\begin{array}{l}\text{0}=\text{(}\frac{\text{1}}{\text{2}}\text{-2)x}+\text{4}\\ \text{0}=\text{-}\frac{\text{3}}{\text{4}}x+4\\ \frac{\text{3}}{\text{4}}x=4\\ x=\frac{16}{3}\\ y=\frac{17}{3}\end{array}}\end{array}$

Function of D(x)is $D(x)=\frac{16}{3}$

(c)

To determine

To calculate: for what value of x D(x)=0

Answer to Problem 49PPE

There is no value of x.

Explanation of Solution

Given information: 

  $D(x)=\frac{16}{3}$

Calculation: 

As the D(x) is a constant function so for any value D(x) will be $\frac{16}{3}$

(d)

To determine

To Explain:The function D(x) 

Answer to Problem 49PPE

It is a independent function.

Explanation of Solution

Given information: 

  $D(x)=\frac{16}{3}$

As the graph of D(x) is a straight line passing through x=3 and parallel to y axis so the value of x in the function D(x) is a constant function and it does not depend on the value of y so it is a independent function.