name date period 10-2 study guide and …...lesson 10-2 pdf pass chapter 10 15 glencoe algebra 1...

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Chapter 10 11 Glencoe Algebra 1

Study Guide and InterventionSimplifying Radical Expressions

Product Property of Square Roots The Product Property of Square Roots and prime factorization can be used to simplify expressions involving irrational square roots. When you simplify radical expressions with variables, use absolute value to ensure nonnegative results.

Simplify √ �� 180 .

√ �� 180 = √ �� 2 � 2 � 3 � 3 � 5 Prime factorization of 180

= √ � 22 � √ � 32 � √ � 5 Product Property of Square Roots

= 2 � 3 � √ � 5 Simplify.

= 6 √ � 5 Simplify.

Simplify √ �� 120a2 · b5 · c4 .

√ �� 120a2 � b5 � c4 = √ �� 23 � 3 � 5 � a2 � b5 � c4

= √ � 22 � √ � 2 � √ � 3 � √ � 5 � √ � a2 � √ �� b4 � b � √ � c4

= 2 � √ � 2 � √ � 3 � √ � 5 � ⎪a⎥ � b2 � √ � b � c2

= 2 ⎪a⎥ b2c2 √ �� 30b

ExercisesSimplify each expression.

1. √ �� 28 2. √ �� 68 3. √ �� 60 4. √ �� 75

5. √ �� 162 6. √ � 3 · √ � 6 7. √ � 2 · √ � 5 8. √ � 5 · √ �� 10

9. √ �� 4a2 10. √ �� 9x4 11. √ �� 300a4 12. √ �� 128c6

13. 4 √ �� 10 � 3 √ � 6 14. √ �� 3x2 � 3 √ �� 3x4 15. √ �� 20a2b4 16. √ �� 100x3y

17. √ �� 24a4b2 18. √ �� 81x4y2 19. √ �� 150a2b2c

20. √ �� 72a6b3c2 21. √ �� 45x2y5z8 22. √ �� 98x4y6z2

10-2

Product Property of Square Roots For any numbers a and b, where a ≥ 0 and b ≥ 0, √ �� ab = √ � a � √ � b .

Example 1

Example 2

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Study Guide and Intervention (continued)

Simplifying Radical Expressions

Quotient Property of Square Roots A fraction containing radicals is in simplest form if no radicals are left in the denominator. The Quotient Property of Square Roots and rationalizing the denominator can be used to simplify radical expressions that involve division. When you rationalize the denominator, you multiply the numerator and denominator by a radical expression that gives a rational number in the denominator.

Simplify √ �� 56 − 45

.

√ �� 56 − 45

= √ �� 4 � 14 − 9 � 5

= 2 � √ �� 14 −

3 � √ �� 15 Simplify the numerator and denominator.

= 2 √ �� 14 −

3 √ � 5 �

√ � 5 −

√ � 5 Multiply by

√ � 5 −

√ � 5 to rationalize the denominator.

= 2 √ �� 70 −

15 Product Property of Square Roots


1. √ � 9

− √ � 18

2. √ � 8

− √ � 24

3. √ �� 100

− √ �� 121

4. √ � 75

− √ � 3

5. 8 √ � 2 −

2 √ � 8 6. √ � 2 −

5 � √ � 6 −

5

7. √ � 3 − 4 � √ � 5 −

2

8. √ � 5 − 7 � √ � 2 −

5

9. √ �� 3a2 −

10b6 10. √ � x

6 −

y4

11. √ �� 100a4 −

144b8 12. √ �� 75b3c6

− a2

13. √ � 4 −

3 - √ � 5 14. √ � 8

− 2 + √ � 3

15. √ � 5 −

5 + √ � 5 16. √ � 8

− 2 √ � 7 + 4 √ � 10

10-2

Quotient Property of Square Roots For any numbers a and b, where a ≥ 0 and b > 0, √ � a − b =

√ � a − √ � b

.

Example

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Word Problem PracticeSimplifying Radical Expressions

1. SPORTS Jasmine calculated the height of her team’s soccer goal to be 15−

√ ⎯⎯

3 feet.

Simplify the expression.

2. NATURE In 2004, an earthquake below the ocean floor initiated a devastating tsunami in the Indian Ocean. Scientists can approximate the velocity (in feet per second) of a tsunami in water of depth d (in feet) with the formula V = √ �� 16d . Determine the velocity of a tsunami in 300 feet of water. Write your answer in simplified radical form.

3. AUTOMOBILES The following formula can be used to find the “zero to sixty” time for a car, or the time it takes for a car to accelerate from a stop to sixty miles per hour.

V = √ �� 2PT − M

V is the velocity (in meters per second).P is the car’s average power (in watts).M is the mass of the car (in kilograms).T is the time (in seconds).

Find the time it takes for a 900-kilogram car with an average 60,000 watts of power to accelerate from stop to 26.82 meters per second (60 miles per hour). Round your answer to the nearest tenth.

4. PHYSICAL SCIENCE When a substance such as water vapor is in its gaseous state, the volume and the velocity of its molecules increase as temperature increases. The average velocity V of a molecule with mass m at temperature T

is given by the formula V = √��3kT−m .

Solve the equation for k.

5. GEOMETRY Suppose Emeryville Hospital wants to build a new helipad on which medic rescue helicopters can land. The helipad will be circular and made of fire resistant rubber.

a. If the area of the helipad is A, write an equation for the radius r.

b. Write an expression in simplified radical form for the radius of a helipad with an area of 288 square meters.

c. Using your calculator, find a decimal approximation for the radius. Round your answer to the nearest hundredth.

r

10-2


Chapter 10 Resource Masters

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CONSUMABLE WORKBOOKS Many of the worksheets contained in the Chapter Resource Masters booklets are available as consumable workbooks in both English and Spanish.

ISBN10 ISBN13Study Guide and Intervention Workbook 0-07-890835-3 978-0-07-890835-4Homework Practice Workbook 0-07-890836-1 978-0-07-890836-1Homework Practice Workbook (Spanish) 0-07-890840-X 978-0-07-890840-8

ANSWERS FOR WORKBOOKS The answers for Chapter 10 of these workbooks can be found in the back of this Chapter Resource Masters booklet.

StudentWorks Plus™ This CD-ROM includes the entire Student Edition text along with the English workbooks listed above.

TeacherWorks Plus™ All of the materials found in this booklet are included for viewing, printing, and editing in this CD-ROM.

Spanish Assessment Masters (ISBN10: 0-07-890843-4, ISBN13: 978-0-07-890843-9) These masters contain a Spanish version of Chapter 10 Test Form 2A and Form 2C.

Copyright © by The McGraw-Hill Companies, Inc. All rights reserved. Permission is granted to reproduce the material contained herein on the condition that such materials be reproduced only for classroom use; be provided to students, teachers, and families without charge; and be used solely in conjunction with the Glencoe Algebra1 program. Any other reproduction, for sale or other use, is expressly prohibited.

Send all inquiries to:Glencoe/McGraw-Hill8787 Orion PlaceColumbus, OH 43240 - 4027

ISBN: 978-0-07-890504- 9MHID: 0-07-890504- 4

Printed in the United States of America.

1 2 3 4 5 6 7 8 9 10 024 14 13 12 11 10 09 08

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Teacher’s Guide to Using the Chapter 10 Resource Masters .........................................iv

Chapter Resources Chapter 10 Student-Built Glossary .................... 1Chapter 10 Anticipation Guide (English) ........... 3Chapter 10 Anticipation Guide (Spanish) .......... 4

Lesson 10-1Square Root Functions Study Guide and Intervention ............................ 5Skills Practice .................................................... 7Practice ............................................................ 8Word Problem Practice ..................................... 9Enrichment ...................................................... 10

Lesson 10-2Simplifying Radical ExpressionsStudy Guide and Intervention .......................... 11Skills Practice .................................................. 13Practice .......................................................... 14Word Problem Practice ................................... 15Enrichment ...................................................... 16

Lesson 10-3Operations with Radical ExpressionsStudy Guide and Intervention .......................... 17Skills Practice .................................................. 19Practice .......................................................... 20Word Problem Practice ................................... 21Enrichment ...................................................... 22

Lesson 10-4Radical EquationsStudy Guide and Intervention .......................... 23Skills Practice .................................................. 25Practice .......................................................... 26Word Problem Practice ................................... 27Enrichment ...................................................... 28Graphing Calculator Activity ............................ 29

Lesson 10-5The Pythagorean TheoremStudy Guide and Intervention .......................... 30Skills Practice .................................................. 32Practice .......................................................... 33Word Problem Practice ................................... 34Enrichment ...................................................... 35Spreadsheet Activity ........................................ 36

Lesson 10-6The Distance and Midpoint FormulasStudy Guide and Intervention .......................... 37Skills Practice .................................................. 39Practice .......................................................... 40Word Problem Practice ................................... 41Enrichment ...................................................... 42

Lesson 10-7Similar TrianglesStudy Guide and Intervention .......................... 43Skills Practice .................................................. 45Practice .......................................................... 46Word Problem Practice ................................... 47Enrichment ...................................................... 48

Lesson 10-8Trigonometric RatiosStudy Guide and Intervention .......................... 49Skills Practice .................................................. 51Practice .......................................................... 52Word Problem Practice ................................... 53Enrichment ...................................................... 54

AssessmentStudent Recording Sheet ............................... 55Rubric for Scoring Extended Response .......... 56Chapter 10 Quizzes 1 and 2 ........................... 57Chapter 10 Quizzes 3 and 4 ........................... 58Chapter 10 Mid-Chapter Test .......................... 59Chapter 10 Vocabulary Test ........................... 60Chapter 10 Test, Form 1 ................................. 61Chapter 10 Test, Form 2A ............................... 63Chapter 10 Test, Form 2B ............................... 65Chapter 10 Test, Form 2C .............................. 67Chapter 10 Test, Form 2D .............................. 69Chapter 10 Test, Form 3 ................................. 71Chapter 10 Extended Response Test ............. 73Standardized Test Practice ............................. 74

Answers ........................................... A1–A33

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Teacher’s Guide to Using the Chapter 10 Resource Masters

The Chapter 10 Resource Masters includes the core materials needed for Chapter 10. These materials include worksheets, extensions, and assessment options. The answers for these pages appear at the back of this booklet.

All of the materials found in this booklet are included for viewing and printing on the TeacherWorks PlusTM CD-ROM.

Chapter Resources

Student-Built Glossary (pages 1–2) These masters are a student study tool that presents up to twenty of the key vocabulary terms from the chapter. Students are to record definitions and/or examples for each term. You may suggest that students highlight or star the terms with which they are not familiar. Give this to students before beginning Lesson 10-1. Encourage them to add these pages to their mathematics study notebooks. Remind them to complete the appropriate words as they study each lesson.

Anticipation Guide (pages 3–4) This master, presented in both English and Spanish, is a survey used before beginning the chapter to pinpoint what students may or may not know about the concepts in the chapter. Students will revisit this survey after they complete the chapter to see if their perceptions have changed.

Lesson ResourcesStudy Guide and Intervention These masters provide vocabulary, key concepts, additional worked-out examples and Check Your Progress exercises to use as a reteaching activity. It can also be used in conjunction with the Student Edition as an instructional tool for students who have been absent.

Skills Practice This master focuses more on the computational nature of the lesson. Use as an additional practice option or as homework for second-day teaching of the lesson.

Practice This master closely follows the types of problems found in the Exercises section of the Student Edition and includes word problems. Use as an additional practice option or as homework for second-day teaching of the lesson.

Word Problem Practice This master includes additional practice in solving word problems that apply the concepts of the lesson. Use as an additional practice or as homework for second-day teaching of the lesson.

Enrichment These activities may extend the concepts of the lesson, offer an historical or multicultural look at the concepts, or widen students’ perspectives on the mathematics they are learning. They are written for use with all levels of students.

Graphing Calculator, TI-Nspire, or Spreadsheet Activities These activities present ways in which technology can be used with the concepts in some lessons of this chapter. Use as an alternative approach to some concepts or as an integral part of your lesson presentation.

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Assessment OptionsThe assessment masters in the Chapter 10 Resource Masters offer a wide range of assessment tools for formative (monitoring) assessment and summative (final) assessment.

Student Recording Sheet This master corresponds with the standardized test practice at the end of the chapter.

Extended Response Rubric This master provides information for teachers and students on how to assess performance on open-ended questions.

Quizzes Four free-response quizzes offer assessment at appropriate intervals in the chapter.

Mid-Chapter Test This 1-page test provides an option to assess the first half of the chapter. It parallels the timing of the Mid-Chapter Quiz in the Student Edition and includes both multiple-choice and free-response questions.

Vocabulary Test This test is suitable for all students. It includes a list of vocabulary words and 11 questions to assess students’ knowledge of those words. This can also be used in conjunction with one of the leveled chapter tests.

Leveled Chapter Tests

• Form 1 contains multiple-choice questions and is intended for use with below grade level students.

• Forms 2A and 2B contain multiple-choice questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Forms 2C and 2D contain free-response questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Form 3 is a free-response test for use with above grade level students.

All of the above mentioned tests include a free-response Bonus question.

Extended-Response Test Performance assessment tasks are suitable for all students. Samples answers and a scoring rubric are included for evaluation.

Standardized Test Practice These three pages are cumulative in nature. It includes three parts: multiple-choice questions with bubble-in answer format, griddable questions with answer grids, and short-answer free-response questions.

Answers• The answers for the Anticipation Guide

and Lesson Resources are provided as reduced pages with answers appearing in red.

• Full-size answer keys are provided for the assessment masters.

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Student-Built Glossary

This is an alphabetical list of the key vocabulary terms you will learn in Chapter 10. As you study the chapter, complete each term’s definition or description. Remember to add the page number where you found the term. Add these pages to your Algebra Study Notebook to review vocabulary at the end of the chapter.

Vocabulary TermFound

on PageDefinition/Description/Example

conjugateKAHN • jih • guht

converse

cosine

Distance Formula

hypotenusehy • PAH • tn • oos

inverse cosine

inverse sine

inverse tangent

legs

(continued on the next page)

10


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Student-Built Glossary (continued)10

Vocabulary TermFound

on PageDefinition/Description/Example

midpoint

Midpoint Formula

radical equations

radical functions

radicandRA • duh • KAND

similar triangles

tangent

trigonometry


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Anticipation GuideRadical Expressions and Triangles

Before you begin Chapter 10

• Read each statement.

• Decide whether you Agree (A) or Disagree (D) with the statement.

• Write A or D in the first column OR if you are not sure whether you agree or disagree, write NS (Not Sure).

STEP 1A, D, or NS

StatementSTEP 2A or D

1. An expression that contains a square root is called a radical expression.

2. It is always true that √ � xy will equal √ � x · √ � y .

3. 1 − √ � 3

is in simplest form because √ � 3 is not a whole number.

4. The sum of 3 √ � 3 and 2 √ � 3 will equal 5 √ � 3 .

5. Before multiplying two radical expressions with different radicands the square roots must be evaluated.

6. When solving radical equations by squaring each side of the equation, it is possible to obtain solutions that are not solutions to the original equation.

7. The longest side of any triangle is called the hypotenuse.

8. Because 52 = 42 + 32, a triangle whose sides have lengths 3, 4, and 5 will be a right triangle.

9. On a coordinate plane, the distance between any two points can be found using the Pythagorean Theorem.

10. The Distance Formula cannot be used to find the distance between two points on the same vertical line.

11. Two triangles are similar only if their corresponding angles are congruent and the measures of their corresponding sidesare in proportion.

12. All right triangles are similar.

After you complete Chapter 10

• Reread each statement and complete the last column by entering an A or a D.

• Did any of your opinions about the statements change from the first column?

• For those statements that you mark with a D, use a piece of paper to write an example of why you disagree.

10

Step 1

Step 2


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Ejercicios preparatoriosExpresiones radicales y triángulos

Antes de comenzar el Capítulo 10

• Lee cada enunciado.

• Decide si estás de acuerdo (A) o en desacuerdo (D) con el enunciado.

• Escribe A o D en la primera columna O si no estás seguro(a) de la respuesta, escribe NS (No estoy seguro(a)).

PASO 1A, D o NS

EnunciadoPASO 2A o D

1. Una expresión que contiene una raíz cuadrada se denomina expresión radical.

2. Siempre es verdadero que √ � xy será igual a √ � x · √ � y .

3. 1 − √ � 3

está en forma reducida porque √ � 3 no es un número entero.

4. La suma de 3 √ � 3 y 2 √ � 3 será igual a 5 √ � 3 .

5. Antes de multiplicar dos expresiones radicales con radicandosdiferentes, se debe evaluar las raíces cuadradas.

6. Cuando se resuelven ecuaciones radicales mediante la elevación al cuadrado de cada lado de la ecuación, es posible obtener soluciones que no son soluciones para la ecuación original.

7. El lado más largo de cualquier triángulo se llama hipotenusa.

8. Debido a que 52 = 42 + 32, un triángulo cuyos lados tienenlongitudes 3, 4 y 5 será un triángulo rectángulo.

9. En el plano de coordenadas, la distancia entre cualesquierados puntos se puede encontrar usando el teorema de Pitágoras.

10. No se puede usar la fórmula de la distancia para calcular ladistancia entre dos puntos en la misma recta vertical.

11. Dos triángulos son semejantes solo si sus ángulos correspondientes son congruentes y las medidas de sus ladoscorrespondientes están en proporción.

12. Todos los triángulos rectángulos son semejantes.

Después de completar el Capítulo 10

• Vuelve a leer cada enunciado y completa la última columna con una A o una D.

• ¿Cambió cualquiera de tus opiniones sobre los enunciados de la primera columna?

• En una hoja de papel aparte, escribe un ejemplo de por qué estás en desacuerdo con los enunciados que marcaste con una D.

Paso 1

Paso 2

10

Capítulo 12 4 Álgebra 1 de Glencoe


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Dilations of Radical Functions A square root function contains the square root of a variable. Square root functions are a type of radical function.In order for a square root to be a real number, the radicand, or the expression under the radical sign, cannot be negative. Values that make the radicant negative are not included in the domain.

Square Root Function

Parent function: f(x) = √ � x

Type of graph: curve

Domain: {x|x ≥ 0}

Range: {y|y ≥ 0}

Graph y = 3 √�x . State the domain and range.

Step 1 Make a table. Choose nonnegative Step 2 Plot points and draw a values for x. smooth curve.

x y

0 0

0.5 ≈ 2.12

1 3

2 ≈ 4.24

4 6

6 ≈ 7.35

y

x

y = 3 x

The domain is {x|x ≥ 0} and the range is {y|y ≥ 0}.

ExercisesGraph each function, and compare to the parent graph. State the domain and range.

1. y = 3 − 2 √ � x 2. y = 4 √ � x 3. y = 5 −

2 √ � x

y

x

y

x

y

x

10-1 Study Guide and InterventionSquare Root Functions

Example

y

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y = x


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Reflections and Translations of Radical Functions Radical functions, like quadratic functions, can be translated horizontally and vertically, as well as reflected across the x-axis. To draw the graph of y = a √ �� x + h , follow these steps.

Graphs of Square Root

Functions

Step 1 Draw the graph of y = +c √ ⎯⎯ x . The graph starts at the origin and passes through the point at (1, a). If a > 0, the graph is in the 1st quadrant. If a < 0, the graph is reflected across the x-axis and is in the 4th quadrant.

Step 2 Translate the graph ⎪c⎥ units up if c is positive and down if c is negative.

Step 3 Translate the graph ⎪h⎥ units left if h is positive and right if h is negative.

Graph y = - √ �� x + 1 and compare to the parent graph. State the domain and range.

Step 1 Make a table of values.

x -1 0 1 3 8

y 0 -1 -1.41 -2 -3

Step 2 This is a horizontal translation 1 unit to the left of the parent function and reflected across the x-axis. The domain is {x | x ≥ 0} and the range is {y | y ≤ 0}.

ExercisesGraph each function, and compare to the parent graph. State the domain and range.

1. y = √ � x + 3 2. y = √ �� x - 1 3. y = - √ �� x - 1

y

x

y

x

y

x

10-1 Study Guide and Intervention (continued)

Square Root Functions

Example

y

x

y = x

y = - x + 1


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Graph each function, and compare to the parent graph. State the domain and range.

1. y = 2 √ � x 2. y = 1 − 2 √ � x 3. y = 5 √ � x

4. y = √ � x + 1 5. y = √ � x - 4 6. y = √ �� x - 1

7. y = - √ �� x - 3 8. y = √ �� x - 2 + 3 9. y = - 1 − 2 √ �� x - 4 + 1

10-1 Skills Practice Square Root Functions

y

x

y

x

y

x

12

8

4

−2 2 4

y

x

y

x

y

x

y

x

y

x

y

x


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Graph each function, and compare to the parent graph. State the domain and range.

1. y = 4 − 3 √ � x 2. y = √ � x + 2 3. y = √ �� x - 3

y

x

y

x

y

x

4. y = - √ � x + 1 5. y = 2 √ ⎯⎯⎯⎯⎯

x - 1 + 1 6. y = - √ ⎯⎯⎯⎯⎯

x - 2 + 2

y

x

y

x

yx

7. OHM’S LAW In electrical engineering, the resistance of a circuit

can be found by the equation I = √ ⎯⎯ P −

R , where I is the current in

amperes, P is the power in watts, and R is the resistance of the circuit in ohms. Graph this function for a circuit with a resistance of 4 ohms.

10-1 PracticeSquare Root Functions

Curr

ent (

ampe

res)

2

3

1

0 20

4

5

Power (watts)40 60 80 100


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10-1 Word Problem PracticeSquare Root Functions

1. PENDULUM MOTION The period T of a pendulum in seconds, which is the time for the pendulum to return to the point of release, is given by the equation T = 1.11 √�L . The length of the pendulum in feet is given by L. Graph this function.

2. EMPIRE STATE BUILDING The roof of the Empire State Building is 1250 feet above the ground. The velocity of an object dropped from a height of h meters is given by the function V = √ �� 2gh , where g is the gravitational constant, 32.2 feet per second squared. If an object is dropped from the roof of the building, how fast is it traveling when it hits the street below?

3. ERROR ANALYSIS Gregory is drawing the graph of y = -5 √ �� x + 1 . He describes the range and domain as {x � x ≥ -1}, { y � y ≥ 0}. Explain and correct the mistake that Gregory made.

4. CAPACITORS A capacitor is a set of plates that can store energy in an electric field. The voltage V required to store E joules of energy in a capacitor with a capacitance of C farads is given

by V = √⎯⎯⎯2E−C .

a. Rewrite and simplify the equation for the case of a 0.0002 farad capacitor.

b. Graph the equation you found in part a.

Volta

ge (v

olts

)

100

150

50

0 2

200

250

300

350

Energy (joules)

4 6 8 10

c. How would the graph differ if you wished to store E + 1 joules of energy in the capacitor instead?

d. How would the graph differ if you applied a voltage of V + 1 volts instead?

Perio

d (s

ec)

2

3

1

0 4

4

5

Pendulum Length (ft)8 12 16 20


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A cubic root function contains the cubic root of a variable. The cubic root of a number x are the numbers y that satisfy the equation y · y · y = x (or, alternatively, y = 3 √ � x ).Unlike square root functions, cubic root functions return real numbers when the radicand is negative.

Graph y = 3 √ � x .

Step 1 Make a table. Step 2 Plot points and draw a smooth curve. x y

−5 −1.71

−3 −1.44

−1 −1

0 0

1 1

3 1.44

5 1.71

y

x

ExercisesGraph each function, and compare to the parent graph.

1. y = 2 3 √ � x 2. y = 3 √ � x + 1 3. y = 3 √ �� x + 1

y

x

y

x

y

x

4. y = 3 √ �� x - 1 + 2 5. y = 3 3 √ �� x - 2 6. y = - 3 √ � x + 3

y

x

y

x

y

x

10-1 Enrichment Cubic Root Functions

Example


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Skills Practice Simplifying Radical Expressions

Simplify each expression.

1. √ �� 28 2. √ �� 40

3. √ �� 72 4. √ �� 99

5. √ � 2 · √ �� 10 6. √ � 5 · √ �� 60

7. 3 √ � 5 · √ � 5 8. √ � 6 · 4 √ �� 24

9. 2 √ � 3 · 3 √ �� 15 10. √ �� 16b4

11. √ �� 81a2d4 12. √ �� 40x4y6

13. √ �� 75m5P2 14. √ � 5 − 3

15. √ � 1 − 6

16. √ � 6 − 7 · √ � 1 −

3

17. √ �� q −

12

18. √ �� 4h −

5

19. √ �� 12 − b2

20. √ �� 45 − 4m4

21. 2 − 4 + √ � 5

22. 3 − 2 - √ � 3

23. 5 − 7 + √ � 7

24. 4 − 3 - √ � 2

10-2


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Practice Simplifying Radical Expressions

Simplify.

1. √ � 24 2.

√ � 60

3. √ �� 108 4. √ � 8 � √ � 6

5. √ � 7 � √ � 14 6. 3 √ � 12 � 5 √ � 6

7. 4 √ � 3 � 3 √ � 18 8. √ �� 27tu3

9. √ �� 50p5 10. √ �� 108x6y4z5

11. √ �� 56m2n4p5 12. √ � 8

− √ � 6

13. √ � 2 − 10

14. √ � 5 −

32

15. √ � 3 − 4 � √ � 4 −

5

16. √ � 1 − 7 � √ � 7 −

11

17. √ � 3k

− √ � 8

18. √ � 18 − x3

19. √ ��

4y −

3y2 20. √ �� 9ab −

4ab4

21. 3 − 5 - √ � 2

22. 8 − 3 + √ � 3

23. 5 − √ � 7 + √ � 3

24. 3 √ � 7 −

-1 - √ � 27

25. SKY DIVING When a skydiver jumps from an airplane, the time t it takes to free fall a

given distance can be estimated by the formula t = √ �� 2s − 9.8

, where t is in seconds and s is

in meters. If Julie jumps from an airplane, how long will it take her to free fall 750 meters?

26. METEOROLOGY To estimate how long a thunderstorm will last, meteorologists can use

the formula t = √ �� d3 −

216 , where t is the time in hours and d is the diameter of the storm in

miles.

a. A thunderstorm is 8 miles in diameter. Estimate how long the storm will last. Give your answer in simplified form and as a decimal.

b. Will a thunderstorm twice this diameter last twice as long? Explain.

10-2


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Enrichment

Squares and Square Roots From a Graph

The graph of y = x2 can be used to find the squares and square roots of numbers.To find the square of 3, locate 3 on the x-axis. Then find its corresponding value on the y-axis.The arrows show that 32 = 9.To find the square root of 4, first locate 4 on the y-axis. Then find its corresponding value on the x-axis. Following the arrows on the graph, you can see that √

⎯⎯ 4 = 2.

A small part of the graph at y = x2 is shown below. A 1:10 ratio for unit length on the y-axis to unit length on the x-axis is used.

Find √ �� 11 .

The arrows show that √ ⎯⎯⎯

11 = 3.3 to the nearest tenth.

ExercisesUse the graph above to find each of the following to the nearest whole number.

1. 1.52 2. 2.72 3. 0.92

4. 3.62 5. 4.22 6. 3.92

Use the graph above to find each of the following to the nearest tenth.

7. √ ⎯⎯⎯

15 8. √ ⎯⎯

8 9. √ ⎯⎯

3

10. √ ⎯⎯

5 11. √ ⎯⎯⎯

14 12. √ ⎯⎯⎯

17

1 2

1110

20

3 43.3O x

y

x

y

O

10-2

Example


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Study Guide and InterventionOperations with Radical Expressions

Add or Subtract Radical Expressions When adding or subtracting radical expressions, use the Associative and Distributive Properties to simplify the expressions. If radical expressions are not in simplest form, simplify them.

Simplify 10 √ � 6 - 5 √ � 3 + 6 √ � 3 - 4 √ � 6 .

10 √ � 6 - 5 √ � 3 + 6 √ � 3 - 4 √ � 6 = (10 - 4) √ � 6 + (-5 + 6) √ � 3 Associative and Distributive Properties

= 6 √ � 6 + √ � 3 Simplify.

Simplify 3 √ �� 12 + 5 √ �� 75 .

3 √ �� 12 + 5 √ �� 75 = 3 √ �� 22 · 3 + 5 √ �� 52 · 3 Simplify.

= 3 · 2 √ � 3 + 5 · 5 √ � 3 Simplify.

= 6 √ � 3 + 25 √ � 3 Simplify.

= 31 √ � 3 Distributive Property


1. 2 √ � 5 + 4 √ � 5 2. √ � 6 - 4 √ � 6

3. √ � 8 - √ � 2 4. 3 √ � 75 + 2 √ � 5

5. √ � 20 + 2 √ � 5 - 3 √ � 5 6. 2 √ � 3 + √ � 6 - 5 √ � 3

7. √ � 12 + 2 √ � 3 - 5 √ � 3 8. 3 √ � 6 + 3 √ � 2 - √ � 50 + √ � 24

9. √ � 8a - √ � 2a + 5 √ � 2a 10. √ � 54 + √ � 24

11. √ � 3 + √ � 1 − 3 12. √ � 12 + √ � 1 −

3

13. √ � 54 - √ � 1 − 6 14. √ � 80 - √ � 20 + √ �� 180

15. √ � 50 + √ � 18 - √ � 75 + √ � 27 16. 2 √ � 3 - 4 √ � 45 + 2 √ � 1 − 3

17. √ �� 125 - 2 √ � 1 − 5 + √ � 1 −

3 18. √ � 2 −

3 + 3 √ � 3 - 4 √ � 1 −

12

10-3

Example 1

Example 2


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Operations with Radical Expressions

Multiply Radical Expressions Multiplying two radical expressions with different radicands is similar to multiplying binomials.

Multiply (3 √ � 2 - 2 √ � 5 )(4 √ �� 20 + √ � 8 ).

Use the FOIL method.

(3 √ � 2 - 2 √ � 5 )(4 √ �� 20 + √ � 8 ) = (3 √ � 2 )(4 √ �� 20 ) + (3 √ � 2 )( √ � 8 ) + (-2 √ � 5 )(4 √ �� 20 ) + (-2 √ � 5 )( √ � 8 )

= 12 √ �� 40 + 3 √ �� 16 - 8 √ �� 100 - 2 √ �� 40 Multiply.

= 12 √ �� 22 · 10 + 3 · 4 - 8 · 10 - 2 √ �� 22 · 10 Simplify.

= 24 √ �� 10 + 12 - 80 - 4 √ �� 10 Simplify.

= 20 √ �� 10 - 68 Combine like terms.


1. 2( √ � 3 + 4 √ � 5 ) 2. √ � 6 ( √ � 3 - 2 √ � 6 )

3. √ � 5 ( √ � 5 - √ � 2 ) 4. √ � 2 (3 √ � 7 + 2 √ � 5 )

5. (2 - 4 √ � 2 )(2 + 4 √ � 2 ) 6. (3 + √ � 6 ) 2

7. (2 - 2 √ � 5 ) 2 8. 3 √ � 2 ( √ � 8 + √ � 24 )

9. √ � 8 ( √ � 2 + 5 √ � 8 ) 10. ( √ � 5 - 3 √ � 2 )( √ � 5 + 3 √ � 2 )

11. ( √ � 3 + √ � 6 ) 2 12. ( √ � 2 - 2 √ � 3 ) 2

13. ( √ � 5 - √ � 2 )( √ � 2 + √ � 6 ) 14. ( √ � 8 - √ � 2 )( √ � 3 + √ � 6 )

15. ( √ � 5 - √ � 18 )(7 √ � 5 + √ � 3 ) 16. (2 √ � 3 - √ � 45 )( √ � 12 + 2 √ � 6 )

17. (2 √ � 5 - 2 √ � 3 )( √ � 10 + √ � 6 ) 18. ( √ � 2 + 3 √ � 3 )( √ � 12 - 4 √ � 8 )

10-3

Example


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Skills PracticeOperations with Radical Expressions

10-3


1. 7 √ � 7 - 2 √ � 7 2. 3 √ �� 13 + 7 √ �� 13

3. 6 √ � 5 - 2 √ � 5 + 8 √ � 5 4. √ �� 15 + 8 √ �� 15 - 12 √ �� 15

5. 12 √ � r - 9 √ � r 6. 9 √ �� 6a - 11 √ �� 6a + 4 √ �� 6a

7. √ �� 44 - √ �� 11 8. √ �� 28 + √ �� 63

9. 4 √ � 3 + 2 √ �� 12 10. 8 √ �� 54 - 4 √ � 6

11. √ �� 27 + √ �� 48 + √ �� 12 12. √ �� 72 + √ �� 50 - √ � 8

13. √ �� 180 - 5 √ � 5 + √ �� 20 14. 2 √ �� 24 + 4 √ �� 54 + 5 √ �� 96

15. 5 √ � 8 + 2 √ �� 20 - √ � 8 16. 2 √ � 13 + 4 √ � 2 - 5 √ � 13 + √ � 2

17. √ � 2 ( √ � 8 + √ � 6 ) 18. √ � 5 ( √ � 10 - √ � 3 )

19. √ � 6 (3 √ � 2 - 2 √ � 3 ) 20. 3 √ � 3 (2 √ � 6 + 4 √ � 10 )

21. (4 + √ � 3 ) (4 -

√ � 3 ) 22. (2 - √ � 6 )2

23. ( √ � 8 + √ � 2 ) ( √ � 5 + √ � 3 ) 24. ( √ � 6 + 4 √ � 5 )(4 √ � 3 - √ � 10 )


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PracticeOperations with Radical Expressions


1. 8 √ � 30 - 4 √ � 30 2. 2 √ � 5 - 7 √ � 5 - 5 √ � 5

3. 7 √ �� 13x - 14 √ �� 13x + 2 √ �� 13x 4. 2 √ �� 45 + 4 √ � 20

5. √ � 40 - √ � 10 + √ � 90 6. 2 √ � 32 + 3 √ � 50 - 3 √ � 18

7. √ � 27 + √ � 18 + √ �� 300 8. 5 √ � 8 + 3 √ � 20 - √ � 32

9. √ � 14 - √ � 2 − 7 10. √ � 50 + √ � 32 - √ � 1 −

2

11. 5 √ � 19 + 4 √ � 28 - 8 √ � 19 + √ � 63 12. 3 √ � 10 + √ � 75 - 2 √ � 40 - 4 √ � 12

13. √ � 6 ( √ � 10 + √ � 15 ) 14. √ � 5 (5 √ � 2 - 4 √ � 8 )

15. 2 √ � 7 (3 √ � 12 + 5 √ � 8 ) 16. (5 - √ � 15 ) 2

17. ( √ � 10 + √ � 6 ) ( √ � 30 - √ � 18 ) 18. ( √ � 8 + √ � 12 ) ( √ � 48 + √ � 18 )

19. ( √ � 2 + 2 √ � 8 )(3 √ � 6 - √ � 5 ) 20. (4 √ � 3 - 2 √ � 5 )(3 √ � 10 + 5 √ � 6 )

21. SOUND The speed of sound V in meters per second near Earth’s surface is given by

V = 20 √ �� t + 273 , where t is the surface temperature in degrees Celsius.

a. What is the speed of sound near Earth’s surface at 15°C and at 2°C in simplest form?

b. How much faster is the speed of sound at 15°C than at 2°C?

22. GEOMETRY A rectangle is 5 √ � 7 + 2 √ � 3 meters long and 6 √ � 7 - 3 √ � 3 meters wide.

a. Find the perimeter of the rectangle in simplest form.

b. Find the area of the rectangle in simplest form.

10-3


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-5-10

5

5

10

10

15

20Ski slope

y

xO

Word Problem PracticeOperations with Radical Expressions

1. ARCHITECTURE The Pentagon is the building that houses the U.S. Department of Defense. Find the approximate perimeter of the building, which is a regular pentagon. Leave your answer as a radical expression.

2. EARTH The surface area of a sphere with radius r is given by the formula 4πr2. Assuming that Earth is close to spherical in shape and has a surface area of about 5.1 × 108 square kilometers, what is the radius of Earth to the nearest ten kilometers?

3. GEOMETRY The area of a trapezoid is found by multiplying its height by the average length of its bases. Find the area of deck attached to Mr. Wilson’s house. Give your answer as a simplified radical expression.

4. RECREATION Carmen surveyed a ski slope using a digital device connected to a computer. The computer model assigned coordinates to the top and bottom points of the hill as shown in the diagram. Write a simplified radical expression that represents the slope of the hill.

5. FREE FALL Suppose a ball is dropped from a building window 800 feet in the air. Another ball is dropped from a lower window 288 feet high. Both balls are released at the same time. Assume air resistance is not a factor and use the following formula to find how many seconds t it will take a ball to fall h feet.

t = 1 − 4 √ � h

a. How much time will pass between when the first ball hits the ground and when the second ball hits the ground? Give your answer as a simplified radical expression.

b. Which ball lands first?

c. Find a decimal approximation of the answer for part a. Round your answer to the nearest tenth.

(7r + 12)̊

House

Deck

h = 7 √ � 5 ft

12 √ � 3 ft

6 √ � 3 ft

A (2 √ �� 14 , 5 √ � 7 )

23 √ �� 149 m

B (-2 √ �� 14 , 7 √ � 7 )

10-3


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Enrichment

The Wheel of TheodorusThe Greek mathematicians were intrigued by problems of representing different numbers and expressions using geometric constructions.Theodorus, a Greek philosopher who lived about 425 B.C., is said to have discovered a way to construct the sequence √ � 1 , √ � 2 , √ � 3 , √ � 4 , . . ..The beginning of his construction is shown. You start with an isosceles right triangle with sides 1 unit long.

Use the figure above. Write each length as a radical expression in simplest form.

1. line segment AO 2. line segment BO

3. line segment CO 4. line segment DO

5. Describe how each new triangle is added to the figure.

6. The length of the hypotenuse of the first triangle is √ � 2 . For the second triangle, the length is √ � 3 . Write an expression for the length of the hypotenuse of the nth triangle.

7. Show that the method of construction will always produce the next number in the sequence. (Hint: Find an expression for the hypotenuse of the (n + 1)th triangle.)

8. In the space below, construct a Wheel of Theodorus. Start with a line segment 1 centimeter long. When does the Wheel start to overlap?

1

1

1

1

O A

B

CD

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Study Guide and InterventionRadical Equations

Radical Equations Equations containing radicals with variables in the radicand are called radical equations. These can be solved by first using the following steps.

Solve 16 = √ � x − 2 for x.

16 = √ � x −

2 Original equation

2(16) = 2( √ � x

− 2 ) Multiply each side by 2.

32 = √ � x Simplify.

(32)2 = ( √ � x )2 Square each side.

1024 = x Simplify.

The solution is 1024, which checks in the original equation.

Solve √ �� 4x - 7 + 2 = 7.

√ �� 4x - 7 + 2 = 7 Original equation

√ �� 4x - 7 + 2 - 2 = 7 - 2 Subtract 2 from each side.

√ �� 4x - 7 = 5 Simplify.

( √ �� 4x - 7 )2 = 52 Square each side.

4x - 7 = 25 Simplify.

4x - 7 + 7 = 25 + 7 Add 7 to each side.

4x = 32 Simplify.

x = 8 Divide each side by 4.

The solution is 8, which checks in the original equation.

Exercises Solve each equation. Check your solution.

1. √ � a = 8 2. √ � a + 6 = 32 3. 2 √ � x = 8

4. 7 = √ �� 26 - n 5. √ �� -a = 6 6. √ �� 3r2 = 3 ± √ �

7. 2 √ � 3 = √ � y 8. 2 √ �� 3a - 2 = 7 9. √ �� x - 4 = 6

10. √ �� 2m + 3 = 5 11. √ �� 3b - 2 + 19 = 24 12. √ �� 4x - 1 = 3

13. √ �� 3r + 2 = 2 √ � 3 14. √ � x − 2 = 1 −

2 15. √ � x −

8 = 4

16. √ �� 6x2 + 5x = 2 17. √ � x − 3 + 6 = 8 18. 2 √ �� 3x −

5 + 3 = 11

10-4

Example 1 Example 2

Step 1 Isolate the radical on one side of the equation.Step 2 Square each side of the equation to eliminate the radical.


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Radical Equations

Extraneous Solutions To solve a radical equation with a variable on both sides, you need to square each side of the equation. Squaring each side of an equation sometimes produces extraneous solutions, or solutions that are not solutions of the original equation. Therefore, it is very important that you check each solution.

Solve √ �� x + 3 = x - 3.

√ �� x + 3 = x - 3 Original equation

( √ �� x + 3 )2 = (x - 3)2 Square each side.

x + 3 = x2 - 6x + 9 Simplify.

0 = x2 - 7x + 6 Subtract x and 3 from each side.

0 = (x - 1)(x - 6) Factor.

x - 1 = 0 or x - 6 = 0 Zero Product Property

x = 1 x = 6 Solve.

CHECK √ �� x + 3 = x - 3 √ �� x + 3 = x - 3 √ �� 1 + 3 � 1 - 3 √ �� 6 + 3 � 6 - 3 √ � 4 � -2 √ � 9 � 3 2 ≠ -2 3 = 3 �Since x = 1 does not satisfy the original equation, x = 6 is the only solution.

Exercises Solve each equation. Check your solution.

1. √ � a = a 2. √ �� a + 6 = a 3. 2 √ � x = x

4. n = √ �� 2 - n 5. √ �� -a = a 6. √ �� 10 - 6k + 3 = k

7. √ �� y - 1 = y - 1 8. √ �� 3a - 2 = a 9. √ �� x + 2 = x

10. √ �� 2b + 5 = b - 5 11. √ �� 3b + 6 = b + 2 12. √ �� 4x - 4 = x

13. r + √ �� 2 - r = 2 14. √ �� x2 + 10x = x + 4 15. -2 √ � x − 8 = 15

16. √ �� 6x2 - 4x = x + 2 17. √ �� 2y2 - 64 = y 18. √ �� 3x2 + 12x + 1 = x + 5

10-4

Example 1


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Skills PracticeRadical Equations

Solve each equation. Check your solution.

1. √ � f = 7 2. √ �� -x = 5

3. √ �� 5p = 10 4. √ � 4y = 6

5. 2 √ � 2 = √ � u 6. 3 √ � 5 = √ �� -n

7. √ � g - 6 = 3 8. √ �� 5a + 2 = 0

9. √ �� 2t - 1 = 5 10. √ �� 3k - 2 = 4

11. √ �� x + 4 - 2 = 1 12. √ �� 4x - 4 - 4 = 0

13. √ � d

− 3 = 4 14. √ � m −

3 = 3

15. x = √ �� x + 2 16. d = √ �� 12 - d

17. √ �� 6x - 9 = x 18. √ �� 6p - 8 = p

19. √ �� x + 5 = x - 1 20. √ �� 8 - d = d - 8

21. √ �� r - 3 + 5 = r 22. √ �� y - 1 + 3 = y

23. √ �� 5n + 4 = n + 2 24. √ �� 3z - 6 = z - 2

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Practice Radical Equations


1. √ �� -b = 8 2. 4 √ � 3 = √ � x

3. 2 √ � 4r + 3 = 11 4. 6 - √ � 2y = -2

5. √ �� k + 2 - 3 = 7 6. √ �� m - 5 = 4 √ � 3

7. √ �� 6t + 12 = 8 √ � 6 8. √ �� 3j - 11 + 2 = 9

9. √ �� 2x + 15 + 5 = 18 10. √ � 3d − 5 - 4 = 2

11. 6 √ �� 3x − 3 - 3 = 0 12. 6 + √ � 5r −

6 = -2

13. y = √ �� y + 6 14. √ �� 15 - 2x = x

15. √ �� w + 4 = w + 4 16. √ �� 17 - k = k - 5

17. √ �� 5m - 16 = m - 2 18. √ �� 24 + 8q = q + 3

19. √ �� 4t + 17 - t - 3 = 0 20. 4 - √ �� 3m + 28 = m

21. √ �� 10p + 61 - 7 = p 22. √ �� 2x2 - 9 = x

23. ELECTRICITY The voltage V in a circuit is given by V = √ �� PR , where P is the power in watts and R is the resistance in ohms.

a. If the voltage in a circuit is 120 volts and the circuit produces 1500 watts of power, what is the resistance in the circuit?

b. Suppose an electrician designs a circuit with 110 volts and a resistance of 10 ohms. How much power will the circuit produce?

24. FREE FALL Assuming no air resistance, the time t in seconds that it takes an object to

fall h feet can be determined by the equation t = √ � h

− 4 .

a. If a skydiver jumps from an airplane and free falls for 10 seconds before opening the parachute, how many feet does the skydiver fall?

b. Suppose a second skydiver jumps and free falls for 6 seconds. How many feet does the second skydiver fall?

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Word Problem PracticeRadical Equations

1. SUBMARINES The distance in miles that the lookout of a submarine can see is approximately d = 1.22 √ � h , where h is the height in feet above the surface of the water. How far would a submarine periscope have to be above the water to locate a ship 6 miles away? Round your answer to the nearest tenth.

2. PETS Find the value of x if the perimeter of a triangular dog pen is 25 meters.

3. LOGGING Doyle’s log rule estimates the amount of usable lumber (in board feet) that can be milled from a shipment of logs. It is represented by the equation

B = L ( d - 4 − 4 )

2 , where d is the log

diameter (in inches) and L is the log length (in feet). Suppose the truck carries 20 logs, each 25 feet long, and that the shipment yields a total of 6000 board feet of lumber. Estimate the diameter of the logs to the nearest inch. Assume that all the logs have uniform length and diameter.

4. FIREFIGHTING Fire fighters calculate the flow rate of water out of a particular hydrant by using the following formula.

F = 26.9d2 √ � p

F is the flow rate (in gallons per minute), p is the nozzle pressure (in pounds per square inch), and d is the diameter of the hose (in inches). In order to effectively fight a fire, the combined flow rate of two hoses needs to be about 2430 gallons per minute. The diameter of each of the hoses is 3 inches, but the nozzle pressure of one hose is 4 times that of the second hose. What are the nozzle pressures for each hose? Round your answers to the nearest tenth.

5. GEOMETRY The lateral surface area s of a right circular cone, not including the base, is represented by the equation s = πr √ �� r2 + h2 , where r is the radius of the circular base and h is the height of the cone.

a. If the lateral surface area of a funnel is 127.54 square centimeters and its radius is 3.5 centimeters, find its height to the nearest tenth of a centimeter.

b. What is the area of the opening (i.e., the base) of the funnel?

x + 1 m

12 m 10 m

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Enrichment

More Than One Square RootYou have learned that to remove the square root in an equation, you first need to isolate the square root, then square both sides of the equation, and finally, solve the resulting equation. However, there are equations that contain more that one square root and simply squaring once is not enough to remove all of the radicals.

Solve √ �� x + 7 = √ � x + 1.

√ �� x + 7 = √ � x + 1 One of the square roots is already isolated.

( √ �� x + 7 ) 2 = ( √ � x + 1 ) 2 Square both sides to remove the square root.

x + 7 = x + 2 √ � x + 1 Simplify. Use the FOIL method to square right side.

x + 7 - x - 1 = 2 √ � x Simplify.

6 = 2 √ � x Simplify. Isolate the square root term again.

3 = √ � x Divide both sides by 2.

9 = x Square both sides to remove the square root.

Check: Substitute into the original equation to make sure your solution is valid.√ �� 9 + 7 = √ � 9 + 1 Replace x with 9.

√ �� 16 = 3 + 1 Simpify.

4 = 4 � The equation is true, so x = 9 is the solution.

ExercisesSolve each equation.

1. √ �� x + 13 - 2 = √ �� x + 1 2. √ �� x + 11 = √ �� x + 3 + 2

3. √ �� x + 9 - 3 = √ �� x - 6 4. √ �� x + 21 = √ � x + 3

5. √ �� x + 9 + 3 = √ �� x + 20 + 2 6. √ �� x - 6 + 6 = √ �� x + 1 + 5

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Graphing Calculator ActivityRadical Inequalities

The graphs of radical equations can be used to determine the solutions of radical inequalities through the CALC menu.

Solve each inequality.

a. √ �� x + 4 ≤ 3

Enter √ �� x + 4 in Y1 and 3 in Y2 and graph. Examine the graphs. Use

TRACE to find the endpoint of the graph of the radical equation. Use CALC to determine the intersection of the graphs. This interval, -4 to 5, where the graph of y = √ �� x + 4 is below the graph of y = 3, represents the solution to the inequality. Thus, the solution is -4 ≤ x ≤ 5.

b. √ �� 2x - 5 > x - 4Graph each side of the inequality. Find the intersection and trace to the endpoint of the radical graph.

The graph of y = √ �� 2x - 5 is above the graph of y = x - 4 from 2.5 up to 7. Thus, the solution is 2.5 < x < 7.

ExercisesSolve each inequality.

1. 6 - √ �� 2x + 1 < 3 2. √ �� 4x - 5 ≤ 7 3. √ �� 5x - 4 ≥ 4

4. -4 > √ �� 3x - 2 5. √ �� 3x - 6 + 5 ≥ -3 6. √ �� 6 - 3x < x + 16

[-10, 10] scl:1 by [-10, 10] scl:1[-10, 10] scl:1 by [-10, 10] scl:1

[-10, 10] scl:1 by [-10, 10] scl:1[-10, 10] scl:1 by [-10, 10] scl:1

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Study Guide and InterventionThe Pythagorean Theorem

The Pythagorean Theorem The side opposite the right angle in a right triangle is called the hypotenuse. This side is always the longest side of a right triangle. The other two sides are called the legs of the triangle. To find the length of any side of a right triangle, given the lengths of the other two sides, you can use the Pythagorean Theorem.

Pythagorean Theorem

If a and b are the measures of the legs of a right triangle

and c is the measure of the hypotenuse, then c2 = a2 + b2.

C

B

Ab

ac

Find the length of the missing side.

c2 = a2 + b2 Pythagorean Theorem

c2 = 52 + 122 a = 5 and b = 12

c2 = 169 Simplify.

c = √ �� 169 Take the square root of each side.

c = 13The length of the hypotenuse is 13.

ExercisesFind the length of each missing side. If necessary, round to the nearest hundredth.

1. 2. 3. 25

25c

100

110a

40

30c

10-5

12

5

14

84 15 89

5

Example

4. 5. 6.


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The Pythagorean Theorem

Right Triangles If a and b are the measures of the shorter sides of a triangle, c is the measure of the longest side, and c2 = a2 + b2, then the triangle is a right triangle.

Determine whether the following side measures form right triangles.

a. 10, 12, 14Since the measure of the longest side is 14, let c = 14, a = 10, and b = 12.


142 � 102 + 122 a = 10, b = 12, c = 14

196 � 100 + 144 Multiply.

196 ≠ 244 Add.

Since c2 ≠ a2 + b2, the triangle is not a right triangle.b. 7, 24, 25

Since the measure of the longest side is 25, let c = 25, a = 7, and b = 24.


252 � 72 + 242 a = 7, b = 24, c = 25

625 � 49 + 576 Multiply.

625 = 625 Add.

Since c2 = a2 + b2, the triangle is a right triangle.

ExercisesDetermine whether each set of measures can be sides of a right triangle. Then determine whether they form a Pythagorean triple.

1. 14, 48, 50 2. 6, 8, 10 3. 8, 8, 10

4. 90, 120, 150 5. 15, 20, 25 6. 4, 8, 4 √ � 5

7. 2, 2, √ � 8 8. 4, 4, √ �� 20 9. 25, 30, 35

10. 24, 36, 48 11. 18, 80, 82 12. 150, 200, 250

13. 100, 200, 300 14. 500, 1200, 1300 15. 700, 1000, 1300

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Skills PracticeThe Pythagorean Theorem

Find the length of each missing side. If necessary, round to the nearest hundredth. 1. 2.

3. 4.

5. 6.

Determine whether each set of measures can be sides of a right triangle. Then determine whether they form a Pythagorean triple.

7. 7, 24, 25 8. 15, 30, 34

9. 16, 28, 32 10. 18, 24, 30

11. 15, 36, 39 12. 5, 7, √ �� 74

13. 4, 5, 6 14. 10, 11, √ �� 221

250

240

a4

9

c

2933

b

1634

b

1539

a

21

72

c

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PracticeThe Pythagorean Theorem

Find the length of each missing side. If necessary, round to the nearest hundredth.

1. 2. 3.

Determine whether each set of measures can be sides of right triangle. Then determine whether they form a Pythagorean triple.

4. 11, 18, 21 5. 21, 72, 75

6. 7, 8, 11 7. 9, 10, √ �� 161

8. 9, 2 √ �� 10 , 11 9. √ � 7 , 2 √ � 2 , √ �� 15

10. STORAGE The shed in Stephan’s back yard has a door that measures 6 feet high and 3 feet wide. Stephan would like to store a square theater prop that is 7 feet on a side. Will it fit through the door diagonally? Explain.

11. SCREEN SIZES The size of a television is measured by the length of the screen’s diagonal.

a. If a television screen measures 24 inches high and 18 inches wide, what size television is it?

b. Darla told Tri that she has a 35-inch television. The height of the screen is 21 inches. What is its width?

c. Tri told Darla that he has a 5-inch handheld television and that the screen measures 2 inches by 3 inches. Is this a reasonable measure for the screen size? Explain.

124

b1911

a

60

32c

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Word Problem PracticePythagorean Theorem

1. BASEBALL A baseball diamond is a square. Each base path is 90 feet long. After a pitch, the catcher quickly throws the ball from home plate to a teammate standing by second base. Find the distance the ball travels. Round your answer to the nearest tenth.

2. TRIANGLES Each student in Mrs. Kelly’s geometry class constructed a unique right triangle from drinking straws. Mrs. Kelly made a chart with the dimensions of each triangle. However, Mrs. Kelly made a mistake when recording their results. Which result was recorded incorrectly?

3. MAPS Find the distance between Macon and Berryville. Round your answer to the nearest tenth.

4. TELEVISION Televisions are identified by the diagonal measurement of the viewing screen. For example, a 27-inch television has a diagonal screen measurement of 27 inches.

Complete the chart to find the screen height of each television given its size and screen width. Round your answers to the nearest whole number.

Source: Best Buy

5. MANUFACTURING Karl works for a company that manufactures car parts. His job is to drill a hole in spherical steel balls. The balls and the holes have the dimensions shown on the diagram.

a. How deep is the hole?

b. What would be the radius of a ball with a similar hole 7 centimeters wide and 24 centimeters deep?

d

Second Base

Home Plate

90 ft

x cm 13 cm

5 cm

Side Lengths

Student a b c Student a b c

Amy 3 4 5 Fran 8 14 16

Belinda 7 24 25 Gus 5 12 13

Emory 9 12 15

27 in.

1

0

2

3

4

5

6

7

8

1 2 3 4 5 6 7 8 9 10

ten

s o

f m

iles

tens of miles

Carter City

Macon Hamilton

Berryville

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TV size width (in.) height (in.)

19-inch 15

25-inch 21

32-inch 25

50-inch 40


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Enrichment

Pythagorean TriplesRecall the Pythagorean Theorem:In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs.

a2 + b2 = c2

Note that c is the length of the hypotenuse.The integers 3, 4, and 5 satisfy the 32 + 42 = 52

Pythagorean Theorem and can be the 9 + 16 = 25lengths of the sides of a right triangle. 25 = 25Furthermore, for any positive integer n, For n = 2: 62 + 82 = 102

the numbers 3n, 4n, and 5n satisfy the 36 + 64 = 100Pythagorean Theorem. 100 = 100

If three numbers satisfy the Pythagorean Theorem, they are called a Pythagorean triple. Here is an easy way to find other Pythagorean triples.

The numbers a, b, and c are a Pythagorean triple if a = m2 - n2, b = 2mn, and c = m2 + n2, where m and n are relatively prime positive integers and m > n.

Choose m = 5 and n = 2.

a = m2 - n2 b = 2mn c = m2 + n2 Check 202 + 212 � 292

= 52 - 22 = 2(5)(2) = 52 + 22 400 + 441 � 841 = 25 - 4 = 20 = 25 + 4 841 = 841 = 21 = 29

ExercisesUse the following values of m and n to find Pythagorean triples.

1. m = 3 and n = 2 2. m = 4 and n = 1 3. m = 5 and n = 3

4. m = 6 and n = 5 5. m = 10 and n = 7 6. m = 8 and n = 5

AC

B

ac

b

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Spreadsheet ActivityPythagorean Triples

A Pythagorean triple is a set of three whole numbers that satisfies the equation a2 + b2 = c2, where c is the greatest number. You can use a spreadsheet to investigate the patterns in Pythagorean triples. A primitive Pythagorean triple is a Pythagorean triple in which the numbers have no common factors other than 1. A family of Pythagorean triples is a primitive Pythagorean triple and its whole number multiples.

The spreadsheet at the right produces a family of Pythagorean triples.

Step 1 Enter a Pythagorean triple into cells A1, A2, and A3.

Step 2 Use rows 2 through 10 to find 9 additional Pythagorean triples that are multiples of the primitive triple. Format the rows so that row 2 multiplies the numbers in row 1 by 2, row 3 multiplies the numbers in row 1 by 3, and so on.

ExercisesUse the spreadsheet of families of Pythagorean triples.

1. Choose one of the triples other than (3, 4, 5) from the spreadsheet. Verify that it is a Pythagorean triple.

2. Two polygons are similar if they are the same shape, but not necessarily the same size. For triangles, if two triangles have angles with the same measures then they are similar. Use a centimeter ruler to draw triangles with measures from the spreadsheet. Do the triangles appear to be similar?

Each of the following is a primitive Pythagorean triple. Use the spreadsheet to find two Pythagorean triples in their families.

3. (5, 12, 13)

4. (9, 40, 41)

5. (20, 21, 29)

Triples.xls

A1

3

4

56

2

8

9

1011

7

B C3

6

9

12

15

18

21

24

27

30

4

8

12

16

20

24

28

32

36

40

5

10

15

20

25

30

35

40

45

50

The formula in cell A10 isA1 * 10.

Sheet 1 Sheet 2 SI

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Find the distance between the points at (-5, 2) and (4, 5).

d = √ �� (x2 - x1)2 + (y2 - y1)

2 Distance Formula

= √ �� (4 - (-5))2 + (5 - 2)2 (x1, y

1) = (-5, 2), (x

2, y

2) = (4, 5)

= √ �� 92 + 32 Simplify.

= √ �� 81 + 9 Evaluate squares and simplify.

= √ �� 90 The distance is √ �� 90 , or about 9.49 units.

Example 1

Study Guide and InterventionThe Distance and Midpoint Formulas

Distance Formula The Pythagorean Theorem can be used to derive the Distance Formula shown below. The Distance Formula can then be used to find the distance between any two points in the coordinate plane.

Distance FormulaThe distance between any two points with coordinates (x

1, y

1) and (x

2, y

2) is

given by d = √ �� (x2 - x

1)2 + (y

2 -y

1)2

Jill draws a line segment from point (1, 4) on her computer screen to point (98, 49). How long is the segment?

d = √ �� (x2 - x1)2 + (y2 - y1)

2 = √ �� (98 - 1)2 + (49 - 4)2 = √ �� 972 + 452 = √ �� 9409 + 2025 = √ �� 11,434 The segment is about 106.93 units long.

ExercisesFind the distance between the points with the given coordinates.

1. (1, 5), (3, 1) 2. (0, 0), (6, 8) 3. (-2, -8), (7, -3)

4. (6, -7), (-2, 8) 5. (1, 5), (-8, 4) 6. (3, -4), (-4, -4)

7. (-1, 4), (3, 2) 8. (0, 0), (-3, 5) 9. (2, -6), (-7, 1)

10. (-2, -5), (0, 8) 11. (3, 4), (0, 0) 12. (3, -4), (-4, -16)

Find the possible values of a if the points with the given coordinates are the indicated distance apart.

13. (1, a), (3, -2); d = √ � 5 14. (0, 0), (a, 4); d = 5 15. (2, -1), (a, 3); d = 5

16. (1, -3), (a, 21); d = 25 17. (1, a), (-2, 4); d = 3 18. (3, -4), (-4, a); d = √ �� 65

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The Distance and Midpoint Formulas

Midpoint Formula The point that is equidistance from both of the endpoints is called the midpoint. You can find the coordinates of the midpoint by using the Midpoint Formula.

Midpoint Formula

The midpoint M of a line segment with endpoints at (x1, y

1) and (x

2, y

2) is

given by M ( x

1 + x

2

− 2 ,

y1 + y

2

− 2 ) .

Find the coordinates of the midpoint of the segment with endpoints at (–2, 5) and (4, 9).

M ( x1 + x2 −

2 ,

y1 + y2 − 2 ) Midpoint Formula

= M ( -2 + 4 −

2 ,

5 + 9 − 2 ) (x

1, y

1) = (-2, 5) and (x

2, y

2) = (4, 9)

= M ( 2 − 2 , 14 −

2 ) Simplify the numerators.

= M (1, 3) Simplify.

ExercisesFind the coordinates of the midpoint of the segment with the given endpoints.

1. (1, 6), (3, 10) 2. (4, -2), (0, 6) 3. (7, 2), (13, -4)

4. (-1, 2), (1, 0) 5. (-3, -3), (5, -11) 6. (0, 8), (-6, 0)

7. (4, -3), (-2, 3) 8. (9, -1), (3, -7) 9. (2, -1), (8, 7)

10. (1, 4), (-3, 12) 11. (4, 0), (-2, 6) 12. (1, 9), (7, 1)

13. (12, 0), (2, -6) 14. (1, 1), (9, -9) 15. (4, 5), (-2, -1)

16. (1, -14), (-5, 0) 17. (2, 2), (6, 8) 18. (-7, 3), (5, -3)

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Skills PracticeThe Distance and Midpoint Formulas

Find the distance between the points with the given coordinates.

1. (9, 7), (1, 1) 2. (5, 2), (8, -2)

3. (1, -3), (1, 4) 4. (7, 2), (-5, 7)

5. (-6, 3), (10, 3) 6. (3, 3), (-2, 3)

7. (-1, -4), (-6, 0) 8. (-2, 4), (5, 8)


9. (-2, -5), (a, 7); d = 13 a 10. (8, -2), (5, a); d = 3

11. (4, a), (1, 6); d = 5 12. (a, 3), (5, -1); d = 5

13. (1, 1), (a, 1); d = 4 14. (2, a), (2, 3); d = 10

15. (a, 2), (-3, 3); d = √ � 2 16. (-5, 3), (-3, a); d = √ � 5

Find the coordinates of the midpoint of the segment with the given endpoints.

17. (-3, 4), (-2, 8) 18. (5, -6), (7, -9)

19. (4, 2), (8, 6) 20. (5, 2), (3, 10)

21. (12, -1), (4, -11) 22. (-3, -1), (-11, 3)

23. (9, 3), (6, -6) 24. (0, -4), (8, 4)

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PracticeThe Distance and Midpoint Formulas

Find the distance between the points with the given coordinates.

1. (4, 7), (1, 3) 2. (0, 9), (-7, -2)

3. (6, 2), + (4, 1 − 2 ) 4. (-1, 7), + ( 1 −

3 , 6)

5. ( √ � 3 , 3) , (2 √ � 3 , 5) 6. (2 √ � 2 , -1) , (3 √ � 2 , 3)


7. (4, -1), (a, 5); d = 10 8. (2, -5), (a, 7); d = 15

9. (6, -7), (a, -4); d = √ �� 18 10. (-4, 1), (a, 8); d = √ �� 50

11. (8, -5), (a, 4); d = √ �� 85 12. (-9, 7), (a, 5); d = √ �� 29

Find the coordinates of the midpoint of the segment with the given endpoints.

13. (4, -6), (3, -9) 14. (-3, -8), (-7, 2)

15. (0, -4), (3, 2) 16. (-13, -9), (-1, -5)

17. (2, - 1 − 2 ) , (1, 1 −

2 ) 18. ( 2 −

3 , -1) , (2, 1 −

3 )

19. BASEBALL Three players are warming up for a baseball game. Player B stands 9 feet to the right and 18 feet in front of Player A. Player C stands 8 feet to the left and 13 feet in front of Player A.

a. Draw a model of the situation on the coordinate grid. Assume that Player A is located at (0, 0).

b. To the nearest tenth, what is the distance between Players A and B and between Players A and C?

c. What is the distance between Players B and C?

20. MAPS Maria and Jackson live in adjacent neighborhoods. If they superimpose a coordinate grid on the map of their neighborhoods, Maria lives at (-9, 1) and Jackson lives at (5, -4).

x

y

O 4 8

16

12

8

4

-8 -4

10-6


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Word Problem PracticeThe Distance and Midpoint Formulas

1. CHESS Margaret’s last two remaining chess pieces are located at the centers of the squares at opposite corners of the board. If the chessboard is a square with 8-inch sides, about how far apart are the pieces? Round your answer to the nearest tenth.

2. ENGINEERING Todd has drawn a cul-de-sac for a residential development plan. He used a compass to draw the cul-de-sac so that it would be circular. On his blueprint, the center of the cul-de-sac has coordinates (-1, -1) and a point on the circle is (2, 3). What is the radius of the cul-de-sac?

3. LANDSCAPING Randy plotted a triangular patio on a landscape plan for a client. What is the length of fencing he will need along the patio edge that borders the property line? Round your answer to the nearest tenth.

4. UTILITIES The electric company is running some wires across an open field. The wire connects a utility pole at (2, 14) and a second utility pole at (7, -8). If the electric company wishes to place a third pole at the midpoint of the two poles, at what coordinates should the pole be placed?

5. MARCHING BAND The Ohio State University marching band performs a famous on-field spelling of O-H-I-O called “Script Ohio”. Sometimes they must adjust the usual dimensions of the word to fit it into the limited guest band performance area. The diagram below shows part of the adjusted drill chart. Each point represents one band member, and the coordinates are in yards.

a. How far is the drum major from the tuba player who dots the “i”?

b. Carol is the band member at the top left of the first O in Ohio. She is located at (0, 26). How far away is Carol from the tuba player? Round your answer to the nearest tenth.

drum majortuba player(20, 13)

(32, 10)

1

2

3

4

5

6

7

8

9

10 2 3 4 5 6 7 8 9

met

ers

meters

shru

bs

pro

per

ty li

ne

ho

use

tree

patio

(2, 4)

(6, 8)

(5, 1)

10-6


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Enrichment

A Space-Saving MethodTwo arrangements for cookies on a 32 cm by 40 cm cookie sheet are shown at the right. The cookies have 8-cm diameters after they are baked. The centers of the cookies are on the vertices of squares in the top arrangement. In the other, the centers are on the vertices of equilateral triangles. Which arrangement is more economical? The triangle arrangement is more economical, because it contains one more cookie.

In the square arrangement, rows are placed every 8 cm. At what intervals are rows placed in the triangle arrangement?

Look at the right triangle labeled a, b, and c. A leg a of the triangle is the radius of a cookie, or 4 cm. The hypotenuse c is the sum of two radii, or 8 cm. Use the Pythagorean theorem to find b, the interval of the rows.

c2 = a2 + b2

82 = 42 + b2

64 - 16 = b2

√ �� 48 = b 4 √ � 3 = b

b = 4 √ � 3 ≈ 6.93

The rows are placed approximately every 6.93 cm.

Solve each problem.

1. Suppose cookies with 10-cm diameters are arranged in the triangular pattern shown above. What is the interval b of the rows?

2. Find the diameter of a cookie if the rows are placed in the triangular pattern every 3 √ � 3 cm.

3. Describe other practical applications in which this kind of triangular pattern can be used to economize on space.

{b = ?

21 cookies

c

a

{8 cm

20 cookies

10-6


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Study Guide and InterventionSimilar Triangles

Similar Triangles �RST is similar to �XYZ. The angles of the two triangles have equal measure. They are called corresponding angles. The sides opposite the corresponding angles are called corresponding sides.

30°

30°

60°60°

SZ X

Y

R T

Determine whether the pair of triangles is similar. Justify your answer.

Since corresponding angles do not have the equal measures, the triangles are not similar.

50°75°

45°89°R

S

T

X

Y

Z

Determine whether the pair of triangles is similar. Justify your answer.

The measure of ∠G = 180° - (90° + 45°) = 45°. The measure of ∠I = 180° - (45° + 45°) = 90°. Since corresponding angles have equal measures, �EFG ∼ �HIJ.

90°

45° 45°45°

F

G

H

J

I

E

ExercisesDetermine whether each pair of triangles is similar. Justify your answer.

1. 2. 3.

4. 5. 6.

20°

30°

120°

115°45°45°

80°

55°

40°

30°30°

110°

90°

45°

60°60°

30°

30°30°

120°

10-7

Similar Triangles

If two triangles are similar, then the

measures of their corresponding sides

are proportional and the measures of

their corresponding angles are equal.

�ABC ∼ �DEF

AB − DE

= BC − EF

= AC − DF

A

B D

E

F

C

Example 1 Example 2


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Similar Triangles

Find Unknown Measures If some of the measurements are known, proportions can be used to find the measures of the other sides of similar triangles.

INDIRECT MEASUREMENT �ABC ∼ �AED in the figure at the right. Find the height of the apartment building.

Let BC = x.ED − BC

= AD − AC

= 25 − 300

ED = 7, AD = 25, AC = 300

25x = 2100 Find the cross products.

x = 84The apartment building is 84 meters high.

ExercisesFind the missing measures for the pair of similar triangles if �ABC ∼ �DEF.

1. c = 15, d = 8, e = 6, f = 10

2. c = 20, a = 12, b = 8, f = 15

3. a = 8, d = 8, e = 6, f = 7

4. a = 20, d = 10, e = 8, f = 10

5. c = 5, d = 10, e = 8, f = 8

6. a = 25, b = 20, c = 15, f = 12

7. b = 8, d = 8, e = 4, f = 10

8. INDIRECT MEASUREMENT Bruce likes to amuse his brother by shining a flashlight on his hand and making a shadow on the wall. How far is it from the flashlight to the wall?

9. INDIRECT MEASUREMENT A forest ranger uses similar triangles to find the height of a tree. Find the height of the tree.

12 ft

20 ft

x

100 ft

5 in.6 in.

x in.

4 ft

Note: Not drawn to scale

A

c a

b e

f d

C D F

EB

25 mNote: Not drawn to scale

275 m

7 m

A D

E

C

B

x

10-7

Example

7 − x


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Skills PracticeSimilar Triangles

P R

Q

q

r p

S U

T

t

u s

52° 52°

63°

65°F E HJ

KG40°

40°45°

F

E G H

JK

60°

60°

57°60°

XV

W

U

Z

Y

40°

50°

A

D F

E

C

B

10-7

Determine whether each pair of triangles is similar. Justify your answer.

1. 2.

3. 4.

Find the missing measures for the pair of similar triangles if �PQR �STU.

5. r = 4, s = 6, t = 3, u = 2

6. t = 8, p = 21, q = 14, r = 7

7. p = 15, q = 10, r = 5, s = 6

8. p = 48, s = 16, t = 8, u = 4

9. q = 6, s = 2, t = 3 − 2 , u = 1 −

2

10. p = 3, q = 2, r = 1, u = 1 − 3

11. p = 14, q = 7, u = 2.5, t = 5

12. r = 6, s = 3, t = 21 − 8 , u = 9 −

4


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PracticeSimilar Triangles

Determine whether each pair of triangles is similar. Justify your answer.

1. 2.

Find the missing measures for the pair of similartriangles if �ABC ∼ �DEF.

3. c = 4, d = 12, e = 16, f = 8

4. e = 20, a = 24, b = 30, c = 15

5. a = 10, b = 12, c = 6, d = 4

6. a = 4, d = 6, e = 4, f = 3

7. b = 15, d = 16, e = 20, f = 10

8. a = 16, b = 22, c = 12, f = 8

9. a = 5 − 2 , b = 3, f = 11 −

2 , e = 7

10. c = 4, d = 6, e = 5.625, f = 12

11. SHADOWS Suppose you are standing near a building and you want to know its height. The building casts a 66-foot shadow. You cast a 3-foot shadow. If you are 5 feet 6 inches tall, how tall is the building?

12. MODELS Truss bridges use triangles in their support beams. Molly made a model of a truss bridge in the scale of 1 inch = 8 feet. If the height of the triangles on the model is 4.5 inches, what is the height of the triangles on the actual bridge?

D F

E

e

f d

A C

B

b

c a

80°

47°47°

56°

E H

F

GD

C

31° 59°R Q S T

UP

10-7


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Word Problem PracticeSimilar Triangles

1. CRAFTS Layla is wants to buy a set of similar magnets for her refrigerator door. Layla finds the magnets below for sale at a local shop. Which two are similar?

2. EXHIBITIONS The world’s largest candle was displayed at the 1897 Stockholm Exhibition. Suppose Lars measured the length of the shadow it cast at 11:00 A.M. and found that it was 12 feet. Suppose that immediately after this, he measured to find that a nearby 25-foot tent pole cast a shadow 5 feet long. How tall was the world’s largest candle?

3. LANDMARKS The Toy and Miniature Museum of Kansas City displays a miniature replica of George Washington’s Mount Vernon mansion. The miniature house is 10 feet long, 6 feet wide, 8 feet tall, and has 22 rooms. The scale of the model to the original is one inch to one foot. If the roof gable of the miniature has dimensions as shown on the diagram below, what is the height of the roof gable on the original Mount Vernon mansion?

Source: Mount Vernon

4. SURVEYING Surveyors use properties of triangles including similarity and the Pythagorean Theorem to find unknown distances. Use the dimensions on the diagram to find the unknown distance xacross the lake.

5. PUZZLES The figure below shows an ancient Chinese movable puzzle called a tangram. It has 7 pieces that can be reconfigured to produce an endless number of designs and pictures.

Assume that the side length of this tangram square is √�2 cm. Leave your answers as simplified radical expressions.

a. What are the side lengths of triangles 1 and 2?

b. What are the side lengths of triangle 7?

c. What are the side lengths of triangles

3 and 5?

A B C

7cm 6cm

5cm

3cm

4cm

8cm

3 ft

1.9 ft 1.9 ft

height

N P

M

x

Q

80 m 40 m

120 m 0

lake

1

2

3

4

5

67

10-7


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Enrichment

A Curious Construction

Many mathematicians have been interested in ways to construct the number π. Here is one such geometric construction.In the drawing, triangles ABC and ADE are righttriangles. The length of

−−− AD equals the length of

−− AC

and −−

FB is parallel to −−−

EG .The length of

−−− BG gives a decimal approximation

of the fractional part of π to six decimal places.

Follow the steps to find the length of −−

BG . Roundto seven decimal places.

1. Use the length of −−−

BC and the Pythagorean Theorem to find the length of

−− AC .

2. Find the length of −−−

AD .

3. Use the length of −−−

AD and the Pythagorean Theorem to find the length of −−

AE .

4. The sides of the similar triangles FED and DEA are in proportion. So, FE − 0.5

= 0.5 − AE

.Find the length of

−− FE .

5. Find the length of −−

AF .

6. The sides of the similar triangles AFB and AEG are in proportion. So, AF − AE

= AB − AG

. Find the length of

−−− AG .

7. Now, find the length of −−−

BG .

8. The value of π to seven decimal places is 3.1415927. Compare the fractional part of pi with the length of

−−− BG .

7–8

1–2

C

E

DGBA

F

1

10-7


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10-8

Trigonometric Ratios Trigonometry is the study of relationships of the angles and the sides of a right triangle. The three most common trigonometric ratios are the sine, cosine, and tangent.

sine of ∠A = leg opposite ∠A

− hypotenuse

sine of ∠B = leg opposite ∠B

− hypotenuse

sin A = a − c

sin B = b − c

cosine of ∠A = leg adjacent to ∠A

−− hypotenuse

cosine of ∠B = leg adjacent to ∠B

−− hypotenuse

cos A = b − c

cos B = a − c

tangent of ∠A = leg opposite ∠A

−− leg adjacent to ∠A

tangent of ∠B = leg opposite ∠B

−− leg adjacent to ∠B

tan A = a − b

tan B = b − a

Find the values of the three trigonometric ratios for angle A .

Step 1 Use the Pythagorean Theorem to find BC. a2 + b2 = c2 Pythagorean Theorem

a2 + 82 = 102 b = 8 and c = 10

a2 + 64 = 100 Simplify.

a2 = 36 Subtract 64 from both sides.

a = 6 Take the square root of each side.

Step 2 Use the side lengths to write the trigonometric ratios.

sin A = opp

− hyp

= 6 − 10

= 3 − 5 cos A =

adj −

hyp = 8 −

10 = 4 −

5 tan A =

opp −

adj = 6 −

8 = 3 −

4

ExercisesFind the values of the three trigonometric ratios for angle A.

1.

817

2. 3

5

3.

24

7

Use a calculator to find the value of each trigonometric ratio to the nearest ten-thousandth.

4. sin 40° 5. cos 25° 6. tan 85°

Study Guide and InterventionTrigonometric Ratios

Example

a10

8

b

a

c


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10-8

Use Trigonometric Ratios When you find all of the unknown measures of the sides and angles of a right triangle, you are solving the triangle. You can find the missing measures of a right triangle if you know the measure of two sides of the triangle, or the measure of one side and the measure of one acute angle.

Solve the triangle. Round each side length to the nearest tenth.

Step 1 Find the measure of ∠B. The sum of the measures of the angles in a triangle is 180.180° − (90° + 38°) = 52°The measure of ∠B is 52°.

Step 2 Find the measure of −−

AB . Because you are given the measure of the side adjacent to ∠ A and are finding the measure of the hypotenuse, use the cosine ratio.

cos 38° = 13 − c Definition of cosine

c cos 38° = 13 Multiply each side by c.

c = 13 − cos 38° Divide each side by sin 41°.

So the measure of −−

AB is about 16.5.

Step 3 Find the measure of −−−

BC . Because you are given the measure of the side adjacent to ∠ A and are finding the measure of the side opposite ∠ A, use the tangent ratio.

tan 38° = a − 13

Definition of tangent

13 tan 38° = a Multiply each side by 13.

10.2 ≈ a Use a calculator.

So the measure of −−−

BC is about 10.2.

ExercisesSolve each right triangle. Round each side length to the nearest tenth.

1.

9

ab

30°

2.

b

8c

44°

3.

16

cb

56°


Trigonometric Ratios

Example

13

ac

38°


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Find the values of the three trigonometric ratios for angle A.

1.

77

85

2.

159

3. 10

24

4. 15

8

12


5. sin 18° 6. cos 68° 7. tan 27°

8. cos 60° 9. tan 75° 10. sin 9°

Solve each right triangle. Round each side length to the nearest tenth.

11.

1317°

12.

655°

Find m ∠J for each right triangle to the nearest degree.

13.

6

5 14.

19

11

10-8 Skills PracticeTrigonometric Ratios


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Find the values of the three trigonometric ratios for angle A.

1.

72

97

2.

36

15


3. tan 26° 4. sin 53° 5. cos 81°

Solve each right triangle. Round each side length to the nearest tenth.

6.

67°

22

7.

9

29°

Find m∠J for each right triangle to the nearest degree.

8. 11

5

9.

1218

10. SURVEYING If point A is 54 feet from the tree, and the angle between the ground at point A and the top of the tree is 25°, find the height h of the tree.

10-8 PracticeTrigonometric Ratios

25°

54 ft

h


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10-8 Word Problem PracticeTrigonometric Ratios

1. WASHINGTON MONUMENT Jeannie is trying to determine the height of the Washington Monument. If point A is 765 feet from the monument, and the angle between the ground and the top of the monument at point A is 36°, find the height h of the monument to the nearest foot.

765 ft

36°

h

2. AIRPLANES A pilot takes off from a runway at an angle of 20º and maintains that angle until it is at its cruising altitude of 2500 feet. What horizontal distance has the plane traveled when it reaches its cruising altitude?

3. TRUCK RAMPS A moving company uses an 11-foot-long ramp to unload furniture from a truck. If the bed of the truck is 3 feet above the ground, what is the angle of incline of the ramp to the nearest degree?

4. SPECIAL TRIANGLES While investigating right triangle KLM, Mercedes finds that cos M = sin M. What is the measure of angle M?

5. TELEVISIONS Televisions are commonly sized by measuring their diagonal. A common size for widescreen plasma TVs is 42 inches.

42’’ h

h169

a. A widescreen television has a 16:9 aspect ratio, that is, the screen width

is 16 − 9 times the screen height. Use the

Pythagorean Theorem to write an equation and solve for the height h of the television in inches.

b. Use the information from part a to solve the right triangle.

c. What would the measure of angle A be on a standard television with a 4:3 aspect ratio?


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In addition to the sine, cosine, and tangent, there are three other common trigonometric ratios. They are the secant, cosecant, and cotangent.

secant of ∠A = hypotenuse

− leg opposite ∠A

secant of ∠B = hypotenuse

− leg opposite ∠B

sec A = c − a

sec B = c − b

b

a

ccosecant of ∠A =

hypotenuse −

leg adjacent ∠A

cosecant of ∠B = hypotenuse

− leg adjacent ∠B

csc A = c − b

csc B = c − a

cotangent of ∠A = leg adjacent to ∠A

−− leg opposite ∠A

cotangent of ∠B = leg adjacent to ∠B

−− leg opposite ∠B

cot A = a − b

cot B = b − a

Find the secant, cosecant, and cotangent of angle A.

Use the side lengths to write the trigonometric ratios.

sec A = hyp

− opp

= 15 − 9 = 5 −

3 csc A =

hyp −

adj = 15 −

12 = 5 −

4

cot A = adj

− opp

= 12 − 9

= 4 − 3

ExercisesFind the secant, cosecant, and cotangent of angle A.

1.

817

2.

3 5

3.

24

7

4. How does the sine of an angle relate to the angle’s cosecant? How does the cosine of an angle relate to the angle’s secant? How does the cotangent of an angle relate to the angle’s secant?

Use the relations that you found in Exercise 4 and a calculator to find the value of each trigonometric ratio to the nearest ten-thousandth.

5. sec 17° 6. csc 49° 7. cot 81°

10-8 EnrichmentMore Trigonometric Ratios

Example

12

159


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10 Student Recording SheetUse this recording sheet with pages 664–665 of the Student Edition.

Read each question. Then fill in the correct answer.

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. A B C D

Multiple Choice

Record your answer in the blank.

For gridded response questions, also enter your answer in the grid by writing each number or symbol in a box. Then fill in the corresponding circle for that number or symbol.

8. ———————— (grid in)

9. ————————

10. ———————— (grid in)

11. ————————

12. ———————— (grid in)

13. ————————

14. ————————

15. ———————— (grid in)

Extended Response

Short Response/Gridded Response

Record your answers for Question 16 on the back of this paper.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

8.

12.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

7

6

5

4

3

2

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15.

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0

9

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6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

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9

8

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4

3

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0


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Rubric for Scoring Extended Response

General Scoring Guidelines

• If a student gives only a correct numerical answer to a problem but does not show how he or she arrived at the answer, the student will be awarded only 1 credit. All extended response questions require the student to show work.

• A fully correct answer for a multiple-part question requires correct responses for all parts of the question. For example, if a question has three parts, the correct response to one or two parts of the question that required work to be shown is not considered a fully correct response.

• Students who use trial and error to solve a problem must show their method. Merely showing that the answer checks or is correct is not considered a complete response forfull credit.

Exercise 16 Rubric

Score Specific Criteria

4 A correct solution that is supported by well-developed, accurate explanations. The distance between Karen’s school and the park is 25.1 miles. The coordinates of Karen’s house are (0.5, 0.5). The student should show a working knowledge of the Distance and Midpoint formulas.

3 A generally correct solution, but may contain minor flaws in reasoning or computation.

2 A partially correct interpretation and/or solution to the problem.

1 A correct solution with no evidence or explanation.

0 An incorrect solution indicating no mathematical understanding of the concept or task, or no solution is given.

10


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1. Graph y = 2 √ �� x + 1 . State the domain and range.

2. MULTIPLE CHOICE Which expression has a domain of {x | x ≥ 4}?

A y = √ �� x - 4 B y = √ �� x + 4 C y = √ � x - 4 D y = √ � x + 4


3. √ �� 72

4. √ �� 8 x 2 y

5. 3 − 4 + √ � 2

Chapter 10 Quiz 1 (Lessons 10-1 and 10-2)

1.

2.

3.

4.

5.

10

Chapter 10 Quiz 2(Lessons 10-3 and 10-4)

10Simplify each expression.

1. 6 √ �� 45 + 2 √ �� 80 2. 5 √ � 6 - 4 √ �� 10 - √ � 6 + 12 √ �� 10

3. √ �� 10 ( √ � 5 + 3 √ � 2 ) 4. ( √ � 6 - √ � 5 ) ( √ �� 10 + √ � 3 )

5. MULTIPLE CHOICE Find the perimeter of a rectangle with a width 2 √ � 5 + 3 √ �� 11 and a length 3 √ � 5 - √ �� 11 .

A 7 √ �� 55 - 3 B 5 √ � 5 + 2 √ �� 11 C 14 √ �� 55 - 6 D 10 √ � 5 + 4 √ �� 11


6. √ �� 2x + 6 + 6 = 10 7. √ �� c + 2 = c - 4

8. √ �� (11x - 24) = x 9. 3 √ �� (m + 5) - 3 = 6

10. √ �� (3a + 4) = √ �� (12a - 14)

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x


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1. If c is the measure of the hypotenuse of a right triangle,find the missing measure. If necessary, round to the nearest hundredth. b = 3, c = 15, a = ?

2. Determine whether the side measures 7, 9, and 12 form a right triangle.

3. MULTIPLE CHOICE What is the area of triangle MNP? A 29.68 units2 B 19.21 units2

C 153.67 units2 D 307.35 units2

4. Find the distance between the points at (0, 7)and (-5, 13).

5. Find the coordinates of the midpoint of the segment with endpoints (-3, 11) and (-5, 5).

1. Determine whether the pair of triangles is similar. Justify your answer.

2. Find the measures of the missing sides if � ABC ∼ � XYZ. a = 9, c = 15, x = 6, y = 14

3. SHADOWS If a 26-foot tree casts a shadow that is 14 feet long and a nearby tower casts a shadow that is 21 feet long, how tall is the tower?

4. MULTIPLE CHOICE Which is not equal to 1?

A sin 45˚ B tan 45˚ C cos 0˚ D sin 90˚

5. Solve the triangle. Round each side length to the nearest tenth.

1.

2.

3.

4.

5.

16

25

P

M N

X y Z A b C

z x c aY

B

Chapter 10 Quiz 3(Lessons 10-5 and 10-6)

10 Chapter 10 Quiz 4(Lessons 10-7 and 10-8)

10

1.

2.

3.

4.

5.

A

D

BE F

C100˚

100˚ 30˚

50˚

1020°


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Part I Write the letter for the correct answer in the blank at the right of each question.

1. Which expression has a range of { y | y ≥ 2}?

A y = √��x - 2 B y = √��x + 2 C y = √�x - 2 D y = √�x + 2

2. Which expression has a domain of {x | x ≥ 1}?

F y = √��x - 1 G y = √��x + 1 H y = √�x - 1 J y = √�x + 1

For Questions 3–5, simplify each expression.

3. √��288 A 4 √��18 B 2 √��12 C 4 √�6 D 12 √�2

4. √��20 x 3y 2

F 5x|y|2 √�x G 2x|y| √�5x H 2|x|y √�5x J 5|x|y √�2x

5. √��t−18

A √ � t −3 √ � 2

B |t|−18

C 3t−18

D √ � 2t −

6

6. Solve √��5n - 1 - n = 1. F 1, 2 G -1, -2 H 1−

4J 1

7. Solve √ �� (7 - 2b) = √ �� (9 - b)

A 1 − 2 B 2 C - 1 −

2 D -2

Part II


8. √ �� 15 (2 √ � 3 - 4 √ � 5 ) 9. (4 √ � 3 + 5) (4 √ � 3 - 5)

10. √ �� 288 + 3 √ �� 162 11. 6 √ � 5 - 2 √ �� 10 + √ � 5

12. 3 √ �� 50 - 2 √ �� 72 + √ �� 24

For Questions 13 and 14, solve each equation.

13. 2 √ � 5x - 3 = 7 14. √ �� x - 4 = x - 24

15. A square has an area of 90 square inches. The formula for the area A of a square with side length ℓ is A = ℓ2. Find the length of one side of the square.

1.

2.

3

4.

5.

6.

7.

10 Chapter 10 Mid-Chapter Test(Lessons 10-1 through 10-4)

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Choose from the terms above to complete each sentence.

1. If you exchange the hypothesis and conclusion of an if-then statement, the result is the of the statement.

2. The trigonometric ratio equivalent to the leg adjacent to an angle divided by the hypotenuse is the .

3. If a coordinate grid is superimposed on a map, you can find the distance between two places on the map using the

.

4. The equation 8 = 3 √ � d is an example of a .

5. The binomials 5 √ � 3 + 2 √ � 5 and 5 √ � 3 - 2 √ � 5 are .

6. In a right triangle, the side opposite the right angle is the .

7. The expression 7x under the radical symbol in √ � 7x is called the .

8. The two sides of a right triangle that are not the hypotenuse are called .

9. If two triangles have three pairs of corresponding angles with equal measures, the triangles are .

Define each term in your own words.

10. trigonometry

11. tangent

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

conjugates

converse

cosine

Distance Formula

hypotenuse

inverse cosine

inverse sine

inverse tangent

legs

midpoint

Midpoint Formula

radical equation

radical function

radicand

similar triangles

tangent

trigonometry

10 Chapter 10 Vocabulary Test


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Chapter 10 Test, Form 1

Write the letter for the correct answer in the blank at the right of each question.

1. How does the graph of y = √ � x + 2 compare to the parent graph?

A translated up 2 C translated left 2 B translated down 2 D translated right 2

2. Which expression has a domain of {x | x ≥ -1}?

F y = √ �� x + 1 G y = √ �� x - 1 H y = √ � x + 1 J y = √ � x - 1


3. √ �� 90

A 9 √ �� 10 B 10 √ � 9 C 3 √ �� 10 D √ �� 30

4. 3 − 5 - √ � 2

F 15 + 3 √ � 2 −

23 G 15 - 3 √ � 2

− 23

H 15 + 3 √ � 2 J 15 + 3 √ � 2 −

3

5. 6 √ � 5 - 2 √ � 5

A 4 B -12 C -12 √ � 5 D 4 √ � 5

6. 3 √ �� 12 + √ �� 27 - 2 √ �� 20

F 14 √ � 3 - 4 √ � 5 G 3 √ � 3 - √ � 2 H 9 √ � 3 - 4 √ � 5 J 21 √ � 3 - 8 √ � 5

7. √ � 2 ( √ � 6 + 3 √ � 2 )

A 3 √ � 2 + 6 B 6 √ � 2 C 2 √ � 3 + 3 √ � 2 D 2 √ � 3 + 6

8. Solve √ �� 2x - 5 = 3.

F 4 G 7 H -8 J 11 − 2

9. Solve √ �� 2x + 8 = x.

A -2, 4 B 4 C -2 D 2, 4

10. Find the length of the hypotenuse of a right triangle if a = 3 and b = 4.

F 5 G √ � 7 H 25 J 7

11. Determine which side measures form a Pythagorean triple.

A 4, 5, 6 B 3, 4, 5 C 5, 11, 12 D 4, 8, 12

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

10

12. Find the coordinates of the midpoint of the segment with endpoints at (1, 3) and (9, 9).

F (4, 6) G (5, 6) H (8, 6) J (10, 12) 12.


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13.

14.

15.

16.

17.

18.

19.

20.

Chapter 10 Test, Form 1 (continued)

13. Find the distance between the points at (0, 0) and (5, 12).

A 17 B √ �� 17 C √ �� 60 D 13

14. Which pair of triangles is similar?

F H 50˚

50˚

85˚

85˚

G

40˚

40˚

55˚

60˚

J

50˚ 45˚

15. If � ABC ˜ � DEF, and c = 8, f = 4, and b = 12, find e.

A 24 B 8 C 6 D 2 2 − 3

16. What is the length of a diagonal of a rectangle with a length of 8 meters and a width of 6 meters?

F 10 m G 14 m H 48 m J 100 m

17. Determine which side measures form a right triangle.

A 10, 24, 28 B 13, 17, 21 C √ � 3 , √ � 4 , √ � 5 D 5, 12, 13

18. SAILING A 12-foot cable attached to the top of the mast of a sailboat is fastened to a point on the deck 4 feet from the base of the mast. What is the height of the mast?

F 9.56 ft G 22 ft H 11.31 ft J 128 ft

For Questions 19 and 20, the leg adjacent to ∠ A in a right triangle measures 8 units, and the hypotenuse measures 13 units.

19. What is cos A?

A 8 − 13

B 13 − 8 C 38° D 52°

20. What is m∠A?

F 1° G 32° H 38° J 52°

Bonus Simplify √ �� 4 x 2 + 4x + 1 . B:

10

50˚ 50˚

x˚ x˚


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1. How does the graph of y = √ �� x + 3 compare to the parent graph?

A translated up 3 C translated right 3 B translated down 3 D translated left 3

2. Which expression has a domain of {x | x ≥ 2} ?

F y = √ � x + 2 G y = √ � x - 2 H y = √ �� x + 2 J y = √ �� x - 2


3. 5 √ � 3 · 2 √ �� 21

A 70 √ � 3 B 10 √ �� 63 C 49 √ � 3 D 30 √ � 7

4. √ �� x 2 −

12

F x 2 − 2 √ � 3

G |x| √ � 3

− 6 H x −

6 J

|x| −

√ �� 12

5. 5 − √ �� 11 - √ � 6

B 5 √ �� 66 −

66 C √ �� 11 + √ � 6 D 5 √ �� 11 + 5 √ � 6

− 17

6. √ �� 18 - √ �� 54 + 2 √ �� 50

F 13 √ � 2 - 3 √ � 6 G -4 √ � 3 + 4 √ � 5 H -4 √ � 3 - 4 √ � 5 J 8 √ � 2 - 3 √ � 6

7. ( √ �� 14 + √ � 3 ) ( √ � 6 - √ � 7 )

A 2 √ � 5 - √ �� 21 + 3 - √ �� 10 C √ �� 21

B √ �� 21 - 4 √ � 2 D √ �� 21 + √ � 2

8. Solve √ �� 3x - 2 + 4 = 8.

F 12 G 6 H 2 − 3 J 3 −

2

9. Solve √ �� 7a + 32 = a + 2.

A -4 B 7 C -4, 7 D -7, 4

10. A right triangle has one leg that is 7 centimeters. The hypotenuse is 25 centimeters. Find the length of the other leg.

F 15 cm G √ �� 674 cm H 24 cm J 5 √ � 7 cm


A 3, 8, 12 B 5, 9, 11 C 11, 13, 16 D 6, 8, 10

12. Find the distance between the points at (-3, 4) and (2, 7). F √ �� 34 G √ �� 74 H 2 √ �� 30 J √ �� 10

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2.

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7.

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12.

Chapter 10 Test, Form 2A 10

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13. Find the coordinates of the midpoint of the segment with endpointsat the origin and (-6, 8).

A (-12, 16) B (-3, 4) C (3,-4) D (6, -8)

14. Which pair of triangles is similar? F H

G J

15. If � ABC ˜ � DEF, and a = 10, b = 12, d = 6, and f = 6.6, find the measures of the missing sides.

A c = 11, e = 7.2 B c = 7.3, e = 10.9

C c = 4, e = 20 D c = 6.6, e = 8.9

16. What is the length of a diagonal of a rectangle with a length of 9 inches and a width of 3 inches?F 3.5 in. G 9.5 in. H 18 in. J 90 in.

17. Determine which side measures form a right triangle. A 1, 2, 3 B 2, 3, 4 C 3, 4, 5 D 4, 5, 6

18. LADDERS A 16 foot ladder leans against a wall. The base of the ladder is 6 feet from where the wall meets the ground. How far up the wall does the ladder reach?

F 14.8 ft G 12.9 ft H 144 ft J 220 ft

For Questions 19 and 20, the leg opposite to ∠ A in a right trianglemeasures 12 units, and the hypotenuse measures 19 units.

19. What is sin A?

A 12 − 19

B 19 − 12

C 0.775 D 0.815

20. What is m∠ A? F 0.01° G 32° H 39° J 51°

Bonus Find the length of a diagonal of a square if its area is 72 square meters.

13.

14.

15.

16.

17.

18.

19.

20.

50˚ 45˚ 40˚100˚

50˚50˚

50˚ 30˚60˚ 45˚

Chapter 10 Test, Form 2A (continued)

B:

FDCA

B Ec a f d

eb

10


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1. How does the graph of y = √ �� x - 8 compare to the parent graph? A translated up 8 C translated left 8 B translated down 8 D translated right 8

2. Which expression has a range of {y | y ≥ 1}?

F y = √ � x + 1 G y = √ � x - 1 H y = √ �� x + 1 J √ �� x - 1


3. 3 √ � 6 · 5 √ � 2

A 24 √ � 2 B 30 √ � 3 C 45 √ � 2 D 15 √ �� 12

4. √ �� 18 − y

F 3 √ � 2y

− y G 3 √ � 2 − y H 6 − y J 3 √ � 2 − y

5. 3 − √ � 8 + √ � 5

A 1 B 3 √ �� 40 −

40 C 2 √ � 2 - √ � 5 D 6 √ � 2 - 3 √ � 5

− 13

6. 3 √ �� 32 - 2 √ �� 18 + √ �� 54

F 4 √ � 2 - 3 √ � 6 G 2 √ � 6 + 6 √ � 3 H 2 √ � 6 - 6 √ � 3 J 6 √ � 2 + 3 √ � 6

7. ( √ � 7 - √ �� 10 ) ( √ � 5 + √ �� 14 )

A 2 √ � 3 + √ �� 21 - √ �� 15 - 2 √ � 6 C 2 √ � 2 - √ �� 35

B - √ �� 35 D 12 √ � 2 + 3 √ �� 35

8. Solve √ �� 3n + 1 + 3 = 7.

F 13 G 1 − 3 H -1, 1 −

3 J 5

9. Solve √ �� 5x + 39 = x + 3.

A -6, 5 B -6 C 5 D -5, 6

10. A right triangle has one leg that is 8 inches. The hypotenuse is 17 inches. Find the length of the other leg.

F 15 in. G √ �� 353 in. H 9 in. J 2 √ �� 34 in.


A 4, 7, 8 B 9, 12, 15 C 3, 7, 9 D 10, 15, 20

12. Find the distance between the points at (-2, 7) and (-3, -4).

F √ �� 29 G √ �� 82 H √ �� 34 J √ �� 122

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5.

6.

7.

8.

9.

10.

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12.

Chapter 10 Test, Form 2B10


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13. Find the coordinates of the midpoint of the segment with endpoints at the origin and (-6, 4).

A (-12, 8) B (-3, 2) C (3, -2) D (6, -4)

14. Which pair of triangles is similar?

F

55˚ 50˚

H

80˚

60˚40˚

60˚

G 35˚72˚

35˚ 68˚

J 30˚70˚

15. If � ABC ∼ � ABC, and a = 10, c = 12, d = 9, and e = 7.2, find the measures of the missing sides.

A b = 6.5, f = 13.3 C b = 12.5, f = 7.5 B b = 6, f = 11.5 D b = 8, f = 10.8

16. What is the length of a diagonal of a rectangle with a length of 14 inches and a width of 7 inches?F 9.4 in. G 15.7 in. H 49 in. J 245 in.

17. Determine which side measures form a right triangle.A 5, 12, 13 B 6, 13, 14 C 7, 14, 15 D 8, 15, 16

18. LADDERS A 24-foot ladder leans against a wall. The base of the ladder is 9 feet from where the wall meets the ground. How far up the wall does the ladder reach?F 495 ft G 20.9 ft H 81 ft J 22.2 ft

For Questions 19 and 20, the leg opposite to ∠ B in a right triangle measures 15 units, and the hypotenuse measures 28 units.

19. What is sin A?

A 15 − 28

B 28 − 15

C 0.634 D 0.844

20. What is m∠ A?

F 0.01° G 28° H 32° J 58°

Bonus Find the length of a diagonal of a square if its area is 98 square feet.

Chapter 10 Test, Form 2B (continued)

13.

14.

15.

16.

17.

18.

19.

20.

B:

10

F

D

C

A B Ec

ab

f

de


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1. Graph y = √ �� x - 2 + 1 and compare to the parent graph.State the domain and range.

2. State the domain and range of y = -3 √ �� x - 1 + 5.


3. √ �� 24 · √ � 3

4. √ �� 75 y 4 w 3

5. 2 √ � 3 −

√ � 6 - 2

6. √ �� 20 + 2 √ �� 45 + 3 √ �� 80

7. ( √ � 6 + √ � 7 ) ( √ �� 21 - √ � 2 )


8. √ � m = 2 √ � 3

9. √ �� 2a + 14 - 13 = -7

10. 10 + √ �� x - 8 = x

If c is the measure of the hypotenuse of a right triangle, find each missing measure. If necessary, round to the nearest hundredth.

11. a = 6, b = 10, c = ? 12. b = 24, c = 25, a = ?


13. 14, 48, 50 14. 12, 24, 36

For Questions 15 and 16, find the distance between each pair of points whose coordinates are given. Express answers in simplest radical form and as decimal approximations rounded to the nearest hundredth,if necessary.

15. (0, -4), (5, 2) 16. (7, 3), (-4, 11)

17. Find the coordinates of the midpoint of the segment withendpoints (3, -2) and (-5, 6).

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

Chapter 10 Test, Form 2C10

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For Questions 18 and 19, determine whether each pair of triangles is similar. Justify your answer.

18. 50˚

75˚ 75˚

55˚

19. 55˚

40˚

20. Find the measures of the missing sides for the set of measures given if � MET ∼ � CUB. b = 2, e = 9, m = 7, t = 3

21. Find the height of the tree.

2 ft

3 ft

25 ft

22. A boat leaves the harbor and sails 7 miles west and 2 miles north to an island. The next day it travels to a second island 5 miles south and 3 miles east of the harbor. How far is it from the first island to the second island?

23. What is the length of a rectangle if the width is 10 cm and the diagonal is 16 cm?

24. Solve m ∠ J for the right triangle

23

10

to the nearest degree.

25. At a loading dock, a ramp is 55 feet long. The angle theramp makes with the ground is 25°. Find the height reachedby the ramp.

Bonus Solve 8 - 3x = √ �� 4 x 2 + 20 + 8.

Chapter 10 Test, Form 2C (continued)

18.

19.

20.

21.

22.

23.

24.

25.

B:

BUTE

MC

te

bu

cm

10


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1. Graph y = √ �� x + 1 - 3 and compare to the parent graph. State the domain and range.

2. State the domain and range of y = -2 √ �� x + 2 - 1.


3. √ �� 40 · √ � 5

4. √ �� 50 x 3 y 2

5. 5 √ � 2 −

√ �� 10 - 3

6. 2 √ �� 24 + √ �� 54 + 3 √ �� 150

7. ( √ �� 11 - √ � 6 ) ( √ � 2 + √ �� 33 )


8. √ �� 7x - 3 = 5

9. √ �� 4x − 3 - 2 = 0

10. x + 3 = √ �� 3x + 37

If c is the measure of the hypotenuse of a right triangle, find each missing measure. If necessary, round to the nearest hundredth.

11. a = 4, b = 7, c = ? 12. b = 15, c = 17, a = ?


13. 15, 20, 25 14. 16, 20, 30

Find the distance between each pair of points whose coordinates are given. Express answers in simplest radical form and as decimal approximations rounded to the nearest hundredth, if necessary.

15. (-3, 0), (2, 7) 16. (5, 8), (-7, 1)

17. Find the coordinates of the midpoint of the segment with endpoints (7, -1) and (-1, 5).

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

Chapter 10 Test, Form 2D10

y

x

1.


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18.

19.

20.

21.

22.

23.

24.

25.

B:


18. 50˚

35˚

85˚

19. 50˚

50˚

20. Find the measures of the missing sides for the set of measures given if � ABC ∼ � XYZ. b = 15, c = 9, x = 9, z = 6

21. Find the height of the lamppost.

18 ft 9 ft

5 ft

For Questions 22 and 23; round to the nearest hundredth.

22. Mandy leaves her home for a walk. How far is she from her home after walking 2 miles due east and then 5 miles due south?

23. What is the width of a rectangle if the length is 13 cm and the diagonal is 20 cm?

24. Solve m∠J for the right triangle to thenearest degree.

25. At a loading dock, a ramp is 80 feet long. The angle the ramp makes with the ground is 22°. Find the height reached by the ramp.

Bonus Solve 12 + √ �� 5 x 2 + 36 = 12 - 3x.

Chapter 10 Test, Form 2D (continued)10

26

10

Z

X

C

A

BY

c

a

b

x

z

y


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1. Graph y = -2 √ �� x - 2 - 2 and compare to the parent graph.State the domain and range.

2. State the domain and range of y = -8 √ �� x + 4 - 12.


3. √ �� 378 · √ � 6

4. √ �� 5 x 4 −

4 n 5

5. √ � 8 −

2 √ � 5 + √ � 6

6. 5 √ �� 12 + 6 √ � 1 − 3 - 3 √ �� 48

7. (2 √ � 6 + 7 √ � 5 ) (2 √ �� 10 - 5 √ � 3 )

For Questions 8–10, solve each equation. If necessary, leave in simplest radical form.

8. √ �� 3n - 2 + 6 = 10

9. √ �� 5n − 3 + 12 = 7

10. 2x = 6 + √ �� 2 x 2 - 7x + 1

11. Find the length of the hypotenuse of a right triangle if a = √ � 5 and b = 6.

12. Find the width of a rectangle with a diagonal of 12 centimeters and a length of 10 centimeters.

For Questions 13 and 14, determine whether thefollowing side measures form right triangles.

13. 16, 49, 65

14. 5, 9, √ �� 106

15. Find the distance between the points at (-4, 6) and (10, 13).

16. Find the value of a if the midpoint of the segment with endpoints (3, a) and ( a −

2 , 10) is (2.5, 7).

17. Find the perimeter of a square ABCD if two of the vertices are A(8, -14) and B(3, -4).

Chapter 10 Test, Form 3

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

10

y

x

1.


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18. 50˚

50˚45˚

19. 60˚

30˚

2x˚x˚

20. Find the measures of the missing sides for the set of measures given if � XYZ ∼ � ABC. a = 2.4, c = 1.02, x = 12.6, y = 8.4

21. Find the distance across the lake from point A to point B, if � ABC ∼ � EDC.

210 m

65 m 100 m

120 m

E

DC

B

A

80˚

80˚

70˚

70˚

For Questions 22 and 23, round to the nearest hundredth.

22. A diagonal of a rectangle measures 15 cm. The length of the rectangle is 11 cm. What is the height of the rectangle?

23. Hubert left his home heading due east. He walked that way for 4 miles then headed due north for 7 miles. How far away is Hubert from his home?

24. Solve m∠J for the right triangle to

7130

the nearest degree.

25. A freeway on-ramp is 625 feet long. The angle the ramp makes with the ground is 8°. Find the height reached by the on-ramp.

Bonus Simplify 2 √ � 6 - √ � 5 −

√ � 6 + 3 √ � 5 .

Chapter 10 Test, Form 3 (continued)

18.

19.

20.

21.

22.

23.

24.

25.

B:

CZ

A

B

X

YEy

z

x

cb

da

10


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Chapter 10 Extended-Response Test

Demonstrate your knowledge by giving a clear, concise solution to each problem. Be sure to include all relevant drawings and justify your answers. You may show your solution in more than one way or investigate beyond the requirements of the problem.

1. The Product Property of Square Roots and the Quotient Property of Square Roots can be written in symbols as √ �� ab = √ � a · √ � b

and √ � a − b = √ � a

− √ � b

, respectively.

a. Explain the Product Property of Square Roots and discuss any limitations of a and b for this property.

b. Explain the Quotient Property of Square Roots and discuss any limitations of a and b for this property.

c. Discuss any similarities of the two properties.

2. The formula L = √ �� kP represents the relationship between an airplane’s length L in feet and the pounds P its wings can lift, where k is a constant of proportionality calculated for each particular plane.

a. Solve the formula for P. b. For k = 0.12, choose three values for L and calculate the takeoff

weight P for each value. c. For k = 0.08, choose three values for L and calculate the takeoff

weight P for each value. d. Determine whether a larger or smaller constant of

proportionality allows a plane to carry more weight.

3. a. Yoki claims that two triangles with two pairs of corresponding angles equal in measure are similar. Do you agree or disagree? Justify your reasoning.

b. Yoki also claims that two triangles with two pairs of sides equal in measure are similar. Do you agree or disagree? Justify your reasoning.

4. y

xB(0, 0) C(a, 0)

A(a, b)

a. Choose values for a and b, then find the distance between each pair of points.

b. Show that the triangle is a right triangle by using the Pythagorean Theorem.

10


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SCORE

1. Simplify (mt2)(m3)(m2t). (Lesson 8-1)

A m6t3 B m6t2 C m9t2 D m3t2

2. Find (x + 2y)2. (Lesson 7-8)

F x2 + 2xy + 2y2 H x2 + 4xy + 4y2

G x2 + 4y2 J 2x2 + 2xy + 4y2

3. Solve 6r2 - 14r - 15 = 0 by using the Quadratic Formula. Round to the nearest tenth. (Lesson 9-5)

A Ø B {-3.1, 0.8} C {-0.8, 3.1} D {0.8, 3.1}

4. Which binomial is a factor of 15t2 - t - 6? (Lesson 8-4)

F 3t - 2 G 5t - 3 H 3t + 1 J 5t - 6

5. Use the graph to identify two consecutive integers between which a root lies. (Lesson 9-2)

A 1, 2 C -3, -2

B -4, -3 D -2, -1

6. Solve x2 - 14x + 49 = 64. (Lesson 8-6)

F {6, 22} G {-1, 15} H {-15, 1} J {-1, 1}

7. Determine the amount of an investment if $800 is invested at an interest rate of 6.5% compounded monthly for 5 years. (Lesson 9-8)

A $15,223.65 B $34,999.87 C $1096 D $1106.25

8. Which expression cannot be simplified? (Lesson 10-3)

F 5 √ � 8 + 2 √ �� 18 H 2 √ �� 112 + √ �� 63

G 3 √ �� 55 - 4 √ �� 65 J 2 √ �� 45 + 4 √ �� 20

9. Find the length of the hypotenuse of a right triangle if a = 21 and b = 20. (Lesson 10-5)

A 6.4 B 841 C 29 D 41

10. Find the value of a if the distance between the points at (-6, 5) and (a, -7) is 13 units. (Lesson 10-6)

F -11 or -1 G -9.6 or -2.4 H 11 or 1 J 9.6 or 2.4

11. If a 10-foot lightpole casts a 12-foot-long shadow and the nearby library casts a 42-foot-long shadow, how high is the library? (Lesson 10-7)

A 35 ft B 50.4 ft C 28 ft D 40 ft

y

xO

10 Standardized Test Practice(Chapters 1–10)

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. A B C D

8. F G H J

9. A B C D

10. F G H J

11. A B C D

Part 1: Multiple Choice

Instructions: Fill in the appropriate circle for the best answer.


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Standardized Test Practice (continued)

12. What is the equation of the line that passes through (3, 2) and (0, -5)? (Lesson 4-2)

F y = ( 3 − 7 ) x - 5 H y = - ( 3 −

7 ) x + 5

G y = - ( 7 − 3 ) x + 5 J y = ( 7 −

3 ) x - 5

13. Find the slope of the line that passes through (5, 6) and (2, 1). (Lesson 3-3)

A - 5 − 3 B 5 −

3 C - 3 −

5 D 3 −

5

14. What is the length of a diagonal of a rectangle with a length of 10 inches and a width of 6 inches? (Lesson 10-5)

F 4 in. G 16 in. H 11.7 in. J 136 in.

15. Solve the proportion b − 12

= 10 − 15

. (Lesson 2-6)

A 6 B 8 C 4 D 120

16. Jackson’s meal cost $33.40. How much money should he leave for a 15% tip? (Lesson 2-7)

F about $2.00 H about $4.00

G about $3.00 J about $5.00

17. What is the length of a diagonal of a square with an area of 36 inches? (Lesson 10-5)

A 9.4 in. B 8.5 in. C 72 in. D 36 in.

18. Find the discounted price. cookbook: $28discount: 65% (Lesson 2-7)

19. The basic breaking strength b in pounds for a natural fiber line is determined by the formula 900c2 = b, where c is the circumference of the line in inches. What circumference in inches of natural line would have 22,500 pounds of breaking strength?

(Lesson 8-5)

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

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9

8

7

6

5

4

3

2

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9

8

7

6

5

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. . . . .

9

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8

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0

10

12. F G H J

13. A B C D

14. F G H J

15. A B C D

16. F G H J

17. A B C D

Part 2: Gridded Response

Instructions: Enter your answer by writing each digit of the answer in a column box

and then shading in the appropriate circle that corresponds to that entry.


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20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30a.

30b.

20. INVESTMENTS Phyllis invested $12,000, part at 14% annual interest and the remainder at 10%. Last year she earned $1632 in interest. How much money did she invest at each rate? (Lesson 2-9)

21. Write an equation of a line whose slope is -5 and whose y-intercept is 14. (Lesson 4-1)

22. Solve 4 + w ≤ 3 or 5w - 14 > -4. Then graph the solution set. (Lesson 5-4)

23. Use a graph to determine whether the system has no solution, one solution, or infinitely many solutions. 2y - x = 1 and 2y - x = -2 (Lesson 6-1)

24. MANUFACTURING A toy manufacturer makes two typeof model airplanes, jets and biplanes. Each month they can make at most 120 planes. Each jet takes 2 hours to build and each biplane takes 5 hours to build. They use 500 hours or less each month to build model airplanes. Make a graph showing the number of jets and biplanes that can be made each month. (Lesson 5-6)

25. Simplify 21 hk 3 j -2

− - 14 h -5 kj 8

. Assume that the denominator does not

equal zero. (Lesson 7-2)

26. Find the GCF of 18a2bc3 and 54ab3c2. (Lesson 8-1)

27. Find two consecutive odd integers whose product is 255. (Lesson 8-3)

28. Solve x2 + 16x + 64 = 13 by taking the square root of each side. Round to the nearest tenth, if necessary. (Lesson 9-4)

29. A conveyer valued at $12,000 depreciates at a steady rate of 18% per year. What is the value of the conveyer in 6 years? (Lesson 9-8)

30. Matt leaves his house to visit some friends. He drives 11 miles due west and then 9 miles due north to get to Joe’s house. He visits for a while and then drives 6 miles due south and 4 miles due west to get to Mason’s house. (Lesson 10-6)

a. When Matt is at Joe’s house, how far is he from home?

b. When Matt is at Mason’s house, how far is he from Joe’s house?

Standardized Test Practice (continued)

(Chapters 1–10)

-1-2-3-4 0 1 2 3 4

10

Part 3: Short Response

Instructions:Write your answers in the space.

b

jO

20406080

100

100 200 300

120


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Chapter 10 A1 Glencoe Algebra 1

Chapter Resources

Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.

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Lesson 10-1

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y =

√ � x ;

D

= {

x |

x ≥

0};

D

= {

x |

x ≥

0};

D

= {

x |

x ≥

0};

R

= {

y |

y ≥

0}

R

= {

y |

y ≥

0}

R =

{y |

y ≥

0}

10-1

Stud

y G

uide

and

Inte

rven

tion

Sq

uare

Ro

ot

Fu

ncti

on

s

Exam

ple

y

x

y=

x

001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd5

4/15

/08

12:2

5:50

AM

Answers (Anticipation Guide and Lesson 10-1)

A01_A27_ALG1CRMC10_890504.indd A1A01_A27_ALG1CRMC10_890504.indd A1 5/9/08 6:10:52 AM5/9/08 6:10:52 AM

Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

6

G

lenc

oe A

lgeb

ra 1

Ref

lect

ion

s an

d T

ran

slat

ion

s o

f R

adic

al F

un

ctio

ns

Rad

ical

fu

nct

ion

s, l

ike

quad

rati

c fu

nct

ion

s, c

an b

e tr

ansl

ated

hor

izon

tall

y an

d ve

rtic

ally

, as

wel

l as

ref

lect

ed a

cros

s th

e x-

axis

. To

draw

th

e gr

aph

of

y =

a √

��

�

x +

h ,

foll

ow t

hes

e st

eps.

Gra

ph

s o

f S

qu

are

Ro

ot

Fu

nc

tio

ns

Ste

p 1

D

raw

th

e gr

aph

of

y =

+c √

⎯⎯ x . T

he

grap

h s

tart

s at

th

e or

igin

an

d pa

sses

th

rou

gh t

he

poin

t at

(1,

a).

If

a >

0, t

he

grap

h i

s in

th

e 1s

t qu

adra

nt.

If

a <

0, t

he

grap

h i

s re

flec

ted

acro

ss t

he

x-ax

is a

nd

is i

n t

he

4th

qu

adra

nt.

Ste

p 2

T

ran

slat

e th

e gr

aph

⎪c⎥

un

its

up

if c

is

posi

tive

an

d do

wn

if

c i

s n

egat

ive.

Ste

p 3

T

ran

slat

e th

e gr

aph

⎪h

⎥ u

nit

s le

ft i

f h

is

posi

tive

an

d ri

ght

if h

is

neg

ativ

e.

G

rap

h y

= -

√ �

��

x +

1 a

nd

com

par

e to

th

e p

aren

t gr

aph

. Sta

te t

he

dom

ain

an

d r

ange

.

Ste

p 1

M

ake

a ta

ble

of v

alu

es.

x-

1

01

3

8

y

0-

1-

1.4

1-

2-

3

Ste

p 2

T

his

is

a h

oriz

onta

l tr

ansl

atio

n 1

un

it t

o th

e le

ft o

f th

e pa

ren

t fu

nct

ion

an

d re

flec

ted

acro

ss t

he

x-ax

is.

Th

e do

mai

n i

s {x

| x ≥

0}

and

the

ran

ge i

s {y

| y ≤

0}.

Exer

cise

sG

rap

h e

ach

fu

nct

ion

, an

d c

omp

are

to t

he

par

ent

grap

h. S

tate

th

e d

omai

n

and

ran

ge.

1. y

= √

� x +

3

2. y

= √

��

�

x -

1

3. y

= -

√ �

��

x -

1

y

x

y

x

y

x

10-1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Sq

uare

Ro

ot

Fu

ncti

on

s

Exam

ple

y

x

y=

x

y=

- x

+ 1

tran

sla

tio

n o

f y =

√ �

x

up

3 u

nit

s;

D =

{x |

x ≥

0};

R =

{y |

y ≥

3}

tran

sla

tio

n o

f y =

√ �

x

rig

ht

1 u

nit

;D

= {

x |

x ≥

1};

R =

{y |

y ≥

0}

tran

sla

tio

n o

f y =

√ �

x

rig

ht

1 u

nit

an

d r

efl

ecte

d

acro

ss t

he x

-axis

;D

= {

x |

x ≥

1}

R =

{y |

y ≤

0}

001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd6

4/11

/08

9:19

:40

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 10-1

Cha

pte

r 10

7

G

lenc

oe A

lgeb

ra 1

Gra

ph

eac

h f

un

ctio

n, a

nd

com

par

e to

th

e p

aren

t gr

aph

. Sta

te t

he

dom

ain

an

d r

ange

.

1. y

= 2

√ �

x 2.

y =

1 −

2 √

� x

3. y

= 5

√ �

x

dil

ati

on

of

y =

√ ⎯⎯

x ;

dil

ati

on

of

y =

√ ⎯⎯

x ;

d

ila

tio

n o

f y =

√ ⎯⎯

x ;

D =

{x |

x ≥

0},

R =

{ y

| y

≥ 0

}

D

= {

x |

x ≥

0},

R =

{ y

| y

≥ 0

}

D =

{x |

x ≥

0},

R =

{ y

| y

≥ 0

}

4. y

= √

� x +

1

5. y

= √

� x -

4

6. y

= √

��

�

x -

1

tra

ns

lati

on

of

tr

an

sla

tio

n o

f t

ran

sla

tio

n o

f

y =

√ �

x u

p 1

un

it;

y =

√ ⎯⎯

x d

ow

n 4

un

its

;

y =

√ ⎯⎯

x r

igh

t 1

un

it;

D =

{x |

x ≥

0},

R =

{ y

| y

≥ 1

}

D =

{x |

x ≥

0},

R =

{ y

| y

≥ -

4}

D =

{x |

x ≥

1},

R =

{ y

| y

≥ 0

}

7. y

= -

√ �

��

x -

3

8. y

= √

��

�

x -

2 +

3

9. y

= -

1 −

2 √

��

�

x -

4 +

1

tra

ns

lati

on

of

y =

√ � x ;

tr

an

sla

tio

n o

f y =

√ ⎯⎯

x

dia

lati

on

of

y =

√ � x

refl

ec

ted

ri

gh

t 3

un

its

re

fle

cte

d a

cro

ss

rig

ht

2 u

nit

s a

nd

up

a

cro

ss

th

e x

-ax

is

th

e x

-ax

is;

D =

{x |

x ≥

3},

3

un

its

; D

= {

x |

x ≥

2},

tran

sla

ted

rig

ht

4 u

nit

s u

p

R

= {

y |

y ≤

0}

R =

{y |

y ≥

3}

1 u

nit

s;

D =

{x |

x ≥

4},

R

= {

y |

y ≤

1}

10-1

Skill

s Pr

acti

ce

Sq

uare

Ro

ot

Fu

ncti

on

s

y

x

y

x

y

x

12 8 4

−2

24

y

x

y

x

y

x

y

x

y

x

y

x

001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd7

6/12

/08

9:11

:39

PM

Answers (Lesson 10-1)


An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

8

G

lenc

oe A

lgeb

ra 1

Gra

ph

eac

h f

un

ctio

n, a

nd

com

par

e to

th

e p

aren

t gr

aph

. Sta

te t

he

dom

ain

an

d r

ange

.

1. y

= 4 −

3 √

� x

2. y

= √

� x +

2

3. y

= √

��

�

x -

3

y

x

y

x

y

x

4. y

= -

√ �

x +

1

5. y

= 2

√ ⎯⎯

⎯⎯⎯

x -

1 +

1

6. y

= -

√ ⎯⎯

⎯⎯⎯

x -

2 +

2

y

x

y

x

yx

7. O

HM

’S L

AW

In

ele

ctri

cal

engi

nee

rin

g, t

he

resi

stan

ce o

f a

circ

uit

can

be

fou

nd

by t

he

equ

atio

n I

= √

⎯⎯

P

−

R ,

wh

ere

I is

th

e cu

rren

t in

ampe

res,

P i

s th

e po

wer

in

wat

ts, a

nd

R i

s th

e re

sist

ance

of

the

circ

uit

in

oh

ms.

Gra

ph t

his

fu

nct

ion

for

a c

ircu

it w

ith

a

resi

stan

ce o

f 4

ohm

s.

10-1

Prac

tice

Sq

uare

Ro

ot

Fu

ncti

on

s

Current (amperes)

23 1 020

45

Pow

er (w

atts

)40

6080

100

dilati

on

of

y =

√ �

x ;

D =

{x |

x ≥

0},

R =

{y |

y ≥

0}

tran

sla

tio

n o

f y =

√ �

x

up

2 u

nit

;D

= {

x |

x ≥

0},

R

= {

y |

y ≥

2}

tran

sla

tio

n o

f y =

√ �

x

left

3 u

nit

s;

D =

{x

| x

≥ -

3},

R

= {

y |

y ≥

0}

tran

sla

tio

n o

f y =

√ �

x

up

1 u

nit

refl

ecte

d

in t

he x

-axis

; D

= {

x |

x ≥

0},

R =

{y |

y ≤

1}

dilati

on

of

y =

√ �

x

tran

sla

ted

up

1 u

nit

an

d r

igh

t 1 u

nit

;D

= {

x |

x ≥

1},

R =

{y |

y ≥

1}

tran

sla

tio

n o

f y =

√ �

x ;

u

p 2

un

its a

nd

rig

ht

2 u

nit

s,

refl

ecte

d

in t

he x

-axis

; D

= {

x |

x ≥

2},

R =

{y |

y ≤

2}

001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd8

4/15

/08

12:3

0:26

AMCopyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.

NA

ME

DA

TE

PE

RIO

D

Lesson 10-1

Cha

pte

r 10

9

G

lenc

oe A

lgeb

ra 1

10-1

Wor

d Pr

oble

m P

ract

ice

Sq

uare

Ro

ot

Fu

ncti

on

s

1. P

END

ULU

M M

OTI

ON

Th

e pe

riod

T o

f a

pen

dulu

m i

n s

econ

ds, w

hic

h i

s th

e ti

me

for

the

pen

dulu

m t

o re

turn

to

the

poin

t of

rel

ease

, is

give

n b

y th

e eq

uat

ion

T

= 1

.11

√� L

. T

he

len

gth

of

the

pen

dulu

m i

n f

eet

is g

iven

by

L. G

raph

th

is f

un

ctio

n.

2. E

MPI

RE

STA

TE B

UIL

DIN

G T

he

roof

of

the

Em

pire

Sta

te B

uil

din

g is

125

0 fe

et

abov

e th

e gr

oun

d. T

he

velo

city

of

an

obje

ct d

ropp

ed f

rom

a h

eigh

t of

h m

eter

s is

giv

en b

y th

e fu

nct

ion

V =

√ �

�

2gh

, w

her

e g

is t

he

grav

itat

ion

al c

onst

ant,

32

.2 f

eet

per

seco

nd

squ

ared

. If

an o

bjec

t is

dro

pped

fro

m t

he

roof

of

the

buil

din

g,

how

fas

t is

it

trav

elin

g w

hen

it

hit

s th

e st

reet

bel

ow?

ap

pro

xim

ate

ly 2

84 f

t/s

3. E

RR

OR

AN

ALY

SIS

Gre

gory

is

draw

ing

the

grap

h o

f y

= -

5 √

��

�

x +

1 .

He

desc

ribe

s th

e ra

nge

an

d do

mai

n a

s {x

� x

≥ -

1},

{y

� y

≥ 0

}. E

xpla

in a

nd

corr

ect

the

mis

take

th

at G

rego

ry m

ade.

T

he d

om

ain

is a

ctu

ally {

y �

y ≤

0}

becau

se t

he g

rap

h h

as b

een

re

flecte

d a

cro

ss t

he x

-axis

.

4. C

APA

CIT

OR

S A

cap

acit

or i

s a

set

of

plat

es t

hat

can

sto

re e

ner

gy i

n a

n

elec

tric

fie

ld. T

he

volt

age

V r

equ

ired

to

stor

e E

jou

les

of e

ner

gy i

n a

cap

acit

or

wit

h a

cap

acit

ance

of

C f

arad

s is

giv

en

by

V=

√⎯⎯⎯

2E − C.

a.

Rew

rite

an

d si

mpl

ify

the

equ

atio

n f

or

the

case

of

a 0.

0002

far

ad c

apac

itor

.

V =

100 √

⎯⎯ E

b

. Gra

ph t

he e

quat

ion

you

foun

d in

par

t a.

Voltage (volts)

100

150 50 0

2

200

250

300

350

Ener

gy (j

oule

s)

46

810

c.

How

wou

ld t

he

grap

h d

iffe

r if

you

w

ish

ed t

o st

ore

E +

1 jo

ule

s of

en

ergy

in

th

e ca

paci

tor

inst

ead?

tran

sla

tio

n o

f V

= 1

00 √

� E

on

e u

nit

to

th

e l

eft

d

. How

wou

ld t

he

grap

h d

iffe

r if

you

ap

plie

d a

volt

age

of V

+ 1

vol

ts

inst

ead?

tran

sla

tio

n o

fV

= 1

00 √

� E

on

e u

nit

do

wn

Period (sec)

23 1 04

45

Pend

ulum

Len

gth

(ft)

812

1620

001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd9

4/11

/08

9:20

:27

AM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

1

0

Gle

ncoe

Alg

ebra

1

A c

ub

ic r

oot

fun

ctio

n c

onta

ins

the

cubi

c ro

ot o

f a

vari

able

. Th

e cu

bic

roo

t of

a n

um

ber

x ar

e th

e n

um

bers

y t

hat

sat

isfy

th

e eq

uat

ion

y ·

y ·

y =

x (

or, a

lter

nat

ivel

y, y

=

3 √ �

x ).

Un

like

squ

are

root

fu

nct

ion

s, c

ubi

c ro

ot f

un

ctio

ns

retu

rn r

eal

nu

mbe

rs w

hen

th

e ra

dica

nd

is n

egat

ive.

G

rap

h y

=

3 √ �

x

.

Ste

p 1

Mak

e a

tabl

e.

Ste

p 2

Plo

t po

ints

an

d dr

aw a

sm

ooth

cu

rve.

xy

−5

−1

.71

−3

−1

.44

−1

−1

00

11

31

.44

51

.71

y

x

Exer

cise

sG

rap

h e

ach

fu

nct

ion

, an

d c

omp

are

to t

he

par

ent

grap

h.

1. y

= 2

3 √ �

x 2.

y =

3 √ �

x +

1

3. y

= 3

√ �

��

x +

1

y

x

y

x

y

x

d

ilati

on

of

y =

3 √ � x

tra

nsla

tio

n o

f t

ran

sla

tio

n o

f

y

=

3 √ � x

up

1 u

nit

y

=

3 √ � x

left

1 u

nit

4. y

= 3 √

��

�

x -

1 +

2

5. y

= 3

3 √ �

��

x -

2

6. y

= -

3 √ �

x +

3

y

x

y

x

y

x

t

ran

sla

tio

n o

f y =

3 √ � x

dilati

on

of

refl

ecti

on

of

y =

3 √ � x

u

p 2

un

its a

nd

y

= 3 √ � x

tran

sla

ted

a

cro

ss t

he x

-axis

r

igh

t 1 u

nit

r

igh

t 2 u

nit

s

tra

nsla

ted

up

3 u

nit

s

10-1

Enri

chm

ent

Cu

bic

Ro

ot

Fu

ncti

on

s

Exam

ple

001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd10

4/11

/08

9:20

:35

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 10-2

Cha

pte

r 10

1

1

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

Sim

plify

ing

Rad

ical

Exp

ressio

ns

Pro

du

ct P

rop

erty

of

Squ

are

Ro

ots

Th

e P

rod

uct

Pro

per

ty o

f S

qu

are

Roo

ts a

nd

prim

e fa

ctor

izat

ion

can

be

use

d to

sim

plif

y ex

pres

sion

s in

volv

ing

irra

tion

al s

quar

e ro

ots.

W

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an

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he

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Exam

ple

1

Exam

ple

2

001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd11

4/14

/08

1:45

:04

PM

Answers (Lesson 10-1 and Lesson 10-2)

A01_A27_ALG1CRMC10_890504.indd A4A01_A27_ALG1CRMC10_890504.indd A4 6/16/08 5:42:21 PM6/16/08 5:42:21 PM

An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

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1

2

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ncoe

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1

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y G

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and

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Exam

ple

001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd12

4/11

/08

9:20

:55

AM


NA

ME

DA

TE

PE

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001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd13

5/8/

084:

13:4

8P

M



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

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Cha

pte

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26

25. S

KY

DIV

ING

Wh

en a

sky

dive

r ju

mps

fro

m a

n a

irpl

ane,

th

e ti

me

t it

tak

es t

o fr

ee f

all

a

g

iven

dis

tan

ce c

an b

e es

tim

ated

by

the

form

ula

t =

√ �

�

2s

−

9.8 ,

wh

ere

t is

in

sec

onds

an

d s

is

i

n m

eter

s. I

f Ju

lie

jum

ps f

rom

an

air

plan

e, h

ow l

ong

wil

l it

tak

e h

er t

o fr

ee f

all

750

met

ers?

26. M

ETEO

RO

LOG

Y T

o es

tim

ate

how

lon

g a

thun

ders

torm

wil

l la

st, m

eteo

rolo

gist

s ca

n us

e

th

e fo

rmul

a t

= √

��

d3

−

216 ,

whe

re t

is

the

tim

e in

hou

rs a

nd d

is

the

diam

eter

of

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stor

m i

n m

iles

.

a.

A t

hu

nde

rsto

rm i

s 8

mil

es i

n d

iam

eter

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imat

e h

ow l

ong

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stor

m w

ill

last

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ive

you

r an

swer

in

sim

plif

ied

form

an

d as

a d

ecim

al.

b

. W

ill

a th

un

ders

torm

tw

ice

this

dia

met

er l

ast

twic

e as

lon

g? E

xpla

in.

N

o;

it w

ill

last

ab

ou

t 4.4

h,

or

nearl

y 3

tim

es a

s l

on

g.

10-2

ab

ou

t 12.4

s

8 √

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3

−

9

h ≈

1.5

h

2 √

��

3y

−

3 y

001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd14

4/11

/08

9:21

:22

AM


NA

ME

DA

TE

PE

RIO

D

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pte

r 10

15

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ncoe

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m P

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ical

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ressio

ns

1. S

POR

TS J

asm

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calc

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ted

the

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ght

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soc

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goal

to

be 15 − √

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impl

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on.

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2. N

ATU

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In 2

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an

ear

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w

the

ocea

n f

loor

in

itia

ted

a de

vast

atin

g ts

un

ami

in t

he

Indi

an O

cean

. Sci

enti

sts

can

app

roxi

mat

e th

e ve

loci

ty (

in f

eet

per

seco

nd)

of

a ts

un

ami

in w

ater

of

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h d

(i

n f

eet)

wit

h t

he

form

ula

V =

√ �

�

16d

. D

eter

min

e th

e ve

loci

ty o

f a

tsu

nam

i in

30

0 fe

et o

f w

ater

. Wri

te y

our

answ

er i

n

sim

plif

ied

radi

cal

form

.

4

0 √

�

3 f

t/s

3. A

UTO

MO

BIL

ES T

he

foll

owin

g fo

rmu

la

can

be

use

d to

fin

d th

e “z

ero

to s

ixty

” ti

me

for

a ca

r, o

r th

e ti

me

it t

akes

for

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to a

ccel

erat

e fr

om a

sto

p to

six

ty

mil

es p

er h

our. V

= √

��

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−

M

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city

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met

ers

per

seco

nd)

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is

the

car’

s av

erag

e po

wer

(in

wat

ts).

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s th

e m

ass

of t

he

car

(in

kil

ogra

ms)

.T

is

the

tim

e (i

n s

econ

ds).

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ind

the

tim

e it

tak

es f

or a

900

-kil

ogra

m

car

wit

h a

n a

vera

ge 6

0,00

0 w

atts

of

pow

er t

o ac

cele

rate

fro

m s

top

to 2

6.82

m

eter

s pe

r se

con

d (6

0 m

iles

per

hou

r).

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nd

you

r an

swer

to

the

nea

rest

ten

th.

ab

ou

t 5.4

s

4. P

HY

SIC

AL

SCIE

NC

E W

hen

a s

ubs

tan

ce

such

as

wat

er v

apor

is

in i

ts g

aseo

us

stat

e, t

he

volu

me

and

the

velo

city

of

its

mol

ecu

les

incr

ease

as

tem

pera

ture

in

crea

ses.

Th

e av

erag

e ve

loci

ty V

of

a m

olec

ule

wit

h m

ass

m a

t te

mpe

ratu

re T

is g

iven

by

the

form

ula

V=

√�

�3

kT−

m.

Sol

ve t

he

equ

atio

n f

or k

.

k

=m

V 2

− 3T

5. G

EOM

ETR

Y S

upp

ose

Em

eryv

ille

H

ospi

tal

wan

ts t

o bu

ild

a n

ew

hel

ipad

on

wh

ich

med

ic r

escu

e h

elic

opte

rs c

an l

and.

Th

e h

elip

ad w

ill

be c

ircu

lar

and

mad

e of

fir

e re

sist

ant

rubb

er.

a.

If t

he

area

of

the

hel

ipad

is

A, w

rite

an

eq

uat

ion

for

th

e ra

diu

s r.

r

= √

�

A

−

π

b

. Wri

te a

n e

xpre

ssio

n i

n s

impl

ifie

d ra

dica

l fo

rm f

or t

he

radi

us

of a

hel

ipad

w

ith

an

are

a of

288

squ

are

met

ers.

r

= 1

2 √

��

2π

−

π

c.

Usi

ng

you

r ca

lcu

lato

r, f

ind

a de

cim

al

appr

oxim

atio

n f

or t

he

radi

us.

Rou

nd

you

r an

swer

to

the

nea

rest

hu

ndr

edth

.

9.5

7 m

r

10-2

001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd15

4/11

/08

9:21

:27

AM



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

1

6

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

Sq

uare

s a

nd

Sq

uare

Ro

ots

Fro

m a

Gra

ph

Th

e gr

aph

of

y =

x2

can

be

use

d to

fin

d th

e sq

uar

es a

nd

squ

are

ro

ots

of n

um

bers

.T

o fi

nd

the

squ

are

of 3

, loc

ate

3 on

th

e x-

axis

. Th

en f

ind

its

corr

espo

ndi

ng

valu

e on

th

e y-

axis

.T

he

arro

ws

show

th

at 3

2 =

9.

To

fin

d th

e sq

uar

e ro

ot o

f 4,

fir

st l

ocat

e 4

on t

he

y-ax

is. T

hen

fin

d it

s co

rres

pon

din

g va

lue

on t

he

x-ax

is. F

ollo

win

g th

e ar

row

s on

th

e

grap

h, y

ou c

an s

ee t

hat

√ ⎯⎯

4

= 2

.A

sm

all

part

of

the

grap

h a

t y

= x

2 is

sh

own

bel

ow. A

1:1

0 ra

tio

for

un

it l

engt

h o

n t

he

y-ax

is t

o u

nit

len

gth

on

th

e x-

axis

is

use

d.

F

ind

√ �

�

11 .

Th

e ar

row

s sh

ow t

hat

√ ⎯⎯

⎯ 11

= 3

.3

to t

he

nea

rest

ten

th.

Exer

cise

sU

se t

he

grap

h a

bov

e to

fin

d e

ach

of

the

foll

owin

g to

th

e n

eare

st w

hol

e n

um

ber

.

1. 1

.52

2

2. 2

.72

7

3. 0

.92

1

4. 3

.62

13

5.

4.2

2 18

6.

3.9

2 15

Use

th

e gr

aph

ab

ove

to f

ind

eac

h o

f th

e fo

llow

ing

to t

he

nea

rest

ten

th.

7.

√ ⎯⎯

⎯ 15

3.9

8.

√ ⎯⎯

8

2.8

9.

√ ⎯⎯

3

1.7

10. √

⎯⎯

5 2

.2

11.

√ ⎯⎯

⎯ 14

3.7

12

. √

⎯⎯⎯

17 4

.1

12

11 1020

34

3.3

Ox

y

x

y O

10-2

Exam

ple

001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd16

4/11

/08

9:21

:31

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 10-3

Cha

pte

r 10

17

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

Op

era

tio

ns w

ith

Rad

ical

Exp

ressio

ns

Ad

d o

r Su

btr

act

Rad

ical

Exp

ress

ion

s W

hen

add

ing

or s

ubt

ract

ing

radi

cal

expr

essi

ons,

use

th

e A

ssoc

iati

ve a

nd

Dis

trib

uti

ve P

rope

rtie

s to

sim

plif

y th

e ex

pres

sion

s.

If r

adic

al e

xpre

ssio

ns

are

not

in

sim

ples

t fo

rm, s

impl

ify

them

.

S

imp

lify

10

√ �

6

- 5

√ �

3

+ 6

√ �

3

- 4

√ �

6

.

10 √

�

6 -

5 √

�

3 +

6 √

�

3 -

4 √

�

6 =

(10

- 4

) √ �

6

+ (

-5

+ 6

) √ �

3

A

ssocia

tive a

nd D

istr

ibutive P

ropert

ies

= 6

√ �

6

+ √

�

3

Sim

plif

y.

S

imp

lify

3 √

��

12 +

5 √

��

75 .

3 √

��

12 +

5 √

��

75 =

3 √

��

�

22 ·

3 +

5 √

��

�

52 ·

3

Sim

plif

y.

= 3

· 2

√ �

3

+ 5

· 5

√ �

3

S

implif

y.

= 6

√ �

3

+ 2

5 √

�

3

Sim

plif

y.

= 3

1 √

�

3

Dis

trib

utive P

ropert

y

Exer

cise

sS

imp

lify

eac

h e

xpre

ssio

n.

1. 2

√ �

5 +

4 √

� 5 6

√ �

5

2. √

� 6 -

4 √

� 6 -

3 √

�

6

3.

√ �

8 -

√ �

2 √

�

2

4. 3

√ �

75

+

2 √

� 5 1

5 √

�

3 +

2 √

�

5

5.

√ �

20

+ 2

√ �

5 -

3 √

� 5 √

�

5

6. 2

√ �

3 +

√ �

6 -

5 √

� 3 -

3 √

�

3 +

√ �

6

7.

√ �

12

+ 2

√ �

3 -

5 √

� 3 -

√ �

3

8. 3

√ �

6 +

3 √

� 2 -

√ �

50

+ √

�

24

5 √

�

6 -

2 √

�

2

9.

√ �

8a

- √

�

2a +

5 √

�

2a

6 √

��

2a

10.

√ �

54

+ √

�

24

5 √

�

6

11. √

� 3 +

√ �

1 −

3

4 √

�

3

−

3

12

. √

�

12 +

√ �

1 −

3

7 √

�

3

−

3

13. √

�

54 -

√ �

1 −

6

17

√ �

6

−

6

14

. √

�

80 -

√ �

20

+ √

��

180

8 √

�

5

15. √

�

50 +

√ �

18

- √

�

75 +

√ �

27

8

√ �

2 -

2 √

�

3

16. 2

√ �

3 -

4 √

�

45 +

2 √ �

1 −

3

8 √

�

3

−

3

- 1

2 √

�

5

17. √

��

125

- 2

√ �

1 −

5 +

√ �

1 −

3

23

√ �

5

−

5

+ √

�

3

−

3

18.

√ �

2 −

3 +

3 √

� 3 -

4 √ �

1 −

12

√

�

6 +

7 √

�

3

−

3

10-3

Exam

ple

1

Exam

ple

2

001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd17

4/11

/08

9:21

:37

AM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

1

8

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Op

era

tio

ns w

ith

Rad

ical

Exp

ressio

ns

Mu

ltip

ly R

adic

al E

xpre

ssio

ns

Mu

ltip

lyin

g tw

o ra

dica

l ex

pres

sion

s w

ith

dif

fere

nt

radi

can

ds i

s si

mil

ar t

o m

ult

iply

ing

bin

omia

ls.

M

ult

iply

(3

√ �

2

- 2

√ �

5

)(4

√ �

�

20 +

√ �

8

).

Use

th

e F

OIL

met

hod

.

(3 √

�

2 -

2 √

�

5 )(

4 √

��

20 +

√ �

8

) = (3

√ �

2

)(4

√ �

�

20 ) +

(3 √

�

2 )(

√ �

8

) + (-

2 √

�

5 )(

4 √

��

20 ) +

(-2

√ �

5

)( √

�

8 )

= 1

2 √

��

40 +

3 √

��

16 -

8 √

��

100

- 2

√ �

�

40

Multip

ly.

= 1

2 √

��

�

22 ·

10 +

3 ·

4 -

8 ·

10

- 2

√ �

��

22 ·

10

Sim

plif

y.

= 2

4 √

��

10 +

12

- 8

0 -

4 √

��

10

Sim

plif

y.

= 2

0 √

��

10 -

68

Com

bin

e lik

e t

erm

s.

Exer

cise

sS

imp

lify

eac

h e

xpre

ssio

n.

1. 2

( √ �

3 +

4 √

� 5 )

2 √

�

3 +

8 √

�

5

2. √

� 6 ( √

� 3 -

2 √

� 6 )

3 √

�

2 -

12

3.

√ �

5 ( √

� 5 -

√ �

2 )

5 -

√ ��

10

4. √

� 2 (3

√ �

7 +

2 √

� 5 )

3 √

��

14 +

2 √

��

10

5. (

2 -

4 √

� 2 )(2

+ 4

√ �

2 )

- 2

8

6. (3

+ √

� 6 ) 2

15 +

6 √

�

6

7. (

2 -

2 √

� 5 ) 2

24 -

8 √

�

5

8. 3

√ �

2 ( √

� 8 +

√ �

24

) 1

2 +

12

√ �

3

9.

√ �

8 ( √

� 2 +

5 √

� 8 )

44

10

. ( √

� 5 -

3 √

� 2 )( √

� 5 +

3 √

� 2 )

-13

11. (

√ �

3 +

√ �

6 ) 2

9 +

6 √

�

2

12. (

√ �

2 -

2 √

� 3 ) 2

14 -

4 √

�

6

13. (

√ �

5 -

√ �

2 )( √

� 2 +

√ �

6 )

14

. ( √

� 8 -

√ �

2 )( √

� 3 +

√ �

6 )

√

��

10 -

2 +

√ ��

30 -

2 √

� 3

√

� 6 +

2 √

� 3

15. (

√ �

5 -

√ �

18

)(7

√ �

5 +

√ �

3 )

16. (

2 √

� 3 -

√ �

45

)( √

�

12 +

2 √

� 6 )

3

5 +

√ ��

15 -

21 √

��

10 -

3 √

� 6

12 -

6 √

��

15 +

12 √

�

2 -

6 √

��

30

17. (

2 √

� 5 -

2 √

� 3 )( √

�

10 +

√ �

6 )

18. (

√ �

2 +

3 √

� 3 )( √

�

12 -

4 √

� 8 )

4

√ �

2

2 -

22 √

�

6

10-3

Exam

ple

001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd18

4/11

/08

9:21

:45

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 10-3

Cha

pte

r 10

19

Gle

ncoe

Alg

ebra

1

Skill

s Pr

acti

ceO

pera

tio

ns w

ith

Rad

ical

Exp

ressio

ns

10-3

Sim

pli

fy e

ach

exp

ress

ion

.

1. 7

√ �

7

- 2

√ �

7

5 √

�

7

2. 3

√ �

�

13 +

7 √

��

13

10 √

��

13

3. 6

√ �

5 -

2 √

� 5 +

8 √

� 5 12 √

�

5

4. √

��

15 +

8 √

��

15 -

12

√ �

�

15

-3 √

��

15

5. 1

2 √

� r -

9 √

� r 3

√ �

r

6.

9 √

��

6a -

11

√ �

�

6a +

4 √

��

6a

2 √

��

6a

7.

√ �

�

44 -

√ �

�

11

√ �

�

11

8. √

��

28 +

√ �

�

63

5 √

�

7

9. 4

√ �

3

+ 2

√ �

�

12

8 √

�

3

10. 8

√ �

�

54 -

4 √

�

6 2

0 √

�

6

11. √

��

27 +

√ �

�

48 +

√ �

�

12

9 √

�

3

12.

√ �

�

72 +

√ �

�

50 -

√ �

8

9 √

�

2

13. √

��

180

- 5

√ �

5

+ √

��

20

3 √

�

5

14. 2

√ �

�

24 +

4 √

��

54 +

5 √

��

96

36 √

�

6

15. 5

√ �

8

+ 2

√ �

�

20 -

√ �

8

16

. 2 √

�

13 +

4 √

� 2 -

5 √

�

13 +

√ �

2

8

√ �

2 +

4 √

�

5

-

3 √

��

13 +

5 √

�

2

17. √

�

2 ( √

�

8 +

√ �

6

) 4

+ 2

√ �

3

18.

√ �

5 ( √

�

10 -

√ �

3 )

5 √

�

2 -

√ �

�

15

19. √

� 6 (3

√ �

2 -

2 √

� 3 )

6 √

�

3 -

6 √

�

2

20. 3

√ �

3 (2

√ �

6 +

4 √

�

10 )

18 √

�

2 +

12 √

��

30

21. (

4 +

√ �

3

) (4

- √

�

3 )

13

22

. (2

- √

� 6 )

2 10 -

4 √

�

6

23. (

√ �

8 +

√ �

2 ) (

√ �

5 +

√ �

3 )

24. (

√ �

6 +

4 √

� 5 )(4

√ �

3 -

√ �

10

)

3

√ �

�

10 +

3 √

�

6

-

8 √

�

2 +

14 √

��

15

001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd19

5/8/

084:

14:2

1P

M



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

2

0

Gle

ncoe

Alg

ebra

1

Prac

tice

Op

era

tio

ns w

ith

Rad

ical

Exp

ressio

ns

Sim

pli

fy e

ach

exp

ress

ion

.

1. 8

√ �

30

- 4

√ �

30

4

√ ��

30

2. 2

√ �

5 -

7 √

� 5 -

5 √

� 5 -

10

√ �

5

3. 7

√ �

�

13x

- 1

4 √

��

13x

+ 2

√ �

�

13x

-5

√ �

�

13x

4. 2

√ �

�

45 +

4 √

�

20

14

√ �

5

5.

√ �

40

- √

�

10 +

√ �

90

4

√ ��

10

6. 2

√ �

32

+ 3

√ �

50

- 3

√ �

18

14

√ �

2

7.

√ �

27

+ √

�

18 +

√ �

�

300

3 √

�

2 +

13

√ �

3

8. 5

√ �

8 +

3 √

�

20 -

√ �

32

6

√ �

2 +

6 √

�

5

9.

√ �

14

- √

�

2 −

7

6 √

��

14

−

7

10

. √

�

50 +

√ �

32

- √

�

1 −

2

17

√ �

2

−

2

11. 5

√ �

19

+ 4

√ �

28

- 8

√ �

19

+ √

�

63

12. 3

√ �

10

+ √

�

75 -

2 √

�

40 -

4 √

�

12

-

3 √

��

19 +

11 √

�

7

-

√ �

�

10 -

3 √

�

3

13. √

� 6 ( √

�

10 +

√ �

15

) 2

√ ��

15 +

3 √

��

10

14.

√ �

5 (5

√ �

2 -

4 √

� 8 )

-3

√ ��

10

15. 2

√ �

7 (3

√ �

12

+ 5

√ �

8 ) 1

2 √

��

21 +

20

√ �

�

14

16. (

5 -

√ �

15

) 2 40 -

10

√ ��

15

17. (

√ �

10

+ √

� 6 ) (

√ �

30

- √

�

18 )

4 √

�

3

18. (

√ �

8 +

√ �

12

) ( √

�

48 +

√ �

18

) 3

6 +

14

√ �

6

19. (

√ �

2 +

2 √

� 8 )(3

√ �

6 -

√ �

5 )

20

. (4

√ �

3 -

2 √

� 5 )(3

√ �

10

+ 5

√ �

6 )

30 √

�

3 -

5 √

��

10

2 √

��

30 +

30

√ �

2

21. S

OU

ND

The

spe

ed o

f so

und

V i

n m

eter

s pe

r se

cond

nea

r E

arth

’s s

urfa

ce i

s gi

ven

by

V

= 2

0 √

��

�

t +

273

, w

her

e t

is t

he

surf

ace

tem

pera

ture

in

deg

rees

Cel

siu

s.

a.

Wh

at i

s th

e sp

eed

of s

oun

d n

ear

Ear

th’s

su

rfac

e at

15°

C a

nd

at 2

°C i

n s

impl

est

form

?

240 √

�

2 m

/s,

100 √

��

11 m

/s

b

. H

ow m

uch

fas

ter

is t

he

spee

d of

sou

nd

at 1

5°C

th

an a

t 2°

C?

240 √

�

2 -

100 √

��

11 ≈

7.7

5 m

/s

22. G

EOM

ETR

Y A

rec

tang

le is

5 √

�

7 +

2 √

�

3 m

eter

s lo

ng a

nd 6

√ �

7

- 3

√ �

3

met

ers

wid

e.

a.

Fin

d th

e pe

rim

eter

of

the

rect

angl

e in

sim

ples

t fo

rm.

22 √

�

7 -

2 √

�

3 m

b

. F

ind

the

area

of

the

rect

angl

e in

sim

ples

t fo

rm.

190 -

3 √

��

21 m

2

10-3

001_

020_

ALG

1CR

MC

10_8

9050

4.in

dd20

4/11

/08

9:22

:00

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 10-3

Cha

pte

r 10

21

Gle

ncoe

Alg

ebra

1

-5

-10

5

5

10

10

1520S

ki s

lope

y

xO

Wor

d Pr

oble

m P

ract

ice

Op

era

tio

ns w

ith

Rad

ical

Exp

ressio

ns

1. A

RC

HIT

ECTU

RE

Th

e P

enta

gon

is

the

buil

din

g th

at h

ouse

s th

e U

.S.

Dep

artm

ent

of D

efen

se. F

ind

the

appr

oxim

ate

peri

met

er o

f th

e bu

ildi

ng,

w

hic

h i

s a

regu

lar

pen

tago

n. L

eave

you

r an

swer

as

a ra

dica

l ex

pres

sion

.

115 √

��

149 m

2. E

AR

TH T

he

surf

ace

area

of

a sp

her

e w

ith

rad

ius

r is

giv

en b

y th

e fo

rmu

la

4πr2 .

Ass

um

ing

that

Ear

th i

s cl

ose

to

sph

eric

al i

n s

hap

e an

d h

as a

su

rfac

e ar

ea

of a

bou

t 5.

1 ×

108

squ

are

kilo

met

ers,

w

hat

is

the

radi

us

of E

arth

to

the

nea

rest

ten

kil

omet

ers?

6370

km

3. G

EOM

ETR

Y T

he

area

of

a tr

apez

oid

is

fou

nd

by m

ult

iply

ing

its

hei

ght

by t

he

aver

age

len

gth

of

its

base

s. F

ind

the

area

of

dec

k at

tach

ed t

o M

r. W

ilso

n’s

hou

se.

Giv

e yo

ur

answ

er a

s a

sim

plif

ied

radi

cal

expr

essi

on.

6

3 √ �

�

15 ft

2

4. R

ECR

EATI

ON

Car

men

su

rvey

ed a

ski

sl

ope

usi

ng

a di

gita

l de

vice

con

nec

ted

to

a co

mpu

ter.

Th

e co

mpu

ter

mod

el

assi

gned

coo

rdin

ates

to

the

top

and

bott

om p

oin

ts o

f th

e h

ill

as s

how

n i

n t

he

diag

ram

. Wri

te a

sim

plif

ied

radi

cal

expr

essi

on t

hat

rep

rese

nts

th

e sl

ope

of

the

hil

l.

5. F

REE

FA

LL S

upp

ose

a ba

ll i

s dr

oppe

d fr

om a

bu

ildi

ng

win

dow

800

fee

t in

th

e ai

r. A

not

her

bal

l is

dro

pped

fro

m a

low

er

win

dow

288

fee

t h

igh

. Bot

h b

alls

are

re

leas

ed a

t th

e sa

me

tim

e. A

ssu

me

air

resi

stan

ce i

s n

ot a

fac

tor

and

use

th

e fo

llow

ing

form

ula

to

fin

d h

ow m

any

seco

nds

t i

t w

ill

take

a b

all

to f

all

h f

eet.

t =

1 −

4 √

�

h

a. H

ow m

uch

tim

e w

ill

pass

bet

wee

n

wh

en t

he

firs

t ba

ll h

its

the

grou

nd

and

wh

en t

he

seco

nd

ball

hit

s th

e gr

oun

d?

Giv

e yo

ur

answ

er a

s a

sim

plif

ied

radi

cal

expr

essi

on.

b.

Wh

ich

bal

l la

nds

fir

st?

Th

e b

all

dro

pp

ed

fro

m 2

88 f

eet

lan

ds

firs

t.

c. F

ind

a de

cim

al a

ppro

xim

atio

n o

f th

e an

swer

for

par

t a.

Rou

nd

you

r an

swer

to

th

e n

eare

st t

enth

.

( 7r+

12)

˚ Ho

use

Dec

k

h =

7 √

� 5 ft

12 √

�

3 f

t

6 √

�

3 f

t

A (

2 √

��

14 ,

5 √

�

7 )

23 √

��

149 m

B (-

2 √

��

14 ,

7 √

�

7 )

10-3

ab

ou

t 2.8

s

2 √

�

2 s

021_

034_

ALG

1CR

MC

10_8

9050

4.in

dd21

4/11

/08

9:22

:47

AM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

2

2

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

The

Wh

eel o

f Th

eod

oru

sT

he

Gre

ek m

ath

emat

icia

ns

wer

e in

trig

ued

by

prob

lem

s of

re

pres

enti

ng

diff

eren

t n

um

bers

an

d ex

pres

sion

s u

sin

g ge

omet

ric

con

stru

ctio

ns.

Th

eodo

rus,

a G

reek

ph

ilos

oph

er w

ho

live

d ab

out

425

B.C

., is

sa

id t

o h

ave

disc

over

ed a

way

to

con

stru

ct t

he

sequ

ence

√

�

1 ,

√ �

2

, √

�

3 ,

√ �

4

, . .

..T

he

begi

nn

ing

of h

is c

onst

ruct

ion

is

show

n. Y

ou s

tart

wit

h a

n

isos

cele

s ri

ght

tria

ngl

e w

ith

sid

es 1

un

it l

ong.

Use

th

e fi

gure

ab

ove.

Wri

te e

ach

len

gth

as

a ra

dic

al e

xpre

ssio

n i

n s

imp

lest

for

m.

1. l

ine

segm

ent

AO

√ �

1

2.

lin

e se

gmen

t B

O

√ �

2

3. l

ine

segm

ent

CO

√

�

3

4.

lin

e se

gmen

t D

O

√ �

4

5. D

escr

ibe

how

eac

h n

ew t

rian

gle

is a

dded

to

the

figu

re.

Dra

w a

new

sid

e o

f le

ng

th

1 a

t ri

gh

t an

gle

s t

o t

he l

ast

hyp

ote

nu

se.

Th

en

dra

w t

he n

ew

hyp

ote

nu

se.

6. T

he

len

gth

of

the

hyp

oten

use

of

the

firs

t tr

ian

gle

is √

�

2 . F

or t

he

seco

nd

tria

ngl

e, t

he

len

gth

is

√ �

3

. Wri

te a

n e

xpre

ssio

n f

or t

he

len

gth

of

the

hyp

oten

use

of

the

nth

tri

angl

e.

√ �

��

n +

1

7. S

how

th

at t

he

met

hod

of

con

stru

ctio

n w

ill

alw

ays

prod

uce

th

e n

ext

nu

mbe

r in

th

e se

quen

ce. (

Hin

t: F

ind

an e

xpre

ssio

n f

or t

he

hyp

oten

use

of

the

(n +

1)t

h t

rian

gle.

)

√ �

��

��

( √ �

n

)2 +

(1)2

or

√ �

��

n +

1

8. I

n t

he

spac

e be

low

, con

stru

ct a

Wh

eel

of T

heo

doru

s. S

tart

wit

h a

lin

e se

gmen

t 1

cen

tim

eter

lon

g. W

hen

doe

s th

e W

hee

l st

art

to o

verl

ap?

aft

er

len

gth

√ �

�

18

1

1

1

1 OAB

CD

10-3

021_

034_

ALG

1CR

MC

10_8

9050

4.in

dd22

4/11

/08

9:22

:55

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 10-4

Cha

pte

r 10

23

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

Rad

ical

Eq

uati

on

s

Rad

ical

Eq

uat

ion

s E

quat

ion

s co

nta

inin

g ra

dica

ls w

ith

var

iabl

es i

n t

he

radi

can

d ar

e ca

lled

rad

ical

eq

uat

ion

s. T

hes

e ca

n b

e so

lved

by

firs

t u

sin

g th

e fo

llow

ing

step

s.

S

olve

16

= √

�

x

−

2 f

or x

.

16

= √

� x

−

2

Origin

al equation

2(1

6) =

2( √

� x

−

2 ) M

ultip

ly e

ach s

ide b

y 2

.

32

= √

� x

Sim

plif

y.

(3

2)2

= (

√ �

x )2

Square

each s

ide.

10

24 =

x

Sim

plif

y.

Th

e so

luti

on i

s 10

24, w

hic

h c

hec

ks i

n

the

orig

inal

equ

atio

n.

S

olve

√ �

��

4x -

7 +

2 =

7.

√

��

�

4x -

7 +

2 =

7

Origin

al equation

√ �

��

4x -

7 +

2 -

2 =

7 -

2

Subtr

act

2 f

rom

each s

ide.

√

��

�

4x -

7 =

5

Sim

plif

y.

( √

��

�

4x -

7 )2

= 5

2 S

quare

each s

ide.

4x

- 7

= 2

5 S

implif

y.

4x

- 7

+ 7

= 2

5 +

7 A

dd 7

to e

ach s

ide.

4x

= 3

2 S

implif

y.

x

= 8

D

ivid

e e

ach s

ide b

y 4

.

Th

e so

luti

on i

s 8,

wh

ich

ch

ecks

in

th

e or

igin

al

equ

atio

n.

Exer

cise

sS

olve

eac

h e

qu

atio

n. C

hec

k y

our

solu

tion

.

1.

√ �

a

= 8

64

2.

√ �

a

+ 6

= 3

2 676

3.

2 √

� x =

8 1

6

4. 7

= √

��

�

26 -

n -

23

5.

√ �

�

-a

= 6

-36

6.

√ �

�

3r2

= 3

±

√ �

3 ±

√ �

7. 2

√ �

3

= √

� y 12

8.

2 √

��

3a -

2 =

7 6

3

−

4

9. √

��

�

x -

4 =

6 4

0

10. √

��

�

2m +

3 =

5 1

1

11.

√ �

��

3b -

2 +

19

= 2

4 9

12

. √

��

�

4x -

1 =

3 5

−

2

13. √

��

�

3r +

2 =

2 √

�

3 1

0

−

3

14.

√ �

x −

2 =

1 −

2

1

−

2

15.

√ �

x −

8 =

4 1

28

16.

√ �

��

�

6x2

+ 5

x =

2 1

−

2 ,

- 4

−

3

17.

√ �

x −

3 +

6 =

8 1

2

18. 2

√ ��

3x

−

5 +

3 =

11

26 2

−

3

10-4

Exam

ple

1Ex

amp

le 2

Ste

p 1

Is

olat

e th

e ra

dica

l on

on

e si

de o

f th

e eq

uat

ion

.S

tep

2

Squ

are

each

sid

e of

th

e eq

uat

ion

to

elim

inat

e th

e ra

dica

l.

021_

034_

ALG

1CR

MC

10_8

9050

4.in

dd23

4/11

/08

9:23

:01

AM



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

2

4

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(con

tin

ued

)

Rad

ical

Eq

uati

on

s

Extr

aneo

us

Solu

tio

ns

To

solv

e a

radi

cal

equ

atio

n w

ith

a v

aria

ble

on b

oth

sid

es, y

ou

nee

d to

squ

are

each

sid

e of

th

e eq

uat

ion

. Squ

arin

g ea

ch s

ide

of a

n e

quat

ion

som

etim

es

prod

uce

s ex

tran

eou

s so

luti

ons,

or

solu

tion

s th

at a

re n

ot s

olu

tion

s of

th

e or

igin

al e

quat

ion

. T

her

efor

e, i

t is

ver

y im

port

ant

that

you

ch

eck

each

sol

uti

on.

S

olve

√ �

��

x +

3 =

x -

3.

√

��

�

x +

3 =

x -

3

Origin

al equation

( √ �

��

x +

3 )2 =

(x

- 3

)2 S

quare

each s

ide.

x

+ 3

= x

2 -

6x

+ 9

S

implif

y.

0

= x

2 -

7x

+ 6

S

ubtr

act

x and 3

fro

m e

ach s

ide.

0

= (

x -

1)(

x -

6)

Facto

r.

x -

1 =

0

or

x -

6 =

0

Zero

Pro

duct

Pro

pert

y

x

= 1

x

= 6

S

olv

e.

CH

ECK

√ �

��

x +

3

= x

- 3

√

��

�

x +

3 =

x -

3

√

��

�

1 +

3

� 1

- 3

√

��

�

6 +

3 �

6 -

3

√ �

4

� -

2 √

�

9 �

3

2 ≠

-2

3 =

3 �

Sin

ce x

= 1

doe

s n

ot s

atis

fy t

he

orig

inal

equ

atio

n, x

= 6

is

the

only

sol

uti

on.

Exer

cise

sS

olve

eac

h e

qu

atio

n. C

hec

k y

our

solu

tion

.

1.

√ �

a

= a

0,

1

2. √

��

�

a +

6 =

a 3

3.

2 √

� x =

x 0

, 4

4. n

= √

��

�

2 -

n 1

5.

√ �

�

-a

= a

0

6. √

��

��

10 -

6k

+ 3

= k

�

7.

√ �

��

y -

1 =

y -

1 1

, 2

8.

√ �

��

3a -

2 =

a 1

, 2

9.

√ �

��

x +

2 =

x 2

10. √

��

�

2b +

5 =

b -

5 1

0

11.

√ �

��

3b +

6 =

b +

2 1

12

. √

��

�

4x -

4 =

x 2

13. r

+ √

��

�

2 -

r =

2 1

, 2

14

. √

��

��

x2 +

10x

= x

+ 4

8

15. -

2 √ �

x −

8 =

15

�

16. √

��

��

6x2

- 4

x =

x +

2

17.

√ �

��

�

2y2

- 6

4 =

y

18.

√ �

��

��

�

3x

2 +

12x

+ 1

=

x +

5

-

2

−

5 ,

2

8

-4,

3

10-4

Exam

ple

1

021_

034_

ALG

1CR

MC

10_8

9050

4.in

dd24

4/11

/08

9:23

:07

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 10-4

Cha

pte

r 10

25

Gle

ncoe

Alg

ebra

1

Skill

s Pr

acti

ceR

ad

ical

Eq

uati

on

sS

olve

eac

h e

qu

atio

n. C

hec

k y

our

solu

tion

.

1.

√ � f

= 7

49

2.

√ �

�

-x

= 5

-25

3.

√ ��

5p

= 1

0 20

4.

√ �

4y

= 6

9

5. 2

√ �

2

=

√ �

u

8

6. 3

√ �

5

= √

��

-n

-45

7.

√ �

g -

6 =

3 8

1

8. √

��

5a +

2 =

0 ø

9.

√ �

��

2t -

1 =

5 1

3

10.

√ �

��

3k -

2 =

4 6

11. √

��

�

x +

4 -

2 =

1 5

12

. √

��

�

4x -

4 -

4 =

0 5

13.

√ �

d

−

3

= 4

144

14

. √ �

m

−

3 =

3 2

7

15. x

=

√ �

��

x +

2 2

16

. d =

√ �

��

12 -

d 3

17. √

��

�

6x -

9 =

x 3

18

. √

��

�

6p -

8 =

p 2

, 4

19. √

��

�

x +

5 =

x -

1 4

20

. √

��

�

8 -

d =

d -

8 8

21. √

��

�

r -

3 +

5 =

r 7

22

. √

��

�

y -

1 +

3 =

y 5

23. √

��

�

5n +

4 =

n +

2 1

, 0

24

. √

��

�

3z -

6 =

z -

2 5

, 2

10-4

021_

034_

ALG

1CR

MC

10_8

9050

4.in

dd25

5/9/

083:

03:0

0A

M



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

2

6

Gle

ncoe

Alg

ebra

1

Prac

tice

R

ad

ical

Eq

uati

on

sS

olve

eac

h e

qu

atio

n. C

hec

k y

our

solu

tion

.

1.

√ �

�

-b

= 8

-64

2.

4 √

�

3 =

√ �

x 48

3. 2

√ �

4r

+ 3

= 1

1 4

4.

6 -

√ �

2y

= -

2 32

5.

√ �

��

k +

2 -

3 =

7 9

8

6. √

��

�

m -

5 =

4 √

�

3 5

3

7.

√ �

��

6t +

12

= 8

√ �

6

62

8.

√ �

��

3j -

11

+ 2

= 9

20

9.

√ �

��

�

2x +

15

+ 5

= 1

8 77

10

. √ �

3d

−

5

- 4

= 2

60

11. 6

√ ��

3x

−

3 -

3 =

0 1

−

4

12. 6

+ √

�

5r

−

6 =

-2

�

13. y

= √

��

�

y +

6 3

14

. √

��

��

15 -

2x

= x

3

15. √

��

�

w +

4 =

w +

4 -

4,

-3

16

. √

��

�

17 -

k =

k -

5 8

17. √

��

��

5m -

16

= m

- 2

4,

5

18.

√ �

��

�

24 +

8q

= q

+ 3

-3,

5

19. √

��

��

4t +

17

- t

- 3

= 0

2

20. 4

- √

��

��

3m +

28

= m

-1

21. √

��

��

10p

+ 6

1 -

7 =

p -

6,

2

22.

√ �

��

2x2

- 9

= x

3

23. E

LEC

TRIC

ITY

Th

e vo

ltag

e V

in

a c

ircu

it i

s gi

ven

by

V =

√ �

�

PR

, w

her

e P

is

the

pow

er i

n

wat

ts a

nd

R i

s th

e re

sist

ance

in

oh

ms.

a.

If

the

volt

age

in a

cir

cuit

is

120

volt

s an

d th

e ci

rcu

it p

rodu

ces

1500

wat

ts o

f po

wer

, w

hat

is

the

resi

stan

ce i

n t

he

circ

uit

?

b

. Su

ppos

e an

ele

ctri

cian

des

ign

s a

circ

uit

wit

h 1

10 v

olts

an

d a

resi

stan

ce o

f 10

oh

ms.

H

ow m

uch

pow

er w

ill

the

circ

uit

pro

duce

?

24. F

REE

FA

LL A

ssu

min

g n

o ai

r re

sist

ance

, th

e ti

me

t in

sec

onds

th

at i

t ta

kes

an o

bjec

t to

fall

h f

eet

can

be

dete

rmin

ed b

y th

e eq

uat

ion

t =

√ �

h

−

4

.

a.

If

a sk

ydiv

er ju

mps

fro

m a

n a

irpl

ane

and

free

fal

ls f

or 1

0 se

con

ds b

efor

e op

enin

g th

e pa

rach

ute

, how

man

y fe

et d

oes

the

skyd

iver

fal

l?

b

. Su

ppos

e a

seco

nd

skyd

iver

jum

ps a

nd

free

fal

ls f

or 6

sec

onds

. How

man

y fe

et d

oes

the

seco

nd

skyd

iver

fal

l?

10-4

1600 f

t

9.6

oh

ms

1210 w

att

s

576 f

t

021_

034_

ALG

1CR

MC

10_8

9050

4.in

dd26

4/11

/08

9:23

:19

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 10-4

Cha

pte

r 10

27

Gle

ncoe

Alg

ebra

1

Wor

d Pr

oble

m P

ract

ice

Rad

ical

Eq

uati

on

s

1. S

UB

MA

RIN

ES T

he

dist

ance

in

mil

es

that

th

e lo

okou

t of

a s

ubm

arin

e ca

n s

ee

is a

ppro

xim

atel

y d

= 1

.22

√ �

h

, w

her

e h

is

the

hei

ght

in f

eet

abov

e th

e su

rfac

e of

th

e w

ater

. How

far

wou

ld a

su

bmar

ine

peri

scop

e h

ave

to b

e ab

ove

the

wat

er t

o lo

cate

a s

hip

6 m

iles

aw

ay?

Rou

nd

you

r an

swer

to

the

nea

rest

ten

th.

2. P

ETS

Fin

d th

e va

lue

of x

if

the

peri

met

er

of a

tri

angu

lar

dog

pen

is

25 m

eter

s.

x =

8

3. L

OG

GIN

G D

oyle

’s l

og r

ule

est

imat

es t

he

amou

nt

of u

sabl

e lu

mbe

r (i

n b

oard

fee

t)

that

can

be

mil

led

from

a s

hip

men

t of

lo

gs. I

t is

rep

rese

nte

d by

th

e eq

uat

ion

B =

L ( d

- 4

−

4 ) 2 , w

her

e d

is

the

log

d

iam

eter

(in

in

ches

) an

d L

is

the

log

len

gth

(in

fee

t). S

upp

ose

the

tru

ck

carr

ies

20 l

ogs,

eac

h 2

5 fe

et l

ong,

an

d th

at t

he

ship

men

t yi

elds

a t

otal

of

6000

bo

ard

feet

of

lum

ber.

Est

imat

e th

e di

amet

er o

f th

e lo

gs t

o th

e n

eare

st i

nch

. A

ssu

me

that

all

th

e lo

gs h

ave

un

ifor

m

len

gth

an

d di

amet

er.

18 i

n.

4. F

IREF

IGH

TIN

G F

ire

figh

ters

cal

cula

te

the

flow

rat

e of

wat

er o

ut

of a

par

ticu

lar

hyd

ran

t by

usi

ng

the

foll

owin

g fo

rmu

la.

F =

26.

9d2 √

�

p

F

is

the

flow

rat

e (i

n g

allo

ns

per

min

ute

),

p is

th

e n

ozzl

e pr

essu

re (

in p

oun

ds p

er

squ

are

inch

), a

nd

d i

s th

e di

amet

er o

f th

e h

ose

(in

in

ches

). I

n o

rder

to

effe

ctiv

ely

figh

t a

fire

, th

e co

mbi

ned

flo

w r

ate

of t

wo

hos

es n

eeds

to

be a

bou

t 24

30 g

allo

ns

per

min

ute

. Th

e di

amet

er o

f ea

ch o

f th

e h

oses

is

3 in

ches

, bu

t th

e n

ozzl

e pr

essu

re

of o

ne

hos

e is

4 t

imes

th

at o

f th

e se

con

d h

ose.

Wh

at a

re t

he

noz

zle

pres

sure

s fo

r ea

ch h

ose?

Rou

nd

you

r an

swer

s to

th

e n

eare

st t

enth

.

11.2

psi

an

d 4

4.8

psi

5. G

EOM

ETR

Y T

he

late

ral

surf

ace

area

s

of a

rig

ht

circ

ula

r co

ne,

not

in

clu

din

g th

e ba

se, i

s re

pres

ente

d by

th

e eq

uat

ion

s

= π

r √ �

��

r2 +

h2 ,

wh

ere

r is

th

e ra

diu

s of

th

e ci

rcu

lar

base

an

d h

is

the

hei

ght

of

the

con

e.

a. I

f th

e la

tera

l su

rfac

e ar

ea o

f a

fun

nel

is

127

.54

squ

are

cen

tim

eter

s an

d it

s ra

diu

s is

3.5

cen

tim

eter

s, f

ind

its

hei

ght

to t

he

nea

rest

ten

th o

f a

cen

tim

eter

.

b.

Wh

at i

s th

e ar

ea o

f th

e op

enin

g (i

.e.,

the

base

) of

th

e fu

nn

el?

x+

1m

12

m 1

0 m

10-4

11.1

cm

38.5

cm

2

24.2

ft

021_

034_

ALG

1CR

MC

10_8

9050

4.in

dd27

4/11

/08

9:23

:22

AM



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

2

8

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

Mo

re T

han

On

e S

qu

are

Ro

ot

You

hav

e le

arn

ed t

hat

to

rem

ove

the

squ

are

root

in

an

equ

atio

n, y

ou f

irst

nee

d to

iso

late

th

e sq

uar

e ro

ot, t

hen

squ

are

both

sid

es o

f th

e eq

uat

ion

, an

d fi

nal

ly, s

olve

th

e re

sult

ing

equ

atio

n.

How

ever

, th

ere

are

equ

atio

ns

that

con

tain

mor

e th

at o

ne

squ

are

root

an

d si

mpl

y sq

uar

ing

once

is

not

en

ough

to

rem

ove

all

of t

he

radi

cals

.

S

olve

√ �

��

x +

7 =

√ � x

+ 1

.

√

��

�

x +

7 =

√ �

x +

1

One o

f th

e s

quare

roots

is a

lready isola

ted.

( √ �

��

x +

7 ) 2 =

( √ �

x +

1 ) 2

Square

both

sid

es t

o r

em

ove t

he s

quare

root.

x

+ 7

= x

+ 2

√ �

x +

1

Sim

plif

y.

Use t

he F

OIL

meth

od t

o s

quare

rig

ht

sid

e.

x +

7 -

x -

1 =

2 √

� x

Sim

plif

y.

6

= 2

√ �

x S

implif

y.

Isola

te t

he s

quare

root

term

again

.

3

= √

� x

Div

ide b

oth

sid

es b

y 2

.

9

= x

S

quare

both

sid

es t

o r

em

ove t

he s

quare

root.

Ch

eck

: Su

bsti

tute

in

to t

he

orig

inal

equ

atio

n t

o m

ake

sure

you

r so

luti

on i

s va

lid.

√ �

��

9 +

7 =

√ �

9

+ 1

R

epla

ce x

with 9

.

√

��

16 =

3 +

1

S

impify.

4

= 4

�

T

he e

quation is t

rue,

so x

= 9

is t

he s

olu

tion.

Exer

cise

sS

olve

eac

h e

qu

atio

n.

1.

√ �

��

x +

13

- 2

= √

��

�

x +

1

3

2. √

��

�

x +

11

= √

��

�

x +

3 +

2 -

2

3.

√ �

��

x +

9 -

3 =

√ �

��

x -

6

7

4. √

��

�

x +

21

= √

� x +

3

4

5.

√ �

��

x +

9 +

3 =

√ �

��

x +

20

+ 2

16

6.

√ �

��

x -

6 +

6 =

√ �

��

x +

1 +

5

15

10-4

Exam

ple

021_

034_

ALG

1CR

MC

10_8

9050

4.in

dd28

4/11

/08

9:23

:27

AM

Lesson 10-4


NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

29

Gle

ncoe

Alg

ebra

1

Gra

phin

g Ca

lcul

ator

Act

ivit

yR

ad

ical

Ineq

ualiti

es

Th

e gr

aph

s of

rad

ical

equ

atio

ns

can

be

use

d to

det

erm

ine

the

solu

tion

s of

rad

ical

in

equ

alit

ies

thro

ugh

th

e C

AL

C m

enu

.

S

olve

eac

h i

neq

ual

ity.

a.

√ �

��

x +

4 ≤

3

En

ter

√ �

��

x +

4 i

n Y

1 an

d 3

in Y

2 an

d gr

aph

. Exa

min

e th

e gr

aph

s. U

se

TR

AC

E t

o fi

nd

the

endp

oin

t of

th

e gr

aph

of

the

radi

cal

equ

atio

n. U

se

CA

LC

to

dete

rmin

e th

e in

ters

ecti

on

of t

he

grap

hs.

Th

is i

nte

rval

, -4

to 5

, w

her

e th

e gr

aph

of

y =

√ �

��

x +

4

is

belo

w t

he

grap

h o

f y

= 3

, rep

rese

nts

th

e so

luti

on t

o th

e in

equ

alit

y. T

hu

s,

the

solu

tion

is

-4

≤ x

≤ 5

.

b. √

��

�

2x -

5 >

x -

4G

raph

eac

h s

ide

of t

he

ineq

ual

ity.

Fin

d th

e in

ters

ecti

on a

nd

trac

e to

th

e en

dpoi

nt

of t

he

radi

cal

grap

h.

Th

e gr

aph

of

y =

√ �

��

2x -

5 i

s ab

ove

the

grap

h o

f y

= x

- 4

fro

m 2

.5 u

p to

7. T

hu

s,

the

solu

tion

is

2.5

< x

< 7

.

Exer

cise

sS

olve

eac

h i

neq

ual

ity.

1. 6

- √

��

�

2x +

1 <

3

2. √

��

�

4x -

5 ≤

7

3. √

��

�

5x -

4 ≥

4

x

> 4

5

−

4 ≤

x ≤

27

−

2

x ≥

4

4. -

4 >

√ �

��

3x -

2

5. √

��

�

3x -

6 +

5 ≥

-3

6. √

��

�

6 -

3x

< x

+ 1

6

n

o s

olu

tio

n

x

≥ 2

-

10

< x

< 2

[-10

, 10]

scl

:1 b

y [-

10, 1

0] s

cl:1

[-10

, 10]

scl

:1 b

y [-

10, 1

0] s

cl:1

[-10

, 10]

scl

:1 b

y [-

10, 1

0] s

cl:1

[-10

, 10]

scl

:1 b

y [-

10, 1

0] s

cl:1

10-4

Exam

ple

021_

034_

ALG

1CR

MC

10_8

9050

4.in

dd29

5/8/

084:

19:1

9P

M



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

3

0

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

Th

e P

yth

ag

ore

an

Th

eo

rem

The

Pyth

ago

rean

Th

eore

m T

he

side

opp

osit

e th

e ri

ght

angl

e in

a r

igh

t tr

ian

gle

is

call

ed t

he

hyp

oten

use

. Th

is s

ide

is a

lway

s th

e lo

nge

st s

ide

of a

rig

ht

tria

ngl

e. T

he

oth

er

two

side

s ar

e ca

lled

th

e le

gs o

f th

e tr

ian

gle.

To

fin

d th

e le

ngt

h o

f an

y si

de o

f a

righ

t tr

ian

gle,

giv

en t

he

len

gth

s of

th

e ot

her

tw

o si

des,

you

can

use

th

e P

yth

agor

ean

Th

eore

m.

Py

tha

go

rea

n T

he

ore

m

If a

an

d b

are

th

e m

ea

su

res o

f th

e le

gs o

f a

rig

ht

tria

ng

le

an

d c

is t

he

me

asu

re o

f th

e h

yp

ote

nu

se

, th

en

c2 =

a2 +

b2.

CB

Ab

ac

F

ind

th

e le

ngt

h o

f th

e m

issi

ng

sid

e.

c2 =

a2

+ b

2 P

yth

agore

an T

heore

m

c2 =

52

+ 1

22 a

= 5

and b

= 1

2

c2 =

169

S

implif

y.

c =

√ �

�

169

T

ake t

he s

quare

root

of

each s

ide.

c =

13

Th

e le

ngt

h o

f th

e h

ypot

enu

se i

s 13

.

Exer

cise

sF

ind

th

e le

ngt

h o

f ea

ch m

issi

ng

sid

e. I

f n

eces

sary

, rou

nd

to

the

nea

rest

hu

nd

red

th.

1.

2.

3.

5

0

45.8

3

35.3

625

25c

100

110

a

40

30c

10-5

12

5

14

84

1589

5

Exam

ple

4.

5.

6.

1

6.1

2

5.5

7

8

021_

034_

ALG

1CR

MC

10_8

9050

4.in

dd30

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/08

9:23

:39

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 10-5

Cha

pte

r 10

31

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(con

tin

ued

)

Th

e P

yth

ag

ore

an

Th

eo

rem

Rig

ht

Tria

ng

les

If a

an

d b

are

the

mea

sure

s of

th

e sh

orte

r si

des

of a

tri

angl

e, c

is

the

mea

sure

of

the

lon

gest

sid

e, a

nd

c2 =

a2

+ b

2 , th

en t

he

tria

ngl

e is

a r

igh

t tr

ian

gle.

D

eter

min

e w

het

her

th

e fo

llow

ing

sid

e m

easu

res

form

ri

ght

tria

ngl

es.

a. 1

0, 1

2, 1

4S

ince

th

e m

easu

re o

f th

e lo

nge

st s

ide

is 1

4, l

et c

= 1

4, a

= 1

0, a

nd

b =

12.

c2

= a

2 +

b2

Pyth

agore

an T

heore

m

14

2 �

102

+ 1

22 a

= 1

0,

b =

12,

c =

14

196

� 1

00 +

144

M

ultip

ly.

196

≠ 2

44

Add.

Sin

ce c

2 ≠

a2

+ b

2 , th

e tr

ian

gle

is n

ot a

rig

ht

tria

ngl

e.b

. 7,

24,

25

Sin

ce t

he

mea

sure

of

the

lon

gest

sid

e is

25,

let

c =

25,

a =

7, a

nd

b =

24.

c2

= a

2 +

b2

Pyth

agore

an T

heore

m

25

2 �

72

+ 2

42 a

= 7

, b

= 2

4,

c =

25

625

� 4

9 +

576

M

ultip

ly.

625

= 6

25

Add.

Sin

ce c

2 =

a2

+ b

2 , th

e tr

ian

gle

is a

rig

ht

tria

ngl

e.

Exer

cise

sD

eter

min

e w

het

her

eac

h s

et o

f m

easu

res

can

be

sid

es o

f a

righ

t tr

ian

gle.

T

hen

det

erm

ine

wh

eth

er t

hey

for

m a

Pyt

hag

orea

n t

rip

le.

1. 1

4, 4

8, 5

0 yes;

yes

2.

6, 8

, 10

yes;

yes

3.

8, 8

, 10

no

; n

o

4. 9

0, 1

20, 1

50 y

es;

yes

5.

15,

20,

25

yes;

yes

6.

4, 8

, 4 √

�

5 y

es;

no

7. 2

, 2,

√ �

8

yes;

no

8.

4, 4

, √

��

20 n

o;

no

9.

25,

30,

35

no

; n

o

10. 2

4, 3

6, 4

8 n

o;

no

11

. 18,

80,

82

yes;

yes

12

. 150

, 200

, 250

yes;

yes

13. 1

00, 2

00, 3

00 n

o;

no

14

. 500

, 120

0, 1

300

yes;

yes

15.

700

, 100

0, 1

300

no

; n

o

10-5

Exam

ple

021_

034_

ALG

1CR

MC

10_8

9050

4.in

dd31

4/11

/08

9:23

:50

AM



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

3

2

Gle

ncoe

Alg

ebra

1

Skill

s Pr

acti

ceTh

e P

yth

ag

ore

an

Th

eo

rem

Fin

d t

he

len

gth

of

each

mis

sin

g si

de.

If

nec

essa

ry, r

oun

d t

o th

e n

eare

st

hu

nd

red

th.

1.

2.

75

36

3.

4.

30

15.7

5

5.

6.

9.8

5

70

Det

erm

ine

wh

eth

er e

ach

set

of

mea

sure

s ca

n b

e si

des

of

a ri

ght

tria

ngl

e. T

hen

d

eter

min

e w

het

her

th

ey f

orm

a P

yth

agor

ean

tri

ple

.

7. 7

, 24,

25

yes;

yes

8. 1

5, 3

0, 3

4 n

o;

no

9. 1

6, 2

8, 3

2 n

o;

no

10

. 18,

24,

30

yes;

yes

11. 1

5, 3

6, 3

9 yes;

yes

12. 5

, 7,

√ �

�

74 y

es;

no

13. 4

, 5, 6

no

; n

o

14. 1

0, 1

1, √

��

221

yes;

no

25024

0

a4

9

c

2933

b

1634

b

1539a

21

72c

10-5

021_

034_

ALG

1CR

MC

10_8

9050

4.in

dd32

5/9/

083:

07:2

7A

M


NA

ME

DA

TE

PE

RIO

D

Lesson 10-5

Cha

pte

r 10

3

3

Gle

ncoe

Alg

ebra

1

Prac

tice

Th

e P

yth

ag

ore

an

Th

eo

rem

Fin

d t

he

len

gth

of

each

mis

sin

g si

de.

If

nec

essa

ry, r

oun

d t

o th

e n

eare

st h

un

dre

dth

.

1.

2.

3.

68

15.4

9

11.3

1

Det

erm

ine

wh

eth

er e

ach

set

of

mea

sure

s ca

n b

e si

des

of

righ

t tr

ian

gle.

T

hen

det

erm

ine

wh

eth

er t

hey

for

m a

Pyt

hag

orea

n t

rip

le.

4. 1

1, 1

8, 2

1

5. 2

1, 7

2, 7

5

n

o;

no

y

es;

yes

6. 7

, 8, 1

1

7. 9

, 10,

√ �

�

161

n

o;

yes

no

; n

o

8. 9

, 2 √

��

10 ,

11

9. √

�

7 , 2

√ �

2

, √

��

15

y

es;

no

yes;

no

10. S

TOR

AG

E T

he

shed

in

Ste

phan

’s b

ack

yard

has

a d

oor

that

mea

sure

s 6

feet

hig

h a

nd

3 fe

et w

ide.

Ste

phan

wou

ld l

ike

to s

tore

a s

quar

e th

eate

r pr

op t

hat

is

7 fe

et o

n a

sid

e.

Wil

l it

fit

th

rou

gh t

he

door

dia

gon

ally

? E

xpla

in.

No

; th

e g

reate

st

len

gth

th

at

will

fit

thro

ug

h t

he d

oo

r is

√ �

�

45 ≈

6.7

1 f

t.

11. S

CR

EEN

SIZ

ES T

he

size

of

a te

levi

sion

is

mea

sure

d by

th

e le

ngt

h o

f th

e sc

reen

’s

diag

onal

.

a. I

f a

tele

visi

on s

cree

n m

easu

res

24 i

nch

es h

igh

an

d 18

in

ches

wid

e, w

hat

siz

e te

levi

sion

is

it?

30-i

n.

tele

vis

ion

b.

Dar

la t

old

Tri

th

at s

he

has

a 3

5-in

ch t

elev

isio

n. T

he

hei

ght

of t

he

scre

en i

s 21

in

ches

. W

hat

is

its

wid

th?

28 i

n.

c. T

ri t

old

Dar

la t

hat

he

has

a 5

-in

ch h

andh

eld

tele

visi

on a

nd

that

th

e sc

reen

mea

sure

s 2

inch

es b

y 3

inch

es. I

s th

is a

rea

son

able

mea

sure

for

th

e sc

reen

siz

e? E

xpla

in.

No

;if

th

e s

cre

en

measu

res 2

in

. b

y 3

in

., t

hen

its

dia

go

nal

is o

nly

ab

ou

t 3.6

1 i

n.

124

b19

11

a

60

32c

10-5

021_

034_

ALG

1CR

MC

10_8

9050

4.in

dd33

4/11

/08

9:24

:10

AM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

3

4

Gle

ncoe

Alg

ebra

1

Wor

d Pr

oble

m P

ract

ice

Pyth

ag

ore

an

Th

eo

rem

1. B

ASE

BA

LL A

bas

ebal

l di

amon

d is

a

squ

are.

Eac

h b

ase

path

is

90 f

eet

lon

g.

Aft

er a

pit

ch, t

he

catc

her

qu

ickl

y th

row

s th

e ba

ll f

rom

hom

e pl

ate

to a

tea

mm

ate

stan

din

g by

sec

ond

base

. Fin

d th

e di

stan

ce t

he

ball

tra

vels

. Rou

nd

you

r an

swer

to

the

nea

rest

ten

th.

12

7.3

ft

2. T

RIA

NG

LES

Eac

h s

tude

nt

in M

rs.

Kel

ly’s

geo

met

ry c

lass

con

stru

cted

a

un

iqu

e ri

ght

tria

ngl

e fr

om d

rin

kin

g st

raw

s. M

rs. K

elly

mad

e a

char

t w

ith

th

e di

men

sion

s of

eac

h t

rian

gle.

How

ever

, M

rs. K

elly

mad

e a

mis

take

wh

en

reco

rdin

g th

eir

resu

lts.

Wh

ich

res

ult

was

re

cord

ed i

nco

rrec

tly?

3. M

APS

Fin

d th

e di

stan

ce b

etw

een

Mac

on

and

Ber

ryvi

lle.

Rou

nd

you

r an

swer

to

the

nea

rest

ten

th.

76.2

mi

4. T

ELEV

ISIO

N T

elev

isio

ns

are

iden

tifi

ed

by t

he

diag

onal

mea

sure

men

t of

th

e vi

ewin

g sc

reen

. For

exa

mpl

e, a

27-

inch

te

levi

sion

has

a d

iago

nal

scr

een

m

easu

rem

ent

of 2

7 in

ches

.

Com

plet

e th

e ch

art

to f

ind

the

scre

en

hei

ght

of e

ach

tel

evis

ion

giv

en i

ts s

ize

and

scre

en w

idth

. Rou

nd

you

r an

swer

s to

th

e n

eare

st w

hol

e n

um

ber.

Sour

ce: B

est

Buy

5. M

AN

UFA

CTU

RIN

G K

arl

wor

ks f

or a

co

mpa

ny

that

man

ufa

ctu

res

car

part

s.

His

job

is t

o dr

ill

a h

ole

in s

pher

ical

ste

el

ball

s. T

he

ball

s an

d th

e h

oles

hav

e th

e di

men

sion

s sh

own

on

th

e di

agra

m.

a. H

ow d

eep

is t

he

hol

e? 1

2 c

m

b.

Wh

at w

ould

be

the

radi

us

of a

bal

l w

ith

a s

imil

ar h

ole

7 ce

nti

met

ers

wid

e an

d 24

cen

tim

eter

s de

ep?

12.5

cm

d

Seco

nd B

ase

Hom

e Pl

ate

90 ft

x c

m 1

3 cm

5 c

m

Sid

e L

en

gth

s

S

tud

en

t a

b

c

S

tud

en

t a

b

c

A

my

3

4

5

Fra

n

8

14

1

6

B

elin

da

7

2

4

25

G

us

5

12

1

3

E

mo

ry

9

12

1

5

27 in

.

1 02345678

12

34

56

78

910

tens of miles

ten

s o

f m

iles

Car

ter

Cit

y

Mac

on

Ham

ilto

n

Ber

ryvi

lle

10-5

TV

siz

ew

idth

(in

.)h

eig

ht

(in

.)

19

-in

ch

15

12

25

-in

ch

21

14

32

-in

ch

25

20

50

-in

ch

40

30

Fra

n’s

021_

034_

ALG

1CR

MC

10_8

9050

4.in

dd34

6/12

/08

9:12

:44

PM

Lesson 10-5


NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

35

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

Pyth

ag

ore

an

Tri

ple

sR

ecal

l th

e P

yth

agor

ean

Th

eore

m:

In a

rig

ht

tria

ngl

e, t

he

squ

are

of t

he

len

gth

of

the

hyp

oten

use

is

equ

al t

o th

e su

m o

f th

e sq

uar

es o

f th

e le

ngt

hs

of t

he

legs

.

a2

+ b

2 =

c2

N

ote

that

c i

s th

e le

ngt

h

of

th

e h

ypot

enu

se.

Th

e in

tege

rs 3

, 4, a

nd

5 sa

tisf

y th

e

32 +

42

= 5

2

Pyt

hag

orea

n T

heo

rem

an

d ca

n b

e th

e 9

+ 1

6 =

25

len

gth

s of

th

e si

des

of a

rig

ht

tria

ngl

e.

25 =

25

Fu

rth

erm

ore,

for

an

y po

siti

ve i

nte

ger

n,

For

n =

2:

62 +

82

= 1

02

the

nu

mbe

rs 3

n, 4

n, a

nd

5n s

atis

fy t

he

36 +

64

= 1

00P

yth

agor

ean

Th

eore

m.

100

= 1

00

If t

hre

e n

um

bers

sat

isfy

th

e P

yth

agor

ean

Th

eore

m, t

hey

are

cal

led

a P

yth

agor

ean

tri

ple

. Her

e is

an

eas

y w

ay t

o fi

nd

oth

er P

yth

agor

ean

tri

ples

.

Th

e n

um

bers

a, b

, an

d c

are

a P

yth

agor

ean

tri

ple

if a

= m

2 -

n2 ,

b =

2m

n,

and

c =

m2

+ n

2 , w

her

e m

an

d n

are

rel

ativ

ely

prim

e po

siti

ve i

nte

gers

an

d m

> n

.

C

hoo

se m

= 5

an

d n

= 2

.

a =

m2

- n

2 b

= 2

mn

c

= m

2 +

n2

Ch

eck

20

2 +

212

� 2

92

=

52

- 2

2 =

2(5

)(2)

=

52

+ 2

2 40

0 +

441

� 8

41

= 2

5 -

4

= 2

0 =

25

+ 4

84

1 =

841

=

21

= 2

9

Exer

cise

sU

se t

he

foll

owin

g va

lues

of

m a

nd

n t

o fi

nd

Pyt

hag

orea

n t

rip

les.

1. m

= 3

an

d n

= 2

2.

m =

4 a

nd

n =

1

3. m

= 5

an

d n

= 3

5

, 12,

13

8,

15,

17

16,

30,

34

4. m

= 6

an

d n

= 5

5.

m =

10

and

n =

7

6. m

= 8

an

d n

= 5

1

1,

60,

61

51,

140,

149

39,

80,

89

ACB a

c b

10-5

Exam

ple

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd35

4/11

/08

9:24

:55

AM



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

3

6

Gle

ncoe

Alg

ebra

1

Spre

adsh

eet

Act

ivit

yP

yth

ag

ore

an

Tri

ple

sA

Pyt

hag

orea

n t

rip

le i

s a

set

of t

hre

e w

hol

e n

um

bers

th

at s

atis

fies

th

e eq

uati

on a

2 + b

2 = c

2 , w

here

c is

the

gre

ates

t nu

mbe

r. Y

ou c

an u

se a

spr

eads

heet

to

inve

stig

ate

the

patt

erns

in P

ytha

gore

an t

ripl

es. A

pri

mit

ive

Pyt

hag

orea

n

trip

le i

s a

Pyt

hag

orea

n t

ripl

e in

wh

ich

th

e n

um

bers

hav

e n

o co

mm

on f

acto

rs

oth

er t

han

1. A

fam

ily

of P

yth

agor

ean

tri

ple

s is

a p

rim

itiv

e P

yth

agor

ean

tr

iple

an

d it

s w

hol

e n

um

ber

mu

ltip

les.

Th

e sp

read

shee

t at

th

e ri

ght

pro

du

ces

a fa

mil

y of

Pyt

hag

orea

n t

rip

les.

Ste

p 1

E

nte

r a

Pyt

hag

orea

n t

ripl

e in

to c

ells

A1,

A

2, a

nd

A3.

Ste

p 2

U

se r

ows

2 th

rou

gh 1

0 to

fin

d 9

addi

tion

al P

yth

agor

ean

tri

ples

th

at a

re

mu

ltip

les

of t

he

prim

itiv

e tr

iple

. For

mat

th

e ro

ws

so t

hat

row

2 m

ult

ipli

es t

he

nu

mbe

rs i

n r

ow 1

by

2, r

ow 3

mu

ltip

lies

th

e n

um

bers

in

row

1 b

y 3,

an

d so

on

.

Exer

cise

sU

se t

he

spre

adsh

eet

of f

amil

ies

of P

yth

agor

ean

tri

ple

s.

1. C

hoo

se o

ne

of t

he

trip

les

oth

er t

han

(3,

4, 5

) fr

om t

he

spre

adsh

eet.

Ver

ify

that

it

is a

Pyt

hag

orea

n t

ripl

e. S

am

ple

an

sw

er:

Fo

r (6

, 8,

10),

6

2 +

82 =

36 +

64 o

r 100 =

10

2.

2. T

wo

poly

gon

s ar

e si

mil

ar i

f th

ey a

re t

he

sam

e sh

ape,

bu

t n

ot n

eces

sari

ly

the

sam

e si

ze. F

or t

rian

gles

, if

two

tria

ngl

es h

ave

angl

es w

ith

th

e sa

me

mea

sure

s th

en t

hey

are

sim

ilar

. Use

a c

enti

met

er r

ule

r to

dra

w t

rian

gles

w

ith

mea

sure

s fr

om t

he

spre

adsh

eet.

Do

the

tria

ngl

es a

ppea

r to

be

sim

ilar

?S

ee s

tud

en

ts’

wo

rk.;

yes

Eac

h o

f th

e fo

llow

ing

is a

pri

mit

ive

Pyt

hag

orea

n t

rip

le. U

se t

he

spre

adsh

eet

to f

ind

tw

o P

yth

agor

ean

tri

ple

s in

th

eir

fam

ilie

s.

3.

(5, 1

2, 1

3) S

am

ple

an

sw

er:

(10,

24,

26),

(15,

36,

39)

4. (

9, 4

0, 4

1) S

am

ple

an

sw

er:

(18,

80,

82),

(27,

120,

123)

5.

(20,

21,

29)

S

am

ple

an

sw

er:

(40,

42,

58),

(60,

63,

87)

Tri

ple

s.x

ls

A1 3 4 5 62 8 9 10

117

BC

3 6 9

12

15

18

21

24

27

30

4 8

12

16

20

24

28

32

36

40

5

10

15

20

25

30

35

40

45

50

The

form

ula

in c

ell A

10 is

A1 *

10.

Sh

eet

1S

heet

2S

I

10-5

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd36

5/8/

084:

14:4

6P

M

Lesson 10-6


NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

37

Gle

ncoe

Alg

ebra

1

F

ind

th

e d

ista

nce

bet

wee

n t

he

poi

nts

at

(-5,

2)

and

(4,

5).

d =

√ �

��

��

��

��

(x

2 -

x1)

2 +

(y2

- y

1)2

Dis

tance F

orm

ula

=

√ �

��

��

��

��

(4

- (

-5)

)2 +

(5

- 2

)2 (x

1,

y 1)

= (

-5,

2),

(x 2

, y 2

) =

(4,

5)

=

√ �

��

92 +

32

Sim

plif

y.

=

√ �

��

81 +

9

Evalu

ate

square

s a

nd s

implif

y.

=

√ �

�

90

Th

e di

stan

ce i

s √

��

90 ,

or a

bou

t 9.

49 u

nit

s.

Exam

ple

1

Stud

y G

uide

and

Inte

rven

tion

Th

e D

ista

nce a

nd

Mid

po

int

Fo

rmu

las

Dis

tan

ce F

orm

ula

Th

e P

yth

agor

ean

Th

eore

m c

an b

e u

sed

to d

eriv

e th

e D

ista

nce

F

orm

ula

sh

own

bel

ow. T

he

Dis

tan

ce F

orm

ula

can

th

en b

e u

sed

to f

ind

the

dist

ance

be

twee

n a

ny

two

poin

ts i

n t

he

coor

din

ate

plan

e.

Dis

tan

ce

Fo

rmu

laT

he

dis

tan

ce

be

twe

en

an

y t

wo

po

ints

with

co

ord

ina

tes (

x 1,

y 1)

an

d (

x 2,

y 2)

is

giv

en

by d

= √

��

��

��

��

(x

2 -

x1)2

+ (

y 2 -

y 1)2

J

ill

dra

ws

a li

ne

segm

ent

from

poi

nt

(1, 4

) on

her

co

mp

ute

r sc

reen

to

poi

nt

(98,

49)

. H

ow l

ong

is t

he

segm

ent?

d =

√ �

��

��

��

��

(x

2 -

x1)

2 +

(y 2

- y

1)2

=

√ �

��

��

��

��

(9

8 -

1)2

+ (

49 -

4)2

=

√ �

��

�

972

+ 4

52

= √

��

��

��

94

09 +

202

5

=

√ �

��

11,4

34

Th

e se

gmen

t is

abo

ut

106.

93 u

nit

s lo

ng.

Exer

cise

sF

ind

th

e d

ista

nce

bet

wee

n t

he

poi

nts

wit

h t

he

give

n c

oord

inat

es.

1. (

1, 5

), (

3, 1

) 2.

(0, 0

), (

6, 8

) 3.

(-2,

-8)

, (7,

-3)

2

√ �

5 ;

4.4

7

10

√

��

106 ;

10.3

0

4. (

6, -

7), (

-2,

8)

5. (1

, 5),

(-

8, 4

) 6.

(3, -

4), (

-4,

-4)

1

7

√

��

82 ;

9.0

6

7

7. (

-1,

4),

(3,

2)

8. (0

, 0),

(-

3, 5

) 9.

(2, -

6), (

-7,

1)

2

√ �

5 ;

4.4

7

√

��

34 ;

5.8

3

√

��

130 ;

11.4

0

10. (

-2,

-5)

, (0,

8)

11. (

3, 4

), (

0, 0

) 12

. (3,

-4)

, (-

4, -

16)

√ �

�

173 ;

13.1

5

5

√

��

193 ;

13.8

9

Fin

d t

he

pos

sib

le v

alu

es o

f a

if

the

poi

nts

wit

h t

he

give

n c

oord

inat

es a

re t

he

ind

icat

ed d

ista

nce

ap

art.

13. (

1, a

), (

3, -

2); d

= √

�

5

14. (

0, 0

), (

a, 4

); d

= 5

15

. (2,

-1)

, (a,

3);

d =

5

-

1 o

r -

3

3 o

r -

3

-1 o

r 5

16. (

1, -

3), (

a, 2

1); d

= 2

5 17

. (1,

a),

(-

2, 4

); d

= 3

18

. (3,

-4)

, (-

4, a

); d

= √

��

65

-

6 o

r 8

4

-8 o

r 0

10-6

Exam

ple

2

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd37

4/15

/08

8:56

:53

PM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

3

8

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(con

tin

ued

)

Th

e D

ista

nce a

nd

Mid

po

int

Fo

rmu

las

Mid

po

int

Form

ula

Th

e po

int

that

is

equ

idis

tan

ce f

rom

bot

h o

f th

e en

dpoi

nts

is

call

ed

the

mid

poi

nt.

You

can

fin

d th

e co

ordi

nat

es o

f th

e m

idpo

int

by u

sin

g th

e M

idpo

int

For

mu

la.

Mid

po

int

Fo

rmu

la

Th

e m

idp

oin

t M

of

a lin

e s

eg

me

nt

with

en

dp

oin

ts a

t (x

1,

y 1)

an

d (

x 2,

y 2)

is

giv

en

by M

( x 1 +

x2

−

2

, y 1

+ y

2

−

2

) .

F

ind

th

e co

ord

inat

es o

f th

e m

idp

oin

t of

th

e se

gmen

t w

ith

en

dp

oin

ts

at (

–2, 5

) an

d (

4, 9

).

M ( x 1

+ x

2 −

2

, y 1

+ y

2 −

2

) M

idpoin

t F

orm

ula

=

M ( -

2 + 4

−

2 ,

5 + 9

−

2 )

(x1,

y 1)

= (

-2,

5)

and (

x 2,

y 2)

= (

4,

9)

=

M ( 2

−

2 , 14

−

2 )

Sim

plif

y t

he n

um

era

tors

.

=

M (

1, 3

) S

implif

y.

Exer

cise

sF

ind

th

e co

ord

inat

es o

f th

e m

idp

oin

t of

th

e se

gmen

t w

ith

th

e gi

ven

en

dp

oin

ts.

1. (

1, 6

), (

3, 1

0)

2. (4

, -2)

, (0,

6)

3. (7

, 2),

(13

, -4)

(

2,

8)

(2,

2)

(10,

-1)

4. (

-1,

2),

(1,

0)

5. (-

3, -

3), (

5, -

11)

6. (0

, 8),

(-

6, 0

)

(

1,

0)

(1,

-7)

(-

3,

4)

7. (

4, -

3), (

-2,

3)

8. (9

, -1)

, (3,

-7)

9.

(2, -

1), (

8, 7

)

(

1,

0)

(6,

-4)

(5,

3)

10. (

1, 4

), (

-3,

12)

11

. (4,

0),

(-

2, 6

) 12

. (1,

9),

(7,

1)

(

-1,

8)

(1,

3)

(4,

5)

13. (

12, 0

), (

2, -

6)

14. (

1, 1

), (

9, -

9)

15. (

4, 5

), (

-2,

-1)

(

7,

-3)

(5,

-4)

(1,

2)

16. (

1, -

14),

(-

5, 0

)

17. (

2, 2

), (

6, 8

) 18

. (-

7, 3

), (

5, -

3)

(

-2,

-7)

(4,

5)

(-

1,

0)

10-6

Exam

ple

1

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd38

4/11

/08

9:25

:16

AM

Lesson 10-6


NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

39

Gle

ncoe

Alg

ebra

1

Skill

s Pr

acti

ceTh

e D

ista

nce a

nd

Mid

po

int

Fo

rmu

las

Fin

d t

he

dis

tan

ce b

etw

een

th

e p

oin

ts w

ith

th

e gi

ven

coo

rdin

ates

.

1. (

9, 7

), (

1, 1

) 10

2.

(5, 2

), (

8, -

2) 5

3. (

1, -

3), (

1, 4

) 7

4.

(7, 2

), (

-5,

7)

13

5. (

-6,

3),

(10

, 3)

16

6.

(3, 3

), (

-2,

3)

5

7. (

-1,

-4)

, (-

6, 0

) √

��

41 ≈

6.4

0

8. (-

2, 4

), (

5, 8

) √

��

65 ≈

8.0

6

Fin

d t

he

pos

sib

le v

alu

es o

f a

if

the

poi

nts

wit

h t

he

give

n c

oord

inat

es a

re t

he

ind

icat

ed d

ista

nce

ap

art.

9. (

-2,

-5)

, (a,

7);

d =

13

a =

-7 o

r 3

10

. (8,

-2)

, (5,

a);

d =

3 a

= -

2

11. (

4, a

), (

1, 6

); d

= 5

a =

2 o

r 10

12

. (a,

3),

(5,

-1)

; d =

5 a

= 2

or

8

13. (

1, 1

), (

a, 1

); d

= 4

a =

-3 o

r 5

14

. (2,

a),

(2,

3);

d =

10

a =

-7 o

r 13

15. (

a, 2

), (

-3,

3);

d =

√ �

2

a =

-4 o

r -

2

16. (

-5,

3),

(-

3, a

); d

= √

�

5 a

= 2

or

4

Fin

d t

he

coor

din

ates

of

the

mid

poi

nt

of t

he

segm

ent

wit

h t

he

give

n e

nd

poi

nts

.

17. (

-3,

4),

(-

2, 8

) (-

2.5

, 6)

18. (

5, -

6), (

7, -

9) (6

, -

7.5

)

19. (

4, 2

), (

8, 6

) (6

, 4)

20. (

5, 2

), (

3, 1

0) (4

, 6)

21. (

12, -

1), (

4, -

11)

(8,

-6)

22. (

-3,

-1)

, (-

11, 3

) (-

7,

1)

23. (

9, 3

), (

6, -

6) (7

.5,

-1.5

) 24

. (0,

-4)

, (8,

4)

(4,

0)

10-6

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd39

5/8/

084:

14:5

8P

M



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

4

0

Gle

ncoe

Alg

ebra

1

Prac

tice

Th

e D

ista

nce a

nd

Mid

po

int

Fo

rmu

las

Fin

d t

he

dis

tan

ce b

etw

een

th

e p

oin

ts w

ith

th

e gi

ven

coo

rdin

ates

.

1. (

4, 7

), (

1, 3

) 5

2.

(0, 9

), (

-7,

-2)

√ �

�

170 ≈

13.0

4

3. (

6, 2

), +

(4,

1 −

2 )

5

−

2 o

r 2.5

0

4. (-

1, 7

), +

( 1 −

3 ,

6 ) 5

−

3 ≈

1.6

7

5. (

√ �

3

, 3) ,

( 2 √

�

3 , 5

) √

�

7 ≈

2.6

5

6. ( 2

√ �

2

, -1)

, ( 3

√ �

2

, 3)

3 √

�

2 ≈

4.2

4

Fin

d t

he

pos

sib

le v

alu

es o

f a

if

the

poi

nts

wit

h t

he

give

n c

oord

inat

es a

re t

he

ind

icat

ed d

ista

nce

ap

art.

7. (

4, -

1), (

a, 5

); d

= 1

0 a =

-4 o

r 12

8.

(2, -

5), (

a, 7

); d

= 1

5 a =

-7 o

r 11

9. (

6, -

7), (

a, -

4); d

= √

��

18 a

= 3

or

9

10. (

-4,

1),

(a,

8);

d =

√ �

�

50 a

= -

5 o

r -

3

11. (

8, -

5), (

a, 4

); d

= √

��

85 a

= 6

or

10

12

. (-

9, 7

), (

a, 5

); d

= √

��

29 a

= -

14 o

r -

4

Fin

d t

he

coor

din

ates

of

the

mid

poi

nt

of t

he

segm

ent

wit

h t

he

give

n e

nd

poi

nts

.

13. (

4, -

6), (

3, -

9) (3

.5,

- 7

.5)

14. (

-3,

-8)

, (-

7, 2

) (-

5,

-3)

15. (

0, -

4), (

3, 2

) (1

.5,

- 1

) 16

. (-

13, -

9), (

-1,

-5)

(-

7,

-7)

17. (2

, - 1 −

2 ) ,

(1, 1 −

2 ) (

1 1

−

2 ,

0)

18. ( 2

−

3 , -

1 ) , (2

, 1 −

3 ) (1

1

−

3 ,

- 1

−

3 )

19. B

ASE

BA

LL T

hre

e pl

ayer

s ar

e w

arm

ing

up

for

a ba

seba

ll

gam

e. P

laye

r B

sta

nds

9 f

eet

to t

he

righ

t an

d 18

fee

t in

fro

nt

of P

laye

r A

. Pla

yer

C s

tand

s 8

feet

to

the

left

and

13

feet

in

fron

t of

Pla

yer

A.

a.

Dra

w a

mod

el o

f th

e si

tuat

ion

on

th

e co

ordi

nat

e gr

id.

Ass

um

e th

at P

laye

r A

is

loca

ted

at (

0, 0

).

b.

To

the

nea

rest

ten

th, w

hat

is

the

dist

ance

bet

wee

n P

laye

rs A

an

d B

an

d be

twee

n P

laye

rs A

an

d C

?

c.

Wh

at i

s th

e di

stan

ce b

etw

een

Pla

yers

B a

nd

C?

20. M

APS

Mar

ia a

nd

Jack

son

liv

e in

adj

acen

t n

eigh

borh

oods

. If

they

su

peri

mpo

se a

co

ordi

nat

e gr

id o

n t

he

map

of

thei

r n

eigh

borh

oods

, Mar

ia l

ives

at

(-9,

1)

and

Jack

son

li

ves

at (

5, -

4).

x

y

O4

8

16 12 8 4

-8

-4

10-6

ab

ou

t 1.9

6 m

i

20.1

ft;

15.3

ft

17.7

ft

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd40

4/11

/08

9:25

:30

AM

Lesson 10-6


NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

41

Gle

ncoe

Alg

ebra

1

Wor

d Pr

oble

m P

ract

ice

Th

e D

ista

nce a

nd

Mid

po

int

Fo

rmu

las

1. C

HES

S M

arga

ret’s

las

t tw

o re

mai

nin

g ch

ess

piec

es a

re l

ocat

ed a

t th

e ce

nte

rs o

f th

e sq

uar

es a

t op

posi

te c

orn

ers

of t

he

boar

d. I

f th

e ch

essb

oard

is

a sq

uar

e w

ith

8-

inch

sid

es, a

bou

t h

ow f

ar a

part

are

th

e pi

eces

? R

oun

d yo

ur

answ

er t

o th

e n

eare

st t

enth

.

2. E

NG

INEE

RIN

G T

odd

has

dra

wn

a

cul-

de-s

ac f

or a

res

iden

tial

dev

elop

men

t pl

an. H

e u

sed

a co

mpa

ss t

o dr

aw t

he

cul-

de-s

ac s

o th

at i

t w

ould

be

circ

ula

r.

On

his

blu

epri

nt,

th

e ce

nte

r of

th

e cu

l-de

-sac

has

coo

rdin

ates

(-

1, -

1) a

nd

a po

int

on t

he

circ

le i

s (2

, 3).

Wh

at i

s th

e ra

diu

s of

th

e cu

l-de

-sac

?

3. L

AN

DSC

API

NG

Ran

dy p

lott

ed a

tr

ian

gula

r pa

tio

on a

lan

dsca

pe p

lan

for

a

clie

nt.

Wh

at i

s th

e le

ngt

h o

f fe

nci

ng

he

wil

l n

eed

alon

g th

e pa

tio

edge

th

at

bord

ers

the

prop

erty

lin

e? R

oun

d yo

ur

answ

er t

o th

e n

eare

st t

enth

.

4. U

TILI

TIES

Th

e el

ectr

ic c

ompa

ny

is

run

nin

g so

me

wir

es a

cros

s an

ope

n f

ield

. T

he

wir

e co

nn

ects

a u

tili

ty p

ole

at (

2, 1

4)

and

a se

con

d u

tili

ty p

ole

at (

7, -

8). I

f th

e el

ectr

ic c

ompa

ny

wis

hes

to

plac

e a

thir

d po

le a

t th

e m

idpo

int

of t

he

two

pole

s, a

t w

hat

coo

rdin

ates

sh

ould

th

e po

le b

e pl

aced

?

5. M

AR

CH

ING

BA

ND

Th

e O

hio

Sta

te

Un

iver

sity

mar

chin

g ba

nd

perf

orm

s a

fam

ous

on-f

ield

spe

llin

g of

O-H

-I-O

cal

led

“Scr

ipt

Oh

io”.

Som

etim

es t

hey

mu

st

adju

st t

he

usu

al d

imen

sion

s of

th

e w

ord

to f

it i

t in

to t

he

lim

ited

gu

est

ban

d pe

rfor

man

ce a

rea.

Th

e di

agra

m b

elow

sh

ows

part

of

the

adju

sted

dri

ll c

har

t.

Eac

h p

oin

t re

pres

ents

on

e ba

nd

mem

ber,

an

d th

e co

ordi

nat

es a

re i

n y

ards

.

a. H

ow f

ar i

s th

e dr

um

maj

or f

rom

th

e tu

ba p

laye

r w

ho

dots

th

e “i

”?

b.

Car

ol i

s th

e ba

nd

mem

ber

at t

he

top

left

of

the

firs

t O

in

Oh

io. S

he

is

loca

ted

at (

0, 2

6). H

ow f

ar a

way

is

Car

ol f

rom

th

e tu

ba p

laye

r? R

oun

d yo

ur

answ

er t

o th

e n

eare

st t

enth

.

dru

m m

ajo

rtu

ba

pla

yer

(20,

13)

(32,

10)

123456789

10

23

45

67

89

meters

met

ers

shru

bs

property line

house

tree

pat

io

( 2, 4

)

( 6, 8

)

( 5, 1

)

10-6

9.9

in

.

5 u

nit

s

7.1

m

(4.5

, 3)

12.4

yd

23.9

yd

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd41

4/14

/08

1:50

:06

PM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

4

2

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

A S

pace-S

avin

g M

eth

od

Tw

o ar

ran

gem

ents

for

coo

kies

on

a 3

2 cm

by

40 c

m c

ooki

e

shee

t ar

e sh

own

at

the

righ

t. T

he

cook

ies

hav

e 8-

cm d

iam

eter

s af

ter

they

are

bak

ed. T

he

cen

ters

of

the

cook

ies

are

on t

he

vert

ices

of

squ

ares

in

th

e to

p ar

ran

gem

ent.

In

th

e ot

her

, th

e ce

nte

rs a

re o

n t

he

vert

ices

of

equ

ilat

eral

tri

angl

es. W

hic

h

arra

nge

men

t is

mor

e ec

onom

ical

? T

he

tria

ngl

e ar

ran

gem

ent

is m

ore

econ

omic

al, b

ecau

se i

t co

nta

ins

one

mor

e co

okie

.

In t

he

squ

are

arra

nge

men

t, r

ows

are

plac

ed e

very

8 c

m. A

t w

hat

in

terv

als

are

row

s pl

aced

in

th

e tr

ian

gle

arra

nge

men

t?

Loo

k at

th

e ri

ght

tria

ngl

e la

bele

d a,

b, a

nd

c. A

leg

a o

f th

e

tria

ngl

e is

th

e ra

diu

s of

a c

ooki

e, o

r 4

cm. T

he

hyp

oten

use

c i

s th

e su

m o

f tw

o ra

dii,

or 8

cm

. Use

th

e P

yth

agor

ean

th

eore

m t

o fi

nd

b, t

he

inte

rval

of

the

row

s.

c2

= a

2 +

b2

82

= 4

2 +

b2

64

- 1

6 =

b2

√

��

48 =

b

4 √

�

3 =

bb

= 4

√ �

3

≈ 6

.93

Th

e ro

ws

are

plac

ed a

ppro

xim

atel

y ev

ery

6.93

cm

.

Sol

ve e

ach

pro

ble

m.

1. S

upp

ose

cook

ies

wit

h 1

0-cm

dia

met

ers

are

arra

nge

d in

th

e tr

ian

gula

r pa

tter

n s

how

n a

bove

. Wh

at i

s th

e in

terv

al b

of

the

row

s? 8

.66 c

m

2. F

ind

the

diam

eter

of

a co

okie

if

the

row

s ar

e pl

aced

in

th

e tr

ian

gula

r pa

tter

n e

very

3 √

�

3 c

m.

6 c

m

3. D

escr

ibe

oth

er p

ract

ical

app

lica

tion

s in

wh

ich

th

is k

ind

of

tria

ngu

lar

patt

ern

can

be

use

d to

eco

nom

ize

on s

pace

.

Sam

ple

an

sw

er:

packag

ing

can

s

{b

= ?

21 c

ooki

es

ca

{8

cm

20 c

ooki

es

10-6

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd42

4/15

/08

9:01

:46

PM

Lesson 10-7


NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

43

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

Sim

ilar

Tri

an

gle

s

Sim

ilar

Tria

ng

les

�R

ST

is

sim

ilar

to

�X

YZ

. Th

e an

gles

of

the

tw

o tr

ian

gles

hav

e eq

ual

mea

sure

. Th

ey a

re c

alle

d co

rres

pon

din

g an

gles

. Th

e si

des

oppo

site

th

e co

rres

pon

din

g an

gles

are

cal

led

corr

esp

ond

ing

sid

es.

30°

30°

60°

60°

SZ

X Y

RT

D

eter

min

e w

het

her

th

e p

air

of t

rian

gles

is

sim

ilar

. J

ust

ify

you

r an

swer

.

Sin

ce c

orre

spon

din

g an

gles

do

not

hav

e th

e eq

ual

mea

sure

s, t

he

tria

ngl

es a

re

not

sim

ilar

.

50°

75°

45°

89°

R

S

T

X

Y

Z

D

eter

min

e w

het

her

th

e p

air

of t

rian

gles

is

sim

ilar

. Ju

stif

y yo

ur

answ

er.

Th

e m

easu

re o

f ∠

G =

180

° -

(90

° +

45°

) =

45°

. T

he

mea

sure

of

∠I

= 1

80°

- (

45°

+ 4

5°)

= 9

0°.

Sin

ce c

orre

spon

din

g an

gles

hav

e eq

ual

m

easu

res,

�E

FG

∼ �

HIJ

.

90°

45°

45°

45°

F

G

H

J

I

E

Exer

cise

sD

eter

min

e w

het

her

eac

h p

air

of t

rian

gles

is

sim

ilar

. Ju

stif

y yo

ur

answ

er.

1.

2.

3.

Y

es;

co

rresp

on

din

g

No

; co

rresp

on

din

g

Yes;

co

rresp

on

din

g

an

gle

s h

ave e

qu

al

a

ng

les d

o n

ot

have

an

gle

s h

ave e

qu

al

measu

res.

eq

ual

measu

res.

measu

res.

4.

5.

6.

Y

es;

co

rresp

on

din

g

Yes;

co

rresp

on

din

g

No

; co

rresp

on

din

g

an

gle

s h

ave e

qu

al

a

ng

les h

ave e

qu

al

an

gle

s d

o n

ot

have

measu

res.

measu

res.

eq

ual

measu

res.

20°

30°

120°

115°

45°

45°

80°

55°

40°

30°

30°

110°

90°

45°

60°

60°

30°

30°

30°

120°

10-7

Sim

ila

r T

ria

ng

les

If t

wo

tria

ng

les a

re s

imila

r, t

he

n t

he

me

asu

res o

f th

eir c

orr

esp

on

din

g s

ide

s

are

pro

po

rtio

na

l a

nd

th

e m

ea

su

res o

f

the

ir c

orr

esp

on

din

g a

ng

les a

re e

qu

al.

�A

BC

∼ �

DE

F

AB

−

DE =

BC

−

EF =

AC

−

DF

A

BD

E

F

C

Exam

ple

1Ex

amp

le 2

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd43

4/11

/08

9:25

:49

AM



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

4

4

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(con

tin

ued

)

Sim

ilar

Tri

an

gle

s

Fin

d U

nkn

ow

n M

easu

res

If s

ome

of t

he

mea

sure

men

ts a

re k

now

n, p

ropo

rtio

ns

can

be

use

d to

fin

d th

e m

easu

res

of t

he

oth

er s

ides

of

sim

ilar

tri

angl

es.

IND

IREC

T M

EASU

REM

ENT

�A

BC

∼ �

AE

D i

n t

he

figu

re a

t th

e ri

ght.

F

ind

th

e h

eigh

t of

th

e ap

artm

ent

bu

ild

ing.

Let

BC

= x

.E

D

−

BC

= A

D

−

AC

=

25

−

300

ED

= 7

, A

D =

25

, A

C =

30

0

25

x =

210

0 F

ind

th

e c

ross p

rod

ucts

.

x

= 8

4T

he

apar

tmen

t bu

ildi

ng

is 8

4 m

eter

s h

igh

.

Exer

cise

sF

ind

th

e m

issi

ng

mea

sure

s fo

r th

e p

air

of s

imil

ar

tria

ngl

es i

f �

AB

C ∼

�D

EF

.

1. c

= 1

5, d

= 8

, e =

6, f

= 1

0 a =

12;

b =

9

2. c

= 2

0, a

= 1

2, b

= 8

, f =

15

d =

9;

e =

6

3. a

= 8

, d =

8, e

= 6

, f =

7 b

= 6

; c =

7

4. a

= 2

0, d

= 1

0, e

= 8

, f =

10

b =

16;

c =

20

5. c

= 5

, d =

10,

e =

8, f

= 8

a =

25

−

4 ;

b =

5

6. a

= 2

5, b

= 2

0, c

= 1

5, f

= 1

2 d

= 2

0;

e =

16

7. b

= 8

, d =

8, e

= 4

, f =

10

a =

16;

c =

20

8. I

ND

IREC

T M

EASU

REM

ENT

Bru

ce l

ikes

to

amu

se

his

bro

ther

by

shin

ing

a fl

ash

ligh

t on

his

han

d an

d m

akin

g a

shad

ow o

n t

he

wal

l. H

ow f

ar i

s it

fro

m

the

flas

hli

ght

to t

he

wal

l? 5

1.6

in

. o

r 4.3

ft

9. I

ND

IREC

T M

EASU

REM

ENT

A f

ores

t ra

nge

r u

ses

sim

ilar

tr

ian

gles

to

fin

d th

e h

eigh

t of

a t

ree.

Fin

d th

e h

eigh

t of

th

e tr

ee.

60 f

t

12 ft

20 ft

x

100

ft

5 in

.6

in.

x in

.

4 ft

Note

: Not

dra

wn

to s

cale

A

ca

be

fd

CD

F

EB

25 m

Note

: Not

dra

wn

to s

cale

275

m

7 m

ADE

CB

x

10-7

Exam

ple

7 −

x

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd44

4/11

/08

9:26

:04

AM

Lesson 10-7


NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

45

Gle

ncoe

Alg

ebra

1

Skill

s Pr

acti

ceS

imilar

Tri

an

gle

s

PR

Q

q

rp

SU

T

t

us

52°

52°

63

°

65

°F

EH

J

KG

40°

40°

45°

F

EG

H

JK

60°

60°

57°

60°

XV

W

U

Z

Y

40°

50°

A

DF

E C

B

10-7

Det

erm

ine

wh

eth

er e

ach

pai

r of

tri

angl

es i

s si

mil

ar. J

ust

ify

you

r an

swer

.

1.

2.

Y

es;

∠A

= ∠

D =

90°;

∠B

=

No

; ∠

Y =

180°

- (

60°

+ 6

0°)

= 6

0°.

180°

- (

90°

+ 4

0°)

= 5

0°

= ∠

E;

S

ince �

UV

W h

as a

57°

an

gle

, b

ut

∠F =

180°

- (

90°

+ 5

0°)

= 4

0°

=

�X

YZ

do

es n

ot,

co

rresp

on

din

g

∠C

. S

ince t

he c

orr

esp

on

din

g

an

gle

s d

o n

ot

all h

ave e

qu

al

an

gle

s h

ave e

qu

al

measu

res,

m

easu

res,

an

d t

he t

rian

gle

s a

re n

ot

�A

BC

∼ �

DE

F.

sim

ilar.

3.

4.

N

o;

∠F =

180°

- (

45°

+ 4

0°)

=

Yes;

∠G

= 1

80°

- (

65°

+ 5

2°)

= 6

3°

=

95°.

Sin

ce �

HJK

has a

90°

∠

K;

∠J =

180°

- (

63°

+ 5

2°)

= 6

5°

=

an

gle

, b

ut

�E

FG

do

es n

ot,

∠

F;

∠E

= ∠

H =

52°.

Sin

ce t

he

co

rresp

on

din

g a

ng

les d

o n

ot

all

co

rresp

on

din

g a

ng

les h

ave e

qu

al

have e

qu

al

measu

res,

an

d t

he

m

easu

res,

�E

FG

∼ �

HJK

.

tria

ng

les a

re n

ot

sim

ilar.

Fin

d t

he

mis

sin

g m

easu

res

for

the

pai

r of

sim

ilar

tr

ian

gles

if

�P

QR

∼ �

ST

U.

5.

r =

4, s

= 6

, t =

3, u

= 2

p =

12,

q =

6

6.

t =

8, p

= 2

1, q

= 1

4, r

= 7

u =

4,

s =

12

7.

p =

15,

q =

10,

r =

5, s

= 6

t =

4,

u =

2

8.

p =

48,

s =

16,

t =

8, u

= 4

r =

12,

q =

24

9.

q =

6, s

= 2

, t =

3 −

2 ,

u =

1 −

2 r

= 2

, p

= 8

10.

p =

3, q

= 2

, r =

1, u

= 1 −

3 s

= 1

, t

= 2

−

3

11.

p =

14,

q =

7, u

= 2

.5, t

= 5

r =

3.5

, s =

10

12.

r =

6, s

= 3

, t =

21

−

8 , u

= 9 −

4 p

= 8

, q

= 7

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd45

5/8/

084:

15:2

2P

M



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

4

6

Gle

ncoe

Alg

ebra

1

Prac

tice

Sim

ilar

Tri

an

gle

sD

eter

min

e w

het

her

eac

h p

air

of t

rian

gles

is

sim

ilar

. Ju

stif

y yo

ur

answ

er.

1.

2.

Y

es;

∠Q

= ∠

T =

90°;

∠P

=

No

; ∠

C =

180°

- (

47°

+ 8

0°)

= 5

3°.

180°

- (

90°

+ 3

1°)

= 5

9°

= ∠

S;

S

ince �

FG

H h

as a

56°

an

gle

, b

ut

∠U

= 1

80°

- (

90°

+ 5

9°)

= 3

1°

=

�C

DE

do

es n

ot,

co

rresp

on

din

g

∠R

. S

ince t

he c

orr

esp

on

din

g

an

gle

s d

o n

ot

all h

ave e

qu

al

an

gle

s h

ave e

qu

al

measu

res,

m

easu

res,

an

d t

he t

rian

gle

s a

re n

ot

�P

QR

, �

STU

. s

imilar.

Fin

d t

he

mis

sin

g m

easu

res

for

the

pai

r of

sim

ilar

tria

ngl

es i

f �

AB

C ∼

�D

EF

.

3. c

= 4

, d =

12,

e =

16,

f =

8 a

= 6

, b

= 8

4. e

= 2

0, a

= 2

4, b

= 3

0, c

= 1

5 d

= 1

6,

f =

10

5. a

= 1

0, b

= 1

2, c

= 6

, d =

4 e

= 4

.8,

f =

2.4

6. a

= 4

, d =

6, e

= 4

, f =

3 c

= 2

, b

= 8

−

3

7. b

= 1

5, d

= 1

6, e

= 2

0, f

= 1

0 a =

12,

c =

15

−

2

8. a

= 1

6, b

= 2

2, c

= 1

2, f

= 8

d =

32

−

3 ,

e =

44

−

3

9. a

= 5 −

2 ,

b =

3, f

= 11

−

2 ,

e =

7 c

= 3

3

−

14 ,

d =

35

−

6

10. c

= 4

, d =

6, e

= 5

.625

, f =

12

a =

2,

b =

1.8

75

11. S

HA

DO

WS

Su

ppos

e yo

u a

re s

tan

din

g n

ear

a bu

ildi

ng

and

you

wan

t to

kn

ow i

ts h

eigh

t.

Th

e bu

ildi

ng

cast

s a

66-f

oot

shad

ow. Y

ou c

ast

a 3-

foot

sh

adow

. If

you

are

5 f

eet

6 in

ches

ta

ll, h

ow t

all

is t

he

buil

din

g?

12. M

OD

ELS

Tru

ss b

ridg

es u

se t

rian

gles

in

th

eir

supp

ort

beam

s. M

olly

mad

e a

mod

el o

f a

tru

ss b

ridg

e in

th

e sc

ale

of 1

in

ch =

8 f

eet.

If

the

hei

ght

of t

he

tria

ngl

es o

n t

he

mod

el i

s 4.

5 in

ches

, wh

at i

s th

e h

eigh

t of

th

e tr

ian

gles

on

th

e ac

tual

bri

dge?

DF

E

e

fd

AC

B

b

ca

80°

47°

47°

56°

EH

F

GD

C

31°

59°

RQ

STU

P

10-7

121 f

t

36 f

t

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd46

4/11

/08

9:26

:13

AM

Lesson 10-7


NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

47

Gle

ncoe

Alg

ebra

1

Wor

d Pr

oble

m P

ract

ice

Sim

ilar

Tri

an

gle

s

1. C

RA

FTS

Lay

la i

s w

ants

to

buy

a se

t of

si

mil

ar m

agn

ets

for

her

ref

rige

rato

r do

or.

Lay

la f

inds

th

e m

agn

ets

belo

w f

or s

ale

at

a lo

cal

shop

. Wh

ich

tw

o ar

e si

mil

ar?

B a

nd

C

2. E

XH

IBIT

ION

S T

he

wor

ld’s

lar

gest

can

dle

was

dis

play

ed a

t th

e 18

97 S

tock

hol

m

Exh

ibit

ion

. Su

ppos

e L

ars

mea

sure

d th

e le

ngt

h o

f th

e sh

adow

it

cast

at

11:0

0 A.M

. an

d fo

un

d th

at i

t w

as 1

2 fe

et. S

upp

ose

that

im

med

iate

ly a

fter

th

is, h

e m

easu

red

to f

ind

that

a n

earb

y 25

-foo

t te

nt

pole

ca

st a

sh

adow

5 f

eet

lon

g. H

ow t

all

was

th

e w

orld

’s l

arge

st c

andl

e?

3. L

AN

DM

AR

KS

Th

e T

oy a

nd

Min

iatu

re

Mu

seu

m o

f K

ansa

s C

ity

disp

lays

a

min

iatu

re r

epli

ca o

f G

eorg

e W

ash

ingt

on’s

M

oun

t V

ern

on m

ansi

on. T

he

min

iatu

re

hou

se i

s 10

fee

t lo

ng,

6 f

eet

wid

e, 8

fee

t ta

ll, a

nd

has

22

room

s. T

he

scal

e of

th

e m

odel

to

the

orig

inal

is

one

inch

to

one

foot

. If

the

roof

gab

le o

f th

e m

inia

ture

h

as d

imen

sion

s as

sh

own

on

th

e di

agra

m

belo

w, w

hat

is

the

hei

ght

of t

he

roof

ga

ble

on t

he

orig

inal

Mou

nt

Ver

non

m

ansi

on?

Sour

ce: M

ount

Ver

non

4. S

UR

VEY

ING

Su

rvey

ors

use

pro

pert

ies

of t

rian

gles

in

clu

din

g si

mil

arit

y an

d th

e P

yth

agor

ean

Th

eore

m t

o fi

nd

un

know

n

dist

ance

s. U

se t

he

dim

ensi

ons

on t

he

diag

ram

to

fin

d th

e u

nkn

own

dis

tan

ce x

acro

ss t

he

lake

. 8

0 m

5. P

UZZ

LES

Th

e fi

gure

bel

ow s

how

s an

an

cien

t C

hin

ese

mov

able

pu

zzle

cal

led

a ta

ngr

am. I

t h

as 7

pie

ces

that

can

be

reco

nfi

gure

d to

pro

duce

an

en

dles

s n

um

ber

of d

esig

ns

and

pict

ure

s.

Ass

um

e th

at t

he

side

len

gth

of

this

ta

ngr

am s

quar

e is

√� 2

cm

. Lea

ve y

our

answ

ers

as s

impl

ifie

d ra

dica

l ex

pres

sion

s.

a.

Wh

at a

re t

he

side

len

gth

s of

tri

angl

es

1 an

d 2?

b

. Wha

t ar

e th

e si

de le

ngth

s of

tri

angl

e 7?

√ �

2

− 2

cm

, √

�

2

− 2c

m,

1 c

m

c. W

hat

are

th

e si

de l

engt

hs

of t

rian

gles

3 an

d 5?

AB

C

7cm

6cm

5cm

3cm

4cm

8cm

3 ft

1.9

ft 1

.9 ft

heig

ht

NP

M x

Q

80

m 4

0 m

120

m 0

lake

1

2

3

4

5

67

10-7

1

−

2 c

m,

1

−

2 c

m,

√ �

2

−

2 c

m

60 f

t

14 f

t1

cm

, 1

cm

, √

�

2 c

m

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd47

4/14

/08

1:51

:24

PM



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

4

8

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

A C

uri

ou

s C

on

str

ucti

on

Man

y m

ath

emat

icia

ns

hav

e be

en i

nte

rest

ed i

n w

ays

to

con

stru

ct t

he

nu

mbe

r π

. Her

e is

on

e su

ch g

eom

etri

c co

nst

ruct

ion

.In

th

e dr

awin

g, t

rian

gles

AB

C a

nd

AD

E a

re r

igh

ttr

ian

gles

. Th

e le

ngt

h o

f −−

− A

D e

qual

s th

e le

ngt

h o

f −

−

AC

an

d −

−

FB

is

para

llel

to

−−−

EG

.T

he

len

gth

of

−−−

BG

giv

es a

dec

imal

app

roxi

mat

ion

of

th

e fr

acti

onal

par

t of

π t

o si

x de

cim

al p

lace

s.

Fol

low

th

e st

eps

to f

ind

th

e le

ngt

h o

f −

−

BG

. Rou

nd

to s

even

dec

imal

pla

ces.

1. U

se t

he

len

gth

of

−−−

BC

an

d th

e P

yth

agor

ean

T

heo

rem

to

fin

d th

e le

ngt

h o

f −

−

AC

.

A

C =

√ ��

��

1 2 +

( 7

−

8 ) 2

= 1

.3287682

2. F

ind

the

len

gth

of

−−−

AD

.

A

D =

AC

= 1

.3287682

3. U

se t

he

len

gth

of

−−−

AD

an

d th

e P

yth

agor

ean

Th

eore

m t

o fi

nd

the

len

gth

of

−−

AE

.

A

E =

√ ��

��

��

(A

D ) 2

+ ( 1

−

2 ) 2

= 1

.4197271

4. T

he

side

s of

th

e si

mil

ar t

rian

gles

FE

D a

nd

DE

A a

re i

n p

ropo

rtio

n. S

o, F

E

−

0.5 =

0.5

−

AE

.F

ind

the

len

gth

of

−−

FE

.

F

E =

1

−

4(A

E)

= 0

.1760902

5. F

ind

the

len

gth

of

−−

AF

.

A

F =

AE

- F

E =

1.2

436369

6. T

he

side

s of

th

e si

mil

ar t

rian

gles

AF

B a

nd

AE

G a

re i

n p

ropo

rtio

n. S

o,

AF

−

AE

= A

B

−

AG

. F

ind

the

len

gth

of

−−−

AG

.

A

G =

AB

· A

E

−

AF

= A

E

−

AF =

1.1

415929

7. N

ow, f

ind

the

len

gth

of

−−−

BG

.

B

G =

AG

- A

B =

AG

- 1

= 0

.1415929

8. T

he

valu

e of

π t

o se

ven

dec

imal

pla

ces

is 3

.141

5927

. Com

pare

th

e fr

acti

onal

par

t of

pi

wit

h t

he

len

gth

of

−−−

BG

.

0

.1415929 -

0.1

415927 =

0.0

000002,

an

err

or

of

less t

han

1 p

art

in

a m

illio

n

7 – 8

1 – 2

C

E DG

BA

F

1

10-7

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd48

4/15

/08

8:47

:18

PM

Lesson 10-8


NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

49

Gle

ncoe

Alg

ebra

1

10-8

Trig

on

om

etri

c R

atio

s T

rigo

nom

etry

is

the

stu

dy o

f re

lati

onsh

ips

of t

he

angl

es a

nd

the

side

s of

a r

igh

t tr

ian

gle.

Th

e th

ree

mos

t co

mm

on t

rigo

nom

etri

c ra

tios

are

th

e si

ne,

cos

ine,

an

d ta

nge

nt.

sin

e o

f ∠

A =

le

g o

pp

osite

∠A

−

hyp

ote

nu

se

sin

e o

f ∠

B =

le

g o

pp

osite

∠B

−

hyp

ote

nu

se

sin

A =

a −

c

sin

B =

b

−

c

co

sin

e o

f ∠

A =

le

g a

dja

ce

nt

to ∠

A

−

−

h

yp

ote

nu

se

co

sin

e o

f ∠

B =

le

g a

dja

ce

nt

to ∠

B

−

−

h

yp

ote

nu

se

co

s A

= b

−

c

co

s B

= a −

c

tan

ge

nt

of

∠A

=

leg

op

po

site

∠A

−

−

le

g a

dja

ce

nt

to ∠

A

tan

ge

nt

of

∠B

=

leg

op

po

site

∠B

−

−

le

g a

dja

ce

nt

to ∠

B

tan

A =

a −

b

tan

B =

b

−

a

F

ind

th

e va

lues

of

the

thre

e tr

igon

omet

ric

rati

os f

or a

ngl

e A

.

Ste

p 1

U

se t

he

Pyt

hag

orea

n T

heo

rem

to

fin

d B

C.

a2 +

b2

= c

2 P

yth

agore

an T

heore

m

a2 +

82

= 1

02 b

= 8

and c

= 1

0

a2

+ 6

4 =

100

S

implif

y.

a2 =

36

Subtr

act

64 f

rom

both

sid

es.

a =

6

Take t

he s

quare

root

of

each s

ide.

Ste

p 2

U

se t

he

side

len

gth

s to

wri

te t

he

trig

onom

etri

c ra

tios

.

sin

A =

opp

−

hyp

=

6 −

10

= 3 −

5

co

s A

=

adj

−

hyp

=

8 −

10 =

4 −

5

tan

A =

opp

−

adj =

6 −

8 =

3 −

4

Exer

cise

sF

ind

th

e va

lues

of

the

thre

e tr

igon

omet

ric

rati

os f

or a

ngl

e A

.

1.

817

2.

3 5

3.

24

7

sin

A =

15

−

17 ,

co

s A

=

8

−

17 ,

sin

A =

7

−

25 ,

co

s A

= 2

4

−

25 ,

tan

A =

15

−

8

ta

n A

=

7

−

24

Use

a c

alcu

lato

r to

fin

d t

he

valu

e of

eac

h t

rigo

nom

etri

c ra

tio

to t

he

nea

rest

te

n-t

hou

san

dth

.

4.

sin

40°

5.

cos

25°

0.9

063

6. t

an 8

5° 1

1.4

301

Stud

y G

uide

and

Inte

rven

tion

Tri

go

no

metr

ic R

ati

os

Exam

ple

a10 8

0.6

428

b

a

c

sin

A =

4

−

5 ,

co

s A

= 3

−

5 ,

tan

A =

4

−

3

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd49

4/11

/08

9:26

:32

AM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

5

0

Gle

ncoe

Alg

ebra

1

10-8

Use

Tri

go

no

met

ric

Rat

ios

Wh

en y

ou f

ind

all

of t

he

un

know

n m

easu

res

of t

he

side

s an

d an

gles

of

a ri

ght

tria

ngl

e, y

ou a

re s

olvi

ng

the

tria

ngl

e. Y

ou c

an f

ind

the

mis

sin

g m

easu

res

of a

rig

ht

tria

ngl

e if

you

kn

ow t

he

mea

sure

of

two

side

s of

th

e tr

ian

gle,

or

the

mea

sure

of

one

side

an

d th

e m

easu

re o

f on

e ac

ute

an

gle.

Sol

ve t

he

tria

ngl

e. R

oun

d e

ach

sid

e le

ngt

h t

o th

e n

eare

st t

enth

.

Ste

p 1

F

ind

the

mea

sure

of

∠B

. Th

e su

m o

f th

e m

easu

res

of

the

angl

es i

n a

tri

angl

e is

180

.18

0° −

(90

° +

38°

) =

52°

Th

e m

easu

re o

f ∠

B i

s 52

°.

Ste

p 2

F

ind

the

mea

sure

of

−−

AB

. Bec

ause

you

are

giv

en t

he

mea

sure

of

the

side

adj

acen

t to

∠ A

an

d ar

e fi

ndi

ng

the

mea

sure

of

the

hyp

oten

use

, use

th

e co

sin

e ra

tio.

co

s 38

° =

13

−

c

Definitio

n o

f cosin

e

c

cos

38°

= 1

3 M

ultip

ly e

ach s

ide b

y c

.

c

=

13

−

cos

38°

Div

ide e

ach s

ide b

y s

in 4

1°.

So

the

mea

sure

of

−−

AB

is

abou

t 16

.5.

Ste

p 3

F

ind

the

mea

sure

of

−−−

BC

. Bec

ause

you

are

giv

en t

he

mea

sure

of

the

side

adj

acen

t to

∠ A

an

d ar

e fi

ndi

ng

the

mea

sure

of

the

side

opp

osit

e ∠

A, u

se t

he

tan

gen

t ra

tio.

tan

38°

=

a −

13

D

efinitio

n o

f ta

ngent

13 t

an 3

8° =

a

Multip

ly e

ach s

ide b

y 1

3.

10.2

≈ a

U

se a

calc

ula

tor.

So

the

mea

sure

of

−−−

BC

is

abou

t 10

.2.

Exer

cise

sS

olve

eac

h r

igh

t tr

ian

gle.

Rou

nd

eac

h s

ide

len

gth

to

the

nea

rest

ten

th.

1.

9

ab 30

°

2.

b

8c

44°

3.

16cb

56°

∠

B =

60°,

AC

≈ 7

.8,

∠A

= 6

0°,

AC

≈ 7

.7,

∠B

= 3

4°,

AC

≈ 1

9.3

,B

C =

4.5

A

B ≈

11.1

A

B ≈

10.8

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Tri

go

no

metr

ic R

ati

os

Exam

ple

13

ac

38°

035_

054_

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Lesson 10-8


NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

51

Gle

ncoe

Alg

ebra

1

Fin

d t

he

valu

es o

f th

e th

ree

trig

onom

etri

c ra

tios

for

an

gle

A.

1.

7785

2

.

159

sin

A =

7

7

−

85 , c

os A

= 3

6

−

85 ,

tan

A =

77

−

36

sin

A =

4

−

5 , c

os A

= 3

−

5 , t

an

A =

4

−

3

3.

10

24

4

. 15

8

sin

A =

12

−

13 ,

co

s A

=

5

−

13 , t

an

A =

12

−

5

sin

A =

8

−

17 , c

os A

= 1

5

−

17 , t

an

A =

8

−

15

Use

a c

alcu

lato

r to

fin

d t

he

valu

e of

eac

h t

rigo

nom

etri

c ra

tio

to t

he

nea

rest

ten

-th

ousa

nd

th.

5.

sin

18°

0.3

090

6.

cos

68°

0.3

746

7.

tan

27°

0.5

095

8.

cos

60°

0.5

9.

tan

75°

3.7

321

10

. si

n 9

° 0.1

564

Sol

ve e

ach

rig

ht

tria

ngl

e. R

oun

d e

ach

sid

e le

ngt

h t

o th

e n

eare

st t

enth

.

11.

1317

°

12

.

655

°

∠A

= 7

3°,

AB

= 1

3.6

, A

C =

4.0

∠

B =

35°,

AB

= 1

0.5

, BC

= 8

.6

Fin

d m

∠J

for

eac

h r

igh

t tr

ian

gle

to t

he

nea

rest

deg

ree.

13.

6

5

14.

19

11

4

0°

55°

10-8

Skill

s Pr

acti

ceTri

go

no

metr

ic R

ati

os

035_

054_

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1CR

MC

10_8

9050

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dd51

5/9/

083:

12:1

5A

M



An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

5

2

Gle

ncoe

Alg

ebra

1

Fin

d t

he

valu

es o

f th

e th

ree

trig

onom

etri

c ra

tios

for

an

gle

A.

1.

7297

2

.

36

15

sin

A =

65

−

97 ,

co

s A

= 7

2

−

97 ,

tan

A =

65

−

72

sin

A =

36

−

39 ,

co

s A

= 1

5

−

39 ,

tan

A =

36

−

15

Use

a c

alcu

lato

r to

fin

d t

he

valu

e of

eac

h t

rigo

nom

etri

c ra

tio

to t

he

nea

rest

te

n-t

hou

san

dth

.

3.

tan

26°

0.4

877

4.

sin

53°

0.7

986

5.

cos

81°

0.1

564

Sol

ve e

ach

rig

ht

tria

ngl

e. R

oun

d e

ach

sid

e le

ngt

h t

o th

e n

eare

st t

enth

.

6.

67°

22

7.

9

29°

∠B

= 2

3°,

AB

= 2

3.9

, A

C =

9.3

∠

A =

61°,

AB

= 1

0.3

, B

C =

5.0

Fin

d m

∠J

for

eac

h r

igh

t tr

ian

gle

to t

he

nea

rest

deg

ree.

8.

11

5

9.

1218

2

4°

42°

10. S

UR

VEY

ING

If

poin

t A

is

54 f

eet

from

th

e tr

ee, a

nd

the

angl

e be

twee

n t

he

grou

nd

at

poin

t A

an

d th

e to

p of

th

e tr

ee i

s 25

°, f

ind

the

hei

ght

h o

f th

e tr

ee.

2

5.2

ft

10-8

Prac

tice

Tri

go

no

metr

ic R

ati

os

25°

54 ft

h

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd52

4/11

/08

9:27

:17

AM

Lesson 10-8


NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

53

Gle

ncoe

Alg

ebra

1

10-8

Wor

d Pr

oble

m P

ract

ice

Tri

go

no

metr

ic R

ati

os

1. W

ASH

ING

TON

MO

NU

MEN

TJe

ann

ie

is t

ryin

g to

det

erm

ine

the

hei

ght

of t

he

Was

hin

gton

Mon

um

ent.

If

poin

t A

is

765

feet

fro

m t

he

mon

um

ent,

an

d th

e an

gle

betw

een

th

e gr

oun

d an

d th

e to

p of

th

e m

onu

men

t at

poi

nt

A i

s 36

°, f

ind

the

hei

ght

h o

f th

e m

onu

men

t to

th

e n

eare

st f

oot.

765

ft36°

h

2. A

IRPL

AN

ES A

pil

ot t

akes

off

fro

m a

ru

nw

ay a

t an

an

gle

of 2

0º a

nd

mai

nta

ins

that

an

gle

un

til

it i

s at

its

cru

isin

g al

titu

de o

f 25

00 f

eet.

Wh

at h

oriz

onta

l di

stan

ce h

as t

he

plan

e tr

avel

ed w

hen

it

reac

hes

its

cru

isin

g al

titu

de?

3. T

RU

CK

RA

MPS

A m

ovin

g co

mpa

ny

use

s an

11-

foot

-lon

g ra

mp

to u

nlo

ad f

urn

itu

re

from

a t

ruck

. If

the

bed

of t

he

tru

ck i

s 3

feet

abo

ve t

he

grou

nd,

wh

at i

s th

e an

gle

of i

ncl

ine

of t

he

ram

p to

th

e n

eare

st

degr

ee?

4. S

PEC

IAL

TRIA

NG

LES

Wh

ile

inve

stig

atin

g ri

ght

tria

ngl

e K

LM

, M

erce

des

fin

ds t

hat

cos

M =

sin

M.

Wh

at i

s th

e m

easu

re o

f an

gle

M?

5. T

ELEV

ISIO

NS

Tel

evis

ion

s ar

e co

mm

only

si

zed

by m

easu

rin

g th

eir

diag

onal

. A

com

mon

siz

e fo

r w

ides

cree

n p

lasm

a T

Vs

is 4

2 in

ches

.

42’’

h

h16 9

a

. A w

ides

cree

n t

elev

isio

n h

as a

16:

9 as

pect

rat

io, t

hat

is,

th

e sc

reen

wid

th

is 16

−

9

tim

es t

he

scre

en h

eigh

t. U

se t

he

Pyt

hag

orea

n T

heo

rem

to

wri

te a

n

equ

atio

n a

nd

solv

e fo

r th

e h

eigh

t h

of

the

tele

visi

on i

n i

nch

es.

b

. Use

th

e in

form

atio

n f

rom

par

t a

to

solv

e th

e ri

ght

tria

ngl

e.

c

. Wh

at w

ould

th

e m

easu

re o

f an

gle

A b

e on

a s

tan

dard

tel

evis

ion

wit

h a

4:3

as

pect

rat

io?

( 16

−

9

h) 2

+ h

2 =

42

2;

h =

20.6

in

.

wid

th =

36.6

in

., ∠

A =

29°,

∠B

= 6

1°

16°

45°

6869 f

t

556 f

t

37°

035_

054_

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/08

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:30

AM



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pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DA

TE

PE

RIO

D

Cha

pte

r 10

5

4

Gle

ncoe

Alg

ebra

1

In a

ddit

ion

to

the

sin

e, c

osin

e, a

nd

tan

gen

t, t

her

e ar

e th

ree

oth

er c

omm

on t

rigo

nom

etri

c ra

tios

. Th

ey a

re t

he

seca

nt,

cos

ecan

t, a

nd

cota

nge

nt.

se

ca

nt

of

∠A

=

hyp

ote

nu

se

−

leg

op

po

site

∠A

se

ca

nt

of

∠B

=

hyp

ote

nu

se

−

leg

op

po

site

∠B

se

c A

= c −

a

se

c B

= c −

b

b

a

cco

se

ca

nt

of

∠A

=

hyp

ote

nu

se

−

leg

ad

jace

nt

∠A

co

se

ca

nt

of

∠B

=

hyp

ote

nu

se

−

leg

ad

jace

nt

∠B

csc A

= c −

b

csc B

= c −

a

co

tan

ge

nt

of

∠A

=

leg

ad

jace

nt

to ∠

A

−

−

le

g o

pp

osite

∠A

co

tan

ge

nt

of

∠B

=

leg

ad

jace

nt

to ∠

B

−

−

le

g o

pp

osite

∠B

co

t A

= a −

b

co

t B

= b

−

a

Fin

d t

he

seca

nt,

cos

ecan

t, a

nd

cot

ange

nt

of a

ngl

e A

.

Use

th

e si

de l

engt

hs

to w

rite

th

e tr

igon

omet

ric

rati

os.

sec

A =

hyp

−

opp =

15

−

9 =

5 −

3

cs

c A

= h

yp

−

adj =

15

−

12 =

5 −

4

cot

A =

ad

j −

op

p = 12

−

9

= 4 −

3

Exer

cise

sF

ind

th

e se

can

t, c

osec

ant,

an

d c

otan

gen

t of

an

gle

A.

1.

817

2.

35

3.

24

7

sec A

= 1

7

−

8 ,

csc A

= 1

7

−

15

sec A

= 2

5

−

24 ,

csc A

= 2

5

−

27 ,

co

t A

=

8

−

15

co

t A

= 2

4

−

7

4. H

ow d

oes

the

sin

e of

an

an

gle

rela

te t

o th

e an

gle’

s co

seca

nt?

How

doe

s th

e co

sin

e of

an

an

gle

rela

te t

o th

e an

gle’

s se

can

t? H

ow d

oes

the

cota

nge

nt

of a

n a

ngl

e re

late

to

the

angl

e’s

seca

nt?

s

ec A

=

1

−

co

s A

, c

sc A

=

1

−

sin

A , a

nd

co

t A

=

1

−

tan

A

Use

th

e re

lati

ons

that

you

fou

nd

in

Exe

rcis

e 4

and

a c

alcu

lato

r to

fin

d t

he

valu

e of

ea

ch t

rigo

nom

etri

c ra

tio

to t

he

nea

rest

ten

-th

ousa

nd

th.

5.

sec

17°

1.0

457

6.

csc

49°

1.3

250

7.

cot

81°

0.1

584

10-8

Enri

chm

ent

Mo

re T

rig

on

om

etr

ic R

ati

os

Exam

ple

12159

sec A

= 5

−

3 ,

csc A

= 5

−

4 ,

co

t A

=

8

−

15

035_

054_

ALG

1CR

MC

10_8

9050

4.in

dd54

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/08

9:27

:44

AM



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pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass


1.

2.

3.

4.

5.

1.

2.

3.

4.

5.

6.

7.

1.

2.

3.

4.

5.

1.

2.

3.

4.

5.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

(-4, 8)

√ �� 61 or 7.81

C

no

14.70

4 √ � 6 + 8 √ �� 10

12 - 3 √ � 2

− 14

2|x| √ �� 2y

D

F

D

G

D

F

A

39 ft

RP = 27.5, PQ = 29.2, m∠Q = 70º

b = 21, z = 10

Chapter 10 Assessment Answer KeyQuiz 1 (Lessons 10-1 and 10-2) Quiz 3 (Lessons 10-4 and 10-5) Mid-Chapter TestPage 57 Page 58 Page 59 (Lessons 10-1 through 10-3)

Quiz 2 (Lesson 10-3)

Page 57

Quiz 4 (Lesson 10-6)

Page 58 8.

9.

10.

11.

12.

13.

14.

15.

Yes, since

corresponding angles

have equal measures.

26 √ � 5

5 √ � 2 + 6 √ � 5

√ �� 15 - 2 √ � 2

9.49 ft

29

5

3 √ � 2 + 2 √ � 6

7 √ � 5 - 2 √ �� 10

30 √ � 2

23

6 √ � 5 - 20 √ � 3

D

D

5

7

3, 8

2

2

A

6 √ � 2

y

x

D = {x|x ≥ -1}; R = {y|y ≥ 0};


An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass


Chapter 10 Assessment Answer KeyVocabulary Test Form 1Page 60 Page 61 Page 62

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

F

B

G

D

H

D

F

C

F

A 13.

14.

15.

16.

17.

18.

19.

20. J

A

H

D

F

C

B:

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

converse

cosine

Distance Formula

radical equation

conjugates

hypotenuse

radicand

legs

similar triangles

Sample answer: the study of

relationships among the angles

and sides of the triangle.

Sample answer: the

trigonometric ratio equivalent to

the leg opposite to an angle

divided by the leg adjacent to

the angle.

J

B

H

|2x + 1|

D


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ht © G

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ivision o

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om

panies, Inc.

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13.

14.

15.

16.

17.

18.

19.

20.

13.

14.

15.

16.

17.

18.

19.

20.

Chapter 10 Assessment Answer KeyForm 2A Form 2BPage 63 Page 64 Page 65 Page 66

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12. F

D

H

B

G

B

F

C

G

D

J

D

H

A

F

C

G

A

G

B 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12. J

B

F

C

J

C

J

C

F

B

F

D

H

A

J

A

G

D

H

B

12 mB:

14 ftB:


An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

PDF Pass


18.

19.

20.

21.

22.

23.

24.

25.

B:

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17. (-1, 2)

√ �� 185 or 13.60

√ �� 61 or 7.81

no

yes

7

11.66

12

11

12

2 √ �� 14 + 5 √ � 3

20 √ � 5

3 √ � 2 + 2 √ � 3

5y2|w| √ �� 3w

6 √ � 2

≈23.2 ft

67º

≈12.5 cm

√ �� 149 mi or 12.2 mi

40 1 − 2 ft

c = 14 −

3 , u = 6

No, not all corresponding angles have equal measures.

Yes, since corresponding angles have equal measures.

Chapter 10 Assessment Answer KeyForm 2CPage 67 Page 68

-2

D = {x | x ≥ 2}; R = {y | y ≥ 1}D = {x | x ≥ 1}; R = {y | y ≤ 5}

y

x


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om

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18.

19.

20.

21.

22.

23.

24.

25.

B:

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

Chapter 10 Assessment Answer KeyForm 2DPage 69 Page 70

√ �� 193 or 13.89

√ �� 74 or 8.60

no

yes

8

8.06

4

3

4

22 √ � 6

2 √ � y

10 √ � 5 + 15 √ � 2

5|xy| √ �� 2x

10 √ � 2

-3

≈30.0 ft

69º

15.20 cm

5.39 mi

15 ft

a = 27 −

7 , y = 10



(3, 2)

R = {x | x ≥ -1}; D = {y | y ≥ -3}

R = {x | x ≥ -2}; D = {y | y ≤ -1}

y

x


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18.

19.

20.

21.

22.

23.

24.

25.

B:

Chapter 10 Assessment Answer KeyForm 3Page 71 Page 72 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17. 20 √ � 5 units

7 √ � 5 or 15.65

yes

no

2 √ �� 11 cm

7 − 2 , 5

no solution

6

40 √ � 2 - 27 √ �� 15

0

2 √ �� 10 - √ � 3

− 7

x2 √ �� 5n

− 2n3

18 √ � 7

7 √ �� 30 - 27

− 39

87.0 ft

54º

8.06 miles

10.20 cm

136 1 − 2 m

b = 1.6, z = 5.355



4

√ �� 41

D = {x | x ≥ 2}; R = {y | y ≤ -2}

D = {x | x ≥ -4}; R = {y | y ≤ -12}

y

x


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Chapter 10 Assessment Answer KeyPage 73, Extended-Response Test

Scoring Rubric

Score General Description Specific Criteria

4 Superior

A correct solution that

is supported by well-

developed, accurate

explanations

• Shows thorough understanding of the concepts of

simplifying radical expressions, solving radical equations, the Pythagorean Theorem, right triangles, similar triangles, and the distance formula.

• Uses appropriate strategies to solve problems.

• Computations are correct.

• Written explanations are exemplary.

• Graphs are accurate and appropriate.

• Goes beyond requirements of some or all problems.

3 Satisfactory

A generally correct solution,

but may contain minor flaws

in reasoning or computation

• Shows an understanding of the concepts of simplifying radical expressions, solving radical equations, the Pythagorean Theorem, right triangles, similar triangles, and the distance formula.

• Uses appropriate strategies to solve problems.

• Computations are mostly correct.

• Written explanations are effective.

• Graphs are mostly accurate and appropriate.

• Satisfies all requirements of problems.

2 Nearly Satisfactory

A partially correct

interpretation and/or

solution to the problem

• Shows an understanding of most of the concepts of


• May not use appropriate strategies to solve problems.

• Computations are mostly correct.

• Written explanations are satisfactory.

• Graphs are mostly accurate.

• Satisfies the requirements of most of the problems.

1 Nearly Unsatisfactory

A correct solution with no

supporting evidence or

explanation

• Fiinal computation is correct.

• No written explanations or work is shown to substantiate

the final computation.

• Graphs may be accurate but lack detail or explanation.

• Satisfies minimal requirements of some of the problems.

0 Unsatisfactory

An incorrect solution

indicating no mathematical

understanding of the

concept or task, or no

solution is given

• Shows little or no understanding of most of the concepts of


• Does not use appropriate strategies to solve problems.

• Computations are incorrect.

• Written explanations are unsatisfactory.

• Graphs are inaccurate or inappropriate.

• Does not satisfy requirements of problems.

• No answer may be given.


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Chapter 10 Assessment Answer Key Page 73, Extended-Response Test

Sample Answers

1a. The Product Property of Square Roots states that the square root of a product is equal to the product of the square roots of the factors. For this property, a ≥ 0 and b ≥ 0.

1b. The Quotient Property of Square Roots states that the square root of a quotient is equal to the quotient of the square roots of the numerator and denominator. For this property, a ≥ 0 and b > 0.

1c. The student should recognize that both properties state that finding a square root can be done before or after certain other operations.

2a. P = L2 − k

2b. Sample answer: L 25 50 75

P 5208 20,833 46,875

2c. Sample answer:

L 25 50 75

P 7813 31,250 70,313

2d. A smaller constant of proportionality allows a plane to carry more weight.

3a. The student should agree. Since the measures of the three angles in a triangle have a sum of 180°, having two pairs of corresponding angles equal in measure forces the third pair of corresponding angles to also be equal in measure. Thus, by definition the two triangles must be similar.

3b. The student should disagree. Although two triangles that are the same size and same shape will have two pairs of sides equal in measure, having two pairs of sides equal in measure does not guarantee that corresponding sides are proportional. The two right triangles with side lengths of 3, 4, 5, and 4, 5, √ �� 41 respectively are an example.

4a. Sample answer: a = 9, b = 8; Using the Distance Formula, AB = √ �� 145 , AC = 8, BC = 9.

4b. Sample answer: a = 9, b = 8. ( √ �� 145 ) 2 � 92 + 82 145 = 81 + 64

In addition to the scoring rubric found on page A30, the following sample answers may be used as guidance in evaluating extended-response assessment items.


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Chapter 10 Assessment Answer KeyStandardized Test PracticePage 74 Page 75

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

5

9 8

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. 0

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. A B C D

8. F G H J

9. A B C D

10. F G H J

11. A B C D

12. F G H J

13. A B C D

14. F G H J

15. A B C D

16. F G H J

17. A B C D

18.

19.


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20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30a.

30b.

Chapter 10 Assessment Answer KeyStandardized Test PracticePage 76

7.2 mi

14.2 mi

$3648.08

{-4.4, -11.6}

-15 and -17 or 15 and 17

18abc2

3h6k2

− - 2j10

no solution

{w | w ≤ -1 or w > 2} ;

y = -5x + 14

$10,800 at 14%,

$1200 at 10%

-1-2-3-4 0 1 2 3 4

b

jO

20406080

100

100 200 300

120

2j + 5b = 500

j + b = 120


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