/*! 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 14 Evaluate each expression. $$\f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 14

Evaluate each expression. $$\frac{|27-5|}{|5-27|}$$

Short Answer

Expert verified
The expression evaluates to 1.

Step by step solution

01

Calculate the absolute value in the numerator

First, evaluate the expression inside the absolute value in the numerator: \(27 - 5\). This simplifies to \(22\). So, the absolute value \(|27 - 5|\) is simply \(22\), since \(22\) is already positive.
02

Calculate the absolute value in the denominator

Next, evaluate the expression inside the absolute value in the denominator: \(5 - 27\). This simplifies to \(-22\). The absolute value of \(-22\) is \(22\), because the absolute value function converts negative numbers to positive numbers.
03

Simplify the fraction

Now you have the fraction \(\frac{22}{22}\). Divide \(22\) by \(22\) to simplify the fraction to \(1\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding the Numerator
When working with fractions, the top part is called the numerator. It represents how many parts of a whole are being considered. In our exercise, the expression \( |27-5| \) represents the numerator. Here, the operation inside the absolute value, \( 27 - 5 \), simplifies to \( 22 \). Because \( 22 \) is already positive, its absolute value stays as \( 22 \).
Understanding how to handle absolute values within the numerator is crucial. Absolute values always convert a negative result into a positive one, but a positive number remains unchanged.
So always evaluate what's inside the absolute value first, then apply the absolute value, ensuring your numerator is positive.
Exploring the Denominator
The denominator is the bottom part of a fraction, indicating into how many equal parts the whole is divided. Let鈥檚 apply this to our exercise: we evaluate \( |5-27| \) as the denominator. First, calculate \( 5 - 27 \). This operation gives us \( -22 \).
When you apply the absolute value, \( |-22| \), it changes to \( 22 \). Absolute values turn negative numbers positive.
The denominator is always a positive result when absolute values are involved, as they ensure no negative division occurs. This makes calculations straightforward and consistent when dividing quantities by changing negative values to positive.
How to Simplify Fractions
Simplifying fractions means making them as simple and small as possible, reducing them to their most basic form. In our exercise, the fraction becomes \( \frac{22}{22} \) once both the numerator and denominator are computed using absolute values.
  • To simplify, divide the numerator \( 22 \) by the denominator \( 22 \).
  • When both the numerator and denominator are the same, as in \( \frac{22}{22} \), the fraction simplifies to \( 1 \).
Simplifying ensures any fraction is in its easiest form to understand and utilize.
Remember, if the numerator and denominator are identical, their division will always yield \( 1 \). This process not only simplifies calculations but also makes it easier to compare fractions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find an equation of the line with the given slope and \(y\) -intercept. (a) slope \(0 ; y\) -intercept 14 (b) slope \(14 ; y\) -intercept 0

(a) Verify that the point (3,7) is on the circle $$x^{2}+y^{2}-2 x-6 y-10=0$$ (b) Find the equation of the line tangent to this circle at the point \((3,7) .\) Hint: A result from elementary geometry says that the tangent to a circle is perpendicular to the radius drawn to the point of contact.

Graph the equation after determining the \(x\) - and \(y\) -intercepts and whether the graph possesses any of the three types of symmetry described on page 58 $$x=y^{3}-1$$

The set of real numbers satisfying the given inequality is one or more intervals on the number line. Show the interval(s) on a number line. $$|x+5| \geq 2$$

Are the lines \(y=x+1\) and \(y=1-x\) parallel, perpendicular, or neither?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.