/*! 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 36 Find each product. $$ \left(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 36

Find each product. $$ \left(\frac{2}{3}+r\right)\left(\frac{2}{3}-r\right) $$

Short Answer

Expert verified
\[\frac{4}{9} - r^2\]

Step by step solution

01

Recognize the formula

Identify that \[\frac{2}{3}+r\] and \[\frac{2}{3}-r\] is in the form of \[(a + b)(a - b)\], where \[^a = \frac{2}{3}\] and \[^b = r\]. The product of this form is given by the difference of squares formula: \[(a + b)(a - b) = a^2 - b^2\].
02

Substitute the values

Substitute \[\frac{2}{3}\] for \[^a\] and \[^r\] for \[^b\] in the formula \[(a + b)(a - b) = a^2 - b^2\]. This yields: \[(\frac{2}{3} + r)(\frac{2}{3} - r) = \frac{2}{3}^2 - r^2\].
03

Simplify the expression

Now, compute the square of \[\frac{2}{3}\] and simplify: \[\frac{2}{3}^2 = \frac{4}{9}\]. Thus, the expression becomes: \[\frac{4}{9} - r^2\].

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.

algebraic identities
Algebraic identities are special equations that hold true for all values of the variables involved. One of the most useful algebraic identities is the difference of squares, represented by \((a + b)(a - b) = a^2 - b^2\). This identity helps us simplify products of binomials. By recognizing patterns in algebraic expressions, we can use these identities to make our work faster and easier. Understanding algebraic identities is crucial for solving various algebra problems efficiently.
binomial products
Binomial products involve multiplying two terms that have two parts each. For example, in \((\frac{2}{3} + r)(\frac{2}{3} - r)\), each binomial has two terms: one with \(\frac{2}{3}\) and one with \(r\). The product of these binomials can often be simplified using algebraic identities. By recognizing that this product fits the difference of squares identity, we save time and avoid lengthy calculations. This technique is particularly useful in more complex problems, allowing you to focus on simplifying the final expression.
simplifying expressions
Simplifying expressions means making them as simple as possible while keeping the same value. In the given problem, we simplified \((\frac{2}{3} + r)(\frac{2}{3} - r)\) by first recognizing the pattern and then using the difference of squares identity. This allowed us to rewrite the expression as \(\frac{2}{3}^2 - r^2\). Simplifying further, \(\frac{2}{3}^2 = \frac{4}{9}\), leading to the final simplified form: \(\frac{4}{9} - r^2\). Understanding how to simplify expressions both enhances problem-solving skills and prepares you for more advanced math studies.
squares
Squares are one of the fundamental concepts in algebra. Squaring a number means multiplying it by itself, such as \(a^2 = a \times a\). In our example, we needed to compute \(\frac{2}{3}^2\). To square a fraction, we square the numerator and the denominator separately: \(\frac{2}{3}^2 = \frac{2^2}{3^2} = \frac{4}{9}\). Recognizing square relationships helps in applying algebraic identities accurately and efficiently. Knowing how to handle squares is essential in many areas of math, including geometry and calculus.

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

Use scientific notation to calculate the answer to each problem. In theory there are \(1 \times 10^{9}\) possible Social Security numbers. The population of the United States is about \(3 \times 10^{8} .\) How many Social Security numbers are available for each person?

The special product $$ (x+y)(x-y)=x^{2}-y^{2} $$ $$ \text { can be used to perform some multiplication problems. Here are two examples.} $$ $$ \begin{aligned} 51 \times 49 &=(50+1)(50-1) \\ &=50^{2}-1^{2} \\ &=2500-1^{2} \\ &=2499 \end{aligned} \quad | \begin{aligned} 102 \times 98 &=(100+2)(100-2) \\ &=100^{2}-2^{2} \\ &=10,000-4 \\ &=9996 \end{aligned} $$ Once these patterns are recognized, multiplications of this type can be done mentally. Use this method to calculate each product mentally. $$ 20 \frac{1}{2} \times 19 \frac{1}{2} $$

Perform each division. $$ \left(2 x^{3}+x+2\right) \div(x+1) $$

Use scientific notation to calculate the answer to each problem. In September of \(2009,\) the population of the United States was about 307.5 million. To the nearest dollar, calculate how much each person in the United States would have had to contribute in order to make one lucky person a trillionaire (that is, to give that person \(\$ 1,000,000,000,000) .\)

Perform each division. $$ \left(3 a^{2}-11 a+17\right) \div(2 a+6) $$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

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.