/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Evaluate each expression without... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate each expression without using a calculator. $$ \left(\frac{16}{9}\right)^{-1 / 2} $$

Short Answer

Expert verified
The value of \( \left(\frac{16}{9}\right)^{-1 / 2} \) is \( \frac{3}{4} \).

Step by step solution

01

Understand the Negative Exponent

The expression given is \( \left(\frac{16}{9}\right)^{-1 / 2} \). A negative exponent means that we take the reciprocal of the base. Therefore, \( x^{-a} = \frac{1}{x^a} \). So, \( \left(\frac{16}{9}\right)^{-1 / 2} \) becomes \( \left(\frac{9}{16}\right)^{1/2} \).
02

Apply the Square Root

The exponent \( \frac{1}{2} \) means we need to find the square root of the base \( \frac{9}{16} \). So, \( \left(\frac{9}{16}\right)^{1/2} = \sqrt{\frac{9}{16}} \).
03

Calculate the Square Root

The square root of a fraction \( \frac{a}{b} \) is \( \frac{\sqrt{a}}{\sqrt{b}} \). Thus, \( \sqrt{\frac{9}{16}} = \frac{\sqrt{9}}{\sqrt{16}} \).
04

Simplify the Result

The square root of 9 is 3, and the square root of 16 is 4. Therefore, \( \frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4} \).

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Key Concepts

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

Reciprocal
A reciprocal is a fundamental concept in mathematics that involves flipping a fraction. If you have a fraction \( \frac{a}{b} \), its reciprocal is \( \frac{b}{a} \). In simple terms, the numerator becomes the denominator and vice-versa.
A reciprocal is particularly important when dealing with negative exponents. For example, when you encounter a term like \( x^{-a} \), you can think of it as \( \frac{1}{x^a} \), which transforms a multiplication operation into a division.
In the given exercise, the expression \( \left(\frac{16}{9}\right)^{-1 / 2} \) uses the reciprocal concept due to the negative exponent. We take the reciprocal of \( \frac{16}{9} \), and it becomes \( \frac{9}{16} \). This step simplifies the expression, allowing us to proceed with other mathematical operations like square roots.
Square Root
The square root is a special operation in mathematics that finds a number, which when multiplied by itself, yields the original number. The square root of a number \( x \) is symbolically represented as \( \sqrt{x} \). It effectively reverses the squaring process.
When you're dealing with fractions and their square roots, you take the square root of the numerator and the denominator separately.
Using the exercise example, after applying the reciprocal, we have \( \left(\frac{9}{16}\right)^{1/2} \). The exponent \( \frac{1}{2} \) indicates that we need to take a square root. So, the expression becomes \( \sqrt{\frac{9}{16}} \), which is further simplified to \( \frac{\sqrt{9}}{\sqrt{16}} \).
This translates to taking the square root of 9, which is 3, and the square root of 16, which is 4. Therefore, the square root of \( \frac{9}{16} \) is \( \frac{3}{4} \).
Exponent Rules
Exponent rules are a collection of guidelines that help simplify mathematical operations involving powers. An understanding of these rules is essential for simplifying expressions like the one in the exercise.
Here are some key rules:
  • Product of Powers Rule: When multiplying like bases, you can add their exponents \( x^a \times x^b = x^{a+b} \).
  • Power of a Power Rule: If you have a power of a power, you multiply the exponents \( (x^a)^b = x^{a \cdot b} \).
  • Power of a Product Rule: Distribute the exponent to each factor in the product: \( (xy)^a = x^a y^a \).
  • Negative Exponent Rule: Rewrites negative exponents as reciprocals: \( x^{-a} = \frac{1}{x^a} \).
In our exercise, the crucial exponent rule applied is the Negative Exponent Rule. It helps transition \( \left(\frac{16}{9}\right)^{-1/2} \) to \( \left(\frac{9}{16}\right)^{1/2} \) by converting the negative exponent into a positive one through the reciprocal. Mastery of these rules accelerates problem-solving and enhances comprehension.

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