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Chapter 1: Problem 56

(a) plot the points, (b) find the distance between the points, and (c) find the midpoint of the line segment joining the points. $$ (-16.8,12.3),(5.6,4.9) $$

Short Answer

Expert verified
The distance between the points is given by the square-root of the sum of the squares of the differences between corresponding components of the two points. The midpoint of the line segment joining the points is given by the mean of their corresponding components.

Step by step solution

01

Plot the points

To plot the points (-16.8, 12.3) and (5.6, 4.9), place a dot at the relevant positions in a coordinate system where the first number in each pair corresponds to the x-coordinate and the second number to the y-coordinate.
02

Find the distance between the points

To find the distance between the points, the formula \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) is used. Using this on the coordinates (-16.8, 12.3) and (5.6, 4.9), the equation \(\sqrt{(5.6 - (-16.8))^2 + (4.9 - 12.3)^2}\) is obtained. Solving this will give the distance.
03

Find the midpoint of the line segment

To find the midpoint of the line segment, the formula \((x_1 + x_2)/2, (y_1 + y_2)/2\) is used. Applying these formulas to the coordinates (-16.8, 12.3) and (5.6, 4.9) results in \((-16.8 + 5.6)/2, (12.3 + 4.9)/2\). Evaluating these equations yields the coordinates of the midpoint.

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Most popular questions from this chapter

Use the functions given by \(f(x)=x+4\) and \(g(x)=2 x-5\) to find the specified function. $$ g^{-1} \circ f^{-1} $$

The winning times (in minutes) in the women's 400-meter freestyle swimming event in the Olympics from 1948 through 2008 are given by the following ordered pairs. $$\begin{array}{lll} (1948,5.30) & (1972,4.32) & (1996,4.12) \\ (1952,5.20) & (1976,4.16) & (2000,4.10) \\ (1956,4.91) & (1980,4.15) & (2004,4.09) \\ (1960,4.84) & (1984,4.12) & (2008,4.05) \\ (1964,4.72) & (1988,4.06) & \\ (1968,4.53) & (1992,4.12) & \end{array}$$ A linear model that approximates the data is \(y=-0.020 t+5.00,\) where \(y\) represents the winning time (in minutes) and \(t=0\) represents \(1950 .\) Plot the actual data and the model on the same set of coordinate axes. How closely does the model represent the data? Does it appear that another type of model may be a better fit? Explain. (Source: International Olympic Committee)

Use the given value of \(k\) to complete the table for the inverse variation model $$y=\frac{k}{x^{2}}$$ Plot the points on a rectangular coordinate system. $$\begin{array}{|l|l|l|l|l|l|} \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y=\frac{k}{x^{2}} & & & & & \\ \hline \end{array}$$ $$ k=10 $$

Determine whether the variation model is of the form \(y=k x\) or \(y=k / x,\) and find \(k .\) Then write \(a\) model that relates \(y\) and \(x\). $$ \begin{array}{|c|c|c|c|c|c|} \hline x & 5 & 10 & 15 & 20 & 25 \\ \hline y & 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\ \hline \end{array} $$

Use the functions given by \(f(x)=x+4\) and \(g(x)=2 x-5\) to find the specified function. $$ g^{-1} \circ f^{-1} $$
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