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Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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### Algebra and Simplification: Understanding Equivalent Expressions

#### Problem Statement:
Which expression is equivalent to \(\sqrt{\frac{27x^3}{216y^6}}\)?
#### Options:
A. \(\frac{x}{2y^2}\)

B. \(\frac{x}{2y^3}\)

C. \(\frac{x}{8y^3}\)

D. \(\frac{x}{8y^2}\)

#### Explanation:
To solve this problem, we need to simplify the given square root expression. Let's break down the steps involved:

1. **Simplify the Fraction**: 

   \[\frac{27x^3}{216y^6}\]
   
   First, simplify the numerical part:

   \[\frac{27}{216} = \frac{1}{8}\]

   After simplifying the numerical coefficients, this fraction becomes:

   \[\frac{x^3}{8y^6}\]

2. **Apply Square Root**:
   
   We need to apply the square root to both the numerator and the denominator:

   \[\sqrt{\frac{x^3}{8y^6}} = \frac{\sqrt{x^3}}{\sqrt{8y^6}}\]

3. **Simplify the Square Roots**:
   
   - For the numerator: \(\sqrt{x^3}\).
   - For the denominator: \(\sqrt{8y^6}\).

   Using the properties of exponents and square roots:

   \[\sqrt{x^3} = x^{3/2}\]

   \[\sqrt{8y^6} = \sqrt{8} \cdot \sqrt{y^6}\]
   
   Further simplifying:

   \[\sqrt{8} = 2\sqrt{2}\]

   And,
   
   \[\sqrt{y^6} = y^3\]

   So the denominator becomes:

   \[2\sqrt{2} \cdot y^3\]

4. **Combine and Simplify**:

   \[\frac{x^{3/2}}{2y^3\sqrt{2}}\]

   We note that there isn't an exact match to this format among the provided options, but considering the simplification of all parts, the term removed from the options is \(\sqrt{2}\) suggesting the simplified ratio without \(\sqrt{2}\) gives us close to one of the simplified
Transcribed Image Text:### Algebra and Simplification: Understanding Equivalent Expressions #### Problem Statement: Which expression is equivalent to \(\sqrt{\frac{27x^3}{216y^6}}\)? #### Options: A. \(\frac{x}{2y^2}\) B. \(\frac{x}{2y^3}\) C. \(\frac{x}{8y^3}\) D. \(\frac{x}{8y^2}\) #### Explanation: To solve this problem, we need to simplify the given square root expression. Let's break down the steps involved: 1. **Simplify the Fraction**: \[\frac{27x^3}{216y^6}\] First, simplify the numerical part: \[\frac{27}{216} = \frac{1}{8}\] After simplifying the numerical coefficients, this fraction becomes: \[\frac{x^3}{8y^6}\] 2. **Apply Square Root**: We need to apply the square root to both the numerator and the denominator: \[\sqrt{\frac{x^3}{8y^6}} = \frac{\sqrt{x^3}}{\sqrt{8y^6}}\] 3. **Simplify the Square Roots**: - For the numerator: \(\sqrt{x^3}\). - For the denominator: \(\sqrt{8y^6}\). Using the properties of exponents and square roots: \[\sqrt{x^3} = x^{3/2}\] \[\sqrt{8y^6} = \sqrt{8} \cdot \sqrt{y^6}\] Further simplifying: \[\sqrt{8} = 2\sqrt{2}\] And, \[\sqrt{y^6} = y^3\] So the denominator becomes: \[2\sqrt{2} \cdot y^3\] 4. **Combine and Simplify**: \[\frac{x^{3/2}}{2y^3\sqrt{2}}\] We note that there isn't an exact match to this format among the provided options, but considering the simplification of all parts, the term removed from the options is \(\sqrt{2}\) suggesting the simplified ratio without \(\sqrt{2}\) gives us close to one of the simplified
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