SAT Alg and Geom Questions

411 SATALGEBRA AND GEOMETRY QUESTIONS 411 SAT ALGEBRA AND GEOMETRY QUESTIONS ® NEW YORK other LearningExpress products. etc. SAT (Educational test)—Study guides. ISBN 1-57685-560-0 1.learnatest.A14 2006 512. questions. or bulk sales.Copyright © 2006 LearningExpress. Library of Congress Cataloging-in-Publication Data: 411 SAT algebra and geometry questions. LearningExpress (Organization). II.0076—dc22 2006005374 Printed in the United States of America 987654321 ISBN 1-57685-560-0 For information on LearningExpress. LLC. I. New York. Geometry—Examinations.. cm. p. please write to us at: LearningExpress 55 Broadway 8th Floor New York. 2. Published in the United States by LearningExpress. Algebra—Examinations. NY 10006 Or visit us at: www. questions. All rights reserved under International and Pan-American Copyright Conventions. etc.com . 3. Title: Four hundred eleven SAT algebra and geometry questions. QA157. LLC. and Range 55 CHAPTER 9 Angles 63 CHAPTER 10 Triangles 71 CHAPTER 11 Right Triangles 77 CHAPTER 12 Polygons 85 CHAPTER 13 Quadrilaterals 89 CHAPTER 14 Area and Volume 93 v . Domain.Contents INTRODUCTION 1 PRETEST 5 CHAPTER 1 Algebraic Expressions 13 CHAPTER 2 Solving Equations and Inequalities 19 CHAPTER 3 Quadratic Expressions and Equations 25 CHAPTER 4 Factoring and Multiplying Polynomials 33 CHAPTER 5 Radicals and Exponents 37 CHAPTER 6 Sequences 45 CHAPTER 7 Systems of Equations 49 CHAPTER 8 Functions. – CONTENTS – CHAPTER 15 Circles CHAPTER 16 Coordinate Geometry 99 107 POSTTEST 113 ANSWERS 121 vi . 411 SAT ALGEBRA AND GEOMETRY QUESTIONS . . If an explanation doesn’t make sense.Introduction I hate math. Your approach to math. Don’t get discouraged. With the majority of SAT math questions involving algebra or geometry.” So many people approach math with an attitude that dooms them to failure before they even get started. and the most common of them are in this book. This book trains you for the SAT with 411 math questions.” “I’ll never understand math—it’s too hard.” “Math is my worst subject. In this respect. 1 . No such list even exists. believing you are prepared for any question. Every question can be solved using the skills described in the following 16 chapters. all of the math skills that you will encounter on the math portion of the SAT are known. With time and practice. you’ll become comfortable with math and enter the SAT with confidence. but also to provide you with the confidence you require to score highly on the exam. Use these chapters to gain an understanding of the algebra and geometry that’s given you trouble in the past. or any subject. review the related material in the chapter. You can’t study every possible word that may appear on the verbal portion of the SAT. this book is designed not only to give you the algebra and geometry skills you need for the exam. it’s actually easier to study for the math sections of the SAT than the verbal sections. and then try the question again. can make all the difference in the world. However. and every answer is explained. subtract. Chapter 8: Functions. combine like and unlike terms. and analyze the graphs of quadratic equations. Work with negative and fractional exponents. multiply. right. and form algebraic equations from word problems.– INTRODUCTION –  An Over view of This Book This book is divided into an algebra section and a geometry section. Chapter 7: Systems of Equations Solve systems of two equations with two variables using substitution and combination. obtuse. The tests are made up of only algebra and geometry questions. factor and solve quadratic equations. Chapter 5: Radicals and Exponents Add. Use the pretest to identify the topics in which you need improvement. 2 . vertical. each comprised of eight chapters. and combination sequences for the next or missing term. and simplify radicals and terms with exponents. Chapter 2: Solving Equations and Inequalities Use basic arithmetic and cross multiplication to solve single. Domain. find the domain and range of functions. straight. Chapter 3: Quadratic Expressions and Equations Multiply binomials. and alternating angles. and find the roots of polynomials. Chapter 6: Sequences Solve arithmetic. geometric. solve nested functions and evaluate functions with newly defined symbols. factor. Use the posttest to help you identify which topics you may need to review again. and evaluate one variable in terms of another. supplementary. plus a pretest and a posttest.and multi-step equations. and Range Determine if an equation is a function using the vertical line test. The last eight chapters focus on geometry skills: Chapter 9: Angles Recognize and use the properties of acute. and raise exponents to exponents. The first eight chapters focus on algebra skills: Chapter 1: Algebraic Expressions Understand the parts of algebraic expressions. Chapter 4: Factoring and Multiplying Polynomials Multiply. complementary. find the values that make an expression undefined. evaluate expressions using substitution. divide. rationalize denominators and solve equations with radicals. Each chapter builds on the skills reviewed in the chapters that precede it. and similar triangles. the volume of cylinders and rectangular solids (including cubes). and use basic trigonometry to determine the size of angles and sides of right triangles. midpoint. as well as the area of a sector of a circle and the length of an arc of a circle. right. a set of skills is reviewed (with key words defined). In each chapter. find the slope. Chapter 13: Quadrilaterals Learn the differences that distinguish parallelograms. Following the lesson portion of the chapter are 15–30 SAT-caliber questions. Chapter 16: Coordinate Geometry Plot and find points on the coordinate plane. scalene. Chapter 12: Polygons Review the properties of interior and exterior angles of polygons and regular polygons. and squares from each other.  How to Use This Book Start at the beginning. If you can handle the questions in each chapter. including radius and diameter. but you’ll need the algebra skills reviewed in the first eight chapters to solve these problems. and the surface area of solids. and distance of line segments. Chapter 14: Area and Volume Find the area of triangles and rectangles (including squares). The last eight chapters focus on geometry. you’re ready for the SAT! 3 . the questions are progressively more difficult. equilateral. Find the perimeter of polygons and work with similar polygons. rhombuses. In general. including examples of each skill with explanations. There’s a reason the algebra chapters come first—the SAT is filled with algebra questions. and many of the geometry questions you’ll encounter will involve algebra. Related questions are grouped together. rectangles. and use them to find the circumference and area of a circle. Chapter 11: Right Triangles Use the Pythagorean theorem to find the missing side of a triangle. Chapter 15: Circles Review the parts of a circle. obtuse. Review the properties of special right triangles: 45-45-90 (isosceles) right triangles and 30-60-90 right triangles. and the similarities that each share.– INTRODUCTION – Chapter 10: Triangles Recognize and use the properties of interior and exterior angles of triangles and acute. isosceles. – INTRODUCTION – Read and reread each section of this book. Review a book such as SAT Math Essentials or Acing the SAT 2006 by LearningExpress to be sure you’ve got all the skills you need to achieve the best possible math score on the SAT. If you don’t understand an explanation of a geometry question that involves algebra. but you may be confused about the algebra part of the question. but this book does not cover every type of math question you’ll see on the SAT. Algebra and geometry questions comprise the bulk of SAT math. it may be helpful to reread an algebra section—you might have the geometry skill mastered. Good luck! 4 . If you score high on the pretest. of which 15 are the multiple-choice questions and 10 are the gridins. but it will tell you the degree of effort you will need to put forth to accomplish your goal of learning algebra and geometry. or shortcut taught in this book. If you get a low score you may need to take more than 20 minutes a day to work through a lesson.Pretest B efore you begin Chapter 1. However. you have a good foundation and should be able to work your way through the book quickly. check your answers with the answer key at the end of the book. When you are finished. The pretest will answer some of these questions for you. The pretest consists of 25 questions that cover the topics in this book. If you score low on the pretest. you may want to get an idea of what you know and what you need to learn. you come up with the answer yourself instead of choosing from a list of possible answers. Take as much time as you need to do the pretest. 5 . your performance on the pretest will give you a good indication of your strengths and weaknesses. You decide when you fully comprehend the lesson and are ready to go on to the next one. skill. For the grid-ins. don’t despair. This book will take you through the algebra concepts. Keep in mind the pretest does not test all the skills taught in this book. While 25 questions can’t cover every concept. step by step. this is a self-paced program. so you can spend as much time on a lesson as you need. . 12. / / • • • 0 0 0 • 24. • d d d d d 6. 10. / / • • • • a a a a a b b b b b c c c c c d d d d d 18. / / • • • • 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 7 . 8. 7. 5. / / • • • • / / • • • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 / / • • • 21. • 22. / / • • • • 23.– LEARNINGEXPRESS ANSWER SHEET – PRETEST 1. a a a a a b b b b b c c c c c 16. 3. 15. / / • • • • b b b b b c c c c c d d d d d e e e e e 20. 2. e e e e e 17. 9. 4. 14. e e e e e a a a a a 19. 13. / / • • • 0 0 0 • 25. / / • • • • 11. . It has a slope of 2. a. 500. 8. x = 1 or y = 6 1 d. c. but not all rhombuses are rectangles. d. x ≥ 3 1 d. c. 720π cm3 4. x = – 3 or y = –3 e. All squares are rectangles and rhombuses. It has no domain. what is the area of rectangle ABCD? a. The inequality –4(x – 1) ≤ 2(x + 1) is equivalent to 1 a. b. It is not a function. 288π cm3 e. 2a d. e. x = 0 or y = 0 1 1 b. x ≥ –3 1 b. 240π cm3 d. The expression  (6x2 + 20x + 6) is undefined when a. 100. 1 b. 4  57 d. 100 square units e. 205 square units c. x = 3 or y = –3 c. All rhombuses are parallelograms and all parallelograms are rhombuses. Each term in the sequence below is five times the previous term. 144π cm3 c. A circle has an area of 64π ft. If the area of square BCEF is 20 square units. c. Which of the following is true of y = 2? a. 32 ft. All rectangles are rhombuses. 2a 9. d. 20 + 205 square units d. Which of the following statements is true? a. 500  8 b. 32π ft. 8 ft. but not all squares are rhombuses. Side BC of the rectangle is shared with side BC of square BCEF. x ≤ –3 1 c. 48 c. If Kerry stacks 10 compact discs on top of each other. . 72π cm3 b. 16π ft. . b. All squares are parallelograms. A circular compact disc has a diameter of 12 cm and a height of 2 mm. 8π ft. 200 square units 3. What is the value of (b)( (2b)–1 )? a. b2 9 . All rhombuses are squares. What is the circumference of the circle? a. . 20. What is the eighth term in the sequence? 4. 4  58 e. 105 square units b. a–1  5. 4 a2 e. e. x ≤ 3 e.– PRETEST – (9x2 + 9y2)  1. 5  48 7. e. but not all squares are rectangles.2. what is the volume of the stack? a. d. It has a range of 2. Side AB of rectangle ABCD measures 10 units. 2 c. b. x = –1 or y = –6 6. It has no y-intercept. x ≤ 3 2. then the area of triangle length of DE DEF is equal to a.– PRETEST – 13. If the length of one side of a triangle is 4. y = (x – 1)2 – 2 c. e. What is the distance between the points (–7. x = –3. 5 5  (area of triangle ABC). what is the measure of angle ABC? 7b – 4 14.12)? 18. Triangles ABC and DEF are similar. 7º b. Steve draws a polygon with 12 sides. 2 25  (area of triangle ABC). d. B 9x + 5 c. x = 2 b. 63 b. e. a  7 4a  7 a+1  7 4a + 4  7 7a – 4  7 16. 4 17. each of which has a vertex at the center of the hexagon. what is the volume of the cube in cubic centimeters? 4  (area of triangle ABC). If ABCD (shown below) is a parallelogram. If the surface area of a cube is 96 square centimeters. Which of the following parabolas has its turning point in the second quadrant of the coordinate plane? a. then b = a. d. A b. x = –2 d. x = –3. y = –(x + 1)2 – 2 d. 25 1  (area of triangle ABC). x = –6 e. For what values of x is the expression 3x2 – 3x – 18 equal to 0? a. 68º c. What is the sum of the measures of the interior angles of Steve’s polygon? 10 . x = –2 11.–4) and (5. y = –(x + 2)2 + 1 e. Side AB of triangle ABC corresponds to side DE of triangle  is 10 units and the DEF. 243 a. 123 d. x = 3. 78º d. Jennifer makes a hexagon by arranging 6 equilateral triangles. If the length of AB  is 4 units. 112º 12. If a = 4. 12 c. 5 2  (area of triangle ABC). 24 e. x = 6 1 c. c. x = 3. 16x D C 15. b. x = 3. y = (x – 2)2 + 1 10. what is the area in square units of the hexagon? a. 102º e. y = (x + 1)2 – 2 b. Find the value of  bb if b = 36. what is the measure in degrees of arc AD? 20. what is the value of 2w? 2x + 8 B C 7 3  a 24. If the measure of angle DCA is 115 degrees. If the measure of angle OBC is 71 degrees. sides AC and AB of triangle ABC are congruent. If 5 = 6. If 3 + 3 = 7.– PRETEST – b + 13 19. Evaluate (5a2 +  10a ) for a = –2. 5x – 6 22. what is the measure in degrees of angle A? D A A O 71° D C 2w B 3 21. what is the value of x? 25. In the diagram below. b 23. In the diagram below. what is the length of side AC? 11 . Angle A of right triangle ABC measures 60 degrees and angle C of the triangle measures 30 degrees. sides OC and OB of triangle OBC are congruent. If the length of side BC is 163. . 3a.C H A P T E R 1  Algebraic Expressions Variables A variable is a letter or symbol that represents a value. y + 1. including the exponents of these variables.  Like and Unlike Terms The terms x. For instance. The terms 3x and 3x2 are unlike terms. and x2 + 4x + 4 are all algebraic expressions.  Parts of a Term The coefficient of a term is the constant in front of the variable. 4x. because these terms all have the same base. 13 . An algebraic expression is one or more terms that contain a variable. and 8x are like terms. and 4. because the variables in each term have different exponents. The coefficient of the term 4a3 is 4. The expression x2 + 4x + 4 is made up of three terms: x2. 3x. The base of a term consists of the variable or variables that follow the coefficient of the term. and the base of the term is a3. Add the exponents of common variables. 14 . and the difference of the exponents of the b terms is 3. divide the coefficients of the terms and subtract the exponents:  5x = 3x . and the sum of the exponents of a is 2 + 4 = 6. What is the value of 7x + 1 when x = 5? Substitute 5 for x in the expression: 7(5) + 1 = 35 + 1 = 36. The expressions y + 2. The product of 6a2 and 3a4 is 18a6.  Evaluate Expressions Using Substitution To evaluate an expression given the value of the variable or variables in the expression. Next. the difference of the exponents of the a terms is 1. What is the product of 4x2 and 5xy4? The product of the coefficients is (4)(5) = 20. and then divide the variables. subtract the coefficients of the terms: 9a2 – 6a2 = 3a2. To divide 15x3 2 like terms. What is the product of 6a2 and 3a4? The product of the coefficients is (6)(3) = 18. To add like terms. substitute the value for the variable in the expression. Multiply the coefficients of the terms. Unlike terms can be simplified by multiplying or dividing. Subtract the exponents of common variables. Unlike terms are divided in a similar way. To multiply like terms. since 5 = 5. 3a + 3a2. and then multiply the variables. 25a3b5  5a2b2 = 25 5ab3. add the coefficients of the terms: 3x + 4x = 7x. To subtract like terms. What is the value of x2 + 3x + 5 when x = –1? The value of x2 + 3x + 5 when x = –1 is (–1)2 + 3(–1) + 5 = 1 – 3 + 5 = 3. multiply the variables. adding the exponents of common variables: (x2)(xy4) = x3y4. The product of 4x2 and 5xy4 is 20x3y4. and 6x6 – 6x cannot be simplified any further. Divide the coefficients of the terms. multiply the coefficients of the terms and add the exponents: (3b4)(2b4) = 6b8. since 3 + 4 = 7. 4 – x.  Combining Unlike Terms Unlike terms cannot be simplified by adding or subtracting.– ALGEBRAIC EXPRESSIONS –  Combining Like Terms Like terms can be combined using the standard operations. 36 7a 6. 8x4y4 + 6x2y2 a. 24 + r c. 9a + 12a2 – 5a = a. 4ab c. –6 b. x2 +  –2x = 3. When x = 3. When a = –2. 4 e. 2 y2 y 7.  a2 + a = a. –7 7 c. 12a2 + 4a e. 3a2 5. pr = 4. x. r e. The value of y is 3x + 3. 16a b. 3 2 b. 2a + 4b 5a + 7b d. is written in terms of the other variable. If 4a + 8b = 16. 26a2 2. 0 c. For instance. –4 7 d. d. 8 b. 8x2 + y2 + 6x2y2 e. 7 = e. y = 3x + 3 is an equation with two variables. 2x2 – 5x + 3 = a. c. 33 e. 6 d. in which the value of one variable. (5a + 7b)b  (b + 2b) = a. 12a2 – 4a d. The value of a in terms of b is 4 – 2b. e. x and y.– ALGEBRAIC EXPRESSIONS –  Evaluate One Variable in Terms of Another An equation that contains two variables can be written so that the value of one variable is given in terms of the other. –14 b. 2x2 + 4y2 + 6x2y2 d. a 1 c. 24 + 3r 6 d. 3a 1 d. (3a)(4a)  6(6a2) 1 a. When p = 6. 12x2y2 b.  Practice 1. 3y  4 y3  4 (–y2 + y)  4 (y2 – y)  4 (y2 + y)  4 4p(p + r) 8. y. 4a b. 14x2y2 c. 5a + 5b b. (2x2)(4y2) + 6x2y2 = a. 3 e. 15 24 + 4r  r . what is the value of a in terms of b? Isolate a on one side of the equation. When x = –2. a. 16a2 c. Now. Subtract 8b from both sides: 4a = 16 – 8b. divide both sides of the equation by 4: a = 4 – 2b. what is the value of g in terms of h? 1 5 a. –1 1 b. 107b3 + 39 c. 4 – 7b 22 d. 108b3 + 39 e. –1 e. If 4(y + 1) = 10. 3 e. –2 d. c. 1 b.– ALGEBRAIC EXPRESSIONS – 9. what is the value of y in terms of x? b. 72b3 + 108 b. –3 c. d. 5 e. When a = 1 and b = –1. 2 – 3f  f 17. 4 + 3b 19 c. 2y2 + 3y = a. what is the value of a in terms of b? a. If a(3a) – b(4 + a) = –(a2 + ab). –3 16  5 x 15. When x = 2 and y = 3. 4a2 3 13. 2a 1 b.  c+d = a. 4 c. – 4x d. When a = 3. 107b3 + 108 d. If fg + 2f – g = 2 – (f + g). If 2g = 9h – 15. (4a2)(3b3 + a) – b3 = a. 6h – 10 27 45 d. what is the value of b in terms of a? 1 a. 9x 6x2 2 d. If 7a + 20b = 28 – b. 3x 4x 11. –4 b. 2x 4  9 4  3 20  9 21  9 13  3 16. 2 – 2f e. 216 14. ab + + a. f a  b 12. 0 a2 – b2 c. = 4  f d. 2h – 2 e. a2 e. b. When c = 1 and d = 4. what is the value of g in terms of f ? a. 4 – 3b b. 4x 25  4 4 c. e. 2a d. 4 – 7b (cd)2 b+4 10. 2a2 c. 17  4 1 a. 6h – 15 c. 6h + 3 b. 18h – 30 16 . e. 2y2 17 . 4h d. what is the value of x in terms of y? 2 a. h b. 3y 3 b. 2y 4 c.– ALGEBRAIC EXPRESSIONS – 10 (x2y) 18. h2 e. what is the value of y in terms of x? 2 a. x b. If 8x2 – 4y2 + x2 = 0. 9y2 2 d. what is the value of g in terms of h? a. If xy = 5y. 2x d. 4h2 20. If 4g2 – 1 = 16h2 – 1. 2x  c. 3y2 3 e. 2x x2 e. 2 19. 2h c. . ≤. subtracting. isolate the variable on one side of the equals sign by adding. less than or equal to. or greater than or equal to the other side of the inequality. To solve an equation. add 2 to both sides of the equation: x – 2 + 2 = 4 + 2. x = 1.C H A P T E R Solving Equations and Inequalities 2 A n equation is an expression that contains an equals sign. An inequality is an expression that contains one of the following symbols: <. x x To solve the inequality 4 < 8. multiply both sides of the equation by 4: (4)(4) < (4)(8). greater than. multiplying. To solve the equation x – 2 = 4. or dividing. or ≥. 19 . x < 32.  Basic Equations and Inequalities Basic equations and inequalities can be solved with one step. x = 6. One side of an inequality may be less than. >. x ≥ 2. divide both sides of the inequality by 3: 3 ≥ 3. To solve the equation x + 5 = 6. To solve an inequality. 3x 6 To solve the inequality 3x ≥ 6. subtract 5 from both sides of the equation: x + 5 – 5 = 6 – 5. Both sides of an equation are equal. isolate the variable on one side of the inequality symbol using the same operations. 3x + 2 = 8 is an equation. Only reverse the inequality symbol when multiplying or dividing both sides of the inequality by a negative number. what is the value of x? Begin by cross multiplying: (3x)(6) = (2)(18x – 6). Solve the inequality –2x < 10 for x. Now. add 2 to both sides of the inequality: 10x – 2 + 2 > 18 + 2. Both 4 and –4 square to 16. Use combinations of the operations above to isolate the variable on one side of the equals sign or inequality symbol. multiply the numerator of the fraction on the right side of the equation or inequality by the denominator of the fraction on the left side of the equation or inequality. x > –5. 10x > 20. you must reverse the inequality symbol. 3x 18x – 6 If 2 = 6. subtract 4 from both sides of the equation: x2 + 4 – 4 = 20 – 4.– SOLVING EQUATIONS AND INEQUALITIES – If solving an inequality requires multiplying or dividing both sides of the inequality by a negative number. 18x = 36x – 12. To solve the equation x2 + 4 = 20. Multiply the numerator of the fraction on the left side of the equation or inequality by the denominator of the fraction on the right side of the equation or inequality. divide both sides of the inequality by 10: 1 0 > 10 . or take a root of both sides of the equation.  Multi-Step Equations and Inequalities Multi-step equations and inequalities require two or more steps to solve. solve the equation by adding and dividing: 18x + 12 = 36x – 12 + 12 18x + 12 = 36x 18x – 18x + 12 = 36x – 18x 12 = 18x 12 18x  =  18 18 12 2 x = 18 = 3 20 . and set the two products equal to each other. –2x 10  Divide both sides of the inequality by –2: – 2 > –2 . Notice that the inequality symbol has changed direction. x2 = 16. Values of x that are less than –5. Then. This is the solution set for –2x < 10. You may also have to raise both sides of the equation to an exponent. begin solving the equation or inequality by cross multiplying. That is why you must reverse the inequality symbol: x > –5. 10x 20  Then.  Cross Multiplying If one or both sides of an equation or inequality contain fractions. if substituted into –2x < 10. so the solutions to x2 + 4 = 20 are x = 4 and x = –4. would make the left side of the inequality greater than the right side of the inequality. x > 2. To solve the inequality 10x – 2 > 18. a = a. solve the equation for x: 3x – 2 = 2x + 2 3x – 2x – 2 = 2x – 2x + 2 x–2=2 x–2+2=2+2 x=4  Practice 2. 2x + 4. Twice the number is 2x. while words such as half signal division. The fraction  2x + 4 is undefined when x = –2. 1  For what value of x is the fraction  2x + 4 undefined? Set the denominator of the fraction. equal to 0 and solve for x: 2x + 4 = 0 2x + 4 – 4 = 0 – 4 2x = –4 2x –4  =  2 2 1  x = –2. Read carefully for key words. To find when an algebraic fraction is undefined. p ≤ 3 e. 0 c.  Forming Algebraic Equations from Word Problems The SAT will often describe a situation that must be transformed into an algebraic equation in order to be solved. 24 3 a. Words such as twice signal multiplication. If a – 12 = 12. If x is the number. what is the number? Transform the situation into an algebraic expression. p ≥ 3 5 d. Now. then two less than three times a number is 3x – 2. greater than signals addition. Two more than that is 2x + 2. p ≥ 2 21 . set the denominator of the fraction equal to zero and solve the equation for the value of the variable. Phrases such as less than signal subtraction.– SOLVING EQUATIONS AND INEQUALITIES –  Undefined Expressions A fraction is undefined when its denominator is equal to zero. then 1. If 6p ≥ 10. 12 e. 1 d. p ≥ 5 3 b. p ≤ 5 5 c. –12 b. The key word is means equals. If two less than three times a number is two more than twice the number. The phrase two less than three times a number is two more than twice the number can be represented algebraically as 3x – 2 = 2x + 2. x ≥ 2 10x 5x – 10 12. If 8 = 8. If 7 = 3. 3 c. 16 e. 7  5 b. what is the value of a? a. –1 d. 14 e. 8 c. –2 c. x = a. 6 e. What is the solution set for the inequality –8(x + 3) ≤ 2(–2x + 10)? a. If 6x – 4x + 9 = 6x + 4 – 9. 9. 5 d.– SOLVING EQUATIONS AND INEQUALITIES – 8. x ≤ 11 1  8 b. –3 d. 2 d. what is the value of x? a. n > 36 a. x ≥ –11 b. x = 11 d. 2 e. –5 b. –2 17 e. 2 e. 5 k 4. 10 7. 9 3. If 9a + 5 = –22. 2 c. If  w + 8 = – 18 . what is the value of w? a. x ≥ 11 e. 4 e. 8 d. n < –4 c. 3 d. – 2 c. 12 6. x ≤ –11 c. –6 b. –9 7 b. n < 4 b. If  6c + 9 = 15. –1 7 d. what is the value of c? 5. n > –4 e. x ≤ –14 c. 64 3c2 10. 30 22 . what is the value of x? a. – 9 w 6   11. n > 4 d. x ≥ –8 e. If x + 10 = 5. –2 1 c. k = a. x ≤ –8 d. The inequality 3x – 6 ≤ 4(x + 2) is equivalent to a. The inequality –3n < 12 is equivalent to a. 1  2 b. x ≥ –14 b. –9 c. –27 b. b. 4 c. what is the value of g? a. If 8x + 4 = 14. 7 e. 2 a. 8 9x + 5 17. 1  2 20. d. 14 8  3 3  8 19. 45 1 x 2 16. 4 c. 2 d. 5 d. 12 –8  21. –8 b. c. what is the value of a? 30 a. 28 5g g+7  15. what is the value of y? a. 4 e. 6 d. what is the value of –2x? a. If 11c – 7 = 8. 5 b. d. 24 e. 0 1 d. 9 6 8 18. 16 e. 3 b. If 3x – 8 = 2. –3 d. 3 e. 7 d. 6 e. If + 6 = –x – 3. –6 c. 23 1  6 . If g =  g – 1 . –1 1 c. c. –37 a. what is the value of 33c – 21? 15  11 8  3 b. If  –y – 1 = –2y – 3 . –4 b. 1 c. If 9x + 3 = 3x + 9. –12 b. –3 1 b. If 7 = –4.– SOLVING EQUATIONS AND INEQUALITIES – 4a + 4 2 – 3a 8 13. e. For what value of x is the fraction  x – 8 undefined? a. For what value of d is the fraction 6d undefined? 1 a. 16 e. –2 c. what is the value of x + 7? a. 5  8 5  4 5  2 d. 14 3d 22. –6 c. 1 e. 10 e. –7 b. what is the value of 4x + 2? a. 2 d. 0 c. 3 12 3  8 b. what is the value of 8x? 10   14. odd whole numbers is 48. –5 b. If five less than three times a number is equal to 10. 19 c. –8 d. 3 d. 22 2x + 10  25. 0. For what value of y is the fraction  8 – 8y + 4 undefined? a. 20 d. 15 e. The sum of four consecutive. 12 3y – 6  24. 5 c. –18 b. 5. 17 26. –5 29. –2 3 c. 3 b. 4 c. 9 27. 1 e. 21 e. what is the value of the number? 3 a. 9 b. 4 28. If three more than one-fourth of a number is three less than the number. –9 c. even integers is –18. 10 e. 10 d. 0. 2 e. What is the value of the smallest integer? a. 15 e. 25 30. What is the value of the largest integer? a. 2 d. The sum of three consecutive integers is 63. 0. 5. –5. 11 c. 13 d. –1 d. –2 1 b.– SOLVING EQUATIONS AND INEQUALITIES – 2a – 18  23. what is the value of that number? 10 a. 18 b. –12 b. –10 c. 5 25 c. For what value of a is the fraction  6a + 18 – 4a undefined? a. 8 e. 6 d. –6 e. 4 b. The sum of three consecutive. –4 24 . What is the value of the smallest number? a. For what values of x is the fraction  2x(x – 5) undefined? a. (x)(x) = x2. Finally. multiply the inside terms: (x + 1)(x + 2). (1)(x) = x.C H A P T E R 3 A  Quadratic Expressions and Equations quadratic expression is an expression that contains an x2 term. The expressions x2 – 4 and x2 + 3x + 2 are two examples of quadratic expressions. multiply the last terms: (x + 1)(x + 2). each with a different base. multiply the outside terms: (x + 1)(x + 2). 25 . add the four products: x2 + 2x + x + 2 = x2 + 3x + 2. Outside. (x – 2) and (5x + 1) are both binomials. To multiply binomials. The acronym FOIL can help you to remember how to multiply binomials. for example. (1)(2) = 2. (x)(2) = 2x. Then. you must multiply each term by every other term and add the products. (x + 1) and (x + 2). To find the product of (x + 1)(x + 2). Next. Inside. A quadratic equation is a quadratic expression set equal to a value. The equation x2 + 3x + 2 = 0 is a quadratic equation. and Last. When multiplying two binomials. Multiplying Binomials A binomial is an expression with two terms. FOIL stands for First. begin by multiplying the first term in each binomial: (x + 1)(x + 2). The values 2 and 3 for x make x2 – 5x + 6 equal to 0. Find two numbers that multiply to 9. Add 6 to both sides of the equation. the expression x2 – 5x + 6 equals 0. so the factors of 2x2 + 9x + 9 are (2x + 3) and (x + 3). The values that solve a quadratic are the roots of the quadratic. The numbers 1 and 8. Factoring is the reverse of FOIL. What are the factors of 2x2 + 9x + 9? This quadratic will be factored into (2x + a)(x + b). Find two numbers. x = 2. x – 2 = 0. Two times one of those numbers plus the other must equal 9. Only –3 and –2 add to –5. The numbers 1 and 6. x = 3. and solve for x. You can check your factoring by using FOIL: (x + 2)(x + 4) = x2 + 4x + 2x + 8 = x2 + 6x + 8. therefore. That expression can be broken back down into (x + 1)(x + 2) by factoring. you can find its factors by finding two numbers whose product is 8 and whose sum is 6. and –3 and –2 multiply to 6. Given the quadratic x2 + 6x + 8. For what values of x does x2 – 5x = –6? First. and the numbers 3 and 3 multiply to 9. a and b. factor the quadratic and find the values of x that make each factor equal to 0. x2 – 5x + 6 = 0. and. the coefficient of the second term of the quadratic trinomial. 3 and 2. so that the quadratic is equal to 0.– QUADRATIC EXPRESSIONS AND EQUATIONS –  Factoring Quadratic Expressions Multiplying the binomials (x + 1) and (x + 2) creates the quadratic expression x2 + 3x + 2. combine like terms and place all terms on one side of the equals sign. The numbers 1 and 9. A quadratic trinomial (a trinomial is an expression with three terms) that begins with the term x2 can be factored into (x + a)(x + b). Set each factor equal to 0. Find two numbers that multiply to 6 and add to –5. 26 . factor the quadratic. Two times 3 plus 3 is equal to 9. –1 and –6. that multiply to the third value of the trinomial (the constant) and that add to the coefficient of the second value of the trinomial (the x term).  Solving Quadratic Equations Quadratic equations have two solutions. Now. 4 and 2 multiply to 8. combine like terms and place all terms on one side of the equals sign. If either of these factors equals 0. x – 3 = 0. but only 4 and 2 add to 6. The factors of x2 + 6x + 8 are (x + 2) and (x + 4). Then. To solve a quadratic equation. The factors of x2 – 5x + 6 are (x – 3) and (x – 2). these values make x2 – 5x equal to –6. the positive values of x yield the same y values as the negative values of x. The numbers 1 and 10.–c) and a parabola of the form y = x2 + c has its vertex at (0. The y value of the vertex is 3. A parabola of the form y = x2 – c has its vertex at (0.3). the values of x that make the fraction undefined are –5 and –2. –1 and –10. Since the x value is squared. is a parabola. The graph of y = x2 has its minimum at (0. If the denominator of a fraction is a quadratic. the graph is shifted down one unit. Since x + 5 = 0 when x = –5 and x + 2 = 0 when x = –2. x = 2.0). the parabola y = x2 – 1 has its vertex at (0.–1). The x value is increased before it is squared. The factors of x2 + 7x + 10 are (x + 5)(x + 2).c).–c).  Graphs of Quadratic Equations y 4 3 2 1 –4 –3 –2 –1 1 –1 x 2 3 4 –2 –3 –4 y = (x + 1) 2 y 4 3 y = x2 – 1 2 1 –4 –3 –2 –1 1 –1 –2 –3 –4 x 2 3 4 The graph of the equation y = x2.0). It is either the minimum or maximum y value of the graph. shown at left. but only 5 and 2 add to 7.0). After the x term is squared. There are no y values less than 0 on the graph. 2x  For what values of x is the fraction  x2 + 7x + 10 undefined? Find the roots of x2 + 7x + 10. 27 . The graph of y = x2 has its vertex at the point (0. 5 and 2. making the coordinates of the vertex of the parabola (2. and –5 and –2 multiply to 10. The graph of y = x2 can be translated around the coordinate plane. a fraction is undefined when its denominator is equal to zero. A parabola of the form y = (x – c)2 has its vertex at (c.0). What are the coordinates of the vertex of the parabola formed by the equation y = (x – 2)2 + 3? To find the x value of the vertex. set (x – 2) equal to 0. The vertex of a parabola is the turning point of the parabola. x – 2 = 0.0) and a parabola of the form y = (x + c)2 has its vertex at (0. The values that make the quadratic equal to 0 are the values that make the fraction undefined. The expression (x + 1)2 is equal to 0 when x = –1.– QUADRATIC EXPRESSIONS AND EQUATIONS –  Undefined Expressions As you saw in the last chapter. The parabola y = (x + 1)2 has its vertex at (–1. The y value of the vertex of the parabola is equal to the constant that is added to or subtracted from the x squared term. factor the quadratic and set the factors equal to 0. The minimum value of the parabola is when y = 0 (since y = (x + 1)2 can never have a negative value). While the parabola y = x2 has its vertex at (0. x2 – 2x – 1 8. (x – 4)(x – 7) b. What is one factor of x2 – 4? a. 6x2 + 36x – 54 28 . (x – 4) 3. –4 and 12 e. What are the factors of x2 – x – 6? a. (x + 14)(x – 2) d. (x – 3)(x + 2) c. x2 – 36x + 36 7. (x + 3)(x – 2) d. x2 – 12x – 36 d. 4 and –12 d. x2 + 2cx + c2 9. What is the product of (x – 3)(x + 7)? a. x2 + 7x – 21 e. x2 + 1 c. (x + 2) e. (x – 11)(x + 28) 4. x2 – x – 1 d. x2 + c2x2 + c2 d. x2 – 3x – 21 c. x2 – 1 b. What are the roots of x2 – 18x + 32 = 0? a. –4 and 8 c. x2 – 36 c. (x – 1)(x + 6) 2. (x – 6)(x + 1) e. What is the product of (x – 1)(x + 1)? a. 6x2 + 18x – 15 d. x2 + c2 b. x2 + cx2 + c2x + c2 e. –6 and 8 c. What are the factors of x2 – 11x + 28? a. x2 + cx + c2 c. x2 – 12x + 36 e. –4 and –8 b. 2 and 16 10. –4 c.– QUADRATIC EXPRESSIONS AND EQUATIONS –  Practice 1. What is the value of (x + c)2? a. –6 and –8 b. x2 + 36 b. –2 and –16 d. (x – 4)(x + 7) c. What are the roots of x2 + 8x – 48 = 0? a. What is the product of (2x + 6)(3x – 9)? a. 6x2 – 54 c. –4 and –12 5. (x – 1) d. 6x2 – 18x – 15 e. (x – 14)(x + 2) e. 2 and –16 e. x2 – 21 b. 5x2 – 54 b. x2 – x + 1 e. x2 – 21x – 21 6. What is the product of (x – 6)(x – 6)? a. x2 b. x2 + 4x – 21 d. (x – 3)(x – 2) b. 1 d. e. –6 b. –4 c. 6x – 6  17. b. –1 (x + 3)  b. e. e. If x2 – 3x – 30 = 10. d. The fraction  (x2 + 11x + 30) is equivalent to a. The fraction  (x2 + 8x – 33) is equivalent to a. If x2 – x = 12. what is one possible value of x? a. d. 3 d. The fraction  (x2 – x – 56) is equivalent to a. what is one possible value of x? a. –6 b. –4 c. b. 4  11. 6 e. (x + 5)  (x – 9) (x + 5)  (x + 6) (x – 9)  (x + 6) (x – 9)  (x + 5) (x + 5)  (x + 5) 18. –2 d. 5 e. 3 e. b. 11 (x – 3)  (x + 3) (x + 3)  (x + 11) (x – 3)  (x + 11) 29 . –2 c. (x + 2)  (x – 8) (x + 2)  (x + 7) (x + 7)  (x + 2) (x – 4)  (x – 8) (x – 8)  (x – 8) x+5 (x2 – 4x – 45)  13. c.  (x – 3) c. d. e. 10 (x2 – 6x – 16)  12. –10 b. 7 (x2 + 14x + 33)  14. what is one possible value of x? a. what is the value of x? a. c. 5 d. If 4 =  x + 4 . d. –5 b. 1  (x – 9) 1  (x + 9) 1   (x2 – 81) (x + 9)  (x – 9) (x – 9)  (x + 9) 16.– QUADRATIC EXPRESSIONS AND EQUATIONS – (x + 9) 15. c. 8 e. 3 c. If (x – 2)(x + 6) = –16. The fraction  (x2 – 81) is equivalent to a. –2 d. 8. The square of a number is equal to two less than three times the number. –7 c. 0 1 c.–4) c. 4 9 d. 2 b. 2. 2 e. 21 a. 7 20. 1. –1. (–4. –1 d. –6 c. –8. –1. 3 e. What are two possible values of the number? a. –3 b. 1 e. –7 b. 2. 1. 9 (x2 + 8x + 7)  22. –9 b. 8 e. –6. –2 e. –4 c. –7 b. 3 d. 16 19. –1. –2 d. What is the vertex of the parabola whose equation is y = x2 + 4? a. –7. –6 b. 3 (x2)  24. 1. –4 c. 7 d. (0. What is one value that makes the fraction  (9x2 – 1) undefined? a. 6 e. 7 e. 1.0) d.– QUADRATIC EXPRESSIONS AND EQUATIONS – (x – 16)  23. What values make the fraction  (x2 – 8x + 7) undefined? a. –1. what is the value of x? a. (–2. 1. (0. What is one value that makes the fraction (4x – 8)   (x2 + x – 42) undefined? a. –16 b. –1. (2.0) b. If (x – 7)(x – 5) = –1. What values make the fraction  (2x2 – 25x + 72) undefined? 21. –2 d. 6 9 c. 9 (x2 – 36)  25. 7 26. 2 c. What is one value that makes the fraction  (x2 – 16) undefined? a. 2.8) 30 .4) e. –2) d. (–2. y = (x – 5)2 d. (2.18) –4 –3 –2 –1 –1 –2 –3 –4 a. y = x2 – 25 31 x 1 2 3 4 . (–2. y = (x – 1)2 + 2 d. y = (x – 1)2 – 2 b. y = (x – 4)2 – 3 d.–4)? a. y = (x – 3)2 – 4 c.2) c. y = x2 – 7 b. What is the equation of the parabola shown below? y 4 3 2 1 28.2) b. What is the vertex of the parabola whose equation is y = (x + 2)2 + 2? a. Which of the following is the equation of a parabola whose vertex is (5. y = (x + 3)2 – 4 30. y = (x + 4)2 – 5 e. y = x2 + 5 c. y = (x + 5)2 e. y = x2 – 5 b. y = x2 – 3 29. y = (x + 1)2 – 2 c.0)? a. Which of the following is the equation of a parabola whose vertex is (–3.– QUADRATIC EXPRESSIONS AND EQUATIONS – 27.–2) e. (2. y = (x + 1)2 + 2 e. (2. . (x)(8) = 8x (x + 4)(x2 + 6x + 8). 33 . (4)(6x) = 24x (x + 4)(x2 + 6x + 8). Multiplying Polynomials To multiply two polynomials. What is (x + 4)(x2 + 6x + 8)? (x + 4)(x2 + 6x + 8). multiply every term of the first polynomial by every term of the second polynomial. Then.C H A P T E R 4 A  Factoring and Multiplying Polynomials polynomial is an expression with more than one term. a trinomial is a polynomial with three terms. A binomial is a polynomial with two terms. (x)(6x) = 6x2 (x + 4)(x2 + 6x + 8). (4)(8) = 32 Add the products and combine like terms: x3 + 6x2 + 8x + 4x2 + 24x + 32 = x3 + 10x2 + 32x + 32. (x)(x2) = x3 (x + 4)(x2 + 6x + 8). add the products and combine like terms. (4)(x2) = 4x2 (x + 4)(x2 + 6x + 8). each with a different base. (x2 + 5x – 7)(x + 2) = a. x = –7. If the denominator of a fraction is a polynomial. (x + 49)  For what values of x is the fraction  (x3 + 7x2) undefined? Factor the denominator and set each factor equal to 0 to find the values of x that make the fraction undefined.  Practice 2. Set each factor equal to 0 and solve for x: x = 0. –3x3 + 6x – 54 b. Then. The values that make the polynomial equal to 0 are the values that make the fraction undefined. –x3 + 3x2 + 24x c. –3x3 – 3x2 – 54 d. factor it and set the factors equal to 0. factor out the variable x: x(x2 – 9x – 10). and 3x2 + 15x – 108 = 3(x – 4)(x + 9). What are the factors of 3x2 + 15x – 108? First. –3x(x + 6)(x – 9) = a. x = –1. The fraction is undefined when x equals 0 or –7. factor the monomial. x = 0. and x3 – 9x2 – 10x = x(x – 10)(x + 1). the roots of an equation are the values that make the equation true. x = 10. such as x or a constant. –1. x + 7 = 0. a fraction is undefined when its denominator is equal to 0. 2x3 + 10x2 – 14x 1. or quadratic that remains. x3 + 7x2 + 3x – 14 e. x3 + 7x2 + 17x – 14 d. binomial. x2 = 0. first look for a factor common to every term in the polynomial. x + 1 = 0. x3 – 3x2 – 17x – 14 b. What are the roots of x3 – 9x2 – 10x = 0? First. x3 + 5x2 – 7x + 14 c. –3x3 + 9x2 + 162x 34 .  Finding Roots As you saw in the last chapter. After factoring out that value.– FACTORING AND MULTIPLYING POLYNOMIALS –  Factoring Polynomials To factor the kinds of polynomials you will encounter on the SAT. factor the quadratic that remains: x2 + 5x – 36 = (x – 4)(x + 9). The roots of this equation are 0. x3 + 7x2 = x2(x + 7). –3x2 + 6x – 72 e. Then. factor the quadratic that remains: x2 – 9x – 10 = (x – 10)(x + 1). factor out the constant 3: 3(x2 + 5x – 36). x – 10 = 0. and 10.  Undefined Expressions As you saw in previous chapters. 2(x – 8)(x + 12) b. x3 – 8x2 + 27x – 18 d. 0 d. x3 – 10x2 – 9x – 18 e. What are the factors of 2x3 + 8x2 – 192x? a. 9 a. –4. –4. 1 e. x3 – 18 b. (x + 1)  x (x + 1)  (x + 5) (x + 1)   (x2 – 5x) (x + 1)   (x2 + 5) (x + 1)   (x2 + 5x) 1  (x – 8) x  (x – 8) (x + 8)  (x – 8) d. 2x(x + 16)(x – 12) e. 1. (x2 + 7x + 12)  11. x3 – 9x – 18 c. What values make the fraction  (x3 + 3x2 – 4x) undefined? a. –1. 4. The fraction  (x3 – 64x) is equivalent to b. 4 (x2 + 8x)  7. –4. e. b. –1. 6. 5. 3 d. –5 c. c. d. –1 c. 0. b. x(2x – 8)(x + 24) d. (x + 2)  (x – 6) x  (x + 2)(x + 6) 1  (2x – 12) (x + 2)  4x(x – 6) 2x(x + 2)  (x – 6) 10. No values make the fraction undefined. e. 0. 1 c. x + 8 35 . –1 d. The fraction  (4x3 – 16x2 – 48x) is equivalent to a.– FACTORING AND MULTIPLYING POLYNOMIALS – (x2 + 6x + 5)  8. 2x(x – 8)(x + 12) c. x – 8 e. 16x(2x – 1)(2x + 1) (2x2 + 4x)  9. What is a root of x(x – 1)(x + 1) = 27 – x? a. 4 e. 2(x2 – 8x)(x2 + 12x) d. 16(x3 – x) c. 16x(4x – 1) e. –9 b. 16x(4x2) d. c. x3 – 10x2 + 27x – 18 a. 5 e. c. 4(16x3 – 16x) b. 1 b. The fraction  (x3 – 25x) is equivalent to 3. What is one value that makes the fraction (x2 – 25)   (x3 + 125) undefined? a. 0. What are the factors of 64x3 – 16x? a. –25 b. (x – 6)(x – 3)(x – 1) = a. 10 e. 6 e. what is the number? a. If the number is greater than 0. –4 c. 20 15. What is one value that makes the fraction (x2 + 11x + 30)  (4x + 44x2 + 120x) undefined? a. 5 d.– FACTORING AND MULTIPLYING POLYNOMIALS – 14. –1 13. 7 12. 5 c. –6 b. 4 c. what is the number? a. The cube of a number minus twice its square is equal to 80 times the number. If the number is greater than 0. 8 36 . Four times the cube of a number is equal to 48 times the number minus four times the square of the number. 8 d. 4 b. What is largest of these integers? a. 5 d. 3 b. 6 e. 4 c. The product of 3 consecutive positive integers is equal to 56 more than the cube of the first integer. –3 d. –2 e. 3 b. of x 2 is equal to x. or x. you will need to find the square or cube root of a number or variable. The number under the radical symbol is called the radicand. The radical symbol. or second root. 22 + 32 = 52 37 . To add two radicals with the same radicand. The square root.C H A P T E R 5  Radicals and Exponents Radicals In order to simplify some expressions and solve some equations.  Adding and Subtracting Radicals Two radicals can be added or subtracted if they have the same radicand. it is assumed that the radical symbol represents the square root of the number. add the coefficients of the radicals and keep the radicand the same. . signifies the root of a value. If there is no root number given. Both radicals represent the same root. To divide two radicals. Since (8)(8) = 64. since 2 = 5 and  = 5. 64  is equal only to 8. You can simplify the original radical. 38 . multiply the coefficients of the radicals and multiply the radicands. and both radicals have the same radicand. (1015 )  (23) 10 15  = 55. but the square root of a positive number must be a positive number. but you will still have a radical in your answer. since both 8 and –8 square to 64. To multiply two radicals. the product is equal to the value of the radicand alone.  Multiplying Radicals Two radicals can be multiplied whether or not they have the same radicand. since (4)(3) = 12 and (6)(7) = 42 . so the product of 6 and 6 is 6. not –8. (6)(6) = 6. subtract the coefficient of the second radical from the coefficient of the first radical and keep the radicand the same. The equation x2 = 64 has two solutions. the square root. when multiplied by itself. 6. Look for a number that. equals 64. However. Many whole numbers and fractions do not have roots that are also whole numbers or fractions. the square root of 64 is 8.64  = 8. since these radicals have different radicands. divide the coefficients of the radicals and divide the radicands. 3 3 Any radical divided by itself is equal to 1. most radicals cannot be simplified so easily.  Dividing Radicals Two radicals can be divided whether or not they have the same radicand. (46)(37) = 1242 .– RADICALS AND EXPONENTS – To subtract two radicals with the same radicand. 65 – 45 = 25 The expressions 3 + 2 and 3 – 2 cannot be simplified any further. If two radicals of the same root with the same radicand are multiplied. find the square root of 64.  = 1. 3  Simplifying a Single Radical To simplify a radical such as 64 . To remove a radical symbol from one side of an equation. In other words. Multiply the top and bottom of the fraction by the radical in the denominator. Remove a cube root symbol from an equation by cubing both sides of the equation. If x = 6. 16  is a perfect square. To remove a square root. raise both sides of the equation to the power that is equal to the root of the radical. square both sides of the equation. or second root. what is the value of x? To remove the radical symbol from the left side of the equation. For example.  Rationalizing Denominators of Fractions An expression is not in simplest form if there is a radical in the denominator of a fraction.– RADICALS AND EXPONENTS – To simplify a single radical. or base. To remove a cube root. Any value with an exponent of 0 is equal to 1. (3)(3) = 3. you can raise both sides of the equation to a power. 32  = (16 )(2). x0 = 1. the frac4 tion  3 is not in simplest form. Any value with an exponent of 1 is equal to itself. raise both sides of the equation to the third power. 26 = (2)(2)(2)(2)(2)(2). 3 3 x = 6. the positive square root of 16 is 4. x1 = x. this will not change the value of the fraction. one of which is a perfect square. Since any radical multiplied by 4 43 itself is equal to the radicand. The exponent is equal to the number of times a base is multiplied by itself. multiplying. (4)(3) = 43. 100 = 1. subtracting. 3  Solving Equations with Radicals Use the properties of adding. 101 = 10. and simplifying radicals to help you solve equations with radicals. 4 3 3   Multiply  3 by 3 . raise both sides of the equation to the second power. (x)2 = (6)2. Therefore. that power is the exponent of the base. x = 36. x = 27  Exponents When a value. dividing. 39 . such as 32 . or third root.32 = (16 )(2) = 42. 11 = 1. The exponent of the term 42 is two. 10 = 1. is raised to a power. find two factors of the radicand. x = 3. Since 3 = 1. 42 = (4)(4). so the fraction  in simplest form is  3 . and the base of the term is 4. (x)3 = (3)3. Remove a square root symbol from an equation by squaring both sides of the equation. divide the coefficients of the bases and subtract the exponents of the bases. multiply the coefficients of the bases and add the exponents of the bases. 3 42 = (4)3 = 23 = 8 It does not matter if you find the root (represented by the denominator) first. (27x5)  (9x) = 3x4 (2x3)  (8x4) x–1 (xc)  (xd) 1  = 4. For example. which means that x must be raised to the first power (x1 = x) and then the second. x 2 = x. 1 or square. and then raise the result to the power (represented by the numerator). 1  3–3 =  (33) 1  x–2 =  (x2)  Multiplying and Dividing Terms with Exponents To multiply two terms with common bases. or  (4x) = xc – d 40 . (3x2)(7x4) = 21x6 4 (2x–5)(2x3) = 4x–2. root must be taken. or if you find the power first and then take the root. 3 3 = 64 =8 42 = (4)  Negative Exponents A base raised to a negative exponent is equal to the reciprocal of the base raised to the positive value of that exponent. The denominator of the fraction is the root of the base that must be taken. or x2 (xc)(xd) = xc + d To divide two terms with common bases.– RADICALS AND EXPONENTS –  Fractional Exponents An exponent can also be a fraction. The numerator of the fraction is the power to which the base is being raised. the square root of a num1 ber can be represented as x 2. 2 b. a2b2 a. 16x2 a. = b. 2b) 2) ((a )((ab )  ab  a.– RADICALS AND EXPONENTS –  Raising a Term with an Exponent to Another Exponent When a term with an exponent is raised to another exponent. 4 6. 2g  g b. 4x2 c. a5 c. (x3)3 = x9 (xc)d = xcd If the term that is being raised to an exponent has a coefficient. keep the base of the term and multiply the exponents. 16x e. 2) (32x  4. abab = e. y e. ab  d. 2g c. a6 e. 4x8 d. y3 5. ab c. a a b. 3 3) = 2. 3  3 y3  3 d. c. be sure to raise the coefficient to the exponent as well. a5a d. 42x  b. e. 3. = a. 41 –2 5 m3  n5 n5 3 m m6 10 n n7 5 m n10  m6 = = . a9 g  4 4g  (27y3)   2) (27y  m   n 3 a. c. e. g d. (3x2)3 = 27x6 (cx3)4 = c4x12  Practice 3 1. ab   ab  b. a3(a  4 a. d.  256 x3y3 c. 6 c. d. 3pr 1 13. What is the value of (ab)–ab if a = 3 and b = 9? 3 b. 9 14. a7b6 d. d. ((4g2)3(g4))2 = a. 8g4 c. a7 b. (b)4 = a. 3 9. – 3 1 b. 3 y (y)2(x)–2  xy 1 a. 66 e. y5 x3 c. d. What is the value of ((xy)y)x if x = 2 and y = –x? a. x5y5 1  16 d. 9 c. If (p)4 = q–2. 3p2r2 x 10. a12b11 1  1 12. y3 1  9 1  3 = e. 36 d. 1  3 d. 4 e. 36 7. 8g10 e. c. 8g3 b. 3pr  b. 16 42 . and q = – 3. e. what is one possible value of p? 8. 3 b. then a3 = a.– RADICALS AND EXPONENTS – (ab)3 2 4 11. a12b8 e. pr a. –4 1  b. 8g5 d. 8g12 1 a. 1 e. 3pr c. a12 c. xy x3 b. 9  a. 9pr   – 3 (pr) 2 = e. If a3 = 6.  2 c.– RADICALS AND EXPONENTS – 3 15. c. 6 b. 12 d. 6 d. 12 17. c. 24 e. e. 55 e. 43 1  1. what is the value of bb a ( a4 ) ? ( ) a. If n   -1 n 2 n  5 5  1  n4 20.024 1  32 1  16 1  4 1  2 . b. 8 c. 46 e. 384 1  36 1  6 6  6 d. If n = 20. 6 e. n 5 d. 10 d. 23 b. What is the value of (x –y)(2x y)(3y x) if x = 2 and y = –2? a. 5n d. 6  n +5 16. 26 c. 1 4 c. b. what is the value of d? a. If g108 = g . If (cd)2 = 48 and c = 2. what is a value of g? a. what is the value of  (2 5)? n  a. n 19. 25 m = 5 what is the value of m in terms of n? a. If a is positive. 5 55 b. and a2 = b = 4. b. 5n  e. 18. . 10.  Arithmetic Sequences An arithmetic sequence is a series of terms in which the difference between any two consecutive terms in the sequence is always the same. The first term of the sequence is referred to as the first term (not the zeroth term). The difference between any two consecutive terms is always 4. 6. 33. the rule of the sequence will always be given to you.” On the SAT. . Each term in the sequence below is five less than the previous term. . .C H A P T E R 6 Sequences A sequence is a series of terms in which each term in the series is generated using a rule. 14. Each value in the sequence is a called a term. 2. . 23. . . The rule of a sequence could be “each term is twice the previous term” or “each term is 4 more than the previous term. 18. For example. What is the next term in the sequence? 38. 45 . is an arithmetic sequence. 28. . 27. 192. . For example. 50th. Sometimes you will be asked for the 20th. multiplied by 3. 12. take the last term given and subtract 5. Each term in the sequence below is four times the previous term. 43. multiplied by 3. the fourth term in the sequence. 192. 11. . . multiply the second term in the sequence. by 6. What is the value of x? 2. . 95. Therefore. and apply the rule. . 432. 43. but you can represent that term with an expression. . The fourth term in the sequence. 3. 47. 81.  Geometric Sequences A geometric sequence is a series of terms in which the ratio between any two consecutive terms in the sequence is always the same.– SEQUENCES – To find the next term in the sequence. .  Combination Sequences Some sequences are defined by rules that are a combination of operations. . For example. the third term in the sequence. the 100th term of the sequence is equal to 4 raised to one less than 100 (99). To find the value of x. or 100th term of a sequence. since the rule of the sequence is “each term in the sequence is five less than the previous term”. 23 – 5 = 18. . 48 is equal to 42  3. 48. Each term in the sequence is equal to 4 raised to an exponent. The terms in these sequences do not differ by a constant value or ratio. and 192 is equal to 43  3. so the next term in the sequence is 18. 4(43) – 1 = 172 – 1 = 171. (12)(6) = 72. 11. It would be unreasonable in many cases to evaluate that term. is a geometric sequence. 12. 3. Write each term of the sequence as a product. Each term in the sequence below is one less than four times the previous term. 1. 12 is equal to 41  3. 3 is equal to 40  3. multiplied by 3. x. What is the 100th term of the sequence? 3. . . each number in a sequence could be generated by the rule “double the previous term and add one”: 5. The 100th term is equal to 499  3. . is equal to 4 raised to one less than four (3). 23. the value of the exponent is one less than the position of the term in the sequence. . The ratio between any two consecutive terms is always 3—each term is three times the previous term. The third term in the sequence is 72. You can check your answer by multiplying 72 by 6: (72)(6) = 432. since every term is six times the previous term. For each term. 46 . Take the last given term in the sequence. What is the next term in the sequence? 1. . Each term in the sequence below is six times the previous term. 12. 9. 11 1 c. b. . 8. d. . 5 e. . 83. 3. –9 c. a. What is the seventh term of the sequence? 3. What is the ninth term of the sequence? 101. What is the 20th term of the sequence? 1 1 1 . . . 520 e. 7 2 7. 125 25 5 a. 9 b. a. 162 e. 113 e. . c. 113 2 d. 192 c. 517 c. –10 d. Each term in the sequence below is –2 times the previous term. 128  81 64  27 128  27 64  9 128  3 8. –24. What is the value of a + b + c + d? 2. 7. –11 b. 15 1 b. 38 e. a. d. c. 519 d. . –96 b. . –12 c. What is the seventh term of the sequence? 16 18. . 18 6. b. 72. 152 c. . 10 b. 29 d. 4. 4. y. . . . Each term in the sequence below is five times the previous term. –16.536 3. a. What is the value of x – y? 12. . –16 b. . . . Each term in the sequence below is six more than the previous term. 1. 521 47 . . a. z. 12. –2 d. 20 c. a. 92. a. 13. Each term in the sequence below is nine less than the previous term. –4 e. . Each term in the sequence below is seven less than the previous term. 3. 5. . y. –6.– SEQUENCES –  Practice 1 5. 1. 102. 768 e. 74. –6 e. Each term in the sequence below is 3 times the previous term. . 9. 12. 516 b. . What is the value of x + z? x. 384 d. Each term in the sequence below is 3 more than the previous term. 16 1 d. . Each term in the sequence below is 2 more than the previous term. x. . 119 a. What is the eighth term of the sequence? 1 1 6. . 1. . 12 3 2. . a. . . . 3201 d. d. –16 d. a. . –162 b. What is the value of x + y? x. e. . 256. . –1. What is the value of xy? x. . 9.5. e – 2b e. 36.– SEQUENCES – 13. . –80. a. 26 12. a + 2b d. Each term in the sequence below is 16 more than –4 times the previous term. What is the product of the 100th and 101st terms of the sequence? 1. –8 b. . 199 a. f – e 48 . –64. . –32 b. a. 0.900 14. eighth c. –28 c. What is the value of x + y? x. tenth e. c. . . y. –158 e. . 20 d. Each term in the sequence below is –4 times the previous term. Each term in the sequence below is 20 less than five times the previous term. 16 e.328. –24 d. . . Each term in the sequence below is two less than 1  the previous term. e – c b. –159 d. –161 c. 126. –256 b. What is the next term of the sequence? –1. 61. All of the following are equal to the value of d EXCEPT a. –16 9. 28. b + c c. eleventh 11. . ninth d. . a. –5. –64 c. y. –40 b. a. –157 15. 27. –120. –7 c. b. 336. . 39. 3300 e. –5 d. 3. . x. . What term of the sequence 2 will be the first term to be a negative number? 256. . –20 e. Each term in the sequence below is equal to the sum of the two previous terms. . Each term in the sequence below is nine more 1 than 3 the previous term. –53. 64 10. –16 c. . Each term in the sequence below is two less than three times the previous term. . What is the value of y – x? 81. . seventh b. 22 e. f. a. 3200 c. . 7 e. 8 16. y. 3 b. Each term in the sequence below is three times the previous term. y. –17. take one equation and rewrite it so that you have the value of one variable in terms of the other. If the number of equations is greater than or equal to the number of variables. Solve this system of equations for x and y: 2x + 2y = 4 3y – 4x = 13 49 . substitute that expression into the second equation and solve.  Substitution To solve a system of equations using substitution.C H A P T E R 7 Systems of Equations A system of equations is a group of two or more equations in which the variables have the same values in each equation. The SAT typically features a system of two equations with two variables. There are two techniques that can be used to solve systems of equations: substitution and combination (sometimes referred to as elimination). you can find the value of each variable. Then. since the expressions would have involved fractions. substitute this expression for x in the second equation and solve for y: 3y – 4(2 – y) = 13 3y – 8 + 4y = 13 7y – 8 = 13 7y = 21 y=3 Now that you have the value of y. but that would have been more cumbersome. substitute 3 for y in either equation and solve for x: x = 2 – (3) x = –1 The solution to this system of equations is (–1.3). and then add it to the other equation to eliminate a variable. Subtract 2y from both sides of the equation and divide by 2: 2x + 2y = 4 2x + 2y – 2y = 4 – 2y 2x = 4 – 2y 2x (4 – 2y)  =  2 2 x=2–y Now that you have the value of x in terms of y. 2x + 2y = 4 3y – 4x = 13 50 . multiply one equation by a constant. Alternatively.  Combination To solve a system of equations using combination.– SYSTEMS OF EQUATIONS – Take the first equation and rewrite it so that you have the value of x in terms of y (you could also choose to rewrite the equation with the value of y in terms of x). you could have used the second equation to rewrite x in terms of y (or y in terms of x). depending on the equations in the system. There are many ways to approach solving a system of equations. and some methods work more easily that others. The same system of equations you just saw could also be solved using combination. what is the value of b? 5a + 3b = –2 5a – 3b = –38 a. –2 c. what is the value of x? 3(x + 4) – 2y = 5 2y – 4x = 8 a. what is the value of x? 2x + y = 6 y  + 4x = 12 2 a. 0 c. –4 c. 15 51 . and you still would have arrived at the same answer. what is one possible value of y? xy = 32 2x – y = 0 a. 1 d. the same answer found using substitution. 1 d. although one may involve less work than the other. 6 d. you could have used either equation. you found the value of x using the equation 3y – 4x = 13. After find the value of y using combination. Both methods will work. But. the x terms will drop out: (2)(2x + 2y = 4) = 4x + 4y = 8 (4x + 4y = 8) + (3y – 4x = 13) 7y = 21 Divide both sides of the equation by 7. you found the value of x using the equation 2x + 2y = 4. substitute the value of y into either equation to find the value of x: 3(3) – 4x = 13 9 – 4x = 13 –4x = 4 x = –1 After finding the value of y using substitution. 3 e. 13 4. Given the equations below. –8 b.– SYSTEMS OF EQUATIONS – Adding one equation to the other won’t eliminate either variable. –6 b. 4 e. you can use either substitution or combination to find the solution.  Practice 3. if the first equation is multiplied by 2 and then added to the second equation. Now. 2 d. 16 1. 12 e. Given the equations below. –2 b. 6 2. and you can see that y = 3. Given the equations below. 13 e. –2 b. Given the equations below. –1 c. In either example. When faced with a system of equations. –2 b. 8 5. what is one possible value of p? 4pq – 6 = 10 4p – 2q = –14 a. 4 7. 8 11. what is the value of b? 9a – 2(b + 4) = 30 1 42a – 3b = 3 a. 6 d. what is the value of y? 3x + 7y = 19 4y  = 1 x a. what is the value of a? 7(2a + 3b) = 56 b + 2a = –4 a. 13 e. –3 b. what is the value of n? 2(m + n) + m = 9 3m – 3n = 24 a. what is the value of b? b –7a + 4 = 25 b + a = 13 a. –5 b. –10 1 b. Given the equations below. 2 e. b. Given the equations below. Given the equations below. 1 d. –2 d. 16 6. 0 1 d. 4 e. 4 c. –4 c. 4 c. 5 e. 2 e. Given the equations below. Given the equations below. –2 1 b. 1 d. 2 e. –5 b. 10. 10 12. –3 c. 6 1  4 1  2 c. what is one possible value of m? m(n + 1) = 2 m–n=0 a. Given the equations below. 5 e. 6 8. 8 e. 12 d.– SYSTEMS OF EQUATIONS – 9. Given the equations below. – 2 c. Given the equations below. what is the value of y? 1 x + 6y = 7 2 –4x – 15y = 10 a. 3 d. 2 52 . –1 c. 2 b. 2 c. 2 d. 20 d. –7 c. 12 e. Given the equations below. Given the equations below. 5 e. 10 d. what is the value of a + b? 4a + 6b = 24 6a – 12b = –6 a. 2 b. 17 19. –2 c. 3 c. 4 d.– SYSTEMS OF EQUATIONS – c 13. –18 c. 12 e. –12 d. 45 16. 2 d. Given the equations below. 14 17. Given the equations below. 10 14. Given the equations below. –12 b. what is the value of a + b? 1 (a + 3) – b = –6 2 3a – 2b = –5 a. 25 e. 7 e. 5 c. what is the value n of m? m – 6(n + 2) = –8 6n + m = 16 a. 6 18. –10 b. –12 b. –27 b. 5 b. 14 15. Given the equations below. 12 53 . what is the value of d? c–d  – 2 = 0 5 c – 6d = 0 a. 7 d. what is the value of xy? –5x + 2y = –51 –x – y = –6 a. 6 c. –6 e. –3 20. what is the value of x – y? x  – 2y = 14 3 2x + 6y = –6 a. 1 d. what is the value of ab? 10b – 9a = 6 b–a=1 a. 12 e. –7 b. 8 d. –9 1 c. 2 b. Given the equations below. 15 c. 9 e. what is the value of y – x? 3x + 4 = –5y + 8 9x + 11y = –8 a. Given the equations below. what is the value of a b? a  = b + 1 2 3(a – b) = –21 a. 25 e. 49 e. c. 5 c. –1 c. Given the equations below. 4 b. Given the equations below. 1 b. 36 22. d. –24 b. 16 d. 2 24. what is the value of (y – x)2? 9(x – 1) = 2 – 4y 2y + 7x = 3 a. Given the equations below. e. 81 54 4  9 2  3 3  4 4  3 3  2 . 6 25.– SYSTEMS OF EQUATIONS – 21. 0 d. Given the equations below. 23. 4 c. Given the equations below. 1 e. 0 d. 1 e. –2 c. –6 b. what is the value of x – y? x+y  = 8 3 2x – y = 9 a. what is the value of (p + q)2? 8q + 15p = 26 –5p + 2q = 24 a. 25 d. what is the value x of y? 4x + 6 = –3y –2x + 3 = y + 9 a. b. substitute the given input for the variable. the equation is not a function. If you can draw a vertical line through any point of the graph and the line crosses the graph no more than once. draw the graph of the equation.  Vertical Line Test An equation is a function if it passes the vertical line test. The set of elements that make up the possible inputs of a function is the domain of the function. and Range Functions A function is an equation with one input (variable) in which each unique input value yields no more than one output. The set of elements that make up the possible outputs of a function is the range of the function. Domain. If the vertical line crosses the graph more than once for any point. The values for x are the domain of this function. A function commonly takes the form f(x) = x + c. To test if an equation is a function.C H A P T E R 8  Functions. the equation is a function. If f(x) = 5x + 2. f(3) = 5(3) + 2 = 15 + 2 = 17. The values of f(x) are the range of the function. 55 . where x is a variable and c is a constant. what is f(3)? To find the value of a function given an input. This equation fails the vertical line test. it will cross the graph of the equation x = y2 in two places. draw a horizontal line through the graph of the function where f(x). Both x = 2 and x = –2 make f(x) = 4. Horizontal Line Test y 4 3 2 1 –4 –3 –2 –1 1 –1 –2 –3 –4 x 2 3 4 The horizontal line test can be used to determine how many different x values. or inputs. For example. = 4. but more than one x value can return the same f(x) value. There is no value of x that can be substituted into the equation that can yield more than one different value for y. AND RANGE – Is y = x2 a function? y 4 3 2 1 –4 –3 –2 –1 1 –1 x 2 3 4 –2 –3 For every value of x. so it is not a function. the equation x = c. a function cannot have one input return two or more outputs. If a vertical line is drawn through any point on the graph. To find how many values make f(x) = 4. Remember. where c is any constant. but it can have more than one input return the same output. it will cross the graph of y = x2 only once. DOMAIN. the function f(x) = x2 is a function. return the same f(x) value. The equation y = x2 passes the vertical line test. since that is the graph of a vertical line. For example. is not a function for any value of c.– FUNCTIONS. or y. –4 Is x = y2 a function? y 4 3 2 1 x –4 –3 –2 –1 1 –1 2 3 4 –2 –3 –4  For every value of x that is greater than 0. only one y value is returned. A vertical line is not a function. If a vertical line is drawn through any point to the right of the y-axis. Look at the graph of y = x2 at left. 56 . two y values are returned. because no x value can return two or more f(x) values. Always be wary of equations that are written in the form x = rather than y =. so it is a function. so the value of x must not be less than 0. For all values of x greater than or equal to 0. the values that make a part of the function undefined are the values that are NOT in the domain of the function. the value of the function would 2 be 0. f(–6) = 5(–6) + 2 = –30 + 2 = –28. since no value for x can make f(x) = 0. the horizontal line test proves that there are two values for x that make f(x) = 4. What is the domain of the function f(x) = x? The square root of a negative number is an imaginary number. what is f(g(x)) when x = 3? Begin with the innermost function: Find g(x) when x = 3. If any real number can be substituted for x. find g(3). substitute the result of that function for x in f(x). The range of this function is also all real numbers. In other words. Therefore. If x = 4. 2  The function f(x) =  x – 4 has a domain of all real numbers excluding 4. the domain of the function is x ≥ 0. What is the value of g(f(x)) when x = 3? 57 . DOMAIN. In a function. Any real number can be substituted for x in the equation and the value of the function will be a real number. AND RANGE – You can see that the line y = 4 crosses the graph of f(x) = x2 in exactly two places. Then. f(g(x)) = –28 when x = 3.  Nested Functions Given the definitions of two functions.  Range As you just saw. the function will return values greater than or equal to 0. Therefore. In earlier chapters. the range of the function is all real numbers excluding 0.– FUNCTIONS. What is the range of the function f(x) = x? You already found the domain of the function to be x ≥ 0. you can find the result of one function (given a value) and place it directly into another function. the function f(x) = 3x has a domain of all real numbers.  Domain The function f(x) = 3x has a domain of all real numbers. For example. 2  Although the domain of the function f(x) =  x – 4 is all real numbers excluding 4. if f(x) = 5x + 2 and g(x) = –2x. g(3) = –2(3) = –6. 3x can yield any real number. Therefore. which is undefined. you found the values for an expression that make the expression undefined. That expression. the value of n#m = n2 + 2m. is now the term after the # symbol: n#(m + 3n). substitute 17 for x in g(x). Add the term before the # symbol to three times the term after the # symbol. Look again at the definition of the function. the value of m#n = m + 3n. but the question asks for the value of n#m. Now. what is the value of n#m? The definition of the function states that the term before the # symbol should be squared and added to twice the term after the # symbol. a symbol such as # may be given a certain definition. DOMAIN. The definition of the function states that the term before the # symbol should be squared and added to the term after the # symbol. what is the value of m#n when m = 2 and n = –2? Substitute the values of m and n into the definition of the symbol. or you may be asked to find an expression that represents the function. Therefore. such as “m#n is equivalent to m2 + n. Therefore.– FUNCTIONS. Add n to three times (m + 3n): n + 3(m + 3n) = n + 3m + 9n = 3m + 10n. If m#n is equivalent to m2 + 2n. AND RANGE – Start with the innermost function—this time. When m = 2 and n = –2.  Newly Defined Symbols A symbol can be used to represent one or more operations. The definition of the function states that the term before the # symbol should be added to three times the term after the # symbol. g(17) = –2(17) = –34. f(3) = 5(3) + 2 = 15 + 2 = 17. it is f(x). m2 + n = (2)2 + (–2) = 4 – 2 = 2. m + 3n. Watch your variables carefully. If m#n is equivalent to m2 + n. If m#n is equivalent to m + 3n.”You may be asked to find the value of the function given the values of m and n. 58 . what is the value of n#(m#n)? Begin with the innermost function. f(g(x)) = –28 and g(f(x)) = –34. When x = 3. m#n. For a particular question on the SAT. The definition of the function is given for m#n. 5 4. 3 c. AND RANGE –  Practice 3. 5 e. For how many values is f(x) = 3? y y x x a. For how many values is f(x) = 0? 1. For how many values is f(x) = –5? 2.– FUNCTIONS. 1 b. For how many values is f(x) = 2? y y x x a. DOMAIN. The graph of f(x) is shown below. 0 b. 5 e. The graph of f(x) is shown below. 4 a. 2 c. 4 d. 2 b. 3 e. The graph of f(x) is shown below. 8 a. 4 e. 8 59 . 4 d. 0 b. The graph of f(x) is shown below. 3 d. 2 d. 1 c. 2 c. – FUNCTIONS, DOMAIN, AND RANGE – 9. If f(x) = 6x + 4 and g(x) = x2 – 1, what is the value of g(f(x))? a. 6x2 – 2 b. 36x2 + 16 c. 36x2 + 48x + 15 d. 36x2 + 48x + 16 e. 6x3 + 4x2 – 6x – 4 5. The graph of f(x) is shown below. For how many values is f(x) = –2? y x 10. If f(x) = 4 – 2x2 and g(x) = 2x, what is the value of f(g(x))? a. 4 – 16x b. 4 – 8x c. 4 – 4x d. 4 – 2x e. 4 – x a. 0 b. 1 c. 2 d. 3 e. 4 11. If w@z is equivalent to 3w – z, what is the value of (w@z)@z? a. 3w – 2z b. 6w – 2z c. 9w – 4z d. 9w – 3z e. 9w – 2z 6. If f(x) = 2x – 1 and g(x) = x2, what is the value of f(g(–3))? a. –7 b. 2 c. 9 d. 17 e. 49 p 12. If p&q is equivalent to q + pq, what is the value of q&p when p = 4 and q = –2? a. –10 b. –8.5 c. –7.5 d. 4 e. 16 7. If f(x) = 3x + 2 and g(x) = 2x – 3, what is the value of g(f(–2))? a. –19 b. –11 c. –7 d. –4 e. –3 13. If j%k is equivalent to kj, what is the value of k%(j%k)? a. kjk b. jkj c. k( j + k) d. j(k + j) e. kjjk 8. If f(x) = 2x + 1 and g(x) = x – 2, what is the value of f(g(f(3)))? a. 1 b. 3 c. 5 d. 7 e. 11 60 – FUNCTIONS, DOMAIN, AND RANGE – b–a ? 19. Which of the following is true of f(x) = 4x –1 a. The domain of the function is all real num1 bers greater than 4 and the range is all real numbers greater than 0. b. The domain of the function is all real numbers 1 greater than or equal to 4 and the range is all real numbers greater than 0. c. The domain of the function is all real numbers 1 greater than or equal to 4 and the range is all real numbers greater than or equal to 0. d. The domain of the function is all real numbers greater than 0 and the range is all real 1 numbers greater than or equal to 4. e. The domain of the function is all real numbers greater than or equal to 0 and the range is all 1 real numbers greater than or equal to 4.  14. If a?b is equivalent to  a + b , what is the value of a?(a?b) when a = 6 and b = –5? a. –11 7 b. –5 c. –1 d. 1 17 e. 5 15. If x^y is equivalent to y2 – x, what is the value of y^(y^y)? a. y2 – 2x b. y2 – 2y c. y4 – 2xy2 + x2 – y d. y4 – 2y3 + y2 – y e. y4 – y3 + y2 – y 1  16. What is the domain of the function f(x) =  (x2 – 9) ? a. all real numbers excluding 0 b. all real numbers excluding 3 and –3 c. all real numbers greater than 9 d. all real numbers greater than or equal to 9 e. all real numbers greater than 3 and less than –3 –1 20. Which of the following is true of f(x) =  ? –5 x a. The domain of the function is all real numbers excluding 5 and the range is all real numbers less than or equal to 0. b. The domain of the function is all real numbers greater than 5 and the range is all real numbers less than or equal to 0. c. The domain of the function is all real numbers greater than 5 and the range is all real numbers less than 0. d. The domain of the function is all real numbers greater than or equal to 5 and the range is all real numbers less than 0. e. The domain of the function is all real numbers greater than or equal to 5 and the range is all real numbers less than or equal to 0. 17. What is the range of the function f(x) = x2 – 4? a. all real numbers excluding 0 b. all real numbers excluding 2 and –2 c. all real numbers greater than or equal to 0 d. all real numbers greater than or equal to 4 e. all real numbers greater than or equal to –4 –x2 18. Which of the following is true of f(x) = 2? a. The range of the function is all real numbers less than or equal to 0. b. The range of the function is all real numbers less than 0. c. The range of the function is all real numbers greater than or equal to 0. d. The domain of the function is all real numbers greater than or equal to 0. e. The domain of the function is all real numbers less than or equal to 2. 61 – FUNCTIONS, DOMAIN, AND RANGE – 22. Which of the coordinate planes shows the graph of a function that has a range that contains negative values? a. A b. B c. D d. B, D e. A, B, D Use the diagrams below to answer questions 21–25. A B y y x x x2 + y2 = 4 23. Which of the coordinate planes shows the graph of a function that has a domain of all real numbers? a. B b. D c. E d. B, D e. B, E  y = |x| – 3 C D y y x x _ y = x y= 1 x 24. Which of the coordinate planes shows the graph of a function that has the same range as its domain? a. B, C b. C, D c. B, D d. B, E e. D, E E y x y = (x – 3)2 + 1 25. Of the equations graphed on the coordinate planes, which function has the smallest range? a. A b. B c. C d. D e. E 21. Which of the coordinate planes shows the graph of an equation that is not a function? a. A b. B c. C d. D e. A, D 62 The angle at left could be named B. or ABC. because that is its label. The angle could also be named CBA. but the vertex of the angle must always be the middle letter. x. An angle is named by either its vertex alone. 63 . The order in which rays or line segments are named does not matter. Two rays that meet at their endpoints form an angle. its label (if it has one). The vertex of the angle is the place where the two endpoints meet. or by the rays and the vertex that form the angle. because that is its vertex.C H A P T E R 9  Angles Naming an Angle A ray is a line with one endpoint. because it is formed by rays AB and BC. since it is formed by rays BC and BA. A x B C The vertex of an angle determines its name. – ANGLES –  Types of Angles Acute angles are angles that measure less than 90°. Obtuse angles are angles that measure greater than 90°. Right angles are angles that measure exactly 90°. Straight angles measure exactly 180°. It is important to remember that there are 180° in a straight line. Acute Angle  Obtuse Angle Right Angle Straight Angle Complementar y, Supplementar y, and Adjacent Angles 1 1 2 Complementary Angles 2 Supplementary Angles The measures of complementary angles add to 90° and the measures of supplementary angles add to 180°. Two angles that combine to form a right angle. 1 4 Adjacent angles are two angles that share a common vertex and a common ray. The complementary angles shown are also adjacent angles, as are the two supplementary angles. 2 3 Adjacent Angles  Vertical Angles and Alternating Angles 1 4 5 8 2 3 Intersecting lines form vertical angles. When two lines intersect, the angles formed on either side of the vertex (the point of intersection) are called opposite, or vertical, angles. Vertical angles are always congruent, or equal in measure. Angles 1 and 3 are congruent, and angles 2 and 4 are congruent. 6 7 Parallel lines cut by a transversal (a third line that is not parallel to the first two lines) also form vertical angles. Sets of vertical angles are called alternating angles. In the diagram at left, the parallel lines and the transversal form eight 64 – ANGLES – angles. The angles labeled 1 and 3 are vertical angles, as are the angles labeled 5 and 7. All four of these angles are congruent. Notice how these congruent angles alternate back and forth across the transversal as you look down the transversal. These vertical angles are alternating angles. Angles 1, 3, 5, and 7 are all alternating, and therefore, congruent. In the same way, angles 2, 4, 6, and 8 are all alternating, and they are all congruent. Angles 1 and 2 are not alternating, but they do form a line. These two angles are supplementary. Every oddnumbered angle is congruent to every other odd-numbered angle, and every odd-numbered angle is supplementary, and adjacent, to every even-numbered angle. In the same way, every even-numbered angle is congruent to every other even-numbered angle, and every even-numbered angle is supplementary, and adjacent, to every oddnumbered angle.  Practice Use the diagram below to answer questions 1–5. The diagram is not to scale. 3. If the measure of angle 4 is 6x + 20 and the measure of angle 7 is 10x – 40, what is the measure of angle 6? a. 60° b. 70° c. 110° d. 116° e. 120° 1 2 4 3 5 6 8 7 4. Which of the following is NOT true if the measure of angle 3 is 90°? a. Angles 1 and 2 are complementary. b. Angles 3 and 6 are supplementary. c. Angles 5 and 7 are adjacent. d. Angles 5 and 7 are congruent. e. Angles 4 and 8 are supplementary and congruent. 1. If the measure of angle 2 is equal to 12x + 10 and the measure of angle 8 is equal to 7x – 1, what is the measure of angle 2? a. 9° b. 62° c. 108° d. 118° e. cannot be determined 2. If the measure of angle 5 is five times the measure of angle 6, what is the measure of angle 5? a. 30° b. 36° c. 120° d. 130° e. 150° 65 5. If the measure of angle 2 is 8x + 10 and the measure of angle 6 is x2 – 38, what is the measure of angle 8? a. 42° b. 74° c. 84° d. 108° e. 138° – ANGLES – 8. If the measure of angle 3 is 2x + 2 and the measure of angle 4 is 5x – 10, what is the measure of angle 7? a. 14° b. 30° c. 60° d. 90° e. 115° Use the diagram below to answer questions 6–10. Lines AE, BF, GD, and ray OC all intersect at point O (not labeled), and line AE is perpendicular to ray OC. A B G 7 1 6 5 4 F 2 3 9. If angle 1 measures 62° and angle 4 measures 57°, what is the measure of angle 6? a. 33° b. 61° c. 72° d. 95° e. 119° C D E 6. Which of the following pairs of angles are complementary? a. angles 1 and 2 b. angles 2 and 3 c. angles 1 and 4 d. angles 1, 2, and 3 e. angles 1, 6, and 7 10. If the measure of angle 3 is 5x + 3 and the measure of angle 4 is 15x + 7, what is the sum of angles 5 and 6? a. 67° b. 113° c. 134° d. 157° e. cannot be determined 7. Which of the following number sentences is NOT true? a. angle 1 + angle 2 = angle 3 + angle 4 b. angle 1 + angle 2 + angle 3 + angle 7 = angle 4 + angle 5 + angle 6 c. angle 2 + angle 3 = angle 6 d. angle 1 + angle 7 = angle 2 + angle 3 e. angle 2 + angle 3 + angle 4 + angle 5 = 180 66 42° b. 125° c. angles AIE and JIL d. 48° c. 52° e. 128° 67 . 130° d. angles ILH and IKG 15. 120° b. 90° H J B 11. angles AIF and KIF e. Lines EF and GH are parallel to each other and line IJ is perpendicular to lines EF and GH. what is the measure of angle AIE? a. 38° c. A C I E F G D 14. 87° e. what is the measure of angle EIC? a. Which of the following pairs of angles must be congruent? a. 64° b. If angles LIF and EIK are congruent. 54° d. what is the measure of angle KIJ? a. angles KIJ and JIL c. If the measure of angle GKI is 15x – 4 and the measure of angle CIF is x2. what is the measure of angle EIL? a. If the measure of angle JLI is 8x – 4 and the measure of angle JIL is 5x + 3. 144° Use the diagram below to answer questions 11–15. 140° 12. If angle JLB is three and a half times the size of angle LIF. 120° d.– ANGLES – 13. 48° d. 116° c. 135° e. 14° b. angles AIC and KIL b. 124° e. and the measure of angle LIF is 6 greater than the measure of angle JIL. 98° e. 35° c. angle 4 + angle 5 + angle 6 = angle 1 + angle 2 + angle 8 68 . 8° b. angle 4 = angle 3 b. If the measure of angle 1 is 3x + 5 and the measure of angle 6 is 5x – 3. If lines EF and AB are perpendicular. Every angle is greater than 0°. what is the measure of angle 4? a. and the measure of angle 1 is 7x. angle 2 + angle 6 = angle 3 + angle 7 e. the measure of angle 4 is 11x. Lines CD and GH are perpendicular to each other. 90° d. angle 2 = angle 3 c. Which of the following number sentences is true? a. angle 7 = angle 3 e. 88° c. angle 1 + angle 2 + angle 3 = angle 2 + angle 3 + angle 4 d. which of the following is NOT true? a. what is the sum of angles 4 and 5? a. If angle 4 is congruent to angle 7. If the sum of angles 2 and 3 is x2 and the sum of angles 6 and 7 is 10x. 73° e. 17° b. 38° c. cannot be determined H 7 6 F B D 20. C E A 2 3 1 8 G 4 5 19. angle 3 = angle 8 d.– ANGLES – 18. 55° 17. 44° d. and every line goes through point O (not labeled). 80° b. angle 2 + angle 3 + angle 4 = angle 5 + angle 6 + angle 7 b. 52° d. angle 7 = angle 8 16. angle 3 + angle 4 = angle 6 + angle 7 c. 100° Use the diagram below to answer questions 16–20. 45° e. what is the measure of angle 2? a. angle 8 + angle 9 = 180 e. angle 14 + angle 15 + angle 16 = 180 Use the diagram below to answer questions 21–25. 62° c. If the sum of the measures of angles 20 and 14 is 15x + 6. Lines EF and GH are parallel to each other. 124° F H J 25. 133° e. 100° d. what is the measure of angle 10? a. 118° e. 24° b. If the sum of the measures of angles 21 and 5 is x2 + 11 and the measure of angle 3 is 9x + 1. 36° c. Every numbered angle is greater than 0°. and the sum of the measures of angles 20 and 15 is 18x. 90° d. angle 16 + angle 10 = angle 11 + angle 15 c. 45° d. 95° c. and line AB is perpendicular to lines EF and GH. angle 1 + angle 8 = angle 19 + angle 15 + angle 20 d. If the sum of the measures of angles 8 and 9 is 133°. lines CD and IJ are parallel to each other.– ANGLES – 23. 90° 21. If angle 5 = 8x – 4 and angle 22 = 7x + 11. 156° 69 . angle 13 + angle 14 = angle 13 + angle 15 b. what is the measure of angle 3? a. 137° 22. 43° b. A E G D C 5 6 1 2 3 4 7 8 9 10 11 12 14 13 15 16 17 18 20 21 22 19 B I 24. 100° d. 116° e. 92° b. 18° b. 47° c. 54° e. what is the measure of angle 19? a. Which of the following must be true? a. what is the measure of angle 16? a. . CBA. The third angle of triangle ABC at left is equal to 180 – (50 + 60) = 180 – 110 = 70°. and C comprise the triangle.C H A P T E R 10  Triangles Interior Angles The three interior angles of a triangle add up to 180°. BAC. or CAB. This triangle could also be named BCA. 50 x B 60 C 71 . ACB. since the vertices A. starting from any one of the vertices. you can find the measure of the third angle by adding the measures of the first two angles and subtracting that sum from 180. B. The triangle below is named ABC. Triangles are named by their vertices. A If you know the measure of two angles of a triangle. The vertices must be named in order. with each side of the triangle extended. Angle z is also supy z plementary to the angle that measures 50°. The largest angle in triangle ABC measures 80°. is 130 + 110 + 120 = 360°. 60 90 30 E Right Triangle F 72 . Here is another look at triangle ABC. The measure of angle y is 130°. you could also find the measure of angle y by adding the measures of the other two interior angles. Notice that angles y and z are vertical angles—another reason these two angles are u 70 w equal in measure. the triangle is a right triangle. since it is supplementary to the 60° angle (180 – 60 = 120) and because the sum of the other interior angles is 120° (70 + 50 = 120). since it is supplementary to the 70° angle (180 – 70 = 110) and because the sum of the other interior angles is 110° (50 + 60 = 110).  Types of Triangles A If the measure of the largest angle of a triangle is less than 90°. since it and angle BAC are supplementary. Angle ABC. therefore. If you find the measure of one exterior angle at each vertex. The measure of angle w is 120°. the triangle is an acute triangle. 70. it is a right triangle. since these angles form a line. The sum of angles y. Angle y A and the angle that measures 50° are supplementary. The largest angle in triangle DEF measures 90°. 40 80 B 60 Acute Triangle C D If the measure of the largest angle of a triangle is equal to 90°. angle z also measures 130°.– TRIANGLES –  Exterior Angles The exterior angles of a triangle are the angles that are formed outside the triangle. However. it is an acute triangle. u. 60 The measure of an exterior angle is equal to the sum of the two inteB v x C rior angles to which the exterior angle is not adjacent. You already know angle y measures 130°. the sum of these three exterior angles is 360°. Since angle z is also supplementary to the 50° angle. Adjacent interior and exterior angles are supplementary. 50 The measure of angle y is equal to 180 – 50 = 130°. 60. therefore. is equal to the measure of the exterior angle of BAC: 70 + 60 = 130°. plus angle ACB. The measure of angle u is 110°. and w. the measure of angles y and z. 100  Similar Triangles 40 40 Isosceles Triangle 60 60 60 Two triangles are similar if the measures of their corresponding angles are identical. two angles) of a triangle are equal. 30° e. all three angles) of a triangle are equal. 15° c. In scalene triangle XYZ. side YZ is the longest side. which of the following is true of triangle ABC? a. If the measure of angle A of triangle ABC is 5x + 10. and each side of triangle JKL is twice as long as its corresponding side of triangle MNO. 1. If the measure of angle A of triangle ABC is 3x. b. the side opposite the largest angle of the triangle is the longest side. it is an obtuse triangle. 20° d. e. Equilateral Triangle  Practice 2. Triangle ABC is right but not isosceles. If triangles JKL and MNO are similar. If exactly two sides (and therefore. If all three sides (and therefore. since it is the side opposite the largest angle. If no two sides or angles of a triangle are equal. the measure of angle B is x + 10. Triangle ABC is acute but not scalene. therefore. 120 35 H I Obutse Triangle X 85 35 Z 60 Y Scalene Triangle In a triangle. and the measure of angle C is 2x. the triangle is equilateral. Triangle ABC is obtuse but not scalene. In an isosceles triangle. 12° b. the measure of angle B is 5x. and the measure of angle C is 4x. and the side opposite the smallest angle is the shortest side. the sides opposite the equal angles are the equal sides.– TRIANGLES – If the measure of the largest angle of a triangle is greater than 90°. the triangle is an obtuse triangle. d. c. Triangle ABC is acute and scalene. the ratio of the sides of triangle JKL to the sides of triangle MNO is 2:1. The sides of similar triangles can be expressed with a ratio. The lengths of the corresponding sides of similar triangles can be different—it is the measures of the angles of the triangles that make the triangles similar. the angle opposite the right angle is the hypotenuse. what is the value of x? a. the triangle is isosceles. G 25 There are three other types of triangles. the triangle is scalene. 45° 73 . The largest angle in triangle GHI measures 120°. which is always the longest side of the triangle. Triangle ABC is obtuse and scalene. In a right triangle. 105° c. The measure of an angle exterior to angle F of triangle DEF measures 120°. what is the measure of angle exterior to angle A? a. 25° b. b. angle 3 + angle 2 = angle 4 d. 7. 125° 5 B 2 3 6 C 6. 115° d. Triangle DEF is acute. and every angle is greater than 0°. Which of the following number sentences is NOT true? a. 49° c. c. 7° b. Triangle DEF is equilateral. The diagram is not to scale. 124° Use the diagram below to answer questions 6–9.– TRIANGLES – 3. 4 A 1 4. d. e. angle 1 + angle 2 + angle 3 = 180 e. what is the measure of angle 4? a. If the measures of angles A and B of triangle ABC are each 2x + 5 and the measure of angle C is 3x – 5. 102° e. Triangle DEF is obtuse. angle 1 + angle 2 = angle 5 b. 56° d. 110° e. 75° b. angle 4 + angle 5 + angle 6 = 360 5. angle 4 + angle 1 = angle 3 + angle 6 c. If angle F measures 8x. 140° e. 190° 74 . The measure of an angle exterior to angle F of triangle DEF measures 16x + 12. Triangle DEF is not isosceles. If angle 6 measures 115° and angle 2 measures 75°. 55° c. what is the measure of angle F? a. Triangle DEF is scalene. 70° d. Which of the following must be true? a. Triangle ABC is equilateral. Triangle ABC is obtuse. Triangle ABC is right. The sum of the sides of triangle ABC is five times the sum of the sides of triangle DEF. 12. angle 5 measures 8x. 36 c. b. 30 b. If the measure of angle F in triangle DEF is half the sum of the measures of angles D and E. the length of  AC is 6x. 6 d.– TRIANGLES – 8. If the measure of angle 1 is x2 + 1. what is the measure of an angle exterior to F? a. The diagram is not to scale. Triangles JKL and MNO are congruent and equilateral. e. 118° 11. If the length of  BC is x and the length of  EF is 1 x. 48 e. Every side of triangle DEF is greater than 3. If the length of  AB is 10x – 2. 66° d. c. 12 10. cannot be determined A D B C E F 75 . If angle 4 measures 7x + 2. what is the length of DF ? of DF a. Triangles ABC and DEF are similar. which of the following is NOT true? 5 a. Use the diagram below to answer questions 11–13. Triangle ABC is scalene. 52° b. Angle B is congruent to angle E. 1 c. 4 c. and angle 6 measures 8x – 10. 3 b. and the length  is x + 2. 14. e. The length of  AB is 5 times the length of  DE . what is the length of side NO? a. Triangle ABC is isosceles. The length of  DF is 5 the length of  AC . the length of  DE is 2x + 2. 60 e. which of the following is true? a. what is the measure of angle 1? a. 7 b. what is the length of  DF ? a. 60° c. d. 10 e.  and the length of AC is 72. d. 42 d. cannot be determined 13. If the length of  AB is 90. 62° c. 120° e. and every angle is greater than 0°. If side JK measures 6x + 3 and side MN measures x2 – 4. 30° b. 90° d. 42 c. 45 d. b. The measure of angle A is five times the measure of angle D. 108 9. 114° e. and the measure of angle 3 is 6x + 2. the measure of angle 2 is 9x – 7. the length of  DE is 60. angle 8 is greater than angle 7. and the measure of angle 13 is 94°. 125° 76 . 58° c. 94° d. What is the measure of angle 10? a. 60° d. 122° e. Side GH is congruent to side PR. e. and every angle is greater than 0°. Angles 11 and 14 are complementary. which of the following must be true? a. 136° 15.5 times the measure of angle 14. Lines EF and GH are parallel. Side GI is congruent to its corresponding side. 40° c. If angle 3 measures 10x + 15. which of the following is NOT true? a. 55° c. Triangles GHI and PQR are similar and isosceles. The diagram is not to scale. b. 85° e. 100° Use the diagram below to answer questions 16–20. and the measure of angle 13 is 7x – 6. and GH is an isosceles right triangle.– TRIANGLES – 18. 80° e. 43° b. What is the sum of angles 10 and 7? a. If the triangle formed by lines AB. what is the measure of angle 7? a. 137° e. angle 13 = 90° c. The measure of angle 5 is 10x + 2. what is the measure of angle 8? a. angle 3 = angle 12 + angle 13 B 16. 65° d. 86° c. 44° b. and angle 6 measures 3x + 7. CD. angle 10 measures 8x – 3. angle 1 = angle 4 e. the measure of angle 11 is 4x – 4. Angle H is not congruent to angle R. If angles 6 and 7 are congruent. c. its corresponding side in triangle PQR. The measure of angle 11 is twice the measure of angle 14. d. If side HG of triangle GHI is congruent to side PQ. 19. d. and angle 8 is congruent to angle 5. 20° b. angle 2 = angle 9 + angle 12 b. Angle G is congruent to angle R. 40° b. 147° 17. 102° d. A E G C 1 2 5 6 3 4 7 8 10 9 11 12 13 14 D F H 20. Angle G is congruent to angle I. and the measure of angle 8 is 2. C H A P T E R 11  Right Triangles Pythagorean Theorem As you saw in the previous chapter. A c a B b C The Pythagorean theorem states that the sum of the squares of the bases of a triangle is equal to the square of the hypotenuse of the triangle. then a2 + b2 = c2. 77 . For instance. Given the lengths of two sides of a right triangle. b is the length of the other base. The formula can be rewritten to find any of the three sides. 2 a = c2 – b2 and b2 = c2 – a2. If a is the length of one base. and c is the length of the hypotenuse. right triangles are triangles in which one angle measures 90°. the Pythagorean theorem can be used to find the missing side of the right triangle.  AC . c 3 B  C 4 Common Pythagorean Triples The three lengths of a right triangle are often referred to as Pythagorean triples. In every 30-60-90 right triangle. use the Pythagorean theorem. Multiples of these triples are also Pythagorean triples. c2 = 16. 78 . 18 = c2. and 90°. The length of side AC is 5 units. What if each base measured 3 units? Then. you can find that 22 + (23)2 = c2. and 12:16:20 are all multiples of the 3:4:5 triple. and c = 18 . Its angles measure 45°. Another common Pythagorean triple is 5:12:13. 32 + 32 = c2. 6:8:10. You just saw that 3:4:5 is a Pythagorean triple. 9 + 16 = c2. Multiples of the 5:12:13 triple are 10:24:26 and 15:36:39.– RIGHT TRIANGLES – A In triangle ABC at left. a2 + b2 = c2. 45°. 25 = c2. and c = 4. what is the length of the hypotenuse? Using the Pythagorean theorem again. If side BC of triangle ABC at right is 2 units and side AB is 23 units. and c = 8. or 22. the length of the hypotenuse is 2 times the length of the shorter base (the base that is opposite the 30° angle). The length of the hypotenuse 2 times is 2 times the length of a base. Common Right Triangles There are two common types of right triangles often seen on the SAT: the 45-45-90 right triangle and the 30-6090 right triangle. (3)2 + (4)2 = c2. Notice that in both cases. the bases are sides AB and BC. the length of the hypotenuse is 2 times the length of a base of the triangle. A c x B C x A 30 60 B C The 45-45-90 right triangle is an isosceles right triangle. Now. and then add those squares. Its angles measure 30°. 8 = c2. what is the length of the hypotenuse? Using the Pythagorean theorem. If the bases each measure 2 units. 9:12:15. The bases of this right triangle are equal. or 32. The 30-60-90 right triangle is another type of right triangle. 60°. In the same way. c = 25 . c = 5. The length of the longer base (the base opposite the 60° angle) is always 3 times the length of the shorter base.  is the side opposite the right angle. To find AC . the length of a base is  2 the length of the hypotenuse. take the square root of both sides of the equation to find c. 4 + 12 = c2. you can find that 22 + 22 = c2. since the hypotenuse. and 90°. This is always the case with 45-45-90 right triangles. Square sides a and b. The cosine of C is 6 3 equal to the side adjacent to angle C. 5 c. and tangent of angle C? 10 8 B 6 30° 45° 60° 1  2 2   2  3  2 cosine 3   2  2  2 1  2 tangent 3   3 1 3  sine  C AB First. 16 units e. If the length of a base of right triangle DEF is 8 units and the hypotenuse of triangle DEF is 85 units.  BC Be sure to know the sine. and tangent of these common angles: Practice 2. 12 e. and tangent. Cosine = Hypotenuse . find the sine. 4 b. or 3. 4 units b. If the hypotenuse of the triangle is 2x – 3. AC : 10 . 32 units 1. what is the length of the hypotenuse? a. A In triangle ABC. 13 79 . 8 d. what is the length of the other base? a. cosine. or 5. and Tangent = Adjacent . what are the sine. The tangent of  8 4 AB C is equal to  = 6. 8 units c. The sine of an angle is equal to the length of the base opposite the angle divided by the length of the hypotenuse. divide the length of  8 4  by the length of the hypotenuse. Opposite Adjacent Opposite    Use the mnemonic device SOHCATOA: Sine =  Hypotenuse . The tangent of an angle is equal to the length of the base opposite the angle divided by the length of the base adjacent to the angle. cosine. divided by  AC : 10 . The cosine of an angle is equal to the length of the base adjacent to the angle divided by the length of the hypotenuse. is the sine of angle C. The bases of a right triangle measure x – 3 and x + 4. Since  AB is the side opposite angle C. side BC. cosine.– RIGHT TRIANGLES –  Trigonometr y There are three trigonometry functions that will appear on the SAT: sine. 82 units d. or 5. b. what is the length of side PR? a. a. If the longer base of triangle XYZ is three times the length of the shorter base. 102 units c. what is the length of the hypotenuse in terms of the shorter base? a. 4 units b. x2 units e. G E 6. 5. and 15 e. 2 units b. 18 units e. 42 units 5. 3a e. a triangle with sides measuring 3. and 5 c. 30 units 7. 10. x2  units 2 x  units 2 1 x units 2 d. Angle T of right triangle TUV measures 45°. If line segment JK is 16 units long. cannot be determined Use the diagram below to answer questions 9–10. The diagram is not to scale. what is length of the hypotenuse of triangle GHI? a. and 10 d. 334  units c.– RIGHT TRIANGLES – 3. x2 units 4. Which of the following triangles is a multiple of the triangle with sides measuring x. What is the length of a base of the triangle? a. 100 units I A C J H B K D F 9. If base TU measures 10 units. a triangle with sides measuring 1. a triangle with sides measuring 5. and x + 5. 4. when x is greater than 0? a. a triangle with sides measuring 1. what is the length of the hypotenuse of triangle TUV? a. 2. and 3 b. cannot be determined 80 . 10. c. 452 units e. 62 units b. If the lengths of the bases of right triangle GHI are 9 units and 15 units respectively. a10  d. 82 units d. 202 units d. x – 5. a2 b. a2 8. Angle Q of triangle PQR is a right angle. a3 c. what is the length of line segment IK? a. a triangle with sides measuring 5. and 20 e. The hypotenuse of an isosceles right triangle measures x units. 42 units c. 934  units d. Line AB is parallel to line CD. and line GH is perpendicular to lines AB and CD. 2  4 units b. 2  2 units c. 162 units e. If the sine of angle P is equal to the sine of angle R. and side QR measures 4 units. 22 units d. 4. and the measure of angle IKD is 135°.5 units d. 8 A C D B E a. 23 units b. If the measure of side AB is 7 units. Triangle ABC is an equilateral triangle and triangle CDE is a right triangle. 18 units e.– RIGHT TRIANGLES – 10. (3x + 1)3 units d. 4 units c. 14 units c. what is the length of side BC? a. 8 units e. what is the measure of side AC? a. 43 units d. If the sum of sides BC and AC is 12 units. Given right triangle ABC with right angle B. 3  3 units b. 93 units 81 . 252  2 units d. 6x + 2 units  b. Angle C measures 60°. 7 units d. 73 units c. If the hypotenuse of triangle ABC is 6x + 2 units long. 20 units C B 12. 14 units 15. what is the length of side AB? a. (6x + 2)2 units e. If the length of side AB is 9 cm. If line CD is 25 units from line AB. 33 units e. what is the length of side AB? a. angle A is twice the size of angle C. 72 units b. 252 units 14. 12 units b. 3 units c. 5 units b. what is the length of side BD? A Use the diagram below to answer questions 12–14. 16 units d. 12x + 4 units a. The diagram is not to scale. cannot be determined 11. 3x + 1 units c. 123 units e. 14  units e. If the length of side AE is 20 units. 52 units c. what is the length of line segment IK? 13. and line AC measures 2x + 2. b. The diagram is not to scale. cosine of angle A d. c. 12 units 17. what is the length of side BC? a. and x does not equal y. triangle ABC is not isosceles. c.– RIGHT TRIANGLES – x 18. 8  17 8  15 12  17 17  15 15  8 x  3 3  x 3x – 6  2x + 2 (x2 – 2x)  (2x + 2) 2x + 2  3x – 6 82 . 3 units b. a. tangent of angle A e. If the cosine of angle C is y. 0 1 b. d. sine of angle A b. 8 units e. 2 c. If the tangent of angle A is 0. 6 units d. A b. sine of angle C c.75 and the measure of side AB is 4 less than the measure of side AC. what is the tangent of angle A? a. what is the cosine of angle A? Use the diagram below to answer questions 17–20. e. cannot be determined 15 19. 2 16. If the sine of angle A is 17 . what is the tangent of angle A? a. If angle B of isosceles triangle ABC is a right angle. e. 4 units c. 1 d. 3  3 e. which of the x following is also equal to y? a. If line AB measures 3x – 6. d. C B 20. line BC measures x2 – 2x. If angle KMH is three times the size of angle KML and the length of side JK is x6 units. 16 units e. 9 units b. 163  3 units b. the length of side IK is twice the length of side JK. If the length of side KM is 10 units and the length of side LM is 5 units. the length of side IK is 2x + 1. what is the measure of angle KIJ? a. If the length of side LM is 8 units and the sum of angles IKJ and LKM is 60°. then a. 0.5° b. 60° 25. 83 units d. If the tangent of angle JIK is 3. what is the length of side KM? a. 15 units d. 20 units 22. 50° e. the sine of triangle JIK is  2 .– RIGHT TRIANGLES – 23. and the length of side LM is 2x + 2. the length of side IJ is equal to the length of side JK. A J E I F K M G L H 24. 163 units 83 . 3x units d. 82 units c. Use the diagram below to answer questions 21–25. 16 units e. 2 c. 3x2 units e. e. 45° d. 2x3 units c. side IJ is 3 times the length of side JK. the length of side KM is 3x – 1. d. If the length of side IJ is 2x – 2. side LK is 3 times the length of side LM. what is the length of side IK? a. 30° c. 2x6 units B 21. Line EF is parallel to line GH. and line JL is perpendicular to lines EF and GH. 12 units c. The diagram is not to scale. b. x2 units b. what is the length of side LM? a. . however. For example. 85 .C H A P T E R 12  Polygons Types of Polygons A polygon is a closed figure with three or more sides. a heptagon has seven sides. an octagon has eight sides. and a decagon has ten sides. An equilateral triangle is a regular polygon. is not—it is a closed figure. but it does not have at least three sides. so is a square. a nonagon has nine sides. a pentagon has five sides. A circle. A quadrilateral has four sides. a triangle is a polygon. A triangle is a polygon with three sides.  Regular Polygons A regular polygon is an equilateral polygon—all sides of the polygon are the same length. You should be familiar with the names of polygons that have three to ten sides. a hexagon has six sides. Every angle of a regular hexagon 720 measures 6 = 120°. As with similar triangles. The perimeter of a regular polygon is equal to the length of one side of the polygon multiplied by the number of sides of the polygon. Perimeter A 8 B 8 8 F C 8 8 E 8 D The perimeter of a polygon is the sum of the lengths of its sides. Therefore. Regular hexagon ABCDEF has six sides. divide the sum of the interior angles by the number of sides (angles) of the polygon. the lengths of the corresponding sides of similar polygons can be different— it is the measures of the angles of the polygons that make the polygons similar. and each side of polygon ABCD is four times the length of its corresponding side of polygon EFGH.  Similarity A B D C E F H G  Two polygons are similar if the measures of their corresponding angles are identical. and s is the number of sides of the polygon. every side is congruent to every other side.– POLYGONS – A B F C E  D If A is the sum of the interior angles of a polygon. add the lengths of all 12 sides. the sum of its interior angles is 180(6 – 2) = 180(4) = 720°. The sides of similar polygons can be expressed with a ratio. To find the length of a polygon with 12 sides. Since hexagon ABCDEF is a regular polygon. then A = 180(s – 2). Even if the polygon has 100 sides. To find the perimeter of a triangle. add the lengths of each of its 3 sides. If polygons ABCD and EFGH are similar. 86 . To find the measure of an angle of a regular polygon. The perimeter of hexagon ABCDEF is also equal to (6)(8). the ratio of the sides of polygon ABCD to the sides of polygon EFGH is 4:1. The ratio of the perimeter of polygon ABCD to the perimeter of polygon EFGH is also 4:1. its exterior angles will still add up to 360°. since 8 + 8 + 8 + 8 + 8 + 8 = 48. and every angle is congruent to every other angle. The perimeter of hexagon ABCDEF is 48. Exterior Angles The sum of the exterior angles of any polygon is 360°. 10 sides 7. 5 sides d. The polygon has 4 sides. b. c. At least one side of octagon ABCDEFGH is equal to at least one side of octagon STUVWXYZ. If the ratio of the perimeter of octagon ABCDEFGH to the perimeter of octagon STUVWXYZ is 1:1. If  AB  and FG are corresponding sides. 4.– POLYGONS –  Practice 5. The two regular pentagons are congruent. The two octagons are regular octagons. 3 sides b. If x is 3 greater than the number of sides of the polygon. 16y4 e. e. which of the following statements must be true? a. and the length of  AB is 4x + 4. 4 sides c. The two octagons are similar. what is the length of  FG ? a. The polygon is a regular polygon. 12 sides e. If the sum of the interior angles of a polygon is equal to the sum of the exterior angles. 2y c. 10 sides d. How many sides does the polygon have? a. Each side of each pentagon is congruent to every other side of that pentagon. 20x + 20 2. The two regular pentagons are similar. 6. The ratio of  AB to  ST is 1:1. Which of the following is NOT always true? a. 6 sides b. b. 6 sides e. how many sides does the polygon have? a. 87 . e. Pentagons ABCDE and FGHIJ are similar. The sums of the interior angles of each octagon are equal. 10 sides d. It is impossible for these sums to be equal. 4x + 4 d. The sum of the interior angles of each pentagon is 540°. which of the following must be true? a. The sum of the exterior angles of each pentagon is 360°. c. The ratio of each side of pentagon ABCDE to its corresponding side of pentagon FGHIJ is 4:1. 8 sides c. 4y2 d. Andrea draws a polygon with x number of sides. cannot be determined 3. 7 sides c. The sum of the interior angles of a polygon is equal to three times the sum of its exterior angles. 16x + 16 e. d. 12 sides e. y2 b. Heather draws two regular pentagons. The sum of the interior angles of her polygon is 60 times its number of sides. 13 sides 1. If the perimeter of quadrilateral ABCD is equal to 4y2. 2x + 2 c. Quadrilaterals ABCD and EFGH are similar. cannot be determined 8. How many sides does Andrea’s polygon have? a. x + 1 b. 6 sides b. The sum of the interior angles of a polygon is 9x2. d. b. c. The polygon has 2 sides. but not necessarily congruent. what is the perimeter of quadrilateral EFGH? a. The polygon has 6 sides. e. d. These sides are corresponding sides. 20 units b. 54 units d. What is the perimeter of an isosceles right triangle whose hypotenuse is 52 units? a. 2:3 d. 18 units 88 . If the ratio of the perimeter of polygon ABCDE to the perimeter of polygon TUVWX is 4:3. 252 units 11. and these sides are corresponding sides. 6 units b. 5 units b. what is the perimeter of polygon ABCDE? a. If the sum of the interior angles of a regular polygon equals 720°. 72 units e. The ratio of the lengths of a side of regular pentagon ABCDE to the length of a side of regular hexagon PQRSTU is 5:6. If x = 8. 18x2 units b. 120 units 14. what is the ratio of the perimeter of polygon GHIJKL to the perimeter of polygon ABCDEF? a. 10 + 52 units d. 36 units d. Regular polygon ABCDE is similar to regular polygon TUVWX. Side AB is 5x – 1. ? what is the length of BC A 2x 2x + 2 B 15. 15 units b. 152 units e. 15 units e. what is the perimeter of hexagon PQRSTU? a. what is the perimeter of the polygon? a. 18 units c. 108 units 10. and side TU is 4x – 2. 27x2 units 3x – 2 D 3x C a.– POLYGONS – 12. 1:3 b. 15 +2 units c. 3:1 e. 2:1 c. and the length of one side of the polygon is 3x2. 12 units c. 8 units b. If side AB is 12x + 6x and side GH is 8x + 4x. If the perimeter of the figure below is 60 units. 24 units c. 14 units d. 90 units e. what is the length of one side of the polygon? a. 56 units 9. The perimeter of a regular seven-sided polygon is 11x – 4. 10 units c. 3:2 13. 24x12 units e. 16 units e. 13 units d. 18x12 units c. Polygon ABCDEF is similar to polygon GHIJKL. If the perimeter of pentagon ABCDE is 75 units. 24x2 units d. Since the sum of the exterior angles of any polygon is 360°.C H A P T E R 13 A  Quadrilaterals quadrilateral is a polygon with four sides. The interior angles of any quadrilateral add up to 360°. A diagonal of a quadrilateral cuts the quadrilateral into two triangles. Diagonals The lines that connect opposite angles of a quadrilateral are its diagonals. 89 . the sum of the exterior angles of a quadrilateral is 360°. could be a rhombus. The diagonals of a parallelogram bisect each other. Therefore. side IK = side JL. but cannot be a parallelogram. or square. Not only do the diagonals of a rhombus bisect each other. A rhombus is a parallelogram with four congruent sides. Angles A and C are congruent. could be a parallelogram or a rectangle. The opposite sides of a parallelogram are parallel and congruent. Since a rectangle is a parallelogram. rhombus. Sides BC and AD are also parallel and congruent. cannot be a parallelogram. Angle E of rhombus EFGH measures 3x + 5. intersect at 90° angles. All four angles are congruent (all measure 90°) and the diagonals of a rectangle are congruent. In rectangle IJKL at left. what is the measure of angle F? a. could be a rectangle. e. rectangle. Opposite angles are congruent and pairs of angles that are not opposite are supplementary. Also true is that a square is a rectangle with four congruent sides or that a square is a rhombus with four right angles. The diagonals of a square bisect each other. as are angles B and D. Sides AB and CD of parallelogram ABCD are parallel and congruent. angle D is also supplementary to angles B and C. are not necessarily congruent. 80° c. its opposite sides are parallel and congruent. but cannot be a square. line segments AC and BD. but not a square. 90 . A square is a parallelogram with four congruent sides and four right angles. d. 120° e. Practice 2.– QUADRILATERALS –  Types of Quadrilaterals A B E D C Parallelogram F E H G Rhombus I J L K Rectangle M N P O Square  A parallelogram is a quadrilateral that has two pairs of parallel sides. If the measure of angle H is 4x. 100° d. but cannot be a rectangle. His polygon a. 140° 1. The two diagonals themselves. b. Andrew constructs a polygon with four sides and no right angles. line segments AE = EC and line segments BE = ED. Angle A is supplementary to angles B and C. 20° b. could be a parallelogram. they are perpendicular to each other. and are congruent. its opposite sides are parallel and its opposite angles are congruent. A rectangle is a parallelogram with four right angles. c. Since a rhombus is a parallelogram. rectangle. Monica draws a quadrilateral whose diagonals form four right triangles inside the quadrilateral. 7 c. angle DCE = angle ECB e. 5 units b. and the perimeter of the square is 2 m. d. e. but not a rectangle. 6x – 4 d. c. she can form a. angle DAE = angle EDA d. 2x2 – 4x b. 6x + 4 e. e. 24x units d. 8 d. 20x units c. The length of one side of a rhombus is – 6. 6x – 8 11. rhombus. but not a parallelogram. 10 units e. or parallelogram. 8x units e. what is the perimeter of the new. larger square? a. 8x2 units x2 7. If x is the width of the rectangle. must have four congruent sides. b. c. angle ECD = angle ABE 9. these three figures must be three congruent squares. a square. must be a square. angle AEB = angle DEC b. c. 32x units e. If the rectangle and the square share a side. 6 units c. b. DeDe has four identical line segments. 46 e. 8. What is the perimeter of a square with a diagonal measuring 2x2 units? a. all of the above 10. If the perimeter of a parallelogram ABCD is equal to the perimeter of rhombus EFGH and the perimeter of square IJKL. 8 units d. must have congruent diagonals. 4x2 units d. 2x units b. Four squares are joined together to form one large square. a square. The diagonals of rectangle ABCD intersect at E. the rhombus could be a square. If the perimeter of the rhombus is 168 units. d. The length of a rectangle is four less than twice its width. but the parallelogram must not be a square. 3x – 4 c. what is the value of x? a. This quadrilateral a. b. but not a parallelogram. 43 b. a square or a rectangle. the parallelogram must also be a rhombus. The length of a rectangle is four times the length of a square. Using them. 64x units 6. 174  91 . If the perimeter of one of the original squares was 8x units. then a. a rhombus. a square or a rhombus. angle EDC = angle EBA c. but not a rhombus or a rectangle.– QUADRILATERALS – 3. the parallelogram must not be a square or a rhombus. 16x units b. 4x units c. what is the perimeter of the rectangle? a. 5. Which of the following is NOT always true? a. d. what is the perimeter of the rectangle? a. every side of the rhombus must be congruent to every side of the square. e. 20 units 4. must have four congruent angles. 18 units b. 72 units e. 3 units b. 64 units d. 32 units c. what is the length of the longer diagonal? a. 11 units 15. 103 units 92 . the tangent of which is 8. Diagonal AC of rectangle ABCD creates angle ACB. what is the length of a side of the square? a. 144 units 13. 8 units d. If the length of side BC is 8 units. the cosine of which is 13 . 10 units e. 53 units d. 5 units c. 2 units b. 34 units e. what is the perimeter of rectangle ABCD? a. 4 units c. 30 units d. 50 units 12. Angle A of rhombus ABCD measures 120°. Diagonal AC of rectangle ABCD creates angle 12 ACD.– QUADRILATERALS – 14. 25 units c. which of the following could be the perimeter of rectangle ABCD? a. If the perimeter of a square is equal to 5x + 1 and the length of the diagonal of the square is (2x – 2)2. If the lengths of the sides of rectangle ABCD are all integers. 18 units b. 10 units e. If one side of the rhombus is 10 units. where b is the base of the triangle.2 or cm2.C H A P T E R 14  Area and Volume Area The area of a two-dimensional figure is the amount of space within the borders of the figure. The area of the rectangle is equal to lw = (8)(4) = 32 square units. you do not need to memorize them.  Area Formulas for Common Shapes A These formulas will be given to you at the SAT. The area of the 1 1 1 triangle is equal to 2bh = 2(4)(5) = 2(20) = 10 square units. 5 B 4 E 1 C F Area of a Triangle: 2bh. Rectangle EFGH has a length of 8 units and a height of 4 units. Area of a Rectangle: lw. where l is the length of the rectangle. Area is always measured in square units. Triangle ABC has a base of 4 units and a height of 5 units. 4 H 8 G 93 . and h is its height. such as in. and w is its width. 3 or cm3. such as in. The cylinder has a height of 10 units and a radius of 3 units. so multiplying the length. width. and a height of 4 units. and h is its height. The length. Although the volume of a cube can be found using the formula for the volume of a rectangular solid. where s is the length of a side of the square. The surface area of the rectangular solid is equal to 2(6  10) + 2(6  3) + 2(3  10) = 2(60) + 2(18) + 2(30) = 120 + 36 + 60 = 216 square units. where e is the length of an edge of the cube. The volume of the cylinder is equal to πr2h = π(3)2(10) = π(9)(10) = 90π cubic units. width. and height is the same as cubing any one of those measurements. where r is the radius of the cylinder. the volume of a cube could also be described as e3. The rectangular solid has a length of 8 units. The surface area of a rectangular solid is equal to the sum of the areas of the 6 rectangles (3 pairs of congruent rectangles) that comprise the rectangular solid. 3 10 4 2  8 Volume of a Cylinder: πr2h. Volume is always measured in cubic units. where l is the length of the rectangular solid. then the area of the square is (3)2 = 9 square units. two rectangles that measure 6 units by 3 units.– AREA AND VOLUME – Although the area of a square can be found using the formula for the area of a rectangle. 94 . Volume of a Rectangular Solid: lwh. If the length of one side of a square is 3 units. The area of one square of the cube is equal to (5)(5) = 25 square units. Surface Area 5 6 3 10 The surface area of a three-dimensional shape is the sum of the areas of each side of the shape. w is its width. The volume of the rectangular solid is lwh = (8)(2)(4) = 64 cubic units. so you do not need to memorize these either. the surface area of the square is equal to (25)(6) = 150 square units. The rectangular solid is composed of two rectangles that measure 6 units by 10 units. For instance. since the length and width of a square are equal. Since all six squares that comprise the cube are identical. the area of a square could also be described as s2.  Volume The volume of a three-dimensional figure is the amount of space the figure occupies. and two rectangles that measure 3 units by 10 units. a width of 2 units. the surface area of a cube is equal to the sum of the areas of the six squares that comprise the cube. and h is its height. and height of a cube are all the same.  Volume Formulas for Common Shapes These formulas will also be given to you at the SAT. – AREA AND VOLUME –  Practice  of square ABCD is x units. The area of a rectangle is x2 + 7x + 10 square units. four times the area of the old square. b. x + 5 units d. 3x2 square units 2 d. 60 square units b. 3x2 square units 3. 108 square units e. 9x square units D D E C 1 a. 723 square units d. x2 – 2x square units 1 b. If the height of a triangle is half its base. 2x2 square units c. c. twice the area of the old square. and side AE = side EB. If the lengths of the sides of a square are halved. What is the area of the triangle? a. 240 square units e. one-fourth the area of the old square. x2 – 3x square units 2 e. what is the width of the rectangle? a. Triangle DEC is inscribed in rectangle ABCD. An isosceles right triangle has a hypotenuse of x6 units. 4b2 c. x2 – 3x square units 1 c. b. 2x + 4 units e. If the length of AB  is 2 of AD. A B 1 b 2 1 b2 2 E e. 4b 1 b. 3x square units d. d. what is the area of triangle DEC? A F 6. what is the area of the triangle? 1 a. 4. If the length of the rectangle is x + 2 units. 3x2 square units e. 136 square units d. one-half the area of the old square.  what is the size and the length of EF 3 of the shaded area? 1. If side AB = 30 units. e. 255 square units 7. b 2. x + 4 units c. equal to the area of the old square. side EC = 17 units. 363 square units b. d. 72 square units c. 120 square units c. What is the area of an equilateral triangle that has a perimeter of 36 units? a. x2 + 6x + 8 units B C a. the area of the new square is a. 144 square units 95 . 5. x3 square units 3 b. x + 2 units b. .. 5 b.. height = 5 in. radius = 3 in. Terri fills with water 3 of a glass that is 15 cm tall. what is the area of rectangle ABCD? A D B 14. 20π cm3 c. d. height = 5 in.. and the radius of cylinder B is 3 the radius of cylinder A. A cylinder has a volume of 45π in. 20 12. 8 cm c. 7 d. The volume of cylinder B is 9 the volume of cylinder A. c. 10π cm3 b. 200 square units 11.  AC is a diagonal of rectangle ABCD below. If the radius of the glass is 2 cm. 1 b. The height of cylinder B is three times the height 1 of cylinder A. what is the value of x? a. 60π cm3 9. (32x2 + 8x)π d. what is the length of the rectangle? a. 30π cm3 d. 8 e. 15. The volume of cylinder B is 3 the volume of cylinder A. 10.. what is the radius of the cylinder? a. 18 cm e. height = 3 in. 16 cm d. radius = 3 in. The volume of cylinder B is the same as the volume of cylinder A. 16 cm d. 503 square units c. 6 c. The radius of a cylinder is 2x and the height of the cylinder is 8x + 2. (16x2 + 4x)π b. radius = 5 in. 20 cm 13. The volume of cylinder B is 3 times the volume of cylinder A.– AREA AND VOLUME – 2 8. The perimeter of a square is 3x – 4 units. 4 cm b. 1003 square units e. b. (128x3 + 64x2 + 8x)π C a. e. The volume of cylinder B is 9 times the volume of cylinder A. If the volume of the cylinder is 256π cm3. 24 cm e. 50 square units b. If the area of the square is 25 square units. The height of a cylinder is four times the radius of the cylinder. Which of the following statements is true? 1 a. 32 cm 96 . height = 15 in. radius = 9 in. (32x3 + 8x2)π e. If angle ACD is 30° and  AC = 20 units. If the area of the rectangle is 96 cm2. 6 cm b. d. height = 10 in. 100 square units d. 40π cm3 e. radius = 9 in. (16x3 + 4x2)π c.3. What is the volume of the cylinder in terms of x? a. what volume of water is in Terri’s glass? a. c. e. The length of a rectangle is two less than three times its width. 8 cm c. Which of the following could be the radius and height of the cylinder? a. 1 c. What is the surface area of the solid? a. 64 in. 12 in. 12 units 17. 384 in. 132 in.2 1 a. 4x2 b. The area of one face of a cube is 9x square units. 120 square units e. What is the volume of the cube? a. 1 b. 27x cubic units c. Danielle’s cube has a volume of 512 in. 6 units e. c. 16x4 e.2 d. What is the area of one face of her cube? a. d. What is the surface area of her cube? a. The length of a rectangular solid is 6 units. If the width is equal to the height and the volume of the solid is 108 in. 2 units b. 8x3 d. 27xx cubic units d.2 c. the width of solid B is five times the width of solid A. what is the width of the solid? 1 a. 9 in. 6 units e. 2 units c. then 22. The surface area of a rectangular solid is 192 cm2. If the height of the solid is 4 units and the length of the solid is 12 units.072 in. 12 units d. b. The volume of Stephanie’s cube is equal to 64x6. 3 units c. 3 in. 74 square units c. 32x3 97 . The length of a rectangular solid is twice the sum of the width and height of the rectangular solid. The volume of rectangular solid A is equal to the volume of rectangular solid B. 27x3 cubic units 21. 6 in. 2 units b.3. 3 units d.– AREA AND VOLUME – 16. 27x cubic units b. 8 in. 4 units d. 512 in.3. and the height of the solid is 12 units. 110 square units d. and the height of solid A is twice the height of solid B. 27x2 cubic units e. 23. If the length of solid A is three times the length of solid B. If the volume of the solid is 36 cubic units. what is the width of the solid? a. 20. 60 square units b. 19. e. the width of solid B is 5 the width of solid A. e.2 b.2 e. 8x2 c. the width of solid B is six times the width of solid A. A rectangular solid measures 4 units by 5 units by 6 units. the width of solid B is 36 the width of solid A. 3. the width of solid B is 6 the width of solid A. what is the length of the solid? a. 148 square units 18. 22x2 e. and the surface area of the cube is x3 square units. The volume of a cube is x3 cubic units. The width of a rectangular solid is twice the height of the solid. 5 units e. 8x2 b. 4 units d.– AREA AND VOLUME – 24. and the height of the solid is twice the length of the solid. If x is the length of the solid. 6 units 25. What is the value of x? a. what is the surface area of the solid in terms of x? a. 3 units c. 1 unit b. 28x2 98 . 11x2 c. 14x2 d.  Circumference B C 5 O A The circumference of a circle is the distance around the circle. but it is not a polygon. It is a 360° arc in which every point on the arc is the same distance from a single point—the center of the circle.  AB is a diameter of the circle. the larger arc is called the major arc. The circumference of a circle can also be given as πd. to another point on the other side of the circle. The diameter of a circle is the length of a straight line from a point on one side of the circle. where r is the radius of the circle. The circumference of a circle is equal to 2πr. The diameter of a circle is twice the radius. The radius of a circle  is a radius is the distance from the center of the circle to any point on the circle. through the center of the circle. The circle has a radius of 5 units. where d is the diameter of the circle. you do not need to memorize it. and the smaller arc is called the minor arc.C H A P T E R 15 Circles A circle is a closed figure. and its circumference is (2π)(5) = 10π units. 99 . If a circle is comprised of two arcs. The formula for circumference will be given on the SAT. CO of the circle. Therefore. An arc is a curved line that makes up part or all of a circle. its diameter is (2)(5) = 10 units. The circle has a sector whose angle measures 90°. The area of the 90 1   sector is equal to a fraction of that:  360 (100π) = 4 (100π) = 25π square units. measures 360 – 80 = 280°. Arc Length The length of an arc is the distance between two points on a circle.2 or cm2. Major arc AB. 80° O  Area of a Sector B A 90° 10  O The area of a sector is the area of a fraction of a circle. Therefore. Two radii form a central angle and its intercepted arc. since there are 360° in a circle. This formula will also be given to you at the SAT. The arc formed by the two radii is the intercepted arc of the central angle. such as in. which measures 80°. the longer distance from A to B. The circle on the previous page has a radius of 5 units. Therefore. its area is equal to π(5)2 = 25π square units. The length of an arc is equal to a fraction of the circumference of a circle. The length of arc AB is equal to  360 (20π) = 4 (20π) = 5π units. Multiply the area of a circle by the fraction of the circle represented by the sector. Therefore. and the radius of the circle is 10 units. intercepted (minor) arc AB also measures 80°. you do not need to memorize it. multiplied by the circumference of the circle. Minor arc AB 90 1   is formed by two radii that meet at a 90° angle. 100 . The area of the circle is π(10)2 = 100π square units. The length of that arc is equal to the size of the central angle divided by 360. In the circle. The size of the sector is equal to the angle of the sector divided by 360. radii AO and BO form central angle AOB. You saw how the area of a sector is equal to a fraction of the area of a circle. since there are 360° in a circle.  Central Angles B A A central angle is an angle whose vertex is at the center of the circle and whose vertex is formed by two radii. As with polygons. its circumference is 2π(10) = 20π units.– CIRCLES –  Area The area of a circle is the amount of space within the border of the circle. The circle above has a radius of 10 units. A central angle and its intercepted arc are equal in measure. The area of a circle is equal to πr2. area is always measured in square units. – CIRCLES –  Practice Use the diagram below to answer questions 1–5. 14 units d. 3π units b. 98 units B D 1. If the area of the circle is 196π square units.200π units 4. 225π units  is 8 units and the measure of 5. 7 units c. The diagram is not to scale. 90 units b. C A O 3. what  ? is the length of CD 49 a. cannot be determined c. 12π units 50 c. If the length of OA angle AOC is 50°. 15π units b. If the length of  OD is 6 units and the measure of angle COB is 100°. 300° e. 3π units d. what is the measure of arc DB? a. what is the circumference of the circle? a. 225 units e. what is the area of sector AOC? a. what is the length of arc CB? 10 a. 28 units e. 120° d. 600π units e. 9π square units 101 . 60° c. If the length of  AO is 15 units. 5 π 18 10 π 9 32 π 25 80 π 9 square units square units square units square units e. b. 30° b. 1. 30π units d. d. If the measure of angle AOC is 60°. 30 units c. 2. (22x)π units 121 e. If the area of a circle is (121x)π square units. what is the length of arc DB? 1 a. 9 times smaller. 36π units e. What is the area of her circle? a. which sector has an area of 24π square units? a. what is the length of arc AE? a. (11x)π units d. 1π unit c. 22. what is the circumference of the circle? a. sector EOA 7. 3 times smaller. 54π units Use the diagram below to answer questions 6–9. 9 times bigger. e. 12. (16x2 + 9)π square units d. 36π units e. If the radius of the circle above is 12 units. (2x)π units 13. (16x2 + 24x + 9)π square units e. If the radius of the circle is 9 units.– CIRCLES – 9. The diagram is not to scale. (18x2)π square units d. (27x2)π square units e.5π square units c. If the radius of the circle is 15 units. the area of the circle becomes a. (64x2 + 96x + 36)π square units 102 . If the radius of the circle is 27 units. sector DOB c. 225π square units 8. 9π square units b. sector BOC d. 9π units b. 6π units b. 6 times bigger. (11x)π units b. (3x)π square units c. (81x4)π square units 11. 3 times bigger.5x2)π square units b. d. (4. sector AOC e. sector EOD b. b. 2π units d. If the diameter of a circle is 8x + 6. 45π square units e. C A 70° O 110° 80° 60° E 40° B D 6. 25π square units d. 27π units d. (22x)π units c. 12π units c. 54π units 10. what is the area of sector DOB? 30 a. what is the area of the circle? a. If the circumference of a circle triples. (4x + 3)π square units b. c. Jasmin draws a circle with a radius of 9x2. (16x + 12)π square units c. what is the circumference of the circle? 14. 13 O A 16. 4x + 10 c. x π 20 x π 18 x π 10 units units units d. what is the diameter of the circle? a. 2x + 5 b. 180 – 3x 17. one-half of the circumference of the original circle. 23π cm d. 6π cm e. a. what is the measure of arc CA? 15. xπ square units D a. 20xπ square units 19. Which of the following could be the value of x? a. (x2 –  360 )π square units x2  d. c.5 x π 360 square units b.– CIRCLES – 18. (4x + 5)π e. 6x c. two times the circumference of the original circle. ( 360 )π square units 103 . d. The measure of a central angle is 18°. (2x + 5)π d. four times the circumference of the original circle. 3 b. 3π cm c. ( 360 )π square units x3  e. the circumference of the new circle is a. b. 3x b. If a circle has an area of 12π cm2 and a diameter  AB. 2x2 + 10x + 12. If the area of a circle is (4x2 + 20x + 25)π. 43π cm x  c. 9 d. the same as the circumference of the original circle. The radius of Carly’s circle is 2x – 7. If the central angle of a sector is x° and the radius of the circle is x units. what is the length of arc AB? a. If the measure of angle COB below is 3x. If the diameter of a circle is doubled. c. 18xπ square units e. 12x e. If the length of its intercepted arc is xπ units. e. 6xπ d. B C 20. 6 c. and the area of her circle is (16x + 9)π. b. 3π cm b. then the area of the sector is equal to a. 10 e. one-fourth of the circumference of the original circle. x2 – π ft. What is the area of a sector whose central angle measures 120°? 8 a. d. 18x c.2 π e. 54x e.2 d. The diagram is not to scale. (4)π ft. If  OF =  FB . 35° b.. Use the diagram below to answer questions 24–25. 32 π cm2 3 64 π cm2 3 256 π cm2 3 Use the diagram below to answer questions 22–23. A B D C C A O E F D B  in square ABCD is x ft. 110° b. what 24. 3π cm2 c. what is the measure of arc AC? a. 50° d. 10π cm e. 70° e. (2)π ft. If the length of BD is the size of the shaded area? x2 a. The circumference of a circle is 16π cm.2 3x2 22.– CIRCLES – 23. what is the length of diagonal AD? a.2 1 c. the radius of the circle is 6x and arc EF is 60°. If the area of the circle is 25π cm2. what is the perimeter of triangle DOB? a. 12x b. 3π cm2 16 b. 40° c. angle D = angle B. 52π cm c. 102 cm d. 102π cm 104 . The diagram is not to scale. x2 – 4 ft. e. cannot be determined 21. 36x d. 52 cm b.2 25. x2 – 4x2π ft. If angles D and B both equal 70°. what is the size of the shaded area? a. b. 2x2π square units 27. The semicircles to the left and right of the center circle are each exactly half the size of the center circle.5π square units. If the radius of the circle is 4 units. 144 – 18π square units d. 128 – 24π square units d. 324 square units 30. x2π square units d. If the area of the circle is 8x2π. 16x22 – 4x2π b. If the area of one semicircle is 4. 72 – 16π square units b.– CIRCLES – Use the diagram below to answer questions 26–27. what is the size of the shaded area? a. A B O D C C D 26. If the area of the square is 144 square units. A O B Use the diagram below to answer questions 28–30. The diagram is not to scale. 16x22 – 4xπ d. 144 + 18π square units e. 128 – 16π square units e. 162 square units e. 144 + 36π square units e. 112π square units 105 . 32x2 + 4xπ 28. 108 square units c. 32x2 – 4x2π c. 4x2π square units 29. If the length of  AB is x units. The diagram is not to scale. what is the area of the rectangle? a. what is the total area of the figure? a. and the three figures are adjacent within rectangle ABCD. 16x22 + 4x2π e. ABCD is a square. 144 – 36π square units c. 144 – 72π square units b. 72 square units b. 128 – 32π square units c. 144 square units d. (x2π)  16 (x2π)  4π square units square units c. what is the area of the center circle? a. . a point with the coordinates (0. The x-axis and the y-axis meet at the origin. The origin is 0 units from the x-axis and 0 units from the y-axis. The y value of a point is the distance from the x-axis to that point.–1) The graph is an example of the coordinate plane. The x value of a point is the distance from the y-axis to that point. 107 . Point A is 3 units from the x-axis.and y-axes are placed and coordinate points called ordered pairs can be plotted. The points and figures that can be plotted on the plane and the operations that can be performed on them fall under the heading coordinate geometry.5) (6. The y-coordinate is listed second in the coordinate pair. with the coordinates (2.0). Look at the point labeled A. The x-coordinate is listed first in the coordinate pair.3) D C (–3.4) A (2.3).C H A P T E R 16 T Coordinate Geometry he coordinate plane is the grid of boxes on which the x. so its x value is 2. Point A is 2 units from the y-axis. B (–4. so its y value is 3. –3) (–9. Perpendicular lines have slopes that are negative reciprocals of each other. Lines such as x = 3. The slope ((–5) – 4) –9   of line AB is 2. or quadrants. If the y value increases and the x value decreases from one point to another. y = –2. x = –2. The points in quadrant I.– COORDINATE GEOMETRY – What are the coordinates of point B? Point B is –4 units from the y-axis and 4 units from the x-axis. if the y value decreases and the x value increases from one point to another. or y = c. Lines given by the equations y = 3x + 5 and y = 3x – 2 are parallel. The coordinates of point B are (–4. the top right corner of the plane. the bottom left corner of the plane. while its y value is positive. the slope of the line is positive.–10) F H (3. are lines with slopes of 0. Lines such as y = 3. A vertical line has no slope. the slope of the line is negative. This line has a slope of 0. 108 . This line has no slope. have negative values for x and positive values for y.4) (–2. and the points in quadrant IV. a line is formed. where c is any constant. or. the bottom right corner of the plane. or x = c are lines with no slopes.  Slope When two points on the coordinate plane are connected. or. If both the y value and the x value increase from one point to another. Line CD is a vertical line. have positive values for both x and y. while the line given 1 by the equation y = – 3x + 1 is perpendicular to those lines. have positive values for x and negative values for y. Line GH is a horizontal line.–5) G (–3.–10) (9. A horizontal line has a slope of 0. When the equation of a line is written in the form y = mx + b. there is no change in the x values from point C to point D.7) B A D (2. the value of m is the slope of the line. Point B is in quadrant II and its x value is negative. if both the y value and the x value decrease from one point to another. The slope of a line is the difference between the y values of two points divided by the difference between the x values of those two points. Parallel lines have the same slope. The coordinate plane is divided into 4 sections.–5) 8   The slope of the line AB is equal to  (2 – (–2)) = 4 = 2. The points in quadrant II. Point A is in quadrant I and its x and y values are both positive. (5 – (–3)) C (–9.4). The slope of line EF is equal to  (9 – 6) = 3 = –3. the top left corner of the plane.5) E (6. The points in quadrant III. there is no change in the y values from point G to point H. have negative values for both x and y. d) is the other endpoint. the midpoint of the line segment is equal a+b c+d to (2. –10 b.8) and (4. x2 represents the x-coordinate of the second point. 0 d.2).2) = (2. –5 c. –5 1 b.3)? 1 + (–3) 5 + 3 –2 8 Using the midpoint formula.2) = (–1.  Distance To find the distance between two points. What is the slope of this line? a. If (a.4).6) and (7. In other words. 5 e. 1. 5 d. use the formula below. This line has no slope. The endpoints of a line segment are (5. and y2 represents the y-coordinate of the second point: 2 2  D = ((x 2 – x 1) + (y) 2 – y1)  What is the distance between the points (–2. The variable x1 represents the x-coordinate of the first point.4). the midpoint of this line is equal to (2. 5 e.c) is one endpoint of a line segment and (b.–5) and (–5. The endpoints of a line segment are (–3.5) and (–3. What is the slope of this line? a. What is the midpoint of a line segment with endpoints at (1. y1 represents the y-coordinate of the first point.–2)? Substitute these values into the formula: 2 +– –2)) ((–2)  8)2) D = ((4 – ( D = ((4 + 2)2 + (–2 – 8)2) 2 +) D = ((6)  (–10)2 D = (36 00) + 1 D = 136  D = 234   Practice 2. the midpoint of a line segment is equal to the average of the x values of the endpoints and the average of the y values of the endpoints.–5). 10 109 . –5 1 c.– COORDINATE GEOMETRY –  Midpoint The midpoint of a line segment is the coordinates of the point that falls exactly in the middle of the line segment. 0)? a. C B 5. (10.2) e. –10 Use the diagram below to answer questions 9–10. (9. (–8.–2) c. What is the slope of line segment CD? 1 a. This line has no slope.5.2) and (1. What is the midpoint of a line segment with endpoints at (6. 3 d. –5 e. (8.12) c.–4) and (0. What is the midpoint of this line? a. (–1. 7.–8) and (–8. 15 A D 110 . What is the slope of a line segment with endpoints at (–1.4) e.–1) d. (12. (0. –2 d. (9. What is the slope of line segment AB? 1 a. 4 d. –4 1 b.4) e.4) b. 4 e.10)? a. This line has no midpoint. The endpoints of a line segment are (0.–8) b. What is the midpoint of a line segment with endpoints at (0. (0. (4.8)? a. 5 b.2) d.2) d.–4) c. –2 b. (–4. 5 e.4). 1 c. (12. –4 1 c. B D 4. (–4.8) 3.– COORDINATE GEOMETRY – 6.–4) and (15.9) Use the diagram below to answer questions 4–5. (0. –1 c.0) b. C A 8. What is the midpoint of line segment CD? a. (4. 10 units c. What is the distance from the point (0.–1) b.–2) Use the diagram below to answer questions 14–15. What is the distance from the point C to point D? a. 5 111 .5.2) to the point (2. 237  units d. (1. 10 units c. 229  units c. 25 units b. (2. 365  units e. 20 units 12. 52 units b. What is the midpoint of line segment AB? a. 341  units b. 247  units c.–1) e. (4.–4) to the point (4. (1. –3 3 c.0) b.2) c.0) d. 258  units e. 274  units a.4)? a. 213  units b. What is the distance from the point A to point B? a. C B 10. (1. 517  units 14. 43 units e.–6)? a. 163 units 16. 4 units c. What is the distance from the point (–6. 42 units d. 5 e. (0.0) d. 253  units e. 8 units b.6) A D 11. –5 6 d.–1) e. What is the slope of the line given by the equation 5y = –3x + 6? 13. 251  units d.–3) c. (2. –3 5 b. 45 units 15. 413  units e. (4. (2.17)? a.8) to the point (7. 102 units d.– COORDINATE GEOMETRY – 9. 17 units d. What is the distance from the point (3. –4 1 b. What is the midpoint of a line with endpoints at (2x + 3. (x + 1. y = 2x – 10 25. Which of the following lines is parallel to the line given by the equation y = –2x + 4? a. (6x + 1.y + 1) 3 3 5 5 b. (2x + 2. y = –2x + 4 c. If the slope of line A is multiplied by 4.–y)? a. –2 c. –4 d.y) to the point (x. –1 1 b. (x  ) units 2 e. –16 e. (x + y) units b. y = 4x – 5 d. y = 8x + 5 1 c.3y + 6)? a. y = –3x + 2 1 –2x – 4 1 x + 4 2 1 16 c. If the slope of one line is 3. y = –2x – 4 1 b. y = 3x + 3 e. Line A is perpendicular to line B.2y – 2) c. Which of the following is the product of the slopes of perpendicular lines? a. y = –4x + 5 a.– COORDINATE GEOMETRY – 22. Which of the following lines is parallel to the line given by the equation 4y = 6x – 6? 24. 2 e.y – 4) and (10x – 1. (x2 + y2) units 2 + y2 d. y = 6x – 8 e. y = –6x + 8 1 c. y = 2x – 4 e. y = –3x + 8 17.2y + 1) d. 4 1 c. Which of the following lines is perpendicular to 1 the line given by the equation y = –6x + 8? a. y = 2x + 3 e. (8x – 4. y = –6x – 8 1 d.5). y = –4x – 5 1 b. 1 3 a.4y + 2) 19. What is the distance from the point (–x.2y + 10) e. y = 6x + 8 23. y = 4x – 5 1 1 e. y = d. Which of the following lines is perpendicular to the line given by the equation –2y = –8x + 10? a. y = –2x + 6 2 b. (x + y) units  c. 2(x  + y2) units 112 . y = –3x + 3 1 14 d. 0 1 d. y = 3x + 6 3 d. (12x + 2. what is the equation of the other line? a. y = 3x + 2 18. y = –3x + 6 2 c. 4 21. y = 1 b. y = –6x – 8 b. what must the slope of line B be multiplied by in order for the lines to still be perpendicular? 20. Two perpendicular lines intersect at the point (1. If you missed a question because you did not know how to work the problem. Did you miss a question because of an error you made? If you can figure out why you missed the problem. you should look at the questions you missed. When you are finished. of which 15 are the multiple-choice questions and 10 are the grid-ins. compare the results with the pretest. You need a solid foundation in algebra and geometry if you plan to do well on the math section of the SAT! 113 . if any. congratulations! You have profited from your hard work. then you are ready to take the posttest to measure your progress. If you scored better on the posttest than you did on the pretest. take a second look at the questions you missed. Once you know your score on the posttest. Take as much time as you need to complete the posttest. Do you know why you missed the question. While the format of the posttest is similar to that of the pretest.Posttest I f you have completed all lessons in this book. the questions are different. go back to the lessons and spend more time working that type of problem. At this point. or do you need to go back to the lesson and review the concept? If your score on the posttest doesn’t show much improvement. then you understand the concept and just need to concentrate more on accuracy when taking a test. check your answers with the answer key at the end of the book. The posttest has 25 multiple-choice questions covering the topics you studied in this book. . / / • • • • 11. 13. a a a a a b b b b b c c c c c 16. / / • • • • 23. 9. 3. • d d d d d 6. / / • • • • / / • • • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 / / • • • 21. 12. / / • • • • 24.– LEARNINGEXPRESS ANSWER SHEET – POSTTEST 1. / • / • • • • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 115 . 5. / / • • • • a a a a a b b b b b c c c c c d d d d d 18. 7. • 22. 14. / / • • • 25. 10. 15. 8. 4. / / • • • • b b b b b c c c c c d d d d d e e e e e 20. 2. e e e e e a a a a a 19. e e e e e 17. . There are at least eight different values for which f(x) = 1. b. The graphed equation is not a function. (6. Which of the following statements is true of the graph below? 1  2 b. what is the length of arc AB? –4 –3 –2 –1 – 11 x 2 3 4 –2 A –3 –4 72° O a. x – 2 y 1 c. There are no values greater than 4 in the domain of the function. 5π c. c. e.12)? a. if the radius of the circle is 25 units.8) e. What is the midpoint of a line segment with endpoints at (–1. 4 3 2 1 x+6  x–6 2.  x–2 e. The expression  (x2 – 8x + 12) is equivalent to a. 5π b.4) and (13. (12. 3. There are eight y-intercepts for the equation. –4x x+2  d.– POSTTEST – (x2 + 4x – 12)  1. (7.8) c. B 2 a. The range of the function contains no values between –2 and 1.4) d. d.4. 10π d.16) 117 . 50π e. 250π 4. In the diagram below.5) b. (7. (5.5. –4) b. what are the values of a? a. then for the value of a to remain the same. a = –2. if lines L and M are parallel. a = –1. a = 10 b. If a + 3 =  a – 2 . what is the size of the shaded area in terms of r? A B D C a. (5. a = 1. which of the following equations is NOT necessarily true? 4a  7. the value of y must be doubled and the value of z must be halved. Which of the following is equivalent to a1a–2a3b–1b2b–3? a. b + c = e + f e.  (b2) 5 c. 2r – πr2 6.3) d.– POSTTEST – x 5. d. b (a2)  e. Based on the diagram. a + b + c = d + e + f b. e.6) 118 . 9. ab c. a + b + c + d + e + f = 360 8. if x is doubled. (0. or. (4. r2 – πr2 b. r  2 πr2 – 4 πr2 d.6) e. a = 0 L M a b c e d f a. a = –5 d. If the circle inscribed in square ABCD has a radius of r. a = 5 e. (r2 – πr2)  4 10. the value of y must be doubled and the value of z must be doubled. a + c = 180 – e c. a2b2 a d. the value of z must be halved. the value of y must be halved and the value of z must be halved. (3. either the value of y must be halved. Which of the following points is in the solution set of 4y + 6 > 3x + 15? a. a = 2. b.4) c. (4. a + e + c = 180 d. either the value of y must be doubled or the value of z must be doubled. c. a = –5. a. r2 – 4 e. Given yz = a. 1 b. a = –10 c. If every book has a width of 5 in. Which of the following lines is perpendicular to the line 4y + 3x = 12? 1 a. shifted 2 units right and 2 units up.. A circle has a circumference equal to g. Aiden sketches a right triangle and labels it ABC. y = –3x + 4 4 e. If a 2 = 512. what is the area of triangle DEF in square units? b. If the length of the court is 80 ft. What is the quotient of 3x ? 2 2 a. Compared to the graph of y = x2. Triangles ABC and DEF are similar. The box has a length of 16 in. shifted 4 units left and 2 units down. shifted 2 units left and 2 units down. then what is a 3 ? 19. 4x2 + 7x – 1 c. If the area of a sector of a circle is 8π and the angle formed by the two radii of the sector is equal to 80°. e. 3π g2 g2 17. If the area of the circle is tripled. 9x2 + 18x – 1 e.. how many books can fit in the box? 14. 9x2 + 18x d. shifted 2 units left and 2 units up. 4π c. what is the value of  8–4 ? 119 . what is the perimeter of the court in feet? 20. what is the length of the radius of the circle? 12. e. If the area of triangle ABC is 72 square units. y = 3x – 4 16. Bria and Lindsay play tennis on a rectangular court with an area of 2. 4x + 7x b. Each side of ABC is three times the length of its corresponding side of triangle DEF. 20 units x2 + 3x  21.. b.– POSTTEST – (12x4 + 21x3 – 3x2) 15. 3g2 π 4 9g2 π 4 9g2 π2 4 3  13. the graph of y = (x – 2)2 – 2 is a.. d. what is the length of side AC? a. y = 4x – 3 4 d. y = –4x + 3 3 c.000 ft. d.. 36x6 + 62x5 – 9x4 11. 10 units c. Marie is filling a box with books. c. Angle B is 90° and the tangent of angle A is 1.2. y = 3x + 12 3 b. shifted 2 units right and 2 units down. and a height of 8 in. a width of 10 in. what is the new area of the circle in terms of g? a. a length of 8 in.. 2  18. 103 units e. If the length of side AB is 10 units. and a height of 1 in. 102 units d. 5 units b. When x = –4. – POSTTEST – 22. In the diagram below, line CD is a tangent and line EO is a secant. If arc AB = 60° and the radius of the circle is 7 units, what is the length of secant EO? D B 24. Each term in the sequence below is twice the previous term. What is the seventh term of the sequence below? 6, 12, 24, 48, . . . 25. Find the positive value of x that makes the (x2 + 9x)  expression  (x2 + 6x – 27) undefined. A E 23. If –3a – 9b = –6 and 5a + 6b = –8, what is the value of b? O C 120 Answers  Pretest 1. d. The expression is undefined when the denominator of the expression is equal to 0. Factor (6x2 + 20x + 6) into (3x + 1)(2x + 6) and set each factor equal to 0; 3x + 1 = 0, 3x = –1, x = 1 –3; 2x + 6 = 0, 2x = –6, x = –3. 2. c. Each term in the sequence is five times the term before it. The first term in the pattern is 4, or 4  50. The second term in the pattern is 20, or 4  51. Every term in the pattern is 4 times a power of 5. The exponent of 5 is equal to one less than the position of the term in the pattern. Therefore, the exponent of 5 for the eighth term is one less than 8: 4  57. 3. c. The area of a circle is equal to πr2, where r is the radius of the circle. Therefore, the radius of this circle is equal to 64  = 8 ft. The circumference of a circle is equal to 2πr; 2π(8) = 16π ft. 4. b. The equation y = 2 is the equation of horizontal line that crosses the y-axis at (0,2). Horizontal lines have a slope of 0. This line is a function, since it passes the vertical line test: When graphed, a vertical line can be drawn through the graph of y = 2 at any point and that vertical line will cross the graphed line in only one place. The domain of the function is infinite, but all x values yield the same y value: 2. Therefore, the range of y = 2 is 2. 5. d. If the numerator of a fraction contains a term with a negative exponent, move that term to the denominator. In the same way, if the denominator of a fraction contains a term with a negative exponent, move that term to the numerator. Therefore, the given expression b ab is equal to (2 )(2a). The a and b terms cancel, leaving (2)(2) = 4. 6. c. –4(x – 1) = –4x + 4 and 2(x + 1) = 2x + 2. Since –4x + 4 ≤ 2x + 2, subtract 2x from both sides of the inequality and subtract 4 from both sides of the inequality, leaving: –6x ≤ –2. Divide both sides by –6 and change the direction of the 1 inequality sign: x ≥ 3. 7. b. The area of a square is equal to the length of one side squared. Therefore, the length of side BC is equal to the square root of the area of rectangle BCEF: 20  = 25. The area of a rectangle is equal to the product of its length 121 – ANSWERS – 8. a. 9. a. 10. e. 11. b. and width. The length of rectangle ABCD is 10 units and the width is 25 units, since rectangle ABCD shares side BC with square BCEF. Therefore, the area of rectangle ABCD = (10)(25) = 205 square units. The stack of compact discs, or a stack of circles, forms a cylinder. The volume of a cylinder is equal to πr2h, where r is the radius of the cylinder (or one of the circles that forms it) and h is the height of the cylinder. The radius of a disc is 12 half its diameter: 2 = 6 cm. Since each disc has a height of 2 mm, the height of the cylinder is (10)(2) = 20 mm. Convert this measure to centimeters, since the radius of the disc is given in (20 mm) centimeters. 1 = 2 cm. The volume of the 0 cylinder = π(6)2(2) = 72π cm3. A parallelogram is a quadrilateral with two pairs of parallel sides. All rectangles, rhombuses, and squares are parallelograms. A rectangle is a parallelogram with four right angles. All rectangles are parallelograms, but not all parallelograms are rectangles. A rhombus is a parallelogram with four equal sides. All rhombuses are parallelograms, but not all rhombuses are rectangles and not all parallelograms are rhombuses. A square is a parallelogram with four right angles and four equal sides. All squares are rectangles, rhombuses, and parallelograms, but not all rectangles are squares, not all rhombuses are squares, and not all parallelograms are squares. Factor the expression 3x2 – 3x – 18; 3x2 is equal to 3x multiplied by x. Find two numbers that multiply to –18 and whose difference is –3 after one of them is multiplied by 3; 3x2 – 3x – 18 factors into (3x + 6)(x – 3). Set each factor equal to 0 and solve for x; 3x + 6 = 0, 3x = –6, x = –2; x – 3 = 0, x = 3. Since ABCD is a parallelogram, lines AB and CD are parallel to each other and lines AC and BD are parallel to each other. Opposite angles in a parallelogram are equal; therefore, the angle labeled 16x and angle CAB are equal. Since angle CAB 12. a. 13. d. 14. d. 15. e. 122 and the angle labeled 9x + 5 are supplementary, the angle labeled 16x and the angle labeled 9x + 5 are supplementary: 9x + 5 + 16x = 180, 25x = 175, x = 7. Therefore, the measure of the angle labeled 9x + 5 is 9(7) + 5 = 63 + 5 = 68°. Since that angle and angle ABD are alternating angles, their measures are equal. The measure of angle ABD is also 68°. The ratio of a side of DE to a side of AB is 4 to 4 2 10, or 10 , which equals 5. The area of each tri1 angle is equal to 2(base)(height). Since the base 2 and the height of triangle DEF are 5 the base and height of triangle ABC, the area of triangle DEF 4 2 2 will be (5)(5) = 25 the area of triangle ABC. To find the turning point of a parabola, find the value that makes the x term of the equation equal to 0. Then, use that value of x to find the value of y. Three of the given equations have x values that have positive numbers added to them before squaring. For each of these (choices a, c, and d), the value of x must be negative in order for that term of the equation to be equal to 0. For example, x must equal –1 for the x term in choices a and c to equal 0. Therefore, the turning points of these parabolas will be in either the second or third quadrants of the coordinate plane. The equations in choices a and c contain negative constants. These equations will have their turning points below the x-axis, in the third quadrant. Only choice d, y = –(x + 2)2 + 1, will have a turning point with a negative x value and a positive y value, placing the turning point in the second quadrant of the coordinate plane. Rewrite the equation in terms of b. Multiply both sides of the equation by 4, then add 4 to 7b – 4  both sides of the equation: a =  4 , 4a = 7b – 4, 4a + 4 = 7b. Finally, divide both sides by 4a + 4 7; b = 7. The area of a triangle is equal to half the product of its base and height. Each triangle has a base of 4 units. If a line is drawn from one vertex of the triangle to its opposite base, that line would be 22. 144 + 256 = 400. 17. 18. w = 6. then the area of one face of the cube 96 is 6 = 16 cm2. 400  = 20.25 Solve the given equation for w. 50 Angles DCA and ACB form a line. the volume of the cube is 43 = 64 cm3. the sum of the interior angles is equal to (180)(12 – 2) = (180)(10) = 1. 19. Angles COB and AOD are vertical angles. 25. angle AOD is also 38°. Substitute 6 for w in the second 3 3 1 expression. Since the area of a square is equal to the length of one side of the square multiplied by itself. Since the longer base of the right triangle is the height of the equilateral triangle. In triangle ABC. 1. 64 A cube has 6 identical faces. 2w = 2(6) = 4. Since angle AOD is 38°. 2w. are congruent. If the total surface area of a cube is 96 square centimeters. 23. add the two differences and take the square root: (5 – (–7))2 = 122 = 144 and (12 – (–4))2 = 162 = 256. 12x + 48 = 25x – 30. angle OCB is also 71°. 21. The 36 sum of 36 and 13 is 49.800. Angle AOD is a central angle of the circle (its vertex is at the center of the circle). The longer base is 3 times the length of the shorter base. arc AD = 38°. 20.– ANSWERS – 16. 2w is equal to 4. where s is the number of sides of the polygon. Therefore. therefore. 38 Since sides OC and OB of triangle OBC are congruent. 20 To find the distance between two points. square the difference between the x values and square the difference between the y values. the measure of angle A = 180 – 65 – 65 = 50°.800 The sum of the interior angles of a polygon is equal to (180)(s – 2). Therefore. Therefore. ((5(–2)2) + (10 (–2)))-2 7 3 -2 7 3 -2 4 = (5(4) + – 20) = (20 – 20) = (20)-2 = 20 (4)2 = 52 = 25. 0. Since 3 = 4. 180 – 115 = 65°. perpendicular to that base and bisect it. Since there are 180° in a triangle. and the hypotenuse is twice the length of the shorter base. the area of the 1 equilateral triangle is 2(4)(23) = 43. 6 Cross multiply and solve for x: (2x + 8)(6) = (5)(5x – 6). Since the area of one equilateral triangle is 43. therefore. 3 = 4. angle ABC is also 65°. x = 6. 123 32 The base opposite the 30° angle of a 30-60-90 right triangle is the shortest side of the triangle. splitting the equilateral triangle into two congruent. Since sides AC and AB are congruent. leaving (49 )(36 ) = (7)(6) = 42. the area of all six. 78 = 13x. the measures of these angles add to 180°. The angles opposite the congruent sides are congruent. triangle ABC is isosceles. 25 Substitute –2 for a. 3 + 3 = 7. the angles opposite these sides. angles OBC and OCB. is (43)(6) = 243. 180 – DCA = ACB. angle COB is equal to 180 – (71 + 71) = 180 – 142 = 38°. 2w 2w 3 notice that 3 = 4. the length of one edge of the cube is equal to 16  = 4 cm. Since there are 180° in a triangle. Alternatively. Then. The reciprocal of 3. 36  + 13 42 Substitute 36 for b:  3636 . 7 3 24. and the area of the hexagon.  BC is the . Since Steve’s polygon has 12 sides. The measure of the intercepted arc of a central angle is equal to the measure of the central angle. will be equal to the reciprocal of the value 3 2w 2w 1 of 3. The length of the shorter base 1 is half the length of the hypotenuse: 2(4) = 2. and the 36 in the denominator cancels with the 36 in front of 36 . where e is the length of one edge of the cube. The length of the longer base is 3 times that: 23. Subtract 3 from both sides of the equation and then 3 2w 2w multiply both sides by 2. Each face of a cube is a square. 30-6090 right triangles. The volume of a cube is equal to e3. 9x2 = 4y2. b. c. 4x + 4y = 10y. Add the coefficients: 8x2y2 + 6x2y2 = 14x2y2. Subtract 20b from both sides of the equation and divide by 7: 7a + 20b = 28 – b. b. To solve the inequality. f 17. b = a2. 3. g = 6h – 10. fg + 2f – g = 2 – (f + g). x2 = 9y2. c. Divide the b term in the numerator by the 3b in the denomb 1 1 5a + 7b inator. Multiply 3 2 2 both sides of the equation by 3: (3)(2g ) = (9h – 2 15) 3. 16. 2 4x = 6y. 4(y + 1) = 10. 3a2 – 4b – ab = –a2 – ab. e. d. they cannot be combined any further. add 12 to both sides of the equation: a – 12 = 12. 3b = 3. p ≥ 3. 18. y = 2x  124 Chapter 2 1. Isolate y on one side of the equation. fg + 2f – g = 2 – f – g. 7a = 28 – 21b. Then. Isolate a on one side of the equation. c. c. d. 9a + 12a2 – 5a = 12a2 + 4a. divide both sides of the 6p 10 5 inequality by 6: 6p ≥ 10.  Chapter 1 1. 10y = 5y. 19. Isolate g on one side of the equation. Substitute 3 for each instance of x in the expression: 2(3)2 – 5(3) + 3 = 2(9) – 15 + 3 = 18 – 15 + 3 = 6. x = 3y.  (xy2) = 5y. Finally. angle. x . b + 2b = 3b. e. Multiply x the (y + 1) term by 4. a. Substitute 1 for each instance of c and substitute 4 for each instance of d in the expression: ((1)(4))2 42 16  =  = . b. Substitute –2 for each instance of x in the (y2) y y y2 (y2 + y)     . Isolate g on one side of the equation. a = 24. 15. AC. Multiply 2x2 and 4y2 by multiplying the coefficients of the terms: (2x2)(4y2) = 8 x2y2. and add the exponents of the bases: (3a)(4a) = 12a2. expression:  (–2)2) + (–2(–2)) = 4 + 4 = 4 8. The terms 5a and 7b have unlike bases. . Substitute 1 for each instance of a and substitute –1 for each instance of b in the expression: 1 (1)(–1) + –1 + (1)2 – (–1)2 = –1 + (–1) + 1 – 1 = –2. so they can be added. e.  163  3  = 16. 9. since it is opposite the 60° AB is equal to angle. Add the terms in the denominator. Multiply the coefficients of the terms in the numerator. 10(x2y)  20. 13. 10. a(3a) – b(4 + a) = –(a2 + ab). 8x2 – 4y2 4 2 + x2 = 0. and the hypotenuse of the tri is equal to (16)(2) = 32. Isolate x on one side of the equation. Substitute –2 for each instance of a in the –14 14 7(–2)   expression:  = (4 – 2) = – 2 = –7. fg = 2 – 3f. 4a2 – 4b = 0. 4a2 = 4b. 2. 2(3) 3(3) (2)(9) 9 18 9 9 9 9 12.– ANSWERS – longer base. divide the numerator by the denominator. b. 9x2 – 4y2 = 0. c. Substitute 3 for each instance of a in the expression: (4(3)2)(3b3 + (3)) – b3 = (4)(9)(3b3 + 3) – b3 = 36(3b3 + 3) – b3 = 108b3 + 108 – b3 = 107b3 + 108. (1) + (4) 5 5 11. a = 4 – 3b. d. 4. multiply both sides of the equation by y to make it easier to work x 4x with. Isolate y on one side of the equation. y = 3x. a – 12 + 12 = 12 + 12. g2 = 4h2. Divide the coefficients of the terms and subtract the expo(12a2) 1   nents of the bases:  (36a2) = 3 . 4a and 12a2 are not like terms. Substitute 6 for each instance of a in the expres4(6)((6) + r) 24(6 + r) 4(6 + r) 24 + 4r sion: 6 = 6r = r = r. 8x2y2 and 6x2y2 have like bases. 14. y + 4 = 10. a. 5. c. 2x = y2. Isolate g on one side of the equation. 6. Do the same with the terms in the denominator: [6(6a2)] = 36a2. 6 ≥ 6. 2. e. d. Since  BC is 163. g = 2h. 4g2 – 1= 16h2 – 1. Isolate b on one side of the equation. a. a. The r expression cannot be simplified any further. To solve the equation. Subtract the like terms by subtracting the coefficients of the terms: 9a – 5a = 4a. ((–2)2 + (–2)) 7. (5a + 7b)(3) = 3. so they cannot be combined any further. 10x = 5y2. 4g2 = 16h2. Substitute 2 for each instance of x and substitute 3 for each instance of y in the expression: 6(2)2 4(2) (6)(4) 8 24 8 12 8 20 2 +  =  +  =  +  =  +  = . g = 2 – 3f . 2x + 6 = 3 –x – 3. a = 6. 3x = –21. 5. Since x = –7. 16. a. when multiplying or dividing both sides of an inequality by a negative number. 3x –21 Divide by 3 to solve for x: 3 = 3. x = –6. x ≥ –11. cross multiply: (2)(3x – 8) = 9x + 5. divide both sides of the 12 –3n  inequality by –3: –3n < 12. x = –7. x = 14. –2y – 18 = –10. Since x = –6. x + 10 – 10 = 5 – 10. –12y – 18 = –10y – 10. combine like terms on each side of the equation. y = –4. Remember. d. c. look at the relationship between the . 3x = 3. 125 Finally. To solve the equation. w = –2. e. 6x – 16 = 9x + 5.– ANSWERS – 3. subtract 10 from both sides of the equation: x + 10 = 5. 7. To solve the equation. (3c2)  First. a. d. n > –4. Now. 4g = 12. 2(–2x + 10) = –4x + 20. e. divide both sides of the equation 14 4x 14 7 by 4: 4 = 4. subtract 5 from both sides of the equation. First. then divide by 9: 9a + 5 = –22. Cross multiply and solve for y: (6)(–2y – 3) = (10)(–y – 1). numerator and denominator by 3c. First. multiply (x + 2) by 4: 4(x + 2) = 4x + 8. a = –3. –2y = 8. 8x = 8(1) = 8. a. 5g – 5 = g + 7. – 3 > –3 . reduce the fraction  the (6c) by dividing c (3c2)  = . subtract 3x from both sides of the inequality and subtract 8 from both sides of the inequality: 3x – 6 ≤ 4x + 8 3x – 6 – 3x ≤ 4x + 8 – 3x –6 ≤ x + 8 –6 – 8 ≤ x + 8 – 8 x ≥ –14 8. 6x – 16 = 9x + 5. 11. 8 8 Subtract 3x and 3 from both sides of the equa8 8 19 19 tion: 9x + 3 = 3x + 9. Although you could solve the first equation for x and substitute that value into the given expression. b. 5g First. 5a = 30. 3 3 3 x = 1. (8)8 = (8)(8). 6. 9a + 5 – 5 = –22 – 5. 18. e. add 8x to both sides of the inequality and subtract 20 from both sides of the inequality: –8x – 24 ≤ –4x + 20 –8x – 24 + 8x ≤ –4x + 20 + 8x –24 ≤ 4x + 20 –24 – 20 ≤ 4x + 20 – 20 –44 ≤ 4x 10. 13.  2 (6c) subtract 9 from both sides of the equation and then multiply both sides of the equation by 2: c  + 9 = 15 2 c  + 9 – 9 = 15 – 9 2 c  = 6 2 c (2)(2) = (6)(2) c = 12 Cross multiply and solve for w: (w)(18) = (–6)(w + 8). you must reverse the inequality symbol. First. First. subtract 2x from both sides of the equation and add 5 to both sides of the equation: 2x + 9 = 6x – 5 2x – 2x + 9 = 6x – 2x – 5 9 = 4x – 5 9 + 5 = 4x – 5 + 5 14 = 4x Finally. 14. 24w = –48. 6x – 4x = 2x and 4 – 9 = –5. Since x = 1. Then. multiply (x + 3) by –8 and multiply (–2x + 10) by 2: –8(x + 3) = –8x – 24. multiply both sides of k k the equation by 8: 8 = 8. x = 4 = 2. d. x = –5. Add x to both sides of the equation and subtract 1 6 from both sides of the equation. c. 16a + 30 = 21a. k = 64. Multiply both sides of the equa2 2 3 2 tion by 3 to isolate x: (3)(2x) = –9( 3). Now. Cross multiply and solve for x: (10x)(3) = (7)(5x – 10). 19. cross multiply and solve for g: (5)(g – 1) = (1)(g + 7). g = 3. divide both sides of the inequality by 4: 44 4x –4 ≤ 4. 16a + 16 = –14 + 21a. b. 9. Then. –5x = –70. 9a = –27. a. –2x = –2(–6) = 12. 4g – 5 = 7. Cross multiply and solve for a: (4a + 4)(4) = (7)(–2 + 3a). reduce g by canceling g from the numer5g 5 ator and denominator: g = 1. 15. 4. 18w = –6w – 48. 17. To solve the inequality. 2x = –9. d. x + 7 = –7 + 7 = 0. 12. Multiply both 3 3 3 19 19 sides of the equation by 19 : (19 )(3x) = (3)(19 ). Subtract 6x from both sides of the equation and subtract 5 from both sides of the equation. 30x = 35x – 70. Now. To solve the equation. d. a. x = 20. 23. A fraction is undefined when its denominator is equal to 0. Although you could solve the first equation for c and substitute that value into the given expression. A fraction is undefined when its denominator is equal to 0. 25.  126 Chapter 3 1. A fraction is undefined when its denominator is equal to 0. d. c. 2x = 0. Set 6d equal to 0 and solve for d. 3x = 15. If x represents the first odd whole number. 26. and x + 6 represents the fourth. x = 5. the value of 33c – 21 will be three times the value of 11c – 7. odd whole numbers. Create an equation that describes the situation. The sum of the numbers is x + x + 2 + x + 4 + x + 6 = 4x + 12. Create an equation that describes the situation. 4x + 2 is exactly half of 8x + 4. Set that expression equal to 10. d. the inside terms. 24. 3 = x – 3. The sum of the integers is x + x + 1 + x + 2 = 3x + 3. Since x is the first integer. 3x = 60. The sum of the integers is x + x + 2 + x + 4 = 3x + 6. 33c – 21 is exactly three times 11c – 7. –8y + 12 = 0. x + 4 represents the third. Since x is the first number. y = 2. b. multiply the first term of each binomial. d. the outside terms. x = 8. 6a + 18 – 4a = 0. 27. 2a = –18. 29. x = 0 and x – 5 = 0. and solve for x: 3x + 3 = 63. Three more than that is 4x + 3. Set these expressions equal to each other. d = 0. and x + 2 represents the third integer. Create an equation that describes the situation. Therefore. To find the product of two binomials. The fraction is undefined when x = 8. 3 The fraction is undefined when y = 2. x = 5. Since x is the first number. 2 = 7. 30. 3x = –24. and solve for x: 3x – 5 = 10. Set x – 8 equal to 0 and solve for x. a. Therefore. 48. 4x = 36. c. 33c – 21 = 24. add the products. the value of 4x + 2 will 14 be half the value of 8x + 4. Create an equation that describes the situation. even integers. The fraction is undefined when x = 0 or 5. Set 6a + 18 – 4a equal to 0 and solve for a. x + 2 represents the second integer and x + 4 represents the third. 2. the outside terms. equation and the expression. add the products. x = 9. 21. Set this expression equal to the sum of the numbers. If x represents the first even integer. x = –8. it is the smallest of the four consecutive.– ANSWERS – 20. b. and the last terms. 8 – 8y + 4 = 0. x + 2 represents the second. A fraction is undefined when its denominator is equal to 0. then one-fourth the 1 1 number is 4x. and solve for x: 1 3 3 x + 3 = x – 3. (x – 6)(x – 6) = x2 – 6x – 6x + 36 = x2 – 12x + 36. A fraction is undefined when its denominator is equal to 0. the inside terms. The fraction is undefined when a = –9. c. e. 21 is the second integer. If x represents the number. . 63. Set this expression equal to the sum of the integers. 3(8) = 24. Set this expression equal to the sum of the integers. 2a + 18 = 0. 22. look at the relationship between the equation and the expression. –18. and the last terms. To find the product of two binomials. Five less than that is 3x – 5. Three less than the number is x – 3. –8y = –12. 4 4 4 28. Create an equation that describes the situation. 4x + 2 = 7. Then. x + 1 represents the second integer. If x represents the first integer. e. a = –9. (x – 3)(x + 7) = x2 + 7x – 3x – 21 = x2 + 4x – 21. x – 8 = 0. 6d = 0. it is the smallest of the three consecutive. Set each term of the denominator equal to 0 and solve for x. multiply the first term of each binomial. b. Then. x = 8. The fraction is undefined when d = 0. and solve for x: 4x + 12 = 48. and solve for x: 3x + 6 = –18. then 3 times the number is 3x. If x represents the number. 6 = x. c. and 22 is the third (and largest) integer. 2x(x – 5) = 0. Set 8 – 8y + 4 equal to 0 and solve for 3 y. e. Factor the numerator and denominator. To find the factors of a quadratic. the outside terms. Cancel the (x – 8) terms from the numerator and denominator. x = 2. –4 and 12 multiply to –48 and add to 8. 12. (x – 1)(x + 1) = x2 + x – x – 1 = x2 – 1. and the last terms. begin by finding two numbers whose product is equal to the constant of the quadratic. Then. –3 and 2 multiply to –6 and add to –1. the factors of x2 – x – 6 are (x – 3) and (x + 2). Of those numbers. Factor the numerator and denominator. the factors of x2 – 18x + 32 are (x – 2) and (x – 16). To find the factors of a quadratic. multiply the first term of each binomial. find the pair that adds to the coefficient of the x term of the quadratic. The roots of a quadratic are the solutions of the quadratic. the factors of x2 – 81 are (x – 9) and (x + 9). the inside terms. find the pair that adds to the coefficient of the x term of the quadratic. begin by finding two numbers whose product is equal to the constant of the quadratic. begin by finding two numbers whose product is equal to the constant of the quadratic. –2 and –16 multiply to 32 and add to –18. 5. find the pair that adds to the coefficient of the x term of the quadratic. Of those numbers. Factor the quadratic and set each factor equal to 0 to find the roots. Therefore. Of those numbers. and the last terms. b. Set each factor equal to 0 and solve for x: x – 2 = 0. Of those numbers. begin by finding two numbers whose product is equal to the constant of the quadratic. a. the factors of x2 – 4 are (x – 2) and (x + 2). (2x + 6)(3x – 9) = 6x2 – 18x + 18x – 54 = 6x2 – 54. find the pair that adds to the coefficient of the x term of the quadratic.– ANSWERS – 3. leaving 1 in the numerator and (x – 9) in the denominator. leaving (x + 2) in the numerator and (x + 7) in the denominator. the outside terms. 6. 11. the outside terms. Then. find the pair that adds to the coefficient of the x term of the quadratic. Therefore. and the last terms. To find the product of two binomials. d. The numerator factors into (x – 8)(x + 2) and the denominator factors into (x – 8)(x + 7). 8. 127 numbers whose product is equal to the constant of the quadratic. the factors of x2 + 8x – 48 are (x – 4) and (x + 12). add the products. Cancel the (x + 9) terms from the numerator and denominator. Then. Of those numbers. a. Therefore. b. a. The roots of a quadratic are the solutions of the quadratic. Factor the denominator. c. x = 16. To find the factors of a quadratic. 13. add the products. To find the product of two binomials. This quadratic has no x term—the sum of the products of the outside and inside terms of the factors is 0. Therefore. multiply the first term of each binomial. find the pair that adds to the coefficient of the x term of the quadratic. e. Factor the quadratic and set each factor equal to 0 to find the roots. and x + 12 = 0. This quadratic has no x term—the sum of the products of the outside and inside terms of the factors is 0. multiply the first term of each binomial. Set each factor equal to 0 and solve for x: x – 4 = 0. 7. The roots of x2 + 8x – 48 are 4 and –12. To find the factors of a quadratic. the inside terms. c. add the products. Cancel the (x + 5) terms from the numerator and . x = 4. x = –12. and x – 16 = 0. the inside terms. –2 and 2 multiply to –4 and add to 0. begin by finding two 10. The numerator factors into (x + 5)(x – 9) and the denominator factors into (x + 5)(x + 6). –9 and 9 multiply to –81 and add to 0. To find the product of two binomials. To find the factors of a quadratic. the factors of x2 – 11x + 28 are (x – 4) and (x – 7). (x + c)2 = (x + c)(x + c). Therefore. (x + c)(x + c) = x2 + cx + cx + c2 = x2 + 2cx + c2. Therefore. 9. 4. The roots of x2 – 18x + 32 are 2 and 16. begin by finding two numbers whose product is equal to the constant of the quadratic. To find the factors of a quadratic. Of those numbers. b. –4 and –7 multiply to 28 and add to –11. x2 – x – 12 = 0. find the pair that adds to the coefficient of the x term of the quadratic. Therefore. –4 and 3 multiply to –12 and add to –1. 16. set either factor equal to 0 and solve for x: x – 6 = 0. Therefore. begin by finding two numbers whose product is equal to the constant of the quadratic. Combine like terms on one side of the equation and set the expression equal to 0. begin by finding two numbers whose product is equal to the constant of the quadratic. x2 – x = 12 when x equals 4 or –3. To find the factors of a quadratic. 128 . and x – 11 = 0. Set each factor equal to 0 and solve for x: x – 8 = 0. 19. begin by finding two numbers whose product is equal to the constant of the quadratic. x2 – 12x + 36 = 0. Of those numbers. x2 – x = 12. Set each factor equal to 0 and solve for x: x – 4 = 0. find the pair that adds to the coefficient of the x term of the quadratic. leaving (x + 3) in the numerator and (x – 3) in the denominator. and x + 3 = 0. the factors of x2 – 3x – 40 are (x – 8) and (x + 5). Set each factor equal to 0 and solve for x: x – 4 = 0. Use FOIL to multiply the binomials: (x – 7)(x – 5) = x2 – 5x – 7x + 35 = x2 – 12x + 35. x = 6. 15. –8 and 5 multiply to –40 and add to –3. begin by finding two numbers whose product is equal to the constant of the quadratic. leaving (x – 9) in the numerator and (x + 6) in the denominator. Factor the quadratic and set each factor equal to 0 to find the solutions for x. x2 – 3x – 30 = 10 when x equals 8 or –5.– ANSWERS – 14. Therefore. Therefore. Cross multiply: (x + 5)(x + 4) = (4)(6x – 6). 2 and 2 multiply to 4 and add to 4. To find the factors of a quadratic. Combine like terms on one side of the equation and set the expression equal to 0. x = 8.To find the factors of a quadratic. x2 + 4x – 12 = –16. Factor the numerator and denominator. but only on a multiple-choice SAT question. d. d. and x + 5 = 0. x2 + 9x + 20 = 24x – 24. 17. the factors of x2 – x – 12 are (x – 4) and (x + 3). the factors of x2 – 15x + 44 are (x – 4) and (x – 11). denominator. the factors of x2 – 12x + 36 are (x – 6) and (x – 6). x2 + 4x + 4 = 0. x2 – 12x + 35 = –1. x = –2. x2 – 3x – 40 = 0. c. Trial and error (plugging each answer choice into the equation x2 – x = 12) could be used. x2 – 15x + 44 = 0. –4 and –11 multiply to 44 and add to –15. To find the factors of a quadratic. Combine like terms on one side of the equation and set the expression equal to 0. x = 4. –6 and –6 multiply to 36 and add to –12. Cancel the (x + 11) terms from the numerator and denominator. 18. Of those numbers. set either factor equal to 0 and solve for x: x + 2 = 0. Combine like terms on one side of the equation and set the expression equal to 0. find the pair that adds to the coefficient of the x term of the quadratic. x = 11. Of those numbers. Factor the quadratic and set each factor equal to 0 to find the solutions for x. Since both factors are the same. x = –3. Factor the quadratic and set each factor equal to 0 to find the solutions for x. x = –5. Of those numbers. To find the factors of a quadratic. It is important to be able to solve questions like this that could occur in the grid-in section of the SAT. begin by finding two numbers whose product is equal to the constant of the quadratic. find the pair that adds to the coefficient of the x term of the quadratic. The numerator factors into (x + 11)(x + 3) and the denominator factors into (x + 11)(x – 3). e. Therefore. Of those numbers. Factor the quadratic and set each factor equal to 0 to find the solutions for x. b. find the pair that adds to the coefficient of the x term of the quadratic. x2 + 9x + 20 = 24x – 24. Factor the quadratic and set each factor equal to 0 to find the solutions for x. Since both factors are the same. x = 4. the factors of x2 + 4x + 4 are (x + 2) and (x + 2). x2 – 3x – 30 = 10. Use FOIL to multiply the binomials: (x – 2)(x + 6) = x2 + 6x – 2x – 12 = x2 + 4x – 12. Combine like terms on one side of the equation and set the expression equal to 0. e. 4).d). x2 = 9. a. but shifted left 3 units and down 4 units. Of those numbers. A fraction is undefined when its denominator is equal to 0. Therefore. find the pair that adds to the coefficient of the x term of the quadratic. Factor the quadratic in the denominator and set each factor equal to 0 to find the values that make the fraction undefined. A fraction is undefined when its denominator is equal to 0. x = 2. d. Set each factor equal to 0 and solve for x: 2x – 9 = 0. the fraction is undefined. Set the denominator equal to 0 1 and solve for x.–4) is y = (x + 3)2 – 4. 28. This parabola is similar to the parabola y = x2. 25. x = 7. To find the factors of a quadratic. Write an algebraic equation that describes the situation. the vertex of the parabola whose equation is y = (x + 2)2 + 2 is (–2. 26. –1 and –2 multiply to 2 and add to –3. c. This parabola is similar to the parabola y = x2. Therefore.d). To find the factors of a quadratic. the equation of a . but shifted up 4 units. the equation of a parabola whose vertex is (–3. and 2(8) + 9 = 25. b. x2 = 3x – 2. b. the factors of x2 + x – 42 are (x – 6) and (x + 7). Set each factor equal to 0 and solve for x: x – 1 = 0. c. To find the factors of a quadratic.d). x = –4 or 4. begin by finding two numbers whose product is equal to the constant of the quadratic. A fraction is undefined when its denominator is equal to 0. Therefore. find the pair that adds to the coefficient of the x term of the quadratic. and x – 8 = 0. Therefore. x = 1. Therefore. Since the square of the number is equal to two less than three times the number. Factor the quadratic in the denominator and set each factor equal to 0 to find the values that make the fraction undefined.2). 27. 2 A parabola of the form y = x2 + c has its vertex at (0. 22. –1 and –7 multiply to 7 and add to –8. Set each factor equal to 0 and solve for x: x – 6 = 0.c). e. the factors of the quadratic will be (2x + c)(x + d). x2 – 16 = 0. x = 2. x = 8. 29. –8 and –9 multiply to 72. Two less than three times the number is 3x – 2. the factors of x2 – 8x + 7 are (x – 1) 23. and x – 7 = 0. Combine like terms on one side of the equation and set the expression equal to 0. 24. –6 and 7 multiply to –42 and add to 1. x = –7. 21. the factors of x2 – 3x + 2 are (x – 1) and (x – 2). and x – 2 = 0. then x2 is the square of the number. begin by finding two numbers whose product is equal to the constant of the quadratic. 9x2 – 1 = 0. and x + 7 = 0. 1 1 x = –3 or 3. the factors of 2x2 – 25x + 72 are (2x – 9) and (x – 8). A parabola of the form y = (x + c)2 + d has its vertex at (–c. the fraction is undefined. Of those numbers. When x equals 9  or 8. Factor the quadratic in the denominator and set each factor equal to 0 to find the values that make the fraction undefined. but shifted left 2 units and up 2 units. Factor the quadratic and set each factor equal to 0 to find the solutions for x. When x equals 1 or 7. A fraction is undefined when its denominator is equal to 0. find the pair that adds to the coefficient of the x term of the quadratic. where c and d are constants that multiply to 72 and add to –25 after either c or d is multiplied by 2. A fraction is undefined when its denominator is equal to 0. d. begin by finding two numbers whose product is equal to the constant of the quadratic. x2 – 3x + 2 = 0. d. x = 6. This parabola is similar to the parabola y = x2. Therefore. Therefore. the vertex of this parabola is at (0. the fraction is undefined. x = 1. begin by finding two numbers whose product is equal to the constant of the quadratic. x2 = 16. a. When x equals 6 or –7. 129 and (x – 7).– ANSWERS – 20. Of those numbers. x2 = 3x – 2. 9x2 = 1. A parabola of the form y = (x + c)2 + d has its vertex at (–c. Set the denominator equal to 0 and solve for x. Therefore. If x is the number. Since the first term of the quadratic is 2x2. To find the factors of a quadratic. 9 2x = 9. A parabola of the form y = (x + c)2 + d has its vertex at (–c. Set each factor equal to 0 and solve for x: x – 1 = 0. 7. x2 + 6x + 5 = (x + 1)(x + 5). A fraction is undefined when its denominator is equal to 0. x3 – x + x = 27 – x + x. e. x3 + 3x2 – 4x = x(x + 4)(x – 1). and x is the largest common variable. A fraction is undefined when its denominator is equal to 0. c. 64x3 – 16x = 16x(4x2 – 1). x2 + 8x = x(x + 8). and x is the largest common variable. Factor the polynomial in the denominator and set each factor equal to 0 to find the values that make the fraction undefined. x(x – 1) = x2 – x. Multiply (x2 – 9x + 18) by (x – 1): (x2 – 9x + 18)(x – 1) = x3 – 9x2 + 18x – x2 + 9x – 18 = x3 – 10x2 + 27x – 18. Cancel the 2x term in the numerator with the 4x term in denominator. 4. A fraction is undefined when its denominator is equal to 0. (x2)(2) = 2x2. The vertex of this parabola is at (1. or solution. Factor x2 + 4x – 96 into (x – 8)(x + 12). 12. Begin by multiplying the first two terms: –3x(x + 6) = –3x2 – 18x. Add the products and combine like terms: x3 + 5x2 + –7x + 2x2 + 10x + –14 = x3 + 7x2 + 3x – 14. 8. Therefore. x = –4. Multiply (–3x2 – 18x) by (x – 9): (–3x2 – 18x)(x – 9) = –3x3 + 27x2 – 18x2 + 162x = –3x3 + 9x2 + 162x. x – 1 = 0. (–7)(x) = –7x. leaving 2 in the denominator. a. so the factors of 64x3 – 16x are 16x(2x – 1)(2x + 1).–2) is y = (x – 1)2 – 2. 10. b. Factor out 2x from every term: 2x3 + 8x2 – 192x: 2x(x2 + 4x – 96). 64x3 2 Factor out 16x from both terms: 16 x = 4x and –16x   = –1. The cube root of 27 is 3. 4x3 – 16x2 – 48x = 4x(x2 – 4x – 12) = 4x(x – 6)(x + 2). of x(x – 1)(x + 1) = 27 – x is x = 3.d). The largest constant common to each term is 2. 2x2 + 4x = 2x(x + 2). 3. 4x2 – 1 = (2x – 1)(2x + 1). x3 – 25x = x(x2 – 25) = x(x – 5)(x + 5). Factor the polynomial in the denominator and set each factor equal to 0 to find the values that make the fraction undefined. A parabola of the form y = (x + c)2 + d has its vertex at (–c. Next. (–7)(2) = –14. 30. (x2 – x)(x + 1) = x3 + x2 – x2 – x = x3 – x. (5x)(x) = 5x2. Multiply each term of the trinomial by each term of the binomial: (x2)(x) = x3. b. Therefore.–2). (5x)(2) = 10x. x3 + 125 = 0. Begin by multiplying the first two terms: (x – 6) (x – 3) = x2 – 3x – 6x + 18 = x2 – 9x + 18. Cancel the x terms and the (x + 8) terms in the numerator and denominator. so the root. e. the equation of a parabola whose vertex is (1. x3 = 27. x3 – 64x = x(x2 – 64) = x(x – 8)(x + 8). b. 11. but shifted right 1 unit and down 2 units. Add x to both sides of the equation. e. 16x factor 4x2 – 1. x = –5. This parabola is similar to the parabola y = x2. 2. a. 5.– ANSWERS – parabola whose vertex is (5. leaving (x + 1) in the numerator and x(x – 5) = (x2 – 5x) in the denominator.  Chapter 4 1. This parabola is similar to the parabola y = x2. 9.0) is y = (x – 5)2 + 0. Factor the numerator and denominator. 4x3 + 44x2 + 120x = 4x(x2 + 11x + 30) = 130 . The factors of 2x3 + 8x2 – 192x are 2x(x – 8)(x + 12). c. First. x = 0. but shifted right 5 units. Factor the numerator and denominator. leaving 2(x – 6) = 2x – 12 in the denominator. x + 4 = 0. x3 = –125. Set the denominator equal to 0 and solve for x. Cancel the (x + 5) terms in the numerator and denominator. or 1. or y = (x – 5)2. 6. and (1)(–1) = –1. Cancel the (x + 2) terms in the numerator and denominator. 0. The fraction is undefined when x is equal to –4. x3 – x = 27 – x. c. 16 is the largest constant common to 64x3 and 16x. x = 1. Factor the numerator and denominator. d. a. (2x)(2x) = 4x2. leaving 1 in the numerator and (x – 8) in the denominator. multiply the terms on the left side of the equation. 13. x3 – 2x2 – 80x = x(x2 – 2x – 80) = x(x + 8)(x – 10). The given situation is true for the numbers 0. Find the square root of the coefficient and the 2) = 32 2) = x32 variable. 15. 3x = –14. Therefore. Set each factor equal to 0 to find the values of x that make the equation true. The expression is 3y now equal to  . 2) = ((9y 2)  )(3 ). Therefore. since (3y)(3y) = 9y2. d. 4x3 + 4x2 – 48x = 0. one of which is a perfect square. x3 – 2x2 = 80x. 4. x + 8 = 0. x + 6 = 0. plus 56. Factor the denominator into two radicals. d. 3 3 . Factor (a  3) into a 2) = aa square. 32  = (16 )(2) = 42. x3 + 3x2 + 2x = x3 + 56. x = 0. That value is equal to 48 times the number (48x) minus four times the square of the number (4x2). leaving  . It is important to be able to solve questions like this that could occur in the grid-in section. x = 0. This polynomial is equal to the cube 131 of the first integer. 14. The cube root of 27y3 = 3y. Multiply the coefficient of the given expression by aa: (a3)(aa) = a4a. Again. a. then the second integer is (x + 1) and the third integer is (x + 2). 3. If the first integer is x. 2) = 4x2 Therefore. The difference in those values is equal to 80 times the number (80x). Therefore. and factor the polynomial. Set each term equal to 0 to find the values of x that make the 14 equation true. then 5 is the second integer and 6 is the third. Trial and error (plugging each answer choice into the equation x3 – 2x2 = 80x) could be used. If the number is x. If the number is x. –4. integer. b. x – 4 = 0. (32x  . That leaves 2 = 2. Twice the square of the number is 2x2. and 3. x = –8. –8. 3) into two radicals. so factor (a  (a  . The product of these integers is x(x + 1)(x + 2). so factor 4g  into 4g = 2g. 3x + 14 = 0. (x2 + x)(x + 2) = x3 + 2x2 + x2 + 2x = x3 + 3x2 + 2x. since (3y)(3y)(3y) = 27y3. x3. 4x = 0. but only 3 is greater than 0. x = –6. The fraction is undefined when x is equal to –6. x – 3 = 0. x3 – 2x2 – 80x = 0.  Chapter 5 1. but only on a multiple-choice SAT question. x = 4. x3 + 3x2 + 2x = x3 + 56. 4 is a perfect square. Place all terms on one side of the equation. and largest. x + 4 = 0. The given situation is true for the numbers 0. x(x + 1) = x2 + x. and 10. Sim3 plify the fraction by multiplying the numerator 1 3 3 )(  ) =  and denominator by 3: (  3 . Simplify the fraction by dividing the numerator by the denominator. and factor the polynomial. factor 32  into two radicals. Move all terms onto one side of the equation. 3x2 + 2x – 56 = 0. Move all terms onto one side of the equation. a. x = –3. x + 5 = 0. Factor the polynomial: 3x2 + 2x – 56 = (3x + 14)(x – 4). x = 0. a. (32x  (x  . x – 10 = 0. then the cube of the number is x3. –5. x3 – 2x2 = 80x. Next. a2 is a perfect 2. combining like terms. 4x3 = 48x – 4x2. Set each factor equal to 0 to find the values of x that make the equation true. x = 10. x = –4. or 0. Trial and error could also be used for a multiple-choice problem such as this. Cancel the g terms from the 4 numerator and denominator. If 4 is the first of the consecutive integers. x = 3. a. but only 10 is greater than 0. but you are looking for a positive integer. The square root of (27y 9y2 = 3y. 4x3 + 4x2 – 48x = 4x(x2 + x – 12) = 4x(x + 4)(x – 3). 4x = 0. trial and error could be used to find the solution. x = –5. 4x3 = 48x – 4x2. then four times the cube of the number is 4x3. Factor 4g  into two radicals. Cancel the 3y terms from the 3y3 1 numerator and denominator.– ANSWERS – 4x(x + 5)(x + 6). To multiply the pr terms. 1 The numerator of the fraction is now (n2)( n2) and the denominator of the fraction is 1. First. 1 take the square root of 64g10. cross multiply: g(g108 = 3. Multiply the exponents: ( p2 )4 = 1 1 p2. (39) = (3)(3) = 1. find the square root of 9pr. Divide both sides of the equation by 3 1  : g2108  = 3.(2d )(2d ) 16. n y (x)–2 term. 5 and m = n. so 3(pr)2(pr)2 = 3(pr)2 = 3p2r2. d. First. p2 = 9. 1 1 1 13. A fraction with a negative exponent can be 132 negative exponent. b. ) = 3. (  ) =   . Square the numer3 2 m 2 m ator and denominator. cube the 4g2 term. g2 =  . since (8g5)(8g5) = 64g10. (x)–2 = (y)2 = y 2 . Factor each term in the numerator: (a  rewritten as the reciprocal of the fraction with a 2) 2b) = (a 2) = (b ) = ab. (ab)3 = a3b3. raise (a3b3) the fraction b to the fourth power. leaving ab. (  m3 n5 )2 = ( 1  m3 n5 3 n5 ) = m3 . a3 = 62 = 36. c. add the 3 1 1 3 4 exponents. multiply 64g6 by g4. d. 9. Finally. 6.substitute 2 for c. ( m3 –2  ) =( n5 1  mn5 ) . The denominator (pr) has a negative exponent. The n term in the denominator has a 17. c. e. . (64g10)2 = 8g5. nents. since (n– 12 ) 1 = n12. Substitute –3 for q. A term with a negative exponent can be rewritten as the reciprocal of the term with a positive exponent. The expression is now 3 1 3(pr)2(pr)2. by  : ( 6 )(  6 6 6    )2 = 48. (–3)–2 = (–3)2 = 9. 2 + 2 = 2 = 2. Next. e. e. Therefore. Multiply the ba. A fraction with a negative exponent can be rewritten as the reciprocal of the fraction with a positive expo1 nent. First. Next. g2108  15. 1 raised to any power is 1. Divide both sides of the equation 4d 48 by 4: 4 = 4. (a  )((b ) (a positive exponent. g = 6 =  = 6 1 . b. subtract the expo4 1 b2 (a12b12) = a12b8. Next. d = 12. n5 n5 3 1 Therefore. b. (p )4 = (p2 )4. 9pr  = 3  – 2 9pr  = 3pr . since a value raised to the 1 exponent 2 is another way of representing the square root of the value. 7. If y = –x. Substitute 3 for a and 9 for b. To divide b12 by b4. and p = –3 or 3. b4 = b8. p2 = (–3)–2. c. the value of a3 is equal to the 2 4 value of a 3 squared. nm = 5. g2 =  . 4d = 48. x4 x3 5 = 5 . 1 is raised to the power –3. First. Finally. The expression is now abab x2 x divide this fraction by xy. Therefore. Next. g2 = 108 36  108  1 . Simplify the fraction by multiplying it 6 1 6 6 6 )= . Take the square root of both sides of the equa6 1 1 tion to find the value of g: g2 = 16. pr  can 12 be written as (pr) . multiply the two radicals. the coefficients of each radical and multiply the . Substitute 2 for x and –2 1 1  for y: (((2)(–2))–2)2 = ((–4)–2)2 = (16 )2 =  256 . d. then y = –2. look y at the x4 x4 radicands of each radical: (ab)(ba) = abab  . cube the ab term. Add the exponents of the g terms. It can be placed in the numer1 ator with a positive exponent. To multiply terms with like bases. 1 1 12. Therefore. Since (a3)2 = a3. ab  x2    nents of the fractions: ( y 2 )( y 2 ) = ( y 4 ). (a3b3) each exponent of the a and b terms by 4. Cube the constant 4 and multiply the exponent of g (2) by 3: (4g2)3 = 64g6. square x x : ()2 y y = x2 2 . b 4 First. (1)2 = 1. (64g6)(g4) = 64g10.(2d = 4d. xy y 2 4 (y4 )  xy x4 1  = ( y4 )( xy ) = 4 11. keep the base and add 1 1 the exponents: (n2)(n2) = n. c. so it can be rewritten in the numerator with a positive exponent. but the value of the exponent does not matter. First. (b)4 (a12b12) = b . 1  = n2. 10. Multiply fractions in the numerator by adding the expo- y x2 Cancel  terms from the numerator and denomthe ab inator. 14.– ANSWERS – 2b) = 5. Multiply 8. 1. a. The 100th term of the sequence is equal to 3(100 – 1) = 399 and the 101st term in the sequence is equal to 3(99 + 1) = 3100. and (x–y)(2xy)(3yx) = 2(3yx) = 6yx. The fourth term in the sequence is –24. The value of the exponent is one less than the position of the term in the sequence. the value of x is 5 – 7 = –2. is 3 more than 2: 23. The second 2 term. you can check 3 1 3 your answer by repeatedly adding 2. The 20th term of the sequence is equal to 5(20 – 4) = 516. 4. 25 . Alternatively. The fourth term in the sequence is 74. 1 1 –3 –2   The first term. you can check your answer by repeatedly subtracting 9. Since the number of terms is reasonable. –4 10. 102 + 2 = 3 1 1 3 3 1 12. Since the number of terms is reasonable. so the seventh term will be ()3 = 3 3 8 8 16 16 128  times . Therefore. 2 2 20. z = 7 – 6 = 1. e. (x-y)(2xy) = 2. and 3 3 1 2 1 2 d: 23 + 23 + 33 + 33 = 12. d. 132 + 2 = 15. 18. Cancel the 5 terms 5 and multiply the fraction by 10:  (105) = 25 5(10) 50  =  = 25. add the exponents of the terms: (399)(3100) = 3199. you can check your answer by repeatedly multiplying by –2. Substitute 20 for n: (105) = 5  25 20  + 5 20   5  2 20  ( ) = 25   20  (105). 74 – 9 = 65. multiply the first two terms. c. 38 – 9 = 29. 15 + 2 = 162. which is three terms after the fourth term. First. The first term. the ninth term will be 5(9) = 45 less than 74. To multiply two terms with common bases. The second term. so the seventh term will be (–2)3 = –8 times –24. If a2 = b = 4. 16 7. The next 1 1 term in the sequence. Therefore. 133 . and the value of y is –2 – 7 = –9. 1 2. Substitute 2 for a and 4 8 (4)(2) 1 1 44 2 for b: (  ) = (16)2 = (16 )2 = (2)2 = 4. Every term in the sequence is 3 raised to a power. Since each term is nine less than the previous term. d is 1 2  more than c. You are looking for the seventh term. 12 + 2 = 132. (–24)(–2) = 48. is 30. c. Substitute 2 for x and –2 for y: 6(–2)2 = 6(4) = 24. ()() = . 23. You are looking for the ninth term. (48)(–2) = –96. 3. 47 – 9 = 38. You are looking for the eighth term. d. Therefore. every 27 27 3 3 81 2 term in the sequence is 18 times 3 raised to a 2 power.The second term. (2)4  Chapter 6 1. d.is 5 . a. The fourth term in the sequence is 102. is 18  (3)0. In the same way. The first term in the sequence is 2. e. is 31. Since –y + y = 0. Multiply the coefficients of the terms (1 and 2) and add the exponents. 65 – 9 = 56. Since each term is 2 more than the previous term. 8. (–24)(–8) = 192. You must multiply by –2 three times. 56 – 9 = 47.  125 . b 1 2 1 1 is 3 more than a. which is three terms after the fourth term. and x = 16  = –4. y = 1 – 6 = –5 and x = –5 – 6 = –11. 9. 3. You are looking for the seventh term. which is 5 terms after the fourth term. 102 + 6 = 162. 19. z must be 6 less than 7. 3. Every term in the sequence is 5 raised to a power. Since the number of terms is reasonable. y is equal to – 4 = 16. Since each term is 6 more than the previous term. a. Since each term in the sequence is –4 times the –64 previous term. the eighth term will be 3 1 1 1 4(2) = 6 more than 102. The value of the exponent is one less than the position of the term in the sequence. Add the values of a. y is 6 less than z and x is 6 less than y.is 5 . is 18  (3)1. The term that precedes x is 5. b. Therefore. b. 5. c is 3 more than 3: 33. The value of the exponent is four less than the position of the term in the sequence. The fourth term in the sequence is 3. x – y = –2 – (–9) = –2 + 9 = 7. then a = 2. which is four terms 3 after the fourth term. a. 74 – 45 = 29. e.You must multiply by 2 2  three times. 12. The sum of x + z is equal to –11 + 1 = –10. (–96)(–2) = 192. c. The seventh term of the sequence is 64 2 64 128    equal to 18  (3)6 = 18  ( 729 ) = 2  ( 81 ) = 81 . xy = (16)(–4) = –64. The term that follows z is 7. b. 6. The first term.– ANSWERS – 18. and c = a + b. the value of x –4 –4 (24 – 16) 8  is – 4 = –4 = –2. –53. Since the rule of the sequence is each term is two less than three times the previous term. the terms that precede it.25 14. The sixth term is (2) – 2 = 4. However. d = a + 2b. The y terms will drop out.25.125. If e = c + d. d = e – c. Then. Therefore. so the eighth term of the sequence is the first term of the sequence that is a negative number. c. Substitute this expression for x in the second equation and solve for y: 2x – y = 0 32 2(y) – y = 0 64  y 64  y –y=0 =y y2 = 64 y = –8. and you can solve for a: 3x – 2y = –7 + 2y – 4x = 8 –x = 1 x = –1 . 12. to find the 1 value of x. 12 + 9 = 21.5 of the sequence. Since the rule of the sequence is each term is 20 less than five times the previous term. the seventh term is (2) – 2 = 2. d = b + c. b. to find the value of y. Half of this number minus two will yield a negative value. and the first equation becomes 3x – 2y = –7. then subtract 2: (–53)(3) = –159 – 2 = –161. The b terms will drop out. Solve the first equation for x in terms of y.0625 – 2 = 0. In the same way. d = e – c. multiply the last term. In the same way.25 – 2 = 12. the value of x + y = 4 + –20 = –16. Add the two equations together. the value of x + y = –2 + 24 = 22. b. 134 Chapter 7 1. multiply the last term.  14. y = 6 – 2x. e. Since the rule of the sequence is each term is 1 nine more than 3 the previous term.125 6. xy = 32 32. not 2b. the value of y is  5 5 –100 = 5= –20. e = c + d. Therefore. and c = a + b. and that equation can be rewritten as d = f – e. 28. then 1 add 9: (36)(3) = 12. Solve the first equation for y in terms of x. d. Since the rule of the sequence is each term is 16 more than –4 times the previous term. d. 1 the value of y is 21(3) + 9 = 7 + 9 = 16.0625. since a is not equal to b. In the first equation. Continue the sequence. and you can solve for a: 5a + 3b = –2 + 5a – 3b = –38 10a = –40 a = –4 Substitute –4 for a in the first equation and solve for b: 5(–4) + 3b = –2 –20 + 3b = –2 3b = 18 b=6 3. The fifth term is (2) – 2 = 12. subtract 16 from –80 and divide by –4: (–80 – 16) –96  =  = 24. c. 15. Therefore. since c and d are the two terms previous to e. d = b + c.125 – 2 = 4. d. subtract 12 from both sides of the equation.5 is the fourth term 28. Add the two equations together. f = d + e. multiply the (x + 4) term by 3: 3(x + 4) = 3x + 12. by 3.– ANSWERS – 11. b. y = 8 4. a. 36. add 20 to 0 and divide by 5: 5 = 20 (–120 + 20)  = 4. In the same way. Since each term in the sequence below is equal to the sum of the two previous terms. to find the (0 + 20) value of x. Substitute this expression for y in the second equation and solve for x: 6 – 2x  = 12 2 + 4x 3 – x + 4x = 12 3x + 3 = 12 3x = 9 x=3 2. In the same way. then. x = y. by 3. the value of y – x = 16 – 21 = –5. 2x + y = 6. by subtracting c from both sides of the equation. Therefore. d = b + (a + b). 16. 13. d is not equal to e – 2b. 12. x = 4y. multiply the (m + n) term by 2 and add m: 2(m + n) + m = 2m + 2n + m = 3m + 2n. Solve the second equation for a in terms of b. Substitute this expression for q in the first equation and solve for p: 4pq – 6 = 10 4p(2p + 7) – 6 = 10 10. a = 13 – b. The m terms will drop out. Substitute this expression for a in the first equation and solve for b: b –7a + 4 = 25 b –7(13 – b) + 4 = 25 b 7b – 91 + 4 = 25 29b  = 116 4 29b = 464 b = 16 4y 6. and you can solve for b: 2(4. Substitute this expression for c in the first equation and solve for d: c–d  – 2 = 0 5 6d – d  – 2 = 0 5 5d  – 2 = 0 5 d–2=0 d=2 . b = –2a – 4. In the first equation.5a – 3b = 3) = 9a – 6b = 6 9a – 2b = 38 – (9a – 6b = 6) 4b = 32 b=8 9. 11. Substitute this expression for b in the first equation and solve for a: 7(2a + 3(–2a – 4)) = 56 7(2a + –6a – 12) = 56 7(–4a – 12) = 56 –28a – 84 = 56 –28a = 140 a = –5 Multiply the first equation by 8 and add it to the second equation. 135 8p2 + 28p – 6 = 10 8p2 + 28p – 16 = 0 2p2 + 7p – 4 = 0 (2p – 1)(p + 4) = 0 1 2p – 1 = 0. c. Solve the second equation for q in terms of p. m = 1 Solve the second equation for c in terms of d. multiply the (b + 4) term by –2: –2(b + 4) = –2b – 8. m – n = 0. p = 2 p + 4 = 0. b + a = 13. m = –2 m – 1 = 0. b. a. c – 6d = 0. b. Add 8 to both sides of the equation. e. p = –4 Solve the second equation for b in terms of a. and the first equation becomes 9a – 2b = 38. 4p – 2q = –14. c = 6d. b + 2a = –4. The a terms will drop out. x = 1. The x terms will drop out. Substitute this expression for n in the first equation and solve for m: m(n + 1) = 2 m(m + 1) = 2 m2 + m = 2 m2 + m – 2 = 0 (m + 2)(m – 1) = 0 m + 2 = 0. a. d. 2p = 1. Multiply the second equation by 2 and subtract it from the first equation. q = 2p + 7. c. and you can solve for y: 1 8(2x + 6y = 7) = 4x + 48y = 56 4x + 48y = 56 + –4x – 15y = 10 33y = 66 y=2 Solve the second equation for n in terms of m.– ANSWERS – 5. –2q = –4p – 14. n = m. b. and you can solve for n: 3m + 2n = 9 – 3m – 3n = –24 5n = –15 n = –3 8. 13. Solve the second equation for x in terms of y. In the first equation. Substitute this expression for x in the first equation and solve for y: 3x + 7y = 19 3(4y) + 7y = 19 12y + 7y = 19 19y = 19 y=1 7. Subtract the second equation from the first equation. First. multiply (n + 2) by –6: –6(n + 2) = –6n – 12. 14. add 12 to both sides of the equation. The first equation becomes 3x = –5y + 4. the value of m = 10 . the value of a + b = 3 + 2 = 5. The first equation becomes m – 6n = 4. The x terms will drop out. Multiply the first equation by –6 and add it to the second equation. Then. The n terms will drop out. Solve the second equation for y in terms of x. d. –x – y = –6. the value of d = 2 = 6. y = –x + 6. Then. the value of x – y = 12 – (–5) = 12 + 5 = 17. The b terms will drop out. 17.– ANSWERS – Substitute the value of d into the second equation and solve for c: c – 6(2) = 0 c – 12 = 0 c = 12 c 12 Since c = 12 and d = 2. and you can solve for m: m – 6n = 4 + 6n + m = 16 2m = 20 m = 10 Substitute the value of m into the second equation and solve for n: 6n + 10 = 16 6n = 6 n=1 n 1 Since m = 1 and n = 10. d. and you can solve for a: (6a – 12b = –6)  = 3a – 6b = –3 2 4a + 6b = 24 + 3a – 6b = –3 7a = 21 a=3 Substitute the value of a into the first equation and solve for b: 4(3) + 6b = 24 12 + 6b = 24 6b = 12 b=2 Since a = 3 and b = 2. e. simplify the first equation by subtracting 4 from both sides. Add the two equations. 18. First. a. 15. Substitute this expression for y in the first equation and solve for x: –5x + 2(–x + 6) = –51 –5x – 2x + 12 = –51 –7x + 12 = –51 –7x = –63 x=9 Substitute the value of x into the second equation and solve for y: –9 – y = –6 –y = 3 y = –3 Since x = 9 and y = –3. and you can solve for y: –3(3x = –5y + 4) = –9x = 15y – 12 –9x = 15y – 12 + 9x + 11y = –8 11y = 15y – 20 –4y = –20 y=5 Substitute the value of y into the second equation and solve for x: 9x + 11(5) = –8 136 . the value of xy = (9)(–3) = –27. –y = x – 6. The x terms will drop out. c. 16. and you can solve for y: x –6(3 – 2y = 14) = –2x + 12y = –84 –2x + 12y = –84 + 2x + 6y = –6 18y = –90 y = –5 Substitute the value of y into the second equation and solve for x: 2x + 6(–5) = –6 2x – 30 = –6 2x = 24 x = 12 Since x = 12 and y = –5. multiply the equation by –3 and add it to the second equation. Divide the second equation by 2 and add it to the first equation. The x terms will drop out. Substitute this expression for y in the first equation and solve for x: x + 2x – 9  = 8 3 3x – 9  = 8 3 x–3=8 x = 11 Substitute the value of x into the second equation and solve for y: 2(11) – y = 9 22 – y = 9 –y = –13 y = 13 Since x = 11 and y = 13. Then. e. multiply the equation by 2 and add it to the first equation. the value of ab = (–4)(–3) = 12. b – a = 1. and you can solve for q: 3(–5p + 2q = 24) = –15p + 6q = 72 –15p + 6q = 72 + 8q + 15p = 26 14q = 98 q=7 137 . the value of y – x = 5 – (–7) = 5 + 7 = 12. and you can solve for y: 2(–2x – 6 = y) = –4x – 12 = 2y –4x – 12 = 2y + 4x + 6 = –3y –6 = –y y=6 Substitute the value of y into the first equation and solve for x: 4x + 6 = –3(6) 4x + 6 = –18 4x = –24 x = –6 x 6 Since x = –6 and y = 6. y = 2x – 9. b. b. 20. The a terms will drop out. 23. 19. The second equation becomes –2x – 6 = y. and the 1 15 equation becomes 2a – b = –2. the value of y = –6 = –1. –y = –2x + 9. Then. Substitute this expression for b in the first equation and solve for a: 10(a + 1) – 9a = 6 10a + 10 – 9a = 6 a + 10 = 6 a = –4 Substitute the value of a into the second equation and solve for b: b – (–4) = 1 b+4=1 b = –3 Since a = –4 and b = –3. and you can solve for b: 1 15 –6(2a – b = –2) = –3a + 6b = 45 –3a + 6b = 45 + 3a – 2b = –5 4b = 40 b = 10 Substitute the value of b into the second equation and solve for a: 3a – 2(10) = –5 3a – 20 = –5 3a = 15 a=5 Since a = 5 and b = 10. Multiply the second equation by 3 and add it to the first equation. The first equation becomes 2a + 2 – b 3 = –6. c. 21. First. multiply the equation by –6 and add it to the second equation. Subtract 2 from both sides. simplify the first equation by multiplying (a 1 1 3 + 3) by 2. 2x –y = 9. First. b = a + 1. b. 22. The p terms will drop out.– ANSWERS – 9x + 55 = –8 9x = –63 x = –7 Since y = 5 and x = –7. the value of a + b = 5 + 10 = 15. Solve the second equation for b in terms of a. Solve the second equation for y in terms of x. simplify the second equation by subtracting 9 from both sides of the equation. the value of x – y = 11 – 13 = –2. This line touches the graph of f(x) in 3 places. multiply the second equation by –2 and add it to the first equation. there are 3 values for which f(x) = 2. Therefore. the value of (p + q)2 = (–2 + 7)2 = 52 = 25. The y terms will drop out. Then. 25.– ANSWERS – Substitute the value of q into the second equation and solve for p: –5p + 2(7) = 24 –5p + 14 = 24 –5p = 10 p = –2 Since p = –2 and q = 7. y x . Draw a horizontal line across the coordinate plane where f(x) = 3. there is 1 value for which f(x) = 3. add 9 and 4y to both sides of the equation. In other words. a = 2b + 2. First. and you can solve for x: –2(2y + 7x = 3) = –4y – 14x = –6 –4y – 14x = –6 + 9x + 4y = 11 –5x = 5 x = –1 Substitute the value of x into the second equation and solve for y: 2y + 7(–1) = 3 2y – 7 = 3 2y = 10 y=5 Since y = 5 and x = –1. e. Substitute this expression for a in the second equation: 3(2b + 2 – b) = –21 3(b + 2) = –21 3b + 6 = –21 3b = –27 b = –9 Substitute the value of b into the first equation and solve for a: a  = –9 + 1 2 a  = –8 2 a = –16 Since a = –16 and b = –9. =  –9 9 3 a  138 Chapter 8 1. the value of b –16 16 4  =   = . This line touches the graph of f(x) in 1 place. Therefore. Solve the first equation for a in terms of b by multiplying both sides of the equation by 2. graph the line y = 2. the value of (y – x)2 = (5 – (–1))2 = 62 = 36. b. c. Draw a horizontal line across the coordinate plane where f(x) = 2. y x 2. Then. The first equation becomes 9x + 4y = 11. 24. simplify the first equation by multiplying (x – 1) by 9: 9(x – 1) = 9x – 9. d. g(–4) = 2(–4) – 3 = –8 – 3 = –11. b. Begin with the innermost function: find g(–3) by substituting –3 for x in the function g(x): 139 g(–3) = (–3)2 = 9. This line touches the graph of f(x) in 8 places. substitute the result of that function for x in g(x). You are given the value of g(x): g(x) = 2x. . w@z. g(f(x)) = 36x2 + 48x + 15 Begin with the innermost function. 10. there are 5 values for which f(x) = 0. Therefore. Substitute this expression for x in the equation f(x): f(x) = 4 – 2x2 f(2x) = 4 – 2(2x)2 f(2x) = 4 – 2(4x) f(2x) = 4 – 8x Therefore. and the problem becomes (3w – z)@z. there are 8 values for which f(x) = –5. e. Therefore. there are 0 values for which f(x) = –2. Begin with the function in parentheses. Begin with the innermost function: Find f(–2) by substituting –2 for x in the function f(x): f(–2) = 3(–2) + 2 = –6 + 2 = –4. f(9) = 2(9) – 1 = 18 – 1 = 17. Draw a horizontal line across the coordinate plane where f(x) = –5.– ANSWERS – 3. x 6. c. 7. Then. Multiply 3w – z by 3 and subtract z: (3w – z)@z = 3(3w – z) – z = 9w – 3z – z = 9w – 4z. y x 9. since the x-axis is the graph of the line f(x) = 0. 8. Then. y 11. substitute the result of that function for x in g(x). This line does not touch the graph of f(x) at all. b. You are given the value of f(x): f(x) = 6x + 4. which is given as 3w – z. Substitute this expression for x in the equation g(x): g(x) = x2 – 1 g(6x + 4) = (6x + 4)2 – 1 g(6x + 4) = 36x2 + 24x + 24x + 16 – 1 g(6x + 4) = 36x2 + 48x + 15 Therefore. 5. d. then subtract from that product the term after the @ symbol. Again. 4. g(7) = 7 – 2 = 5. Draw a horizontal line across the coordinate plane where f(x) = –2. f(x) = 0 every time the graph touches the x-axis. a. and subtract from it the term after the @ symbol. substitute 5 for x in f(x): f(5) = 2(5) + 1 = 10 + 1 = 11. Begin with the innermost function. Next. d. Finally. f(g(x)) = 4 – 8x The definition of the function states that you must multiply by 3 the term before the @ symbol. The graph of f(x) touches the x-axis in 5 places. Therefore. e. substitute the result of that function for x in f(x). c. Replace the expression w@z with 3w – z. Begin with the innermost function: Find f(3) by substituting 3 for x in the function f(x): f(3) = 2(3) + 1 = 6 + 1 = 7. multiply the term before the @ symbol by 3. when an exponent is raised to an exponent. x + 3 = 0. The definition of the function states that the term after the % symbol should be raised to the power of the term before the % symbol. Substitute the values of a b–a –5 – 6 –11     and b: a + b = 6 + (–5) = 1 = –11. Now evaluate y^(y2 – y). the . f(x) = 02 – 4 = –4. (x – 3)(x + 3) = 0. Since the x term is squared. x – 3 = 0. x > 5. e. The definition of the function states that the term before the ? symbol should be subtracted from the term after the ? symbol. b–a  which is given as  a + b . c. it cannot be equal to 0 either. the value of q&p = p + qp. the smallest value that this term can equal is 0 (when x = 0). When x = 0. k: (kj)k = kjk. There are no x values that can make the function undefined. These values are not in the domain of the function. 16. Since the x term is squared. e. Remember. so the value of x – 5 cannot be negative. and that difference should be divided by the sum of the two terms. Begin with the innermost function: a?b. kj. b. the smallest value that f(x) can have is when x = 0. The square root of a negative value is imaginary. and that quotient should be added to the product of the two q terms. the smallest value of f(x) is the square root of 0. 140 x values that make the fraction undefined. 20. 19. y2 – y. The domain of the function is all real numbers. The value of x – 5 must be greater than 0. y: (y2 – y)2 – y = y4 – y3 – y3 + y2 – y = y4 – 2y3 + y2 – y. so the value of 4x – 1 must be greater than or 1 equal to 0. The domain of f(x) is all real numbers greater than or equal 1 1 to 4. The domain of f(x) is all real numbers excluding 3 and –3.– ANSWERS – 12. x = 3. The range of the function is all real numbers greater than or equal to 0. x –5 > 0. Substitute the value of –11 – 6 –17 17    a again: 6 + (–11) = –5 = 5 . 18. and the problem becomes k%(kj). Set the denominator equal to 0 to find the 17. the largest value that this term can equal is 0 (when x = 0). which is 0. Now evaluate a?–11. Substitute 4 for p and –2 for q in this definition: 2 1 –4 + (–2)(4) = –2 – 8 = –8. Every other x value will result in a negative value for f(x). The range of f(x) is all real numbers greater than or equal to –4. Begin with the innermost function: y^y: Square the second term. and that difference should be divided by the sum of the two terms. so the domain of the function is all real numbers. The range of f(x) is all real numbers less than or equal to 0. 15. The definition of the function states that the term before the ^ symbol should be squared and the term after the ^ symbol should be subtracted from that square. multiply the exponents. b. c. x2 – 9 = 0. Remember. any real number can be substituted for x. The square root of a negative value is imaginary. then made negative. Replace the expression j%k with its given value. a. y: y2 – y. The range of a function is the set of all possible outputs of the function.5. and subtract the first term. Since the denominator of the fraction must always be positive. The definition of the function states that the term before the & symbol should be divided by the term after the & symbol. Therefore. Since the square root is the denominator of a fraction. should be raised to the power of the term before the % symbol. 4x ≥ 1. square the second term. and subtract the first term. 13. The domain of a function is the set of all possible inputs to the function. Therefore. The domain of f(x) is all real numbers greater than 5. and the numerator of the fraction is –1. a. All real numbers can 1  be substituted for x in the function f(x) =  (x2 – 9) excluding those that make the fraction undefined. The term after the % symbol. x ≥ 4. d. Since x must be greater than or equal to 4. All real numbers can be substituted for x in the function f(x) = x2 – 4. x = –3. The range of a function is the set of all possible outputs of the function. kj. 14. the term before the ? symbol should be subtracted from the term after the ? symbol. y. 4x – 1 ≥ 0. Again. x – 5 > 0. the range of the equation y = |x| – 3 is all real numbers greater than or equal to –3. d. The graph of the equation in diagram A is not a function. e. and D extend below the x-axis. this function will be undefined. a. Since the square root of a negative number is imaginary. For each of these x values (inputs). The domain and range of y = (x – 3)2 + 1 are not the same. the range of the equation y = (x – 3)2 + 1 is all real numbers greater than or equal to 1. The equation of the graph in diagram B is y = |x| – 3. 24. The equation of the graph in diagram E is y = (x – 3)2 + 1. and the domain and range of the equation are the same. value of f(x) will always be negative. One divided by a real number (excluding 0) will yield real numbers. The equation of the graph in diagram E is y = (x – 3)2 + 1. Any x value greater than or less than 3 will generate a y value that is greater than 1. The graphs of the equations in diagrams A. 25.– ANSWERS – 21. it is impossible to generate a y value that is less than –3. These equations have negative y values. Therefore. 22. Any x value greater than or less than 3 will generate a y value that is greater than –3. Any real number can be substituted into this equation. there are two y values (outputs). this function will be undefined. any real number can be substituted for x—there are no x values that will generate an undefined or imaginary y value. the graph of the equation in diagram A is not a function. the domain of this equation is all 141 real numbers greater than or equal to 0. no values less than . The equations graphed in diagrams B and D are functions whose ranges contain negative values. If x = 0. Any real number can be substituted into this equation. The range of a function is the set of possible outputs of the function. The domain and range of y = |x| – 3 are not the same. Therefore. any real number can be substituted for x—there are no x values that will generate an undefined or imaginary y value. B. Therefore. The equation in diagram A fails the vertical line test for all x values where –2 < x < 2. it is impossible to generate a y value that is less than 1. Any x value greater than or less than 3 will generate a y value that is greater than –3. Only the functions in diagrams B and E have a domain of all real numbers with no exclusions. The graph of the equation in diagram A is not a function. However. The equation of the graph in diagram B is y = |x| – 3. If x = 0. 23. It fails the vertical line test for all x values where –2 < x < 2. and the domain and range of the equation are the same. The equation of the graph in 1 diagram D is y = x. Find the coordinate planes that show a graph that extends below the x-axis. the set of possible y values that can be generated for the equation is the range of the equation. The equation of the graph in diagram B is y = |x| – 3. e. It fails the vertical line test for all x values where –2 < x < 2. The equation of the graph in diagram C is y = x. The equation of the graph in diagram 1 D is y = x. There are no x values that will generate an undefined or imaginary y value. the range of the 1 equation y = x is all real numbers excluding 0. No value of x will make f(x) = 0. b. A function is an equation in which each unique input yields no more than one output. There are no x values that will generate an undefined or imaginary y value. However. excluding 0. Therefore. The square roots of real numbers greater than or equal to 0 are also real numbers that are greater than or equal to 0. the domain of this function is all real numbers excluding 0. which means that the range of the equation contains negative values. With this equation as well.With this equation as well. Therefore. the domain of this function is all real numbers excluding 0. so the range of f(x) is all real numbers less than 0. the range of the equation y = x is all real numbers greater than 0. In each of the five equations. Therefore. However. x = 12. Notice that replacing x with 12 in the measure of angle 6 also yields 106: (12)2 – 38 = 144 – 38 = 106°. Therefore. Angles 5 and 7 are in fact adjacent. every numbered angle is congruent and supplementary to every other numbered angle. then the measure of angle 2 is 8(12) + 10 = 106°. and 7 must also measure 90°. Therefore. Since x = 9. e. x – 12 = 0. the measure of angle 2 is 12(9) + 10 = 108 + 10 = 118°. and the measure of angle 5 is 5(30) = 150°. 6x = 180. x + 4 = 0.  Chapter 9 1. and smaller than the set of all real numbers excluding 0 (D). 19x + 9 = 180. x = 15. smaller than the set of all real numbers greater than or equal to 0 (C). their measures are equal. If the measure of angle 6 is x. If angle 3 measures 90°.– ANSWERS – –3 can be generated by this equation. the sum of the measures of angles 1 and 2 must be 90°. Therefore. b. Angles 2 and 6 are alternating angles. Therefore. 5. not 90°. c. 4. 4. excluding 0. their measures are equal. Therefore. Perpendicular lines cross at right angles. 6. d. x = –4. the measure of angle 4 is equal to 180 – 90 = 90°. angles AOC and EOC are both 90°. Therefore. a. the range of this equation is all real numbers greater than or equal to 0.6. 2. Angles 3 and 4 are supplementary. 5. However. Since angles 6 and 8 are supplementary. Angles 5 and 6 are supplementary. and 8 are alternating angles. Therefore. If x = 12. Therefore. then the measure of angle 5 is 5x. the equation y = (x – 3)2 + 1 (E). 7. angles 1 and 2 are not complementary—their measures add to 180°. since these angles form a line. 4x = 60. Of the four equations that are functions. x = 9. Angles 2. no values less than 1 can be generated by this equation. The equation of the graph in diagram E is y = (x – 3)2 + 1. x2 – 8x – 48 = (x + 4)(x – 12).and 7 are alternating (vertical) angles. 6. the range of the equation y = (x – 3)2 + 1 is all real numbers greater than or equal to 1. since they share a common vertex and a common ray. Factor x2 – 8x – 48 and set each factor equal to 0. x2 – 8x – 48 = 0. the sum of their measures is 180°. x = 30. Therefore. 19x = 171. Since angles 1 and 2 combine to form angle AOC. 6x + 20 = 10x – 40. since the set of real numbers that are greater than or equal to 1 is smaller than the set of all real numbers greater than or equal to –3 (B). The equation of the graph in diagram C is y = x. Any x value greater than or less than 3 will generate a y value that is greater than 1. angles 2 and 8 are also supplementary. Since angles 6 and 7 are vertical angles. then angles 1. An angle cannot have a negative measure. Since angles 1 and 2 combine to form angle AOC and angles 3 and . since they are alternating angles. The measure of angle 6 is 30°. Therefore.4. Notice that replacing x with 15 in the measure of angle 7 also yields 110: 10(15) – 40 = 150 – 40 = 110. they are complementary angles. Angles 4 and 7 are alternating angles. Since x = 15. Since line AE is perpendicular to ray OC. 5x + x = 180. the range of the 1 equation y = x is all real numbers excluding 0. There is no value for x that can make y = 0. a. the measure of angle 8 is equal to 180 – 106 = 74°. has the smallest range (fewest elements). angle AOC is 90°. the measure of angles 4 and 7 is 6(15) + 20 = 90 + 20 = 110. they are all congruent to each other. Angles 3 and 4 are congruent and supplementary. the measure of angle 6 is also 110°. 8x + 10 = x2 – 38. Since the square roots of negative numbers are imaginary. so the –4 value of x must be discarded. the range of the equation y = |x| – 3 is all real numbers greater than or equal to –3. Every numbered angle measures 90°. The equation of 1 the graph in diagram D is y = x. One divided by a real number (excluding 0) will yield real numbers. 142 3. Angles 7 and 8 are supplementary. Therefore. Since angles 2. Therefore. d. 12x + 10 + 7x – 1 = 180. their measures are equal. Therefore. the sum of angle 1 and angle 7 is not equal to the sum of angle 2 and angle 3. Since angles AIC and KIL are vertical angles. 2. Angles EIK and KIJ are complementary. their measures are equal. Since angles JLI and AIE are alternating angles. Since line AE is perpendicular to ray OC. x = 14. x + x + 6 = 90. 4. and angle 7. Angle 3 measures 33°. the measure of angle 2 is equal to: 90 – 62 = 28°. 2x = 84. the measures of angles 1. 7x – 8 = 90. Therefore. as to angles 4. and 6. too. since its measure is 6 greater than the measure of angle JIL. 3. 14. then the measure of angle LIF is x + 6. 10. x = 7. There are 180° in a triangle. their sum is equal to 90°: 90 – 57 = 33. x = –23. the sum of angles 2 and 3 is supplementary to the sum of angles 7 and 1. their measures are equal. Since angles 4 and 7 are vertical angles. x – 8 = 0. Therefore. therefore. so the measures of angles JLI and JIL must add to 90°: 8x – 4 + 5x + 3 = 90. Angles GOF and BOD are vertical angles. and 6. measures 60°. therefore. Since angles 2 and 3 combine to form angle BOD. with angles 1 and 2 combining to form that angle. Angles 4. and 7 form a line. angle EIC also measures 15x – 4. angle 1 + angle 2 = angle 3 + angle 4. Angle AOC is also a right angle. therefore. 5. Since line AE is perpendicular to ray OC. 20x = 80. angle EOC is a right angle (measuring 90°). is equal to: 33 + 28 = 61°. 20x + 10 = 90. and 6 form a line. 13x – 1 = 90. then 67 + x = 180. since angle 6 is supplementary to the sum of angles 7 and 1. x = 42. 15x – 4 + x2 = 180. Since line AE is perpendicular to ray OC. Angle 6 and angle BOD are vertical angles. Therefore. Angles 3 and 4 combine to form angle EOC. Since angles LIF and EIK are congruent. Since angles EIC and CIF form a line. Angles GKI and EIC are alternating angles. angle EIK is also 48°. their measures are equal. However. Angles LIF and JIL combine to form angle JIF. angles JIF and EIJ are right angles. a. their measures are equal. 2x + 6 = 90. The measure of angle JIL is 42° and the measure of angle LIF is 42 + 6 = 48°. Since angle 6 is angle GOF and angles 2 and 3 combine to form angle BOD. 5. as do the sum of the measures of angles 4. which measures 90°. therefore. 3. angles 2. 15. x2 + 15x – 184 = 0. x = 8. angle EOC is a right angle (measuring 90°). c. their sum is equal to 90°: 5x + 3 + 15x + 7 = 90. If the measure of angle JIL is x. e. d. the measure of angle 6 is equal to the sum of the measures of angles 2 and 3.– ANSWERS – 4 combine to form angle EOC. and the sum of angles 2 and 3 is equal to the measure of angle 6. Therefore. 12. If x is the sum of angles 5 and 6. In the same way. angle EOC is a right angle (measuring 90°). Angles 3 and 4 combine to form angle EOC. 5. Angle 4 is equal to 5(14) – 10 = 70 – 10 = 60°. therefore. Factor and solve for x: (x + 23)(x – 8) = 0. Since line IJ is perpendicular to line GH. and 7 add to 180°. 9. b. In fact. the measure of angle EIC is equal to: 15(8) – 4 = 120 – 4 = 116°. the measure of angle 4 is equal to: 15(4) + 7 = 67°. 13x = 91. and EIL are alternating angles. Angles 1. the measure of angle JLI is 8(7) – 4 = 52°. and x = 113°. 11. a. Angles 3 and 4 combine to form angle EOC. therefore. Angle AIE is also 52°. An angle cannot have a negative value. Since line IJ is perpendicular to line EF. b. their measures add to 90°. x + 23 = 0. ILH. x = 4. angle IJL is a right angle. Therefore. their measures add to 180°. Therefore. their sum is equal to 90°: 2x + 2 + 5x – 10 = 90. b. the sum of their measures is 180°. their measures are equal. and angle 6. 13. these sums must both equal 90°. these angles are 143 . their measures are equal. which measures 90° each. Since angles EIL and LIF form a line. Angles JLB. 7x = 98. 2. the measure of BOD. and 5 form a line. so disregard the negative value of x. so angle KIJ is 90 – 48 = 42°. therefore. 8. so the sum of the measures of those angles is also 180°. 3. angle COG is a right angle. 23. Since vertical angles have equal measures. 8(15) – 4 = 120 – 4 = 116°. Lines AB and EF are not known to be perpendicular. and there are 180° in a line. Angles 2 and 6 are vertical angles. In the same way. Angles 5. and 8. so their measures are also equal. Using these numbers. Therefore. and angles 4 and 8 measure 50°. since angles 4 and 7 are congruent and angles 3 and 7 are congruent. The measure of angle 3 is equal to: 180 – 43 = 137°. then angle JLB is 3. Therefore. angles 1 and 5 are vertical angles. and 6 is equal to the sum of angles 1. 8x – 4 = 7x + 11. Since angles 4 and 7 are congruent. 12. nothing is known about angles 1. (x + 13)(x – 13) = 0. and 6. angle 3 is equal to angle 7. 16. the measure of angle 4 is 11(5) = 55°. their measures are equal. and 6 measure 45°. 5. 90 + angle 9 = 133°. so angle 9 measures 43°. and 20 also add to 180°. the sum of their measures is 180°. Therefore. Angle AOE is made up of angles 2 and 3. and angles 1 and 2 are complementary. a. c. Since angle 3 and angle 10 are vertical angles. 3x + 5 + 5x – 3 = 90. d. 7. since 90 + 90 = 180°. e. 3. angle 8 measures 90°. Angles 3 and 9 form a line. x = 10. If angle LIF is x. Since lines EF and AB are perpendicular. 4. x2 + 11 = 180. and x = 13 (disregard the negative value of x. since no angles measure 0°. angle 10 is also 118°. 5. 2. and 11 are alternating angles. Since lines CD and GH are perpendicular to each other. Angles 21. 24. 20. The measure of angles 5. x – 4 = 11. 2. then the measure of angle 3 is 9(13) + 1 = 117 + 1 = 118°. and angles 2 and 6 are vertical angles. the sum of their measures is 180°. 21. c. 4. x – 10 = 0. and 22 are alternating (vertical) angles. Angles 3 and 7 could measure 40°. Therefore. 8x + 2 = 90. their measures are equal. since every numbered angle is greater than 0).– ANSWERS – 16. Notice that you could also substitute 15 into the measure of angle 22: 7(15) + 11 = 105 + 11 = 116°. 3. 2. x = 5. Angles 1 and 8 are both right angles. Angles 3 and 7 are vertical angles. supplementary and add to 180°. and x + 3.5x = 180. 18x = 90. the sum of angles 4. the measure of angle 2 also 5x – 3. and 8 are all congruent. 17. Since lines EF and AB are perpendicular. and the sum of angles 2 and 3 is (10)2 = 100. 144 . Test the other answer choices with different possible values. 12. so their measures are equal. If x = 13. 133 – 90 = 43. angles 3 and 4 must be congruent. angles 7 and 8 must be congruent. the sum of angles 4 and 5 is equal to 180 – 100 = 80°. x cannot be 0. Since angles 2. However. 15. Angles 1. x2 = 169. angles 21 and 5 must be supplementary. x = 40. 22. In addition to be equal. Therefore. the measures of angles 1 and 4 must add to 90°. e. Since angle 11 and angle 5 are supplementary (they combine to form a line). In fact. Angles 3. and the sum of angles 2 and 3 is equal to the sum of angles 6 and 7: x2 = 10x. 4. d. x(x – 10) = 0.5x. 2. 8x = 88. and 22 is 116°. x = 15. Angle 1 + angle 8 = angle 19 + angle 15 + angle 20 = 180°. b. Angles 4 and 8 are also vertical angles. only the number sentence in choice e holds true. they are also supplementary. Angles 2 and 6 are vertical angles and angles 3 and 7 are vertical angles. 16. and angles 4 and 8 are congruent. Angles 4 and 8 are vertical angles. and 5 form a line. since these angles form a line. e. their measure are equal. angle 2 is equal to angle 6. and 4 form a line. Therefore. 19. Angles 19. the measure of angle LIF is 40° and the measure of angle EIL is 3. since angles 3 and 4 are complementary and angles 7 and 8 are complementary (since lines CD and GH are perpendicular) all four angles measure 45°. 17. x2 – 10x = 0. Therefore.5(40) = 140°.5x = 180. 11x + 7x = 90. while angles 1. 18. 5. It cannot be stated that angle 2 = angle 3. x = 11. Since the measure of angles 2 and 3 add to 90°. angle AOE measures 90°. The measure of angle 2 is 5(11) – 3 = 55 – 3 = 52°. x2 + 15x – 184 = 0. Angles D and E cannot be congruent to each other without also being congruent to angle F. 3x + 4x + 5x = 180. so the measure of angle 1 is equal to: 180 – 114 = 66°. Therefore. and the measure of angle C is 2(20) = 40. x2 + 15x – 4 = 180. 9. Since angles 14 and 19 are vertical angles. the sum of angles 1. triangle ABC is obtuse and scalene. 24x = 168. Angles D and E could each measure 60°. the sum of angles 2 and 3 is equal to the measure of angle 4: 75 + 65 = 140°. 33x = 264. However. x = 8. and the sum of the measures of interior angles D and E is 120°. d. the sum of these exterior angles is 360°. Since there are 180° in a triangle. triangle DEF cannot be isosceles. Since x = 7. Therefore. x2 + 1 + 9x – 7 + 6x + 2 = 180. Angles 3 and 6 are supplementary. Therefore. making triangle DEF equilateral. An exterior angle and its adjacent interior angle are supplementary. 2x + 5 + 2x + 5 + 3x – 5 = 180. angle 4 measures 7(16) + 2 = 112 + 2 = 114. 5. x = 16. or equilateral. scalene. 2. the measure of . making triangle DEF obtuse and scalene. obtuse. c. 145 7. the measure of angle F = 8(7) = 56°. a. the measure of interior angle F is 60°. Since there are 180° in a triangle. The measures of the angles of a triangle add to 180°. Since lines AB and GH are perpendicular. An angle and its exterior angle are supplementary. 33x – 174 = 90. if either angles D or E measure 60°. and x = 20. 24x + 12 = 180. e. the other must also measure 60°. 23x = 368. 8x + 20 = 180. Therefore. a. The measures of the angles of a triangle add to 180°.  6. x = 7. 8. and x = 15. Therefore. since angle 4 is an exterior angle that is not adjacent to 2 or 3. The measure of angle A is 5(20) + 10 = 110. e. x = 8 (disregard the negative value of x since an angle cannot have a negative measure). Therefore. 8x + 16x + 12 = 180. 7x = 175. If one exterior angle is taken from each vertex of a triangle. (x – 8)(x + 23) = 0. the measure of angle 5 is equal to the sum of the measures of angles 1 and 3. It is true that the sum of angles 4 and 1 is equal to the sum of angles 3 and 6. and x = 25. angle 20 measures 90° and the sum of the measures of angles 14 and 15 is 90°. the measure of an angle exterior to A is 180 – 55 = 125°. 5x + 10 + x + 10 + 2x = 180. 4. since both pairs of angles form lines. the measure of angle B is (20) + 10 = 30. b. Therefore. making triangle DEF acute and equilateral. angle 14 = 15x + 6 – 90 and angle 15 = 18x – 90. Therefore. Since the sum of angles 14 and 15 is 90°. angle 19 also measures 36°. Chapter 10 1. Since the largest angle of triangle ABC is greater than 90° and no two angles of the triangle are equal in measure. but not isosceles. triangle DEF can be acute. but these angles could also measure 100° and 20° respectively. b. Therefore. (15x + 6 – 90) + (18x – 90) = 90. 7x + 2 + 8x + 8x – 10 = 23x – 8 = 360. 3. Since an angle and its exterior angle are supplementary. The measures of the angles of a triangle add to 180°. Therefore. so the statement in choice e is also true. and 3 is equal to 180°. The measure of an exterior angle is equal to the sum of the measures of the interior angles to which the exterior angle is not adjacent. Since an angle exterior to angle F is 120°. Therefore. Angle F measures 60°. The measure of an exterior angle is equal to the sum of the measures of the interior angles to which the exterior angle is not adjacent. 8x = 160. The sum of the measures of one exterior angle from each vertex of a triangle is 360°. Therefore. It is also true that the sum of angles 2 and 3 is equal to the measure of angle 4. e. 2.– ANSWERS – 25. 7x + 5 = 180. not angles 1 and 2. 12x = 180. and the measure of angle 14 is equal to: 126 – 90 = 36. the sum of angles 20 and 14 is 15(8) + 6 = 126. c. the measure of angle 3 = 180 – 115 = 65. The measure of angle A is 2(25) + 5 = 50 + 5 = 55. If the measure of angle F is x. If the measure of angle 14 is x. and the measure of angle 3 is 6(8) + 2 = 48 + 2 = 50°. If x = 1. The ratio of side 1 BC to side EF is 1: 5. d. and each side 1 of triangle DEF is 5 the length of its corresponding side in triangle ABC. 10x + 2 = 4x – 4 + 7x – 6. Therefore. Each side of triangle GHI is congruent to its corresponding side. (10x – 2)(x + 2) = (6x)(2x + 2). 10x + 15 = 11x + 4. Since there are 180° in a triangle. x = 43. since angle 5 and the combination of angles 10 and 11 are alternating angles. Since vertical angles are equal. and 14 form a line. 17. their measures are equal. Therefore. 3x = 180. Therefore.  2x + 2 = x + 2 . 2 = x. Since triangles JKL and MNO are congruent and equilateral. not five times its measure. angle 1 is (8)2 + 1 = 64 + 1 = 65. x = 40. the measures of their corresponding sides are in the same ratio. 60 = x. Since angles 8 and 11 are supplementary. 11. There are 180° in a triangle. or 5:1. 14. 2x = 86. x2 – 3x + 2 = 0.5x. Since the measure of angle 7 is 43 and angles 7 and 8 are supplementary. every side of triangle JKL is congruent every other side of triangle JKL and congruent to every side of triangle MNO. Therefore. Every length of triangles JKL and MNO is equal to 6(7) + 3 = 42 + 3 = 45. angle A is congruent to angle D. d. Since angles 3 and 7 are supplementary. The length of side DF is 4. d. Disregard the negative value of x. The measures of corresponding sides of similar triangles are in the same ratio. 19. Corresponding angles of similar triangles are congruent. the measure of an angle exterior to F is equal to 2(60) = 120°. 11. 3x = 144. d. angle 3 is equal to the sum of angles 6 and 10: 10x + 15 = 8x – 3 + 3x + 7. 10x + 2 = 11x – 10. Angles 10 and 13 are vertical angles. However. each side of triangle ABC is five times the length of its corresponding side in triangle DEF. Therefore. angle 14 = 40° and angle 11 = 2(40) = 80°. 146 . 4. (x – 7)(x + 1) = 0. Sides HG and PQ are corresponding sides that are congruent. 13. angle 5 is equal to the sum of angles 10 and 7: 10(12) + 2 = 120 + 2 = 122°. Therefore. b. 2x + 2. x = 12. then the measure of angle 11 is 2x and the measure of angle 8 is 2. since a side of a triangle cannot be negative. and angles 6 and 7 are congruent. 6x + 3 = x2 – 4. the measure of angle 2 is 9(8) – 7 = 72 – 7 = 65. The measure of angle 3 is 10(11) + 15 = 110 + 15 = 125. 15. then the sum of the measures of angles D and E is 2x. x = 7. and x = 60. Therefore. Therefore.– ANSWERS – 10. The ratio of side AB to side DE is equal to the ratio of side AC to 90 72 3 72 side DF. the measure of angle 7 is 180 – 125 = 55°. triangle ABC is isosceles. 18. a. 2 + 2 = 4. 10x2 – 2x + 20x – 4 = 12x2 + 12x. The measures of corresponding sides of similar triangles are in the same ratio. The length of  DF is 48. Therefore. the measure of angle 8 is equal to 180 – 43 = 137°. x = 11. b. angle 10 is also 94°. Since the triangles are similar.5x = 180. The measure of an exterior angle is equal to the sum of the interior angles to which the exterior angle is not adjacent. Therefore. every side of triangle DEF is greater than 3. Angle 5 and the sum of angles 10 and 11 are equal. 2x2 – 6x + 4 = 0. Therefore. the measure of angle 10 is equal to 180 – (40 + 80) = 180 – 120 = 60°. x = 48. x2 – 6x – 7. Since the measure of an exterior angle is equal to the sum of the interior angles to which the exterior angle is not adjacent. c. 94 + x + x = 180. Angles 10 and 13 are vertical angles. x = 2. x = 1. Since exactly two of the angles of triangle ABC are equal. 12. c. The ratio of side AB to side DE is equal to the ratio of side AC to 6x 10x – 2   side DF. 16. x must be equal to 2.5x = 180. side GI is congruent to its corresponding side. (x – 2)(x – 1) = 0. Since angles 10. b. x + 2x = 180. then side DF = (1) + 2 = 3. the ratio of side HG to side PQ is 1:1. therefore. Since the measure of an exterior angle is equal to the sum of the measures of the interior angles to which the exterior angle is not adjacent. The measures of corresponding sides of similar triangles are in the same ratio. Since angles 8 and 5 are congruent. 8. x + 5 > x > x – 5. 20. the length of the hypotenuse is 2(8) – 3 = 16 – 3 = 13. 4. and since angles 7 and 12 are alternating angles. Use the Pythagorean theorem: 92 + 152 = x2. x = 16 (x cannot equal –16 since the side of a triangle cannot be negative). 15. and 20 + 5 = 25. x = 306  = 334 . 5. The length of side IJ is 147 . you must add the squares of the base lengths. x2 = 256. If angle Q is the right angle of the triangle. If the shorter base is a. 81 + 225 = x2. not 135. c. angle 6 must be the other 45° angle of the triangle.– ANSWERS – 20. x2 – 7x – 8 = 0. Therefore. angle 7 is one of the acute. the sum of angles 9 and 12 is 45 + 45 = 90. angle JIK is also 45° (180 – (90 + 45) = 45). Use the Pythagorean theorem: (x – 3)2 + (x + 4)2 = (2x – 3)2. b. x2 – 6x + 9 + x2 + 8x + 16 = 4x2 – 12x + 9. Using the Pythagorean theorem. If the measure of one base angle of a right triangle is 45°. Therefore. Since the hypotenuse of the triangle is x. since a side of a triangle cannot be negative. Therefore. x(x – 20) = 0. Use the Pythagorean theorem: 82 + x2 = (85)2. a. The hypotenuse of the given triangle is x + 5. Therefore. Since angle IKD is 135°. since x cannot be 0 or negative (the side of a triangle cannot be 0 or negative). The hypotenuse of a 45-45-90 right triangle is equal to 2 times the length of a base. and 5(5) = 25. otherwise. 2x2 – 10x + 25 = x2 + 10x + 25. 25. angles 8 and 7 would be congruent. the measure of angle 2 is 180 – 45 = 135°. (x)2 + (x – 5)2 = (x + 5)2. the measure of angle IKJ = 180 – 135 = 45. d. Disregard the negative value of x. or a 45-45-90 right triangle. d. so angle 2 is not equal to angle 9 + angle 12. d. 6. a2 + 9a2 = c2. PQR is an isosceles right triangle. Since x = 8. since x cannot be 0. since these lines are perpendicular. 2x2 + 2x + 25 = 4x2 – 12x + 9. IJK. Use the Pythagorean theorem: a2 + (3a)2 = c2. congruent angles of the right triangle. Since angle 6 is 45° and angles 2 and 6 are supplementary. x2 – 20x = 0. e. (x – 8)(x + 1) = 0. and x = 20. are a multiple of the triangle with sides 3. and it is supplementary to angle IKJ. Since the length of a base is 10. 2. There is no information provided about the angles of triangle IJK. therefore. the hypotenuse of the triangle is 102 units. However. e. so you cannot determine if this is a certain type of right triangle. The hypotenuse of an isosceles right triangle is equal to 2 times the length of a base. c = a10 . x = 8. 10. If the sine of P is equal to the sine of R. 20 – 5 = 15. which means that it measures 45°. Therefore. TUV is an isosceles right triangle. Therefore. The hypotenuse of an isosceles right triangle is equal to 2 times the length of a base. 3. since 3(5) = 15. and the 2 base of the triangle is equal to  2 times the length of the hypotenuse. x2 + x2 – 10x + 25 = x2 + 10x + 25. If angle 8 is greater than angle 7. then side QR is a base of the right triangle. Since angle IKJ = 45°. 2x2 – 14x – 16 = 0. The sides of the triangle. b. 5. Therefore. Since angles 6 and 9 are alternating angles. 64 + x2 = 320. a. then the longer base is 3a. b. 4. the sides of the triangle measure 20. angle 7 cannot be the right angle of the triangle. the length of side IK cannot be determined from the information provided. the bases of the triangle each x2 measure  2 units. x2 = 306. the hypotenuse of the triangle is 42 units. only the length of one base is given. the measure of angle 9 is 45°. 10a2 = c2. triangle IJK is an isosceles right triangle. 9. then the measure of the other base angle must also be 45° (180 – (90 + 45) = 45). 7. To find the hypotenuse of a right triangle. 4(5) = 20. Lines GH and CD form a right angle. then angle P is equal to angle R. the measure of angle 12 is also 45°. Since the length of a base is 4.  Chapter 11 1. making angle 10 the right angle of the triangle. – ANSWERS – 11. e. 12. d. 13. c. 14. c. 25 units, since line CD is 25 units from line AB. The hypotenuse of an isosceles right triangle is equal to 2 times the length of a base. Since the length of a base, IJ, is 25, the hypotenuse, IK, is 252 units. If angle A is twice the measure of angle C, then let angle C = x and let angle A = 2x; x + 2x + 90 = 180, 3x = 90, x = 30. The measure of angle C is 30° and the measure of angle A is 2(30) = 60°. Therefore, this is a 30-60-90 right triangle. Side AB is the shorter base, since it is the side opposite the smallest angle. The hypotenuse of a 30-60-90 right triangle is twice the length of the shorter base. Therefore, the measure of AC is 2(7) = 14 units. Since angle B of triangle ABC is a right angle and angle C is 60°, angle A must be 30° (180 – (90 + 60) = 30). Therefore, this is a 30-60-90 right triangle. Side AB is the longer base, since it is the side opposite the larger base angle. The measure of the larger base of a 30-60-90 right triangle is 3 times the length of the shorter base. Therefore, the length of side BC is equal to the length 9 93  =  = 33 of side AB divided by 3:   3  3 units. Since angle B of triangle ABC is a right angle and angle C is 60°, angle A must be 30° (180 – (90 + 60) = 30). Therefore, this is a 30-60-90 right triangle. Side AB is the longer base, since it is the side opposite the larger base angle. The hypotenuse of a 30-60-90 right triangle is twice the length of the shorter base. Therefore, the measure of side BC is (6x + 2)  = 3x + 1 units. The measure of the larger 2 base of a 30-60-90 right triangle is 3 times the length of the shorter base. Therefore, the length of side AB is (3x + 1)3 units. Since angle B of triangle ABC is a right angle and angle C is 60°, angle A must be 30° (180 – (90 + 60) = 30). Therefore, this is a 30-60-90 right triangle. Side AB is the longer base, since it is the side opposite the larger base angle. The hypotenuse of a 30-60-90 right triangle is twice the length of the shorter base. Therefore, if the length of side BC is x, then the length of side AC is 2x, and their sum is equal to 3x; 3x = 12, x = 4. The length of side BC is 4 units, and the length of side AB is 43 units, since the longer base of a 30-60-90 right triangle is 3 times the length of the shorter base. 15. b. Side AB of equilateral triangle ABC is 8 units; therefore, side AC is also 8 units. Since side AE is 20 units, side CE must be 20 – 8 = 12 units. Angle ACB is 60°, since every angle of an equilateral triangle is 60°. Angles ACB and ECD are vertical angles, so angle ECD also measures 60°. Since angle ECD is 60°, angle CED is 30°, and triangle CDE is a 30-60-90 right triangle. Side CD is half the length of side CE, since it is the shorter base of the 30-60-90 right triangle. Therefore, 12 side CD is equal to 2 = 6. Line segment BC is 8 units, since it is a side of equilateral triangle ABC, of which every side measures 8 units. The length of side BD is equal to the sum of side BC and side CD: 8 + 6 = 14 units. 16. c. Since triangle ABC is an isosceles right triangle, the measures of both bases are equal. Therefore, the tangent of either angle A or angle C is the tangent of 45°, which is 1, since any number divided by itself is 1. 17. a. The tangent of an angle is equal to the side opposite the angle divided by the side adjacent to the angle; BC AB (x2 – 2x) x(x – 2) x    = (3x + 6) = 3(x – 2) = 3 . 18. a. The sine of one base angle of a right triangle is equal to the cosine of the other base angle. Therefore, the sine of angle C is equal to the cosine of angle A, and the cosine of angle C is equal to the sine of angle A. Why? If the cosine of x angle C is y, then the length of side BC is x and the length of side AC is y. Since the sine of angle A is equal to the side opposite angle A, side BC, divided by the hypotenuse, side AC, the sine of x angle A is also y. 15 19. a. If the sine of angle A is 17, then the length of side BC is 15 and the length of side AC is 17, since the sine of an angle is equal to the side 148 – ANSWERS – opposite the angle divided by the hypotenuse. Use the Pythagorean theorem to find the length  (AB)  2 + 152 = 172, (AB)  2 + 225 = 289, of AB:  2 = 64,  (AB) AB = 8. Since the cosine of an angle is equal to the side adjacent to the angle divided by the hypotenuse, the cosine of angle A is equal 8 AB to AC = 17 . 3 20. e. The tangent of angle A is equal to 0.75, or 4. Since the tangent of A is the measure of side BC divided by side AB, the ratio of side BC to side AB is 3:4. To find the relative length of side AC,  2, use the Pythagorean theorem: 32 + 42 = (AC) 2 2  , 25 = (AC)  , 9 + 16 = (AC) AC = 5. The ratio of side BC to side AB to side AC is 3:4:5. Since the length of side AB is 4 less, not 1 less, than the length of side AC, multiply each number in the  = 3(4) = 12 units, AB  = 4(4) = 16 ratio by 4: BC  units, and AC = 5(4) = 20 units. If the length of  is 16 units and the length of AC  is 20 units, AB  is 4 less than AC,  and the tangent of angle A AB 12  is is still 0.75, since 16 = 0.75. The length of BC 12 units.  = 10 and LM  = 5, the cosine of angle 21. e. Since KM 1 1 LM LMK is KM = 2. The cosine of a 60° angle is 2, so the measure of angle LMK is 60°. Angles LMK and KIJ are alternating angles, so their measures are equal. Therefore, angle KIJ is also 60°. 22. d. Angles IKJ and LKM are vertical angles; their measures are equal. Therefore, each angle is 60 1 equal to 2 = 30°. The sine of a 30° angle is 2. Since the measure of angle LKM is 30°, the  divided by the length of KM  is length of LM 1 LM 1 8 1  equal to 2: KM = 2, KM = 2, KM = 16 units. KMH is 3y, then 4y = 180, y = 45°. The measure of angle KML is 45°. Angle JIK is also 45°, since KML and JIK are alternating angles. Since JIK is 45° and KJI is 90°, angle IKJ is 45° (180 – (45 + 90) = 45). Therefore, triangle IJK is a 4545-90 right triangle. The hypotenuse of a 4545-90 right triangle is 2 times the length of either base: (x6)(2) = x12  = x43 = 2x3 units. 25. d. Angles IKJ and LKM are vertical angles; their measures are equal. Since the sine of an angle is equal to the length of the side opposite the angle divided by the length of the hypotenuse, the sine of angle IKJ is equal to IJ IK and the sine of angle LM LKM is equal to KM .Two equal angles have equal 2x – 2 2x + 2 IJ LM  2  sines; IK = KM , 2x +1 =  3x – 1 , 6x – 6x – 2x + 2 = 4x2 + 2x + 4x + 2, 6x2 – 8x + 2 = 4x2 + 6x + 2, 2x2 – 14x = 0, x2 – 7x = 0, x(x – 7) = 0, x = 7 (x cannot be 0 since the length of a line cannot be neg would be –2 units). If x = ative—if x were 0, IJ  = 2(7) + 2 = 14 + 2 = 7, then the length of LM 16 units.  23. a. The tangent of a 60° angle is 3; therefore, the measure of angle JIK is 60°. Angles JIK and LMK are alternating angles; their measures are equal. Angle LMK is 60° and its tangent is 3.  the side oppoTherefore, the measurec of LK,  site angle LMK, is 3 times the length of LM, the side adjacent to angle LMK. 24. b. Angles KMH and KML are supplementary angles; their measures add to 180°. If the measure of angle KML is y and the measure of angle 149 Chapter 12 1. b. Regular pentagons are equilateral; every side is equal in length. Therefore, every angle is equal in size, and every regular pentagon is similar to every other regular pentagon. However, regular pentagons are not necessarily congruent. One regular pentagon could be ten times the size of another. Heather’s regular pentagons may not be congruent. 2. b. The sum of the exterior angles of any polygon is 360°. Therefore, the sum of the interior angles of this polygon is (360)(3) = 1,080. The sum of the interior angles of a polygon is equal to 180(s – 2), where s is the number of sides of the polygon; 1,080 = 180(s – 2), 1,080 = 180s – 360, 1,440 = 180s, 144 = 18s, s = 8. The polygon has 8 sides. – ANSWERS – 3. a. The sum of the interior angles of a polygon is equal to 180(s – 2), where s is the number of sides of the polygon. If x is the number of sides of the polygon, and the sum of the interior angles is 60 times that number, then 60x = 180(x – 2), 60x = 180x – 360, 120x = 360, x = 3. Andrea’s polygon has three sides. 4. c. The sum of the exterior angles of any polygon is 360°. The sum of the interior angles of a polygon is equal to 180(s – 2), where s is the number of sides of the polygon. If these sums are equal, then 180(s – 2) = 360, 180s – 360 = 360, 180s = 720, and s = 4. The polygon has 4 sides. 5. b. The sum of the interior angles of a polygon is equal to 180(s – 2), where s is the number of sides of the polygon. Since the number of sides is three less than x and the sum of the interior angles is 9x2, 180(x – 3 – 2) = 9x2, 180x – 900 = 9x2, 9x2 – 180x + 900 = 0, x2 – 20x + 100 = 0, (x – 10)(x – 10) = 0, x = 10. Therefore, the number of sides of the polygon is (10) – 3 = 7 sides.  to FG  is 4:1. Therefore, 4 = 6. a. The ratio of AB 1 4x + 4  = 4x + 4, FG  = x + 1. , 4( FG) FG 7. e. Although it is known that ABCD and EFGH are similar, the ratio of their corresponding sides is not provided, nor is the ratio of their perimeters. Therefore, the perimeter of EFGH cannot be determined. 8. d. The number of sides of a polygon determines the sum of the interior angles of the polygon. The sum of the interior angles of any octagon is 180(8 – 2) = 180(6) = 1,080°. These two octagons are not necessarily similar or congru to ent, although they could be. The ratio of AB  could be 1:1, but there could also be no ratio ST between the sides of ABCDEFGH and STUVWXYZ. No one side of ABCDEFGH must equal one side of STUVWXYZ.  and GH  are corresponding sides of 9. c. Since AB  to AB  is equal similar polygons, the ratio of GH to the ratio of the perimeter of GHIJKL to the 8x + 4x 10. e. 11. d. 12. b. 13. e. 12x   perimeter of ABCDEF; GH:AB = 1 2x + 6x = 18x = 2  = 2:3. 3  and TU  are corresponding sides of simSince AB  to TU  is equal to ilar polygons, the ratio of AB the ratio of the perimeter of ABCDE to the 5x – 1 4  TU =   perimeter of TUVWX; AB: 4x – 2 = 3 , 15x – 3 = 16x – 8, –3 = x – 8, x = 5. Therefore, the length  is 5(5) – 1 = 25 – 1 = 24 units. Since of AB ABCDE is a regular polygon with 5 sides, each of the 5 sides measures 24 units. Therefore, the perimeter of ABCDE is (5)(24) = 120 units. The perimeter of ABCD is equal to the sum of its sides: 2x + 2x + 2 + 3x + 3x – 2 = 60, 10x = 60,  is 3(6) – 2 = x = 6. Therefore, the length of BC 18 – 2 = 16 units. If x = 8, then the perimeter of the figure is 11(8) – 4 = 88 – 4 = 84. Every side of a regular polygon is equal in length. Therefore, the length 84 of one side of this seven-sided polygon is 7 = 12 units. If the perimeter of a regular pentagon is 75 units, then the length of one side of the penta75 gon is 5 = 15 units. The ratio of a length of ABCDE to a length of PQRSTU is 5:6. Therefore, 5 if x is the length of a side of PQRSTU, then 6 = 15 , 5x = 90, and x = 18 units. Since the length of x one side of the regular hexagon is 18, the perimeter of the hexagon is (6)(18) = 108 units. 14. c. The hypotenuse of an isosceles right triangle is equal to 2 times the length of a base of the triangle. If the hypotenuse is 52, then the (52) length of each base is  = 5. Therefore, the (2) perimeter of the triangle is 5 + 5 + 52 = 10 + 52 units. 15. a. The sum of the interior angles of a polygon is equal to 180(s – 2), where s is the number of sides of the polygon; 720 = 180(s – 2), 720 = 180s – 360, 1,080 = 180s, 108 = 18s, s = 6. Since the polygon is regular, has 6 sides, and the length of one side is 3x2, the perimeter of the polygon is 6(3x2) = 18x2 units. 150 e. 151 7. The measure of angle H is 4(25) = 100. Every side of a rhombus is equal in length. 8. It does not contain a right angle. since these quadrilaterals each have four right angles. and  = 64. these four line segments could be connected at right angles to form a square. you must have four identical line segments. Since a square is a type of rectangle and a type of rhombus. e. a. The perimeter of ABCD is equal to AB 64 + 8 + 64 + 8 = 144 units. and the sides of a square are all congruent to each other. Therefore. 7x + 5 = 180. c. The tangent of angle ACB is equal to the length  divided by the length of BC. angle F is also 100°. both of which are types of parallelograms. the width of 1 1 the rectangle is 2 and the length is 4(2) = 2. since the 4 angles of a square are right angles. 11. their measures are supplementary: 3x + 5 + 4x = 180. These angles are only equal if ABCD is a square. Therefore. 6. large square is comprised of two sides from each of the four squares (the remaining four sides are now within the large square). the perimeter of the square is equal to 2x + 2x + 2x + 2x = 8x units. any of these four types of quadrilaterals could be formed. then 2x – 4 is the length of the rectangle.– ANSWERS –  Chapter 13 1. 3. Since opposite sides of a rectangle are congruent. and not all rectangles are squares. the perimeter of the rectangle is equal to 2x – 4 + x + 2x – 4 + x = 6x – 8. Angles E and H are consecutive angles in rhombus EFGH. Andrew’s polygon has four sides. x = 25. then AB if the length of BC 8 = 8. this polygon could be a parallelogram or a rhombus. b. therefore. Since nothing about the angles is stated in the question. every side of the rhombus must be congruent to every side of the square. Since the new. . 4. the measure of a side of the square is equal to the 2x2 measure of the diagonal divided by 2 :  2 = 2x units. The diagonal of a square is the hypotenuse of an isosceles right triangle. it is not always true that angle DCE = angle ECB. 4x = 2. Therefore. these angles combine to form right angle DCB of the rectangle. However. x = 48 = 16 3 = 43. and only the opposite angles of a rhombus are necessarily congruent. the length of a side of each small 8x square is 4 = 2x units. so it must be a quadrilateral. The perimeter of each small square is 8x units. x2 – 6 = 42. e. Angles DCE and ECB are complementary angles. large square is equal to: 4(2x + 2x) = 4(4x) = 16x units. so it cannot be a rectangle or a square. a.  Therefore. a. If a quadrilateral has perpendicular diagonals. 5. of AB  is 8 units. but its diagonals are not necessarily congruent. If the length of the square 1 is x. Therefore. Therefore. Since the width of the rectangle is equal to the length of the square and the length of the rectangle is 4 times the length of the square. since these are four-sided polygons that do not necessarily contain a right angle. However. Since the rhombus and the square have the same perimeter and both figures have four congruent sides. 10. a. ABCD is a rectangle. If x is the width of the rectangle. 12. then x + x + x + x = 2. x2 = 48. Therefore. 2. x = 2. Since opposite angles of a rhombus are congruent and angles H and F are opposite angles. Therefore. In order to form a square or a rhombus. every side of the rhombus and every side of the square is equal to exactly one-fourth of the perimeter. it must have four congruent sides. the perimeter of the rectangle is 1 1 equal to: 2 + 2 + 2 + 2 = 5 units. If the square and the rectangle share a side.A rhombus has four congruent sides. 7x = 175. d. d. then the width of the rectangle is equal to the length of the square. b. 9. The diagonals of a rhombus are perpendicular. One 168 side of this rhombus is equal to 4 = 42 units. the perimeter of the new. and the measure of the longer base is 53 units. 15. 10 units. The hypotenuse is the side of the rhombus. Since angle A measures 120°. Use the Pythagorean theorem to find the  ( AD)  2 + 122 = 132. The hypotenuse of an isosceles right triangle is 2 times the length of a base of the triangle. the perime2 ter of the square. The cosine could have been given in reduced  could be 24. AD  = 5 units. Since these right triangles are 30-60-90 right 10 triangles. and triangle EBC is a right triangle. ABCD is a rectangle. a. Therefore. The diagonals of a rhombus divide the rhombus into 4 congruent right triangles. the measure of the shorter base is 2 = 5 units. The area of a triangle is equal to 1 bh. b. Therefore. DC nate sides of rectangles are congruent. The length of one base of the isosceles right triangle is its height. since the 4 angles of a square are right angles. Choice d is the only answer choice that is a multiple of 34. 5x + 1. and the form. The diagonal of a square is the hypotenuse of an isosceles right triangle.  = 30 and AE  = EB. since this line is perpendicular to the base. x6 Therefore. which means that angle B is a right angle. 3x = 9. The length of this line is equal to the length of the longer base of a 30-60-90 right triangle. The longer base is half the length of the longer diagonal. is equal to 2x – 2 + 2x – 2 + 2x – 2 + 2x – 2 = 8x – 8.  152 Chapter 14 1. However. x = 3. Therefore. e. since the length of AC length of every side of ABCD is an integer. the length of the longer diagonal is 2(53) = 103 units. The bases of each triangle are made up of half the shorter diagonal (opposite the 30° angle) and half the longer diagonal (opposite the 60° angle). (BC)  2 = 64. a base of this triangle measures  2 = x3 units. The length of a side of the square is equal to 2(3) – 2 = 6 – 2 = 4 units. therefore. If the length of = 169. b. the area of triangle 1 1 DEC = 2(30)(8) = 2(240) = 120 square units. Therefore. Diagonal AC along with sides AD and DC form a right triangle.  14. and the sides of a square are all congruent to each other. If a line is drawn from the vertex of an angle of the triangle to its opposite base. the area of this triangle is equal 2 1 1 1 to 2(b(2b)) = 4b2. 8x – 8 = 5x + 1.– ANSWERS – 13. b. and every angle of an equilateral triangle is 60°. 225 + Pythagorean theorem: 152 + (BC)  2 = 289. 2. 3. ( AD)  2 + 144 length of AD: 2  = 25. and a 60° angle (the angle formed by the diagonal bisecting the 120° angle of the rhombus). b. angle D also measures 120° and angles B and C each measure 60°. so the length of one side of this triangle 36 is 3 = 12 units. The cosine of ACD is equal to the length of CD divided by the length of diagonal AC. The base of triangle DEC is 30 units and the height of triangle DEC is 8 units. so the area of 1 3 the triangle is 2(x3)(x3) = 2x2 square units. the measure of a side of the square is equal to the measure of the diagonal divided by 2: (2x – 2)2  = 2x – 2 units.  then EB  = 30 = 15 Since AB 2  can be found using the units. 4. since alter= 30 units. then the length of BC  is 12 units. this line represents the height of the triangle. If the base the triangle is b. Since AB (BC)  also equals 30 units. then the height of the 1 triangle is 2b. BC  = 8 units. Since the length of CD  is also 12 units. The perimeter of length of AB ABCD is 5 + 12 + 5 + 12 = 34 units. This line cuts the triangle into two identical 30-60-90 right triangles. the only possible perimeters are multiples of 34. Notice that this is only one possible perimeter of ABCD. a 30° angle (the angle formed by the diagonal bisecting the 60° angle of the rhombus). d. The length of BC  2 = 172. which is 3 times the length of the shorter base. Every side of an equilateral triangle is equal in length. The shorter . The length of CD  could be 26. Each right triangle is made up of a right angle. the units. ( AD)   is also 5 AD is 5 units. 3 The area of a rectangle is equal to its length times its width. Since the length of every side of a square is the same. Only the values of the radius and height given in choice a hold true in the formula: π(3)2(5) = π(9)(5) = 45π in. the area of the 1 equilateral triangle is 2(12)(63) = (6)(63) = 363 square units. x2 + 7x + 10 can be factored into (x + 2) (x + 2)(x + 5). Therefore. Therefore. the length of the rectangle is equal to 3(6) – 2 = 18 – 2 = 16 cm. the area of the triangle is 1 2 1 equal to: 2(x)(3)x = 3x2 square units. Since the hypotenuse is 12 units. c. 2 Since the height of the triangle is 3x and the base of the triangle is x. w is the 153 . 13. where r is the radius of the cylinder and h is the height of the cylinder. 3x = 24. Therefore. r = 4. e. (3x + 16)(x – 6) = 0. Angle ACD is the 30° angle of 30-60-90 right 1 triangle ACD. 1 the size of the shaded area is equal to x2 – 3x2 = 2 x2 square units. base is equal to half the length of the hypotenuse. 7. The volume of a rectangular solid is equal to lwh. the area of the square is x2. if the length of a side of a square is x. 3x2 – 2x – 96 = 0. 6. it is the  the longer shorter side of the right triangle. 11. where r is the radius of the cylinder and h is the height 2 of the cylinder. 2  = 10 units. The area of the new square. The volume of a cylinder is equal to πr2h. a. 2(AD) = 20. 2 the height of the water is 3(15) = 10 cm. d. The volume of this cylinder is equal to π(2x)2(8x + 2) = π4x2(8x + 2) = (32x3 + 8x2)π. 15. the width of the rectangle is equal to its area divided by its length: (x2 + 7 x + 10) . a. 14. Set 3x – 4 equal to 20 and solve for x: 3x – 4 = 20. The sine of 30° is 2. is one-fourth the area of the old square. The volume of a cylinder is equal to πr2h. DC. since each side of the square has a length of x units.3. therefore. side of the triangle. then the volume of cylinder B is π(3r)2(3h) 1 1 1 = π(9r2)(3h) = π3r2h. is 3 times the length of  AD: 103. then the area of the square becomes 1 1 1 1 (2x)(2x) = 4x2. 3x2 – 2x = 96. 4r3 = 256.  = AD 20 . Cancel the (x + 2) terms from the numerator and denominator of the fraction. and x = 6 (disregard the negative value of x since the width of rectangle cannot be negative. The area of a square is equal to the length of one side of the square multiplied by itself. where r is the radius of the cylinder and h is the height of the cylinder. 12. Therefore. where l is the length of the solid. Therefore. 10. The area of the square is x2 square units. d. If the width of the rectangle is x. 16. the length of the shorter base is 6 units and the height is 63 units. x2. the length of one side of this square is equal to the square root of the area: 25 =5 units. the volume of water is equal to: π(2)2(10) = 40π cm3. The radius of the cylinder is 4 cm. x = 8. Since the area of a rectangle is equal to its length times its width. the perimeter of this square is (4)(5) = 20 units. The volume of a cylinder is equal to πr2h. r3 = 64. b. 4x2. the area of ABCD = (10)(103) = 1003 square units. 9.– ANSWERS – 5. The volume of a cylinder is equal to πr2h. Since the height is 4 times the radius. Since Terri’s glass is only 3 full. The volume of a cylinder is equal to πr2h. where r is the radius of the cylinder and h is the height of the cylinder. then the length of the rectangle is 3x – 2. e. (x)(3x – 2) = 96. Since the sine of an angle is equal to the side opposite the angle 1  divided by the hypotenuse. d. Since the area of a rectangle is equal to its length times its width. AD  is the side of the right trianAD gle opposite the 30° angle. c. The width of the rectangle is x + 5 units. the volume of this cylinder is equal to πr2(4r) = 256π. d. 8. Since x = 6. which is 3 the volume of cylinder A. If the sides of the square are halved. The area of a square is equal to the length of one side of the square multiplied by itself. where r is the radius of the cylinder and h is the height of the cylinder. If the volume of cylinder A is 1 πr2h. Therefore. a. e. then the length of solid A is 3l and the height of solid A is 2h. and height. the area of one face of the cube is (8)(8) = 64 square units. the length of a side of this square (and edge of the cube) is equal to 9x . 23. if w1 represents the width of solid A and w2 represents the width of solid B. width. Therefore. the height as well). 21. 24. and two faces measure 5 units by 6 units. (6)(12)(w) = 36. The volume of a cube is equal to the product of its length. w is the width of the solid. are all 3x units). the area of any one face of the 154 . The width of the solid is 3 units. A rectangular solid has 6 rectangular faces. then the length of the solid is equal to 2(x + x). Since every edge of a cube is equal in length and the volume of a cube is equal to e3. Therefore. If the width and height of the solid are each 3 in. Since the surface area of the solid is 192 cm2. which is equal to 4x2. width. Therefore. Two faces measure 4 units by 5 units. since (4x2)(4x2)(4x2) = 64x6. 20. then the length of the solid is 2(3 + 3) = 2(6) = 12 in. and height. two faces measure 4 units by w units. or 2(2x) = 4x. the measure of one edge of Stephanie’s cube is equal to the cube root of 64x6. Since every length and width of the cube is 4x2. If x represents the width (and therefore. If w is the width of the solid. e. and h is the height of the solid. the width of solid B. a. and h is the height of the cube. which means that w2. 72w = 36. Since the volumes of the solids are equal. b. the surface area of the solid is equal to 2(4  12) + 2(4  w) + 2(12  w) = 96 + 8w + 24w = 96 + 32w. and h is the height of the solid. which in this case. 6w1 = w2. and height. since (x)(x)(x) = x3. or 3x units. and h is the height of the solid. and two faces measures 12 units by w units. w is the width of the solid. which is equal to 8. the area of any one face of the cube is (4x2)(4x2) = 16x4. d. the measure of one edge of the cube is equal to the cube root of x3. where l is the length of the solid. 18. the volume of the cube is equal to (3x)3 = 27xx cubic units. Since the length. then (3l)(2h)(w1) = (l)(h)(w2). One face of a cube is a square. width. where e is the length of an edge (or lwh. The surface area of a solid is the sum of the areas of each side of the solid. then two faces measure 4 units by 12 units. width. Since the length. and x = 3. e. A cube has six faces. and height of a cube are identical in measure. The volume of a cube is equal to the product of its length. x3 = 27. width. so the total surface area of the cube is equal to (64)(6) = 384 square units. The volume of a rectangular solid is equal to lwh. l is the length of the cube. width of the solid. w = 3. c. and height of a cube are identical in measure. 96 + 32w = 192. w is the width of the cube. the surface area of the solid is equal to 2(4  5) + 2(4  6) +2(5  6) = 2(20) + 2(24) + 2(30) = 40 + 48 + 60 = 148 square units. If l is the length of solid B and h is the height of solid B. Since every length and width of the cube is 8 units. width. The area of one face of the cube is equal to the product of the length and width of that face. and 1 w = 2 units. 4x3 = 108. The area of one face of the cube is equal to the product of the length and width of that face. Since the length. A rectangular solid has 6 rectangular faces. Therefore. 19. The volume of a cube is equal to the product of its length. Therefore. 22. (4x)(x)(x) = 108. 32w = 96. The area of one face of the cube is equal to the product of the length and width of that face. which is equal to x. where l is the length of the solid. and height of a cube are identical in measure. Since every length and width of the cube is x. The volume of a rectangular solid is equal to lwh. The area of a square is equal to the length of one side of the square multiplied by itself.– ANSWERS – 17. The surface area of a solid is the sum of the areas of each side of the solid. the measure of one edge of Danielle’s cube is equal to the cube root of 512. since (8)(8)(8) = 512. c. two faces measure 4 units by 6 units. is equal to 6 times the width of solid A. e. a. the surface area of the solid is equal to 2(x  2x) + 2(x  4x) + 2(2x  4x) = 2(2x2) + 2(4x2) + 2(8x2) = 4x2 + 8x2 + 16x2 = 28x2. e. The area of a circle is equal to πr2. 10. b. Therefore. so the total surface area of the cube is equal to 6x2 square units.  360 (225π) = 9 (225π) = 25π square units. The area of the circle is equal to π(152) = 225π. A rectangular solid has 6 rectangular faces. their measures are equal. where r is the radius of the circle. The area of a circle is equal to πr2. The area of this circle is π(8)2 = 64π square units. 5 = 24π. 6x2 = x3. Therefore. It is a diameter. c. It is given that the surface area of the square is x3 square units. The measure of line CD is (2)(14) = 28 units. Divide both sides by x2. through the center of the circle. x = 60. 8. The area of the circle is now 9 times bigger. 7. and the value of x is 6. where r is the radius of the circle. CD the circle. Therefore. therefore.  6. r = 14. Angles AOC and DOB are vertical angles. Two faces of the solid measure x units by 2x units. The area of the circle is equal to π(122) = 144π. The length of an arc is equal to the circumference of the circle multiplied by the fraction of the circle that the arc covers. angle DOB is also 60°. The area of a circle is equal to πr2. Since the angle of the sector is 40 1   40°.  360 (12π) = 18 (12π) = 3 π units. The surface area of a solid is the sum of the areas of each side of the solid. The circumference of the circle is equal to (2π)(27) = 54π units. The circumference of the circle is equal to (2π)(9) = 18π units. e. The length of arc CB is equal to the size of central angle COB divided by 360. 36 9 9 The area of a sector is equal to the area of the circle multiplied by the fraction of the circle that the sector covers. If the area of the sector is equal to 24π. The length of an arc is equal to the circumference of the circle multiplied by the fraction of the circle that the arc covers. x 2x   then  360 (144π) = 24. If x is the length of the solid. which means 80  that the length of the arc is  360 of the circum80 2   ference of the circle:  360 (54π) = 9 (54π) = 12π units. sector EOD is the sector with an area of 24π square units. the circumference of the 100 5 10    circle is 12π.– ANSWERS – 50 cube is (x)(x) = x2. 2. and the measure of a diameter is twice the measure of a radius. c. the radius of this circle is equal to the square root of 121x. multiplied by the circumference of the circle. b. 155  sector is equal to a fraction of that:  360 (64π) = 5 5 80 (64π) = (16π) = π square units. 11. Therefore. Arc DB measures 60°. then 2x is the height of the solid and 4x is the width of the solid. d. that means the radius of the circle has tripled. and the angle of the sector is x. The circumference of a circle is equal to 2πr. The central angle of arc AE is 80°. 4.  is a line from one side of r2 = 196. The intercepted arc of a central angle is equal in measure to the central angle. which means that 40  the length of the arc is  360 of the circumference 40 1   of the circle:  360 (18π) = 9 (18π) = 2π units. The central angle of arc DB is 40°. a. If the circumference of a circle triples. and two faces measures 2x units by 4x units. two faces measure x units by 4x units. where r is the radius of the circle. Since the radius of the circle is 6 units. the circumference of this circle is equal to (2π)(15) = 30π units. A cube has six faces. 25. The circumference of a circle is 2πr. 3. Therefore. Therefore. The area of a circle is equal to πr2. c. πr2 = 196π. Angle EOD is 60°. Chapter 15 1. to the other side of the circle. 5. the area of Jasmin’s circle is equal to π(9x2)2 = (81x4)π square units. The area of the 9. The area of a sector is equal to the area of the circle multiplied by the fraction of the circle that the sector covers. d. or . 2x = 120. 12. The area of the circle has gone from πr2 to π (3r)2 = π9r2. where r is the radius of the circle. b. where r is the radius of the circle. d. (5)(5) = 25. Therefore. 4x2 – 14x – 14x + 49 = 16x + 9. The area of a sector is equal to the area of the circle multiplied by the fraction of the circle that the sector covers. so the radius of this cir16π cle is 2π = 8 cm. The area of a circle is equal to πr2. Factor this trinomial into 2 identical factors. The measure of central angle DOB is 60°. b. 360 360 18. Therefore. is 60°. Therefore. d. then angle DOB is equal to 180 – (70 + 70) = 180 – 140 = 40°. c. where r is the radius of the circle. Therefore. of a circle. The area of a circle is equal to πr2. Since angle COA is a central angle and CA is its intercepted arc. 15. or two times its original size. if c is 18  the circumference of the circle. and (2x)(5) + (2x)(5) = 20x. 14. If angles D and B are 70°. their measures are equal. π(2x – 7)2 = (16x + 9)π. so the circumference of this circle is 2π(23) = 43π cm. Therefore. since the measure of its intercepted arc. The area of a circle is equal to πr2. Angles COB and COA form a line. they are supplementary angles. the radius of this circle is equal to the square root of (4x2 + 20x + 25). the circumference becomes π2d. where r is the radius of the circle. The circumference of a circle is 2πr. the circumference of a circle is equal to πd. The length of the arc is 180 1  equal to  π) = 2(43π) = 23π cm. to isolate c: (1 8 )xπ = ( 18 ) 360 c. e. arc AC also measures 40°. Therefore. the measure of arc AB is 180°. Therefore. 156 . which means that the sum of angles D and B is 180 – 60 = 120. 360 (43 21. The area of a circle is equal to πr2. since that would make the radius equal to –5. That fraction is x  equal to  the area of the sector is equal to 360 . d. x = 10. where r is the radius of the circle. 4x2 – 44x + 40 = 0. The area of a sector is equal to the area of the circle multiplied by the fraction of the circle that 120  the sector covers. e. where r is the radius of the circle. 23. d. The area of a circle is equal to πr2. since there are 180° in a triangle. x cannot be equal to 1. x2 – 11x + 10 = 0. 2r is equal to the diameter. 360 Multiply both sides of the equation by 1 8 . the measure of angle COA is equal to 180 – 3x. Therefore. then xπ =  360 c. so x x3 2 (πx ) = ()π square units. The radius of this circle is equal (8x + 6) to 2 = 4x + 3. There are 180° in a triangle. The length of an arc is equal to the circumference of the circle multiplied by the fraction of the circle that the arc covers. 16. The circumference of a circle is 2πr. Angles DOB and AOC are vertical angles. so the diameter of this circle is 2(2x + 5) = 4x + 10. or 360 360 18   20. so the area of this circle is (8)2π = 64π cm2. The circumference of a circle is equal to 2πr. Therefore. b. (2x)(2x) = 4x2. Since the radius of a circle is half the diameter of a circle. 17. The diameter of a circle is twice its radius. (x – 1)(x – 10) = 0. 20. 13. the area of this circle is equal to π(4x + 3)2 = π(16x2 + 12x + 12x + 9) = (16x2 + 24x + 9)π square units. Therefore. (4x2 + 20x + 25) = (2x + 5)(2x + 5). e. and the square root of (4x2 + 20x + 25) is 2x + 5. and a radius cannot have a negative length. c. EF. 19. 120  so the area of the sector is equal to  360 (64π) = 1 64 2  (64π) =  π cm . so the circumference of this circle is equal to 2π(11x) = (22x)π units. and the measure of an intercepted arc of a central angle is equal to the measure of the central angle. If the diameter is doubled. d.– ANSWERS – 11x units. That fraction is equal to  360 . the measure of CA is also 180 – 3x. where r is the radius of the circle. 3 3 22. c = 20xπ square units. Since angle AOC is also 40°. A diameter divides a circle into two 180° arcs. The area of a circle is equal to πr2. The radius of a circle is equal to half the diameter of a circle. The circumference of a circle is equal to 2πr. the area of this circle is πx2. the radius of this circle is 12  cm or 23 cm. d. all three sides of the triangle measure 12x.2. b. and the length of a side of the square. 25.  is equal to the length of 4 radii 28. Since every angle in the triangle meas2 ures 60°. since the area of a circle is equal to πr2. Therefore. is 2(2x2) = 4x2. Since the  is the length of 2 radii. The shaded area is the difference between the area of the square and the x2 1 area of the circle: x2 – (4)π = x2 – 4x2π ft. If the area of one semicircle is 4. 26. the length of   is the length of AB is 4(4) = 16 units. which means that the radius of the circle is equal to 1 x ft. Since the area of a circle is equal to πr2. where r is the radius of the circle. its length is 2(4) = 8 units. If the radius of the circle is 2x2. The total area of the figure is equal to the area of the square plus half the area of the circle (since the other half of the circle is within the area of the square): 36π 144 + (2) = 144 + 18π square units. the length of diagonal AD is 102 cm. Therefore. then the radius of the circle is 5 cm. then the diameter of the circle. c.5π square units. the area of this square is equal to (4x2)2 = 32x2. a. The perimeter of the triangle is 12x + 12x + 12x = 36x. where l is the length of the rectangle and w is the width of the rectangle. 29. each angle measures 120  = 60°.5π) = 9π square units. The area of a square is equal to the length of one of its sides squared. and the length of a side of the square. b. The length of AB of the center circle. The shaded area is equal to the area of the square minus half the (8x2π) area of the circle: 32x2 – 2 = 32x2 – 4x2π. The area of a rectangle is lw. c. The area of a rectangle is lw. 27. Since a side of the square is equal to the diameter of the circle. The  is a radius of the cirradius of the circle is 6x. Since these angles are equal. or 4(3) = 12 units. The diagonal of a square is 2 times the length of one of its sides. 30. If the area of the circle is 8x2π. where r 2 is the radius of the circle.2. The area of the circle is equal to 16π square units. If the radius of the circle is 5 cm. where r is the radius of the circle. then the diameter of the circle. Since this is an equilateral triangle. The area of the circle is 62π = 36π square units. the radius of the x center circle is 4. where l is the length of the rectangle and w is the width of the rectangle. The area of a square is equal to the length of one of its sides squared. Therefore. DOB is an equilateral triangle. The area of this circle is 1 x2 equal to (2x)2π = (4)π ft. 157 . a. and it is equal in length to FB. Since the area of a circle is equal to πr2. the area of the square is x2 ft. The length of  is equal to the length of four radii of the cirAB cle placed end to end. the radius of the circle is equal to 9 = 3 units. is 2(5) = 10 cm. Therefore.2. The diameter of the circle is equal to the length of a side of the square. then the area of a whole circle is 2(4. where r is the radius of the circle. Therefore. then the radius of the circle is 2x2. The area of this rectangle is (12)(6) = 72 square units. since the area of a circle is equal to πr2. the cle. the diameter is 12 units and the radius of the circle is 6 units. If the area of the square is 144 square units. where r is the radius of the circle. Subtract the area of the circles from the area of the rectangle to find the size of the shaded area: 128 – 2(16π) = 128 – 32π square units. where r is the radius of the circle. There is one whole circle and two half circles within the rectangle—a total of 2 circles. since the area of a circle is equal to πr2. x (x2π) the area of the circle is equal to π(4)2 = 16 square units. The area of a circle is equal to πr2. the length length of BC  is 2(3) = 6 units. If the area of the circle is 25π cm2. The area of this rectangle is (16)(8) = 128 square units.– ANSWERS – 24. length of side OB of triangle DOB is 2(6x) = 12x. x. The length of AB  is the of BC length of 4 radii. OF  Therefore. Since BC two radii. then the area of one side of the square is 144  = 12 units. a. The midpoint of a line is equal to the average of the x values of the endpoints and the average 0 + 0 –4 + 4 of the y values of the endpoints: (2. The coordinates of point C are (–3. To find the distance between two points.2) = (10. The coordinates of point A are (–5.–2) = (2. 8. To find the distance between two points.–6). The slope of a line is the difference between the y values of two points divided by the difference between –4 – 1 –5   the x values of those two points:  5 – ( –5) = 10 1 = –2. The midpoint of a line is equal to the average of the x values of the endpoints and the average of the y values of –5 + 9 –4 + 2 4 2 the endpoints: (2.6) and the coordinates of point D are (5.) = (1. c.5.1) and the coordinates of point B are (5. use the distance formula: 2 2 D = ((x  2 – x 1) + (y) 2 – y1)  2 +4)) 2) D = ((9 –5)) – ( (2 – (– 2 2 D = (14  + (6) ) D = (196  + 36) D = 232  = 458  = 258  units 15.–4). c.0). To find the distance between two points. b.2).6) and the coordinates of point D are (5.2) = (–4. use the distance formula: 2 2 D = ((x  2 – x 1) + (y) 2 – y1)  2 + (4) 2) D = ((4 ) – 0 – (–4) 2 2 D = (4  + 8) D = (16 4) + 6 D = 80  = 16 5 = 45 units 13.–6). The slope of a line is the difference between the y values of two points divided by the difference 4–6  between the x values of those two points:  7 – (–3) –2 1 = 10 = –5. a. To find the distance between two points. The slope of a line is the difference between the y values of two points divided by the difference between –7 – 8 –15   the x values of those two points:  –1 – 2 = –3 = 5. The coordinates of point A are (–5.2). 2 2 2 11. 9.– ANSWERS –  Chapter 16 1. 2. 10. The slope of a line is the difference between the y values of two points divided by the difference between the x values of those two points: 0 – 5 – (–5)  =  = 0. The coordinates of point C are (–3.2). 6. d. d. 5. The coordinates of point C are (2.–4).0). use the distance formula: D = ((x  )2 + (y2) – y1)2 2 – x1 2 D = ((5 –3)) – ( +– ((–6)  6)2) 2 + ( 2) D = (8 –12) 158 . The midpoint of a line segment is equal to the average of the x values of the endpoints and the average of the y values of the endpoints: 0 +(–8) –8 + 0 –8 –8 (2. use the distance formula: 2 2 D = ((x  2 – x 1) + (y) 2 – y1)  2 + ((– D = ((7 ) – 36) – 8)2) 2 + ( 2) D = (4 –14) D = (16 96) + 1 D = 212  = 453  = 253  units 14. e. d. – 10 –5 –5 3. use the distance formula: D = ((x  )2 + (y2) – y1)2 2 – x1 2 2) D = ((2 –6)) – ( +) (17 – 2 2 + ( 2) D = (8 15) D = (64 25) + 2 D = 289  = 17 units 12. b.8) and the coordinates of point D are (–1. d.–4) and the coordinates of point B are (9.2) = (2.–4) and the coordinates of point B are (9.–7). The slope of a line is the difference between the y values of two points divided by the difference between the x values of those two points: 10 – 2 8  =  = 4.2) = 0 0 (2. The coordinates of point A are (–5. d. d. 7. a.2) = (2.2) = (2.–1). To find the distance between two points. the endpoints: ( 2 .2) = (0. 1 – (– 1) 2 4. The midpoint of a line segment is equal to the average of the x values of the endpoints and the average of the y values of the endpoints: 6 + 15 –4 + 8 21 4 (2. c. The midpoint of a line is equal to the average of the x values of the endpoints and the average of the y values of –3 + 5 6 + (–6) 2 0  ) = (. When an equation is written in slope-intercept form (y = mx + b). 19. Since the graphed equation passes the vertical line test. 4y = 6x – 6 is equivalent to y = 2x – 2. it becomes 4m. 20. it is a function. 21. This leaves  x–6. a. If the slope of line A is multiplied by 4. you must multiply by 4. if the slope 1 of line A is m. 3 6 5y = –3x + 6 is equivalent to y = –5x + 5. c. c. Divide both sides of the equation by 4 to put the given equation in slope-intercept form (y = 3 3 mx + b). The 3 slope of this line is the coefficient of x. the slope of line B is –m. To find the distance between two points. The negative reciprocal of –6 is 6. –2y = –8x + 10 is equivalent to y = 4x – 5. When an equation is written in slope-intercept form (y = mx + b). d. The equation of the other line is y = 1 –3x + b. 17. Therefore. 2πr. Divide both sides of the given equation by –2 to put the equation in slope-intercept form (y = mx + b). e. therefore. To change 1 the original slope of line B. The function extends beyond the line x = 4. multiplied by the measure of the angle formed by the two radii that 72  intercept the arc. e. 1 The negative reciprocal of 4m is –4m. 23. The slope of this line is the coefficient of x. b. if the slope 1 of one line is m.5) to find the y-intercept. use the distance formula: 2 2 D = ((x  2 – x 1) + (y) 2 – y1)  2 + D = ((x –x)) – ( ((–y) – y)2) 2 ) 2) D = ((2x) + (–2y 2+ 2) D = (4x 4y 2 2 D = 4(x y + ) 2 D = 2(x  + y2) 22. Only choice 3 d is a line with a slope of 2. a. Therefore. e. b = 3. 1 1 –4m. 3. the slope of the line is m. The slopes of perpendicular lines are negative reciprocals of each other. ) = (6x + 1. The slopes of perpendicular lines are negative reciprocals of each other. The slopes of perpendicular lines are negative reciprocals of each other. the equation of the 1 16 line is y = –3 + 3. The slope of the line y = 1 1 1 –6x + 8 is –6. 3 The slope of this line is the coefficient of x. The slopes of perpendicular lines are negative reciprocals of each other. 2. . 2 2 2 24. d. The length of an arc is equal to the circumference of the circle. The function could contain more than 8 values for which f(x) = 1 if more of the coordinate plane was visible. where b is the y-intercept of the line. 5 = 1 16 –3(1) + b. Factor the numerator and denominator. f(x).2y + 1). –5. –m. there are values greater than 4 in the domain of the function. Therefore. The negative reciprocal of 4 1 1 is –4. c. The slope of the line y = –2x + 4 is –2. Only choice a is a line with a slope of –2. c. Therefore. divided by 360: 2π(25)( 360 ) = 1 50π(5) = 10π. The slopes of perpendicular lines are negative reciprocals of each other. y – 4 + 3y + 6 12x + 2 4y + 2 ) = (. Divide both sides of the equation by 5 to put the equation in slope-intercept form (y = mx + b). The term (x – 2) is common to the numerator and denominator. Use the point (1. Graph the line y = 1. to this new slope. the slope of the line is m. 1 The product of these slopes is (m)(–m) = –1. is equal to 1. Only choice b is a line with a slope of –4. The graph of this line crosses the graphed equation in 8 places. 18. Parallel lines have identical slopes. 4. so it can be x+6  canceled. The slope of the other 1 line is –3. there are at least 8 different values for which the function. Parallel lines have identical slopes. 2.– ANSWERS – D = (64 44) + 1 D = 208  = 16 13  = 413  units 16. the slope of the other line is –m. 25.  159 Posttest 1. Only choice e is a line with a slope of 6. (x2 + 4x – 12) = (x – 2)(x + 6) and (x2 – 8x + 12) = (x – 2)(x – 6). The midpoint of a line is equal to the average of the x values of the endpoints and the average of 2x + 3 + 10x – 1 the y values of the endpoints: (2 . b. The area of a square is equal to the length of one side squared. 30 > 27. 12. and f are both 180°. Either the value of y is doubled. Since there are 180° in a line and 180° in a triangle. 4y = –3x + 12. so the equation has only one y-intercept. the denominator of the fraction must be doubled in order for the value of the fraction to remain the x 2x same. The midpoint of the line is (6. a + e + c = 180. (0. g = 2πr. The length of a side of the square is equal to the diameter of the circle. e. . a = –1. a = 10. so their measures are equal. Plug each answer choice into the inequality. which is 2r. 2yz. since 3 is the negative reciprocal 3 4 of –4. The only line given that has a slope of 3 is choice e. 13. Since the turning point of y = x2 is (0. Therefore. find the value that makes the x term of the equation equal to 0. express the area of the original circle in terms of g. a + b + c + d + e + f = 360. a2 + 3a – 10 = 12a. However. And. 7. b + c is not equal to e + f. divide the difference of (4r2 – πr2) πr2 the areas by 4: 4 = r2 – 4. To find the turning point of a parabola. Then. The difference between the area of the square and the area of the circle is 4r2 – πr2. a2 – 9a – 10 = 0. a + 1 = 0. 6. The radius in terms of g is equal to g . c. a1a–2a3 = a2. the sum of a. Since the radius of this circle is r. a = f and c = d. 5. First. use that value of x to find the value of y. relative to the graph of y = x2. and c form a line and angles d. represents (2y)z or (2z)y. 3 y = –4x + 3. Factor this quadratic and solve for a: (a + 1)(a – 10) = 0. To find the midpoint of the line segment. Cross multiply: (a + 5)(a – 2) = (3)(4a). a.3). the area of the square is equal to (2r)(2r) = 4r2. Since the area is tripled. Angles b and e are vertical angles. since d + e + f = 180. 24 + 6 > 12 + 15. Therefore.0). the slope of a line perpendicular to y = 3 4 4 –4x + 3 is 3x. rewrite the equation in slope-intercept form (y = mx + b). since a + b + c = 180 and b = e. the new area of the 3g2 circle is equal to 3 times this quantity: 4π. since those are pairs of alternating angles. b.– ANSWERS – 4. a – 10 = 0. the entire graph) has been shifted 2 units right and 2 units down. The x term of y = (x – 2)2 – 2 will be 0 when x = 2. If that were the case. the turning point of the graph of y = (x – 2)2 – 2 (and therefore. since (2 – 2)2 = 0. When multiplying like bases. Angles a. b. This area represents the area between the circle and the square. d. The graphed equation crosses the y-axis at only one point. 8. Although b = e. making the coordinates of the turning point (2. the denominator would be equal to (2y)(2z) = 4yz. There are many y values between –2 and 1. choice d is correct. keep the base and add the exponents: Since 1 + (–2) + 3 = 2. where r is the radius of the circle. and c and the sum of d.–2). a + c = 180 – b. d. The area of a circle is equal to πr2. If the numerator of a fraction is doubled. Also. First. The slopes of perpendicular lines are negative reciprocals of each other. 9. and the value of the fraction would be halved. 4(6) + 6 > 3(4) + 15. e. find the average of the x-coordinates and the average of the y-coordinates of the endpoints of the line (–1 + 13) 12 (4 + 12) 16 segment: 2 = 2 = 6. e. d. c is not equal to f. a + c = 180 – e. Therefore. 2 = 2 = 8. or the value of z is doubled. The y-coordinate of that point is equal to (2 – 2)2 – 2 = –2. 11. Since only one of these 10. Notice that both values are not doubled. Since –1 + 2 + (–3) = –2. d. yz = 2yz . b–1b2b–3 = b–2. Therefore. e. 160 four regions is shaded. Since a + b + c = 180. The denominator.8). and f are the angles of a triangle. as the range of the function shown extends from –3 to 4. a. the area of this circle is πr2. 4y + 3x = 12. The area of a circle is equal to πr2. where r is the radius of the circle. b. Since the inequality 30 > 27 is true. Replace r 2π g g with 2π. (a2)  (a2)(b–2) =  (b2) . The circumference of a circle is equal to 2πr. The area of the circle is equal to: π(2π)2 2 g = 4π. 12x4 + 21x3 – 3x2 2  = 4x + 7x – 1. 14 If a radius is drawn to the point where a tangent touches a circle. 3x2 = 7x. The 2 square of 4 is 16. 210 The area of a sector of a circle is equal to the area of a circle. 36 2 2 (8π)(8π) = r . relative to 1 9 DEF. since the tangent is equal to the length of the opposite base divided by the length of the adjacent base. –15b + 10 + 6b = –8. Therefore.5 Substitute –4 for x:  8–4 8–4 12 – 1. Therefore. Since the length of each  is 102 units. Now. 6 17.280 can fit in the box: 4 0 = 32. 16 19. Since the 7 1 1 1  cosine of 60° = 2. 2 = EO . Since  AO is opposite the 30° angle of the triangle. The cube root 3 1 of a2= a2 and the cube root of 512 is equal 1 to 8. 3x2 = –1. The area of a rectangle is equal to its length multiplied by its width. and segment EA of tangent CD form a right triangle. The hypotenuse of an isosceles right triangle is equal to 2 times the length of one of the bases. b = 2. 8 18. The volume of the box is equal to (16)(10)(8) = 1. is 2(3b)(3h) = 2bh. πr2.– ANSWERS – 14. Therefore. Divide each term in the numerator by 3x2. triangle ABC is an isosceles right triangle. EO.000 is equal to: 8 0 = 25 ft. The hypotenuse of a 30-60-90 right triangle is twice the length of the shorter base.  EO = 14. w is the width. If each side of ABC is three times the length of its corresponding side of DEF. Divide the volume of the box by the volume of one book to find how many books 1. the length of AC 15. 8π = ( 360 )πr . –9b = –18.5 = 10. . the area of DEF is equal to one-ninth of 72 the area of ABC: 9 = 8 square units. Alternatively. its measure is equal to the measure of its intercepted arc. or nine times the 1 size of the area of DEF (2bh). Substitute this value for a in the second equation: 5(–3b + 2) + 6b = –8.000. where l is the length of the prism. secant EO. (–4)2 + 3(–4) 16 – 12  =  = 21. a = –3b + 2. 23. The volume of a rectangular prism is equal to lwh. divided by the hypotenuse. 32 First. Therefore. 10. find the volume of one book. The cosine of angle O is equal to the adjacent side. 2 Use the first equation to write a in terms of b: –3a – 9b = –6. Therefore. 22. Divide the coefficient of each term by 3. 36 = r . then the area of ABC. 20. First. where b is the base of the triangle and h is the height. it is the shorter base of the triangle. square both sides of the 1 equation to find the value of a: a(2)(2) = 82.that means the lengths of the bases of the triangle are equal. The volume of one book is equal to: (8)(5)(1) = 40 cubic inches. multiplied by the angle formed by the two radii of the sector 80 2  divided by 360. then the width of the court 161 2.5. 643 = 16. b. The cube root of 64 is 4. so the length of EO is equal to (2)(7) = 14.280 cubic inches. base is 10 units. therefore. if the length of the court is 80 and the area of the court is 2. substitute the value of a into 2 2 the second expression. 12x4 21x3 –3x2 2     3 x2 = 4x . Radius AO. a3= 643. and h is the height. (2)EO = 7. If the tangent of an angle in a right triangle is 1. Since angle AOB is a central angle. take the cube (third) root of both sides of the given equation. c. r = 6. Since a2 = 8. AO. and subtract 2 from the exponent of each x term. the measure of angle AOB is 60°. The perimeter of a rectangle is equal to twice its length plus twice its width. you can recognize right triangle EAO is a 30-60-90 right triangle. 2 16. a = 64. Therefore. a right angle is formed. Therefore. 1 The area of a triangle is equal to 2bh. the perimeter of the court is equal to: (2)(80) + (2)(25) = 160 + 50 = 210 ft. since (4)(4)(4) = 64. –3a = 9b – 6. . The exponent of 2 is equal to one less than the position of the term in the sequence. (x2 + 6x – 27) = (x + 9)(x – 3). the sixth term is (96)(2) = 192. Therefore. each term is equal to 6 times a power of 2. The positive value of x that makes the expression undefined is 3. x = 3. x – 3 = 0. x + 9 = 0. the 25. Alternatively. Therefore. 384 Each term is twice the previous term. and the seventh term is (192)(2) = 384. the fifth term in the pattern is (48)(2) = 96.– ANSWERS – 24. the second term is equal to 6(21) = 12. The first term is equal to (6)(20) = 6. 162 3 seventh term is equal to (6)(27 – 1) = (6)(26) = (6)(64) = 384. The expression is undefined when the denominator is equal to 0. x = –9. Factor the denominator to find the values of x that make the denominator equal to 0. – NOTES – . – NOTES – . – NOTES – . – NOTES – . . com to sign in. Start practicing your SAT math skills online right away! Once you’ve logged on.LearningExpressFreeOffer. Use your individualized access code found on the scratch card and go to www. you have FREE access to: ■ ■ ■ ■ 54 practice questions covering ALL types of math questions found on the SAT including algebra I and II. 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