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Use \(<,>,\) or \(=\) to make each statement true. a. \(\sqrt{3}+\sqrt{7} \underline{?} \sqrt{3+7}\) b. \(\sqrt{3^{2}+2^{2}} \underline{?} 5\) c. \(\sqrt{5^{2}-4^{2}} \geq 2\)

Short Answer

Expert verified
a. > b. < c. \geq \.

Step by step solution

01

Simplify the left-hand side of (a)

First, compute the value of \(\backslash sqrt{3}+\backslash sqrt{7}\). The approximate values are \(\backslash sqrt{3} \approx 1.732\) and \(\backslash sqrt{7} \approx 2.646\). Therefore, \(\backslash sqrt{3} + \backslash sqrt{7} \approx 4.378\).
02

Simplify the right-hand side of (a)

Next, compute the value of \(\backslash sqrt{3+7}\). This simplifies to \(\backslash sqrt{10} \approx 3.162\).
03

Compare values for (a)

Compare \(\backslash sqrt{3}+\backslash sqrt{7}\) and \(\backslash sqrt{10}\). Since \(\backslash sqrt{3}+\backslash sqrt{7} \approx 4.378\) is greater than \(\backslash sqrt{10} \approx 3.162\), the correct inequality is \(\backslash sqrt{3}+\backslash sqrt{7} > \backslash sqrt{10}\).
04

Simplify the left-hand side of (b)

First, compute \(\backslash sqrt{3^{2}+2^{2}}\). This simplifies to \(\backslash sqrt{9+4} = \backslash sqrt{13}\).
05

Compare \(\backslash sqrt{13}\) to 5 for (b)

Approximate the value of \(\backslash sqrt{13} \approx 3.606\). Since \(\backslash sqrt{13}\) is less than 5, the correct inequality is \(\backslash sqrt{13} < 5\).
06

Simplify the left-hand side of (c)

First, compute \(\backslash sqrt{5^{2} - 4^{2}}\). This simplifies to \(\backslash sqrt{25 - 16} = \backslash sqrt{9} = 3\).
07

Compare \(\backslash sqrt{9}\) to 2 for (c)

Since 3 is greater than 2, the inequality \(\backslash sqrt{5^{2} - 4^{2}} \backslash geq 2\) is true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Square Roots
Understanding square roots is essential for solving many algebra problems. A square root of a number is a value that, when multiplied by itself, gives the original number. For example, the square root of 9 is 3 because 3 multiplied by 3 equals 9.

In notation, the square root of a number is represented as \( \sqrt{n} \). It is important to know that every positive number has two square roots: a positive and a negative (e.g., \( \sqrt{9} = \pm 3 \)). Generally, when we talk about square roots, we refer to the positive root unless otherwise stated.

Note: Square roots of non-perfect squares (like 7) result in irrational numbers with non-repeating decimal sequences. We often approximate such values for simplicity. For example, \( \sqrt{7} \approx 2.646 \).
  • Always simplify inside the radical first before taking the square root.
  • Use approximate values for quick comparisons but be aware of the exact irrational values.
Inequality Comparisons
Inequality comparisons involve determining the relationship between two expressions. The main symbols used are:
  • \( < \) less than
  • \( > \) greater than
  • \( = \) equal to
  • \( \leq \) less than or equal to
  • \( \geq \) greater than or equal to
These symbols are used to compare the values on the left and right sides of the inequality. For example, \( 3 > 2 \) means 3 is greater than 2.

When working with square roots and other expressions, it is important to first simplify both sides of the inequality or equation before comparing them.

In example (a), we simplified each side separately: \( \sqrt{3} + \sqrt{7} \approx 4.378 \) and \( \sqrt{10} \approx 3.162 \). We then compared the simplified results: \( 4.378 > 3.162 \), leading to \( \sqrt{3} + \sqrt{7} > \sqrt{10} \).
  • Always simplify and approximate where necessary before making comparisons.
  • If both sides involve square roots, try approximating the values.
  • Double-check calculations to ensure accuracy in comparisons.
Simplifying Expressions
Simplifying expressions makes it easier to solve equations and inequalities. Simplification involves breaking down complex expressions into their simplest forms.

Let's see how this applies:
In example (b), we started with \( \sqrt{3^{2} + 2^{2}} \). We simplified the inside of the square root: \( 3^{2} = 9 \) and \( 2^{2} = 4 \). Therefore, \( 3^{2} + 2^{2} = 9 + 4 = 13 \), resulting in \( \sqrt{13} \approx 3.606 \).

Similarly, in example (c), we dealt with \( \sqrt{5^{2} - 4^{2}} \). We simplified inside the square root: \( 5^{2} = 25 \) and \( 4^{2} = 16 \). Thus, \( 5^{2} - 4^{2} = 25 - 16 = 9 \), giving us \( \sqrt{9} = 3 \).
  • Always perform operations inside parentheses or roots first.
  • Combine like terms where possible before taking square roots.
  • Simplify step-by-step to avoid errors and to make comparisons easier.
In summary, simplify expressions as much as possible to make analysis and comparison straightforward.

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Most popular questions from this chapter

Rewrite the following equations using logs instead of exponents. Estimate a solution for \(x\) and then check your estimate with a calculator. Round the value of \(x\) to three decimal places. a. \(10^{x}=153\) b. \(10^{x}=153,000\) c. \(10^{x}=0.125\) d. \(10^{x}=0.00125\)

Simplify each expression by removing all possible factors from the radical, then combining any like terms. a. \(2 \sqrt{50}+12 \sqrt{8}\) b. \(3 \sqrt{27}-2 \sqrt{75}\) c. \(10 \sqrt{32}-6 \sqrt{18}\) d. \(2 \sqrt[3]{16}+4 \sqrt[3]{54}\)

Radio waves, sent from a broadcast station and picked up by the antenna of your radio, are a form of electromagnetic (EM) radiation, as are microwaves, X-rays, and visible, infrared, and ultraviolet light. They all travel at the speed of light. Electromagnetic radiation can be thought of as oscillations like the vibrating strings of a violin or guitar or like ocean swells that have crests and troughs. The distance between the crest or peak of one wave and the next is called the wavelength. The number of times a wave crests per minute, or per second for fast-oscillating waves, is called its frequency. Wavelength and frequency are inversely proportional: the longer the wavelength, the lower the frequency, and vice versa-the faster the oscillation, the shorter the wavelength. For radio waves and other \(\mathrm{EM}\), the number of oscillations per second of a wave is measured in hertz, after the German scientist who first demonstrated that electrical waves could transmit information across space. One cycle or oscillation per second equals 1 hertz \((\mathrm{Hz})\). For the following exercise you may want to find an old radio or look on a stereo tuner at the AM and FM radio bands. You may see the notation \(\mathrm{kHz}\) beside the AM band and MHz beside the FM band. AM radio waves oscillate at frequencies measured in the kilohertz range, and FM radio waves oscillate at frequencies measured in the megahertz range. a. The Boston FM rock station WBCN transmits at \(104.1 \mathrm{MHz}\). Write its frequency in hertz using scientific notation. b. The Boston AM radio news station WBZ broadcasts at 1030 \(\mathrm{kHz}\). Write its frequency in hertz using scientific notation. The wavelength \(\lambda\) (Greek lambda) in meters and frequency \(\mu\) (Greek mu) in oscillations per second are related by the formula \(\lambda=\frac{c}{\mu}\) where \(c\) is the speed of light in meters per second. c. Estimate the wavelength of the WBCN FM radio transmission. d. Estimate the wavelength of the WBZ AM radio transmission. e. Compare your answers in parts (c) and (d), using orders of magnitude, with the length of a football field (approximately 100 meters).

Justify the following rule for exponents. Consider the case of \(n \geq m\) and assume \(m\) and \(n\) are integers \(>0 .\) If \(a\) is any nonzero real number, then $$ \frac{a^{n}}{a^{m}}=a^{(n-m)} $$

Write each of the following in standard decimal form: a. \(7.23 \cdot 10^{5}\) d. \(1.5 \cdot 10^{6}\) b. \(5.26 \cdot 10^{-4}\) e. \(1.88 \cdot 10^{-4}\) c. \(1.0 \cdot 10^{-3}\) f. \(6.78 \cdot 10^{7}\)

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