/*! 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 27 Express each repeating decimal a... [FREE SOLUTION] | 91Ó°ÊÓ

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Express each repeating decimal as a fraction. $$0 . \overline{432}$$

Short Answer

Expert verified
The repeating decimal \(0.\overline{432}\) is equivalent to \(\frac{16}{37}\).

Step by step solution

01

Identify the repeating part of the decimal

The decimal given is \(0.\overline{432}\), meaning the digits \(432\) repeat indefinitely.
02

Set up the equation to represent the repeating decimal

Let \(x = 0.\overline{432}\). This implies \(x = 0.432432432...\).
03

Multiply the equation to shift the decimal

Since \(432\) is 3 digits long, multiply both sides by \(1000\) to move the repeating block before the decimal point: \(1000x = 432.432432...\).
04

Set up a subtraction equation to eliminate the repeat

Subtract the original equation \(x = 0.432432432...\) from the new one. This gives: \(1000x - x = 432.432432... - 0.432432...\).
05

Solve for x

Simplify the subtraction: \(999x = 432\), thus \(x = \frac{432}{999}\).
06

Simplify the fraction

Find the greatest common divisor (GCD) of \(432\) and \(999\). The GCD is \(9\). Therefore, \(x = \frac{432 \div 9}{999 \div 9} = \frac{48}{111}\). Further simplification gives \(x = \frac{16}{37}\) as \(\frac{16}{37}\) is in its lowest terms.

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Key Concepts

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

Fractions
Fractions are a way of representing parts of a whole or a division between two quantities. They consist of a numerator, which is the top part of the fraction, and a denominator, which is the bottom part. For example, in the fraction \( \frac{3}{4} \), "3" is the numerator, and "4" is the denominator, indicating that you have 3 parts out of 4 total parts.

When converting repeating decimals to fractions, the goal is to express the endless decimal sequence as a ratio of two integers. Repeating decimals, like \( 0.\overline{432} \), repeat a certain block of numbers indefinitely. To transform this into a fraction, we can use algebraic techniques to isolate the repeating part and express it as a fraction.

Understanding fractions and how they relate to other numeric forms is crucial in mathematics because they offer precision in measurement and calculations that decimals may not provide.
Simplifying Fractions
Simplifying fractions involves reducing them to their simplest form. This means making the numerator and denominator as small as possible while still retaining the same value.

For example, the fraction \( \frac{432}{999} \) can be simplified by dividing both the numerator and the denominator by their greatest common factor. In the example from the exercise, dividing both by 9 gives \( \frac{48}{111} \).

Further simplification occurs if there's another common divisor. Continuing the simplification shows that \( 16 \) is a factor of both 48 and 111, allowing the fraction to be simplified to \( \frac{16}{37} \).

It is important to simplify fractions for ease of use, clarity, and sometimes to satisfy requirements in various academic and real-world contexts. To consider a fraction truly simplified, the numerator and denominator should be as small as possible with no common divisors between them except 1.
Greatest Common Divisor
The Greatest Common Divisor (GCD), also known as the greatest common factor (GCF), is the largest positive whole number that evenly divides two or more integers without leaving a remainder. Identifying the GCD is essential when simplifying fractions because it helps reduce them to their simplest form.

In the solution example, to simplify \( \frac{432}{999} \), we first determined that the GCD of 432 and 999 is 9. By dividing both numbers by their GCD, we reduce the fraction to its simplest form, \( \frac{48}{111} \), and then further to \( \frac{16}{37} \).

Finding the GCD can be done through different methods, such as listing out the factors of each number and identifying the largest common one, or by using the Euclidean algorithm, which involves a series of division steps.

Using the GCD in mathematical operations ensures that fractions are efficiently simplified, making them more straightforward to use in calculations and comparisons.

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Most popular questions from this chapter

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An important model that is used in population biology and ecology is the Ricker model. The Canadian biologist William E. Ricker introduced this model in his paper Stock and Recruitment (Journal of the Fisheries Research Board of Canada, \(11(1954) 559-623\) ). For information on Ricker himself, see the web page The general form of the Ricker model that we will use here is defined by a recursive sequence of the form \(P_{0}=\) initial population at time \(t=0\) \(P_{t}=r P_{t-1} e^{-k P_{t-1}} \quad\) for \(t \geq 1,\) and where \(r\) and \(k\) are positive constants (a) Suppose that the initial size of a population is \(P_{0}=300\) and that the size of the population at the end of year \(t\) is given by$$P_{t}=5 P_{t-1} e^{-P_{t-1} / 1000} \quad(t \geq 1)$$ Use a graphing utility to compute the population sizes through the end of year \(t=5 .\) (As in Example 5, round the final answers to the nearest integers.) Then use the graphing utility to draw the population scatter plot for \(t=0,1, \ldots, 5 .\) Describe in complete sentences how the size of the population changes over this period. Does the population seem to be approaching an equilibrium level? (b) Using a graphing utility, compute the sizes of the population in part (a) through the end of the year \(t=20\) and draw the corresponding scatter plot. Note that the population seems to be approaching an equilibrium level of \(1609(\text { or } 1610)\) (c) Determine the equilibrium population algebraically by solving the following equation for \(P_{t-1} .\) For the final answer, use a calculator and round to the nearest integer. $$P_{t-1}=5 P_{t-1} e^{-P_{t-1} / 1000}$$

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