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Chapter 1: Problem 35

Find the distance between the points. $$ \left(\frac{1}{2}, \frac{4}{3}\right),(2,-1) $$

Short Answer

Expert verified
The distance between the two points is \(1.5\).

Step by step solution

01

Identify The Coordinates

Recognize the given coordinates. Here, the first point is \((\frac{1}{2}, \frac{4}{3})\) designated as \((x_1, y_1)\) and the second point is \((2, -1)\) designated as \((x_2, y_2)\).
02

Substitute Into The Distance Formula

Insert the given values into the distance formula: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).This will give: \(\sqrt{(2 - \frac{1}{2})^2 + (-1 - \frac{4}{3})^2}\).
03

Simplify The Expression

Simplify the expression within the square root to obtain: \(\sqrt{(1.5)^2+(-\frac{7}{3})^2} = \sqrt{2.25 + \frac{49}{9}}\).
04

Final Calculation

After unifying the denominators and simplifying the expression, you get: \(\sqrt{\frac{81}{36}} = \frac{9}{6} = 1.5\).

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Most popular questions from this chapter

Use the given value of \(k\) to complete the table for the direct variation model $$y=k x^{2}$$ Plot the points on a rectangular coordinate system. $$\begin{array}{|l|l|l|l|l|l|} \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y=k x^{2} & & & & & \\ \hline \end{array}$$ $$ k=\frac{1}{2} $$

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(a) find the inverse function of \(f\), (b) graph both \(f\) and \(f^{-1}\) on the same set of coordinate axes, (c) describe the relationship between the graphs of \(f\) and \(f^{-1}\), and (d) state the domain and range of \(f\) and \(f^{-1}\). $$ f(x)=\frac{4}{x} $$

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