Chapter 6: Problem 87
Write each rational expression in lowest terms. $$ \frac{m^{2}-1}{1-m} $$
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! 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}
Learning Materials
Features
Discover
Chapter 6: Problem 87
Write each rational expression in lowest terms. $$ \frac{m^{2}-1}{1-m} $$
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
Simplify by starting at "the bottom" and working upward. $$ 7-\frac{3}{5+\frac{2}{4-2}} $$
involve factoring by grouping (Section 5.1) and factoring sums and differences of cubes (Section 5.4). Write each rational expression in lowest terms $$ \frac{1-8 r^{3}}{8 r^{2}+4 r+2} $$
Match each multiplication problem in Column I with the correct product in Column II. (a) \(\frac{5 x^{3}}{10 x^{4}} \cdot \frac{10 x^{7}}{4 x}\) (b) \(\frac{10 x^{4}}{5 x^{3}} \cdot \frac{10 x^{7}}{4 x}\) (c) \(\frac{5 x^{3}}{10 x^{4}} \cdot \frac{4 x}{10 x^{7}}\) (d) \(\frac{10 x^{4}}{5 x^{3}} \cdot \frac{4 x}{10 x^{7}}\) A. \(\frac{4}{5 x^{5}}\) B. \(\frac{5 x^{5}}{4}\) C. \(\frac{1}{5 x^{7}}\) D. \(5 x^{7}\)
Multiply. Write each answer in lowest terms. $$ \frac{t-4}{8} \cdot \frac{4 t^{2}}{t-4} $$
Which complex fraction is equivalent to \(\frac{2-\frac{1}{4}}{3-\frac{1}{2}} ?\) Answer this question without showing any work, and explain your reasoning. A. \(\frac{2+\frac{1}{4}}{3+\frac{1}{2}}\) B. \(\frac{2-\frac{1}{4}}{-3+\frac{1}{2}}\) C. \(\frac{-2-\frac{1}{4}}{-3-\frac{1}{2}}\) D. \(\frac{-2+\frac{1}{4}}{-3+\frac{1}{2}}\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.