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Chapter 1: Problem 44

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

Short Answer

Expert verified
The expression \(\left(\frac{16}{9}\right)^{-1 / 2}\) evaluates to \(\frac{3}{4}\).

Step by step solution

01

Understand Negative Exponentiation

First, let's recall that a negative exponent, like \((-1/2)\), indicates the reciprocal of the base raised to the corresponding positive exponent. Thus, \\(\left(\frac{16}{9}\right)^{-1/2} = \frac{1}{\left(\frac{16}{9}\right)^{1/2}}\).
02

Simplify the Reciprocal

Since \\(\left(\frac{16}{9}\right)^{1/2}\) is under the reciprocal, we can treat it as \(\sqrt{\frac{16}{9}}\), which is equivalent to simplifying the square root of a fraction.
03

Apply the Square Root to the Fraction

Calculate the square root of the fraction by taking the square root of the numerator and the denominator separately:\[\sqrt{\frac{16}{9}} = \frac{\sqrt{16}}{\sqrt{9}}.\]
04

Evaluate the Square Roots

The square root of 16 is 4, and the square root of 9 is 3, so:\[\frac{\sqrt{16}}{\sqrt{9}} = \frac{4}{3}.\]
05

Finalize the Evaluation

Substitute back the result into the reciprocal:\[\left(\frac{16}{9}\right)^{-1/2} = \frac{1}{\frac{4}{3}}.\]Then, find the reciprocal of \\(\frac{4}{3}\) to get the final answer, which is \\(\frac{3}{4}\).

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Key Concepts

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

Fractional Exponents
Fractional exponents provide a wonderful way to express roots. The numerator of a fractional exponent indicates the power, while the denominator signifies the root. For instance, in the expression \(a^{m/n}\), \(m\) is the power and \(n\) is the root.
For this exercise, when we see the exponent \(-1/2\), we need to understand two parts:
  • The negative sign, which tells us to find the reciprocal.
  • The fraction \(1/2\), which means we're looking for the square root.
Therefore, the expression \(\left(\frac{16}{9}\right)^{-1/2}\) becomes the reciprocal of the square root of \(\frac{16}{9}\), simplifying the process of finding the solution by breaking it down into clearer operations.
Square Roots
Square roots are a fundamental aspect of managing fractional exponents. They simplify expressions by finding a number which, when multiplied by itself, gives the original value. In our example, when we calculate the square root of a fraction like \(\frac{16}{9}\), it is important to apply the square root to both the numerator and the denominator separately.
This results in:
  • \(\sqrt{16} = 4\)
  • \(\sqrt{9} = 3\)
Thus, \(\sqrt{\frac{16}{9}} = \frac{4}{3}\).
Understanding how to split and simplify using square roots makes working with fractions and roots much more manageable.
Reciprocals
Reciprocals are integral when working with negative exponents and help to simplify complex expressions. A reciprocal of a number is simply\(\frac{1}{x}\), where \(x\) is any given number or expression.
For example, when tasked with simplifying \(\frac{1}{\frac{4}{3}}\), you take the reciprocal of \(\frac{4}{3}\). This operation turns the fraction upside down, leading to \(\frac{3}{4}\).
Reciprocals therefore play a key role in transforming expressions to their simplest forms when negative exponents are involved. They help step through the calculations and lead to clear and concise final results.

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Most popular questions from this chapter

GENERAL: Impact Velocity If a marble is dropped from a height of \(x\) feet, it will hit the ground with velocity \(v(x)=\frac{60}{11} \sqrt{x}\) miles per hour (neglecting air resistance). Use this formula to find the velocity with which a marble will strike the ground if it is dropped from the height of the tallest building in the United States, the 1776 -foot One World Trade Center in New York City.

Should \(\sqrt{9}\) be evaluated as 3 or \(\pm 3 ?\)

When defining \(x^{m / n}\), why did we require that the exponent \(\frac{\mathrm{m}}{\mathrm{u}}\) be fully reduced? [Hint: \((-1)^{2 / 3}=(\sqrt[3]{-1})^{2}=1,\) but with an equal but unreduced exponent you get \((-1)^{4 / 6}=(\sqrt[6]{-1})^{4}\). Is this defined?]

ALLOMETRY: Heart Rate It is well known that the hearts of smaller animals beat faster than the hearts of larger animals. The actual relationship is approximately (Heart rate) \(=250(\text { Weight })^{-1 / 4}\) where the heart rate is in beats per minute and the weight is in pounds. Use this relationship to estimate the heart rate of: A 625 -pound grizzly bear.

Write each expression in power form \(a x^{b}\) for numbers \(a\) and \(b\). $$ \sqrt[3]{\frac{8}{x^{6}}} $$
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