Chapter 9.4, Problem 1GP

a.

To determine

To state whetherrelationship between the number of seconds and the radius of circle 1 is direct variation.

Answer to Problem 1GP

The relationship between the number of seconds and the radius of circle 1 is inverse variation.

Explanation of Solution

Given:

The graph of circle 1 with two points (2,5) and (6,4)

Concept Used:

Slope of a line is given by $m=\frac{y_2-y_1}{x_2-x_1}$

If the slope of the line is positive, then it is direct variation.

If the slope of the line is negative, then it is inverse variation.

Calculation:

Here,

  $\begin{array}{l}{x}_1=2,x_2=6\\ y_1=5,y_2=4\end{array}$

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{4-5}{6-2}\\ =-\frac{1}{4}\\ =-0.25\end{array}$

Since, m is negative

Therefore,

The relationship between the number of seconds and the radius of circle 1 is inverse variation.

b.

To determine

To state whether relationship between the number of seconds and the radius of circle 2 is direct variation.

Answer to Problem 1GP

The relationship between the number of seconds and the radius of circle 2 is direct variation.

Explanation of Solution

Given:

The graph of circle 2 with two points (2,3) and (4,6)

Concept Used:

Slope of a line is given by $m=\frac{y_2-y_1}{x_2-x_1}$

If the slope of the line is positive, then it is direct variation.

If the slope of the line is negative, then it is inverse variation.

Calculation:

Here,

  $\begin{array}{l}{x}_1=2,x_2=4\\ y_1=3,y_2=6\end{array}$

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ \text{}=\frac{6-3}{4-2}\\ \text{}=\frac{3}{2}\\ \text{}=1.50\end{array}$

Since, m is positive.

Therefore,

The relationship between the number of seconds and the radius of circle 2 is direct variation.