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Chapter 2: Problem 16

find the distance between each pair of points. If necessary, round answers to two decimals places. $$ \left(\frac{7}{3}, \frac{1}{5}\right) \text { and }\left(\frac{1}{3}, \frac{6}{5}\right) $$

Short Answer

Expert verified
The distance between the points \((\frac{7}{3}, \frac{1}{5})\) and \((\frac{1}{3}, \frac{6}{5})\) is approximately \(2.24\).

Step by step solution

01

Identify the coordinates

The first point is \((\frac{7}{3}, \frac{1}{5})\) and the second point is \((\frac{1}{3}, \frac{6}{5})\). Therefore, \(x_1=\frac{7}{3}\), \(y_1=\frac{1}{5}\), \(x_2=\frac{1}{3}\), and \(y_2=\frac{6}{5}\).
02

Substitute the coordinates into the distance formula

The formula is \(\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}\). Substituting the coordinates into the formula yields: \(\sqrt{{(\frac{1}{3}-\frac{7}{3})}^2+{(\frac{6}{5}-\frac{1}{5})}^2}\)
03

Simplify the expression

Simplify each subtraction inside the square roots: \(\sqrt{{(-2)}^2+1^2} = \sqrt{4+1} = \sqrt{5}\).
04

Convert the radical to decimal form

\(\sqrt{5}\) is approximately \(2.23607\). Round this value to two decimal places, yielding \(2.24\).

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Most popular questions from this chapter

Exercises \(98-100\) will help you prepare for the material covered in the first section of the next chapter. In Exercises \(98-99,\) solve each quadratic equation by the method of your choice. Use the graph of \(f(x)=x^{2}\) to graph \(g(x)=(x+3)^{2}+1\)

Exercises \(129-131\) will help you prepare for the material covered in the next section. Use point plotting to graph \(f(x)=x+2\) if \(x \leq 1\)

What is the graph of a function?

complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$ x^{2}-2 x+y^{2}-15=0 $$

determine whether each statement makes sense or does not make sense, and explain your reasoning. My graph of \((x-2)^{2}+(y+1)^{2}=16\) is my graph of \(x^{2}+y^{2}=16\) translated two units right and one unit down.
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