Chapter 10 Resource Masters Bothell, WA • Chicago, IL • Columbus, OH • New York, NY
Chapter 10 Resource Masters
Bothell, WA • Chicago, IL • Columbus, OH • New York, NY
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Copyright © by The McGraw-Hill Companies, Inc.
All rights reserved. The contents, or parts thereof, may be reproduced in print form for non-profit educational use with Glencoe Algebra 1, provided such reproductions bear copyright notice, but may not be reproduced in any form for any other purpose without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, network storage or transmission, or broadcast for distance learning.
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ISBN: 978-0-07-660284-1MHID: 0-07-660284-2
Printed in the United States of America.
1 2 3 4 5 6 7 8 9 DOH 16 15 14 13 12 11
CONSUMABLE WORKBOOKS Many of the worksheets contained in the Chapter Resource Masters booklets are available as consumable workbooks in both English and Spanish.
MHID ISBNStudy Guide and Intervention Workbook 0-07-660292-3 978-0-07-660292-6Homework Practice Workbook 0-07-660291-5 978-0-07-660291-9
Spanish VersionHomework Practice Workbook 0-07-660294-X 978-0-07-660294-0
Answers For Workbooks The answers for Chapter 10 of these workbooks can be found in the back of this Chapter Resource Masters booklet.
ConnectED All of the materials found in this booklet are included for viewing, printing, and editing at connected.mcgraw-hill.com.
Spanish Assessment Masters (MHID: 0-07-660289-3, ISBN: 978-0-07-660289-6) These masters contain a Spanish version of Chapter 10 Test Form 2A and Form 2C.
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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 10-6Trigonometric RatiosStudy Guide and Intervention .......................... 37Skills Practice .................................................. 39Practice ........................................................... 40Word Problem Practice .................................... 41Enrichment ...................................................... 42
AssessmentStudent Recording Sheet ............................... 43Rubric for Scoring Extended Response .......... 44Chapter 10 Quizzes 1 and 2 ............................ 45Chapter 10 Quizzes 3 and 4 ............................ 46Chapter 10 Mid-Chapter Test ........................... 47Chapter 10 Vocabulary Test ............................. 48Chapter 10 Test, Form 1 .................................. 49Chapter 10 Test, Form 2A ............................... 51Chapter 10 Test, Form 2B ............................... 53Chapter 10 Test, Form 2C ............................... 55
Chapter 10 Test, Form 2D ............................... 57Chapter 10 Test, Form 3 .................................. 59Chapter 10 Extended Response Test .............. 61Standardized Test Practice .............................. 62
Answers ........................................... A1–A31
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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, printing, and editing at connectED.mcgraw-hill.com.
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. Sample 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.
• Full-size answer keys are provided for the assessment masters.
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Chapter 10 1 Glencoe Algebra 1
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 PageDefi nition/Description/Example
conjugateKAHN • jih • guht
converse
cosine
hypotenusehy • PAH • tn • oos
inverse cosine
inverse sine
inverse tangent
legs
(continued on the next page)
10
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Chapter 10 2 Glencoe Algebra 1
Student-Built Glossary (continued)10
Vocabulary TermFound
on PageDefi nition/Description/Example
radical equations
radical functions
radicandRA • duh • KAND
tangent
trigonometry
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Chapter 10 3 Glencoe Algebra 1
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 missing measures of a triangle can be found if the measure of one of its sides is known.
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. Las medidas ausentes de un triángulo pueden ser encontradas si la medida de uno de sus lados es conocida.
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 10 4 Álgebra 1 de Glencoe
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Chapter 10 5 Glencoe Algebra 1
10-1
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 radicand 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
Study Guide and InterventionSquare Root Functions
Example
y
x
y = x
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Chapter 10 6 Glencoe Algebra 1
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 + k, follow these steps.
Graphs of Square Root
Functions
Step 1 Draw the graph of y = a √ ⎯⎯ 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 ⎪k⎥ units up if k is positive and down if k 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 ≥ -1} 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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Chapter 10 7 Glencoe Algebra 1
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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Chapter 10 8 Glencoe Algebra 1
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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Chapter 10 9 Glencoe Algebra 1
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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Chapter 10 10 Glencoe Algebra 1
A cube root function contains the cube root of a variable. The cube roots 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, cube root functions return real numbers when the radicand is negative.
Graph y = 3 √ � x .
Step 1 Make a table. Round to the nearest Step 2 Plot points and draw a hundredth. 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 Cube Root Functions
Example
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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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Chapter 10 12 Glencoe Algebra 1
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
Factor 56 and 45.
= 2 � √ �� 14 −
3 � √ � 5 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
ExercisesSimplify each expression.
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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Chapter 10 13 Glencoe Algebra 1
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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Chapter 10 14 Glencoe Algebra 1
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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Chapter 10 15 Glencoe Algebra 1
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 2010, an earthquake below the ocean floor initiated a devastating tsunami in Sumatra. Scientists can approximate the velocity V in feet per second of a tsunami in water of depth d 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, and 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 60 miles per hour, which is 26.82 meters per second. 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 a calculator, find a decimal approximation for the radius. Round your answer to the nearest hundredth.
r
10-2
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Chapter 10 16 Glencoe Algebra 1
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 10:1 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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Chapter 10 17 Glencoe Algebra 1
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 Factor 12 and 75.
= 3 · 2 √ � 3 + 5 · 5 √ � 3 Simplify.
= 6 √ � 3 + 25 √ � 3 Multiply.
= 31 √ � 3 Distributive Property
Exercises Simplify each expression.
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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Chapter 10 18 Glencoe Algebra 1
Study Guide and Intervention (continued)
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.
ExercisesSimplify each expression.
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
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Chapter 10 19 Glencoe Algebra 1
Skills PracticeOperations with Radical Expressions
10-3
Simplify each expression.
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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Chapter 10 20 Glencoe Algebra 1
PracticeOperations with Radical Expressions
Simplify each expression.
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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Chapter 10 21 Glencoe Algebra 1
-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 S of a sphere with radius r is given by the formula S = 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 A ball is dropped from a building window 800 feet above the ground. A heavier 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.
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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Chapter 10 22 Glencoe Algebra 1
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
10-3
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Chapter 10 23 Glencoe Algebra 1
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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Chapter 10 24 Glencoe Algebra 1
Study Guide and Intervention (continued)
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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Chapter 10 25 Glencoe Algebra 1
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
10-4
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Chapter 10 26 Glencoe Algebra 1
Practice Radical Equations
Solve each equation. Check your solution.
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?
10-4
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Chapter 10 27 Glencoe Algebra 1
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 L of a right circular cone is the surface area not including the area of the base. The lateral surface area is represented by L = πr √ ��� r2 + h2 , where r is the radius of the base and h is the height.
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 of the funnel, that is, the base of the cone?
x + 1 m
12 m 10 m
10-4
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Chapter 10 28 Glencoe Algebra 1
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 each side to remove the square root.
x + 7 = x + 2 √ � x + 1 Simplify. Use the FOIL method to square the right side.
x + 7 - x - 1 = 2 √ � x Isolate the square root term again.
6 = 2 √ � x Simplify.
3 = √ � x Divide each side by 2.
9 = x Square each side 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
10-4
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Chapter 10 29 Glencoe Algebra 1
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 - 4 Graph 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
10-4
Example
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Chapter 10 30 Glencoe Algebra 1
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 missing length.
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 = 13 Simplify.
The 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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Chapter 10 31 Glencoe Algebra 1
Study Guide and Intervention (continued)
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 each set of measures can be sides of a right triangle.
a. 10, 12, 14Since the greatest measure is 14, let c = 14, a = 10, and b = 12.
c2 = a2 + b2 Pythagorean Theorem
142 � 102 + 122 a = 10, b = 12, c = 14
196 � 100 + 144 Multiply.
196 ≠ 244 Add.
Since c2 ≠ a2 + b2, segments with these measures cannot form a right triangle.b. 7, 24, 25
Since the greatest measure is 25, let c = 25, a = 7, and b = 24.
c2 = a2 + b2 Pythagorean Theorem
252 � 72 + 242 a = 7, b = 24, c = 25
625 � 49 + 576 Multiply.
625 = 625 Add.
Since c2 = a2 + b2, segments with these measures can form 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
10-5
Example
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Chapter 10 32 Glencoe Algebra 1
Skills PracticeThe Pythagorean Theorem
Find each missing length. 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
10-5
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Chapter 10 33 Glencoe Algebra 1
PracticeThe Pythagorean Theorem
10-5
Find each missing length. If necessary, round to the nearest hundredth.
1. 2. 3.
Determine whether each set of measures can be sides of a 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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Chapter 10 34 Glencoe Algebra 1
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
10-5
TV size width (in.) height (in.)
19-inch 15
25-inch 21
32-inch 25
50-inch 40
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Chapter 10 35 Glencoe Algebra 1
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 = 25 �Furthermore, 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 positive integers 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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Chapter 10 36 Glencoe Algebra 1
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 primitive Pythagorean triple into cells A1, B1, and C1.
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.xlsA
1
3456
2
891011
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
10-5
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Chapter 10 37 Glencoe Algebra 1
Study Guide and InterventionTrigonometric Ratios
10-6
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 each side.
a = 6 Take the positive 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°
Example
a10
8
b
a
c
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Chapter 10 38 Glencoe Algebra 1
10-6
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 right 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 Defi nition of cosine
c cos 38° = 13 Multiply each side by c.
c = 13 − cos 38° Divide each side by cos 38°.
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
Defi nition 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°
Study Guide and Intervention (continued)
Trigonometric Ratios
Example
13
ac
38°
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Chapter 10 39 Glencoe Algebra 1
Find the values of the three trigonometric ratios for angle A.
1.
77
85
2.
159
3. 10
24
4. 15
8
Use a calculator to find the value of each trigonometric ratio to the nearest ten-thousandth.
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
10-6 Skills PracticeTrigonometric Ratios
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Chapter 10 40 Glencoe Algebra 1
Find the values of the three trigonometric ratios for angle A.
1.
72
97
2.
36
15
Use a calculator to find the value of each trigonometric ratio to the nearest ten-thousandth.
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.
10-6 PracticeTrigonometric Ratios
25°
54 ft
h
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Chapter 10 41 Glencoe Algebra 1
10-6 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 truck bed 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. To the nearest degree what would the measure of angle BAC be on a standard television with a 4:3 aspect ratio?
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Chapter 10 42 Glencoe Algebra 1
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 adjacent ∠A
secant of ∠B = hypotenuse
− leg adjacent ∠B
sec A = c − b
sec B = c − a
b
a
c
cosecant of ∠A = hypotenuse
− leg opposite ∠A
cosecant of ∠B = hypotenuse
− leg opposite ∠B
csc A = c − a
csc B = c − b
cotangent of ∠A = leg adjacent to ∠A
−− leg opposite ∠A
cotangent of ∠B = leg adjacent to ∠B
−− leg opposite ∠B
cot A = b − a
cot B = a − b
Find the secant, cosecant, and cotangent of angle A.
Use the side lengths to write the trigonometric ratios.
sec A = hyp
− adj
= 15 − 12
= 5 − 4 csc A =
hyp − opp = 15 −
9 = 5 −
3
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 tangent?
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°
10-6 EnrichmentMore Trigonometric Ratios
Example
12
159
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Chapter 10 43 Glencoe Algebra 1
8.
12.
10 Student Recording SheetUse this recording sheet with pages 670–671 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
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
10.
15.
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
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Chapter 10 44 Glencoe Algebra 1
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 Specifi c 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 Pythagorean Theorem.
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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Chapter 10 45 Glencoe Algebra 1
SCORE
1. Graph y = 2 √ ��� x + 1 . State the domain and range.
2. MULTIPLE CHOICE Which function has a domain of {x | x ≥ 4}?
A y = √ ��� x - 4 B y = √ ��� x + 4 C y = √ � x - 4 D y = √ � x + 4
Simplify each expression.
3. √ �� 72
4. √ �� 8 x 2 y
5. 3 − 4 + √ � 2
Chapter 10 Quiz 1 (Lessons 10-1 and 10-2)
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
Solve each equation. Check your solution.
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
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
1.
2.
3.
4.
5.
y
x
SCORE
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Chapter 10 46 Glencoe Algebra 1
SCORE
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
Determine whether the following set of measures form a Pythagorean triple.
4. 48, 64, 82
5. √ � 4 , √ � 6 , √ �� 10
Use the triangle for Questions 1–3.
1. Find x.
2. Find the values of the three trigonometric ratios for angle B.
3. Find the values of the three trigonometric ratios for angle A.
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.
16
25
P
M N
Chapter 10 Quiz 3(Lesson 10-5)
10 Chapter 10 Quiz 4(Lesson 10-6)
10
1.
2.
3.
4.
5.
10
Q
R P20°
SCORE
1.
2.
3.
4.
5.
10 6
A
B
Cx
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Chapter 10 47 Glencoe Algebra 1
SCORE
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
Simplify each expression.
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.
10 Chapter 10 Mid-Chapter Test(Lessons 10-1 through 10-4)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
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Chapter 10 48 Glencoe Algebra 1
SCORE
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 the lengths of two sides of a right triangle are given, the third side can be found 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. When you find all unknown measures of the sides and angles of a right triangle, you are .
Define each term in your own words.
10. trigonometry
11. tangent
conjugates
converse
cosine
hypotenuse
inverse cosine
inverse sine
inverse tangent
legs
Pythagorean Theorem
radical equation
radical function
radicand
solving the triangle
tangent
trigonometry
10 Chapter 10 Vocabulary Test
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
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Chapter 10 49 Glencoe Algebra 1
SCORE 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
For Questions 3–7, simplify each expression.
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
12. Find m∠A to the nearest tenth if cos A = 3 − 5 .
F 0.9° G 31.0° H 36.9° J 53.1°
10
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
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Chapter 10 50 Glencoe Algebra 1
13. Determine which set of measures can be the lengths of the sides of a right triangle.
A 2, 3, 5 B 4, 6, 7 C 10, 12, 13 D 1, √ � 3 , 2
14. What is the equation of the graph?
F y = √ ��� x + 2 + 1 H y = √ ��� x + 1 + 2
G y = √ ��� x - 2 + 1 J y = √ ��� x - 1 + 2
15. Simplify 2 √ � x • 5 √ � x • 3 √ � x .
A 30 √ � x B 30 x 2 √ � x C 30 ⎪x⎥ √ � x D 30 x 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 .
Chapter 10 Test, Form 1 (continued)10
13.
14.
15.
16.
17.
18.
19.
20.
B:
y
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Chapter 10 51 Glencoe Algebra 1
SCORE
Write the letter for the correct answer in the blank at the right of each question.
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
For Questions 3–7, simplify each expression.
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
11. Determine which side measures form a right triangle.
A 3, 8, 12 B 5, 9, 11 C 11, 13, 16 D 6, 8, 10
12. Find m∠B to the nearest tenth if sin B = 1 − 3 .
F 0.5° G 18.4° H 19.5° J 70.5°
Chapter 10 Test, Form 2A 10
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
A 1
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Chapter 10 52 Glencoe Algebra 1
13. Determine which set of measures can be the lengths of the sides of a right triangle.
A 3, 6, 9 B 1, 1, √ � 2 C 3, 3, 4 D 1, 3, 7
14. What is the equation of the graph?
F y = √ ��� x + 1 - 1 H y = √ ��� x + 1 + 1
G y = √ ��� x - 1 - 1 J y = √ ��� x - 1 + 1
15. Simplify 2 √ � y • 5 √ � y • 2 √ � y .
A 20 ⎪y⎥ √ � y B 20 √ � y C 20 y 2 √ � y D 20 y 3
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.
Chapter 10 Test, Form 2A (continued)
B:
13.
14.
15.
16.
17.
18.
19.
20.
10
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Chapter 10 53 Glencoe Algebra 1
SCORE
Write the letter for the correct answer in the blank at the right of each question.
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
For Questions 3–7, simplify each expression.
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.
11. Determine which side measures form a right triangle.
A 4, 7, 8 B 9, 12, 15 C 3, 7, 9 D 10, 15, 20
12. Find m∠B to the nearest tenth if sin B = 2 − 7 .
F 1.3° G 16.6° H 23.1° J 73.4°
Chapter 10 Test, Form 2B10
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
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Chapter 10 54 Glencoe Algebra 1
13. Determine which set of measures can be the lengths of the sides of a right triangle.
A 2, 2, 8 B 12, 14, 20 C 4, 16, 256 D 11, 60, 61
14. What is the equation of the graph?
F y = √ ��� x - 3 - 2 H y = √ ��� x - 2 - 3
G y = √ ��� x + 2 - 3 J y = √ ��� x + 3 - 2
15. Simplify √ ��� 72 x 3 y .
A 6 ⎪x⎥ √ �� 2x y B 6 ⎪x⎥ √ � 2y C 6 ⎪x⎥ √ � y D 12 ⎪x⎥ √ � x y
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 ∠ A 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)10
B:
13.
14.
15.
16.
17.
18.
19.
20.
y
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Chapter 10 55 Glencoe Algebra 1
SCORE
1. Graph y = √ ��� x - 2 + 1.State the domain and range.
2. State the domain and range of y = -3 √ ��� x - 1 + 5.
Simplify each expression.
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 )
Solve each equation. Check your solution.
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 = ?
Determine whether the following side measures form right triangles.
13. 14, 48, 50
14. 12, 24, 36
Use a calculator to find the value of each trigonometric ratio to the nearest ten-thousandth.
15. sin 48°
16. cos 7°
17. tan 71°
Chapter 10 Test, Form 2C10
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
y
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Chapter 10 56 Glencoe Algebra 1
For Questions 18 and 19, find the values of the three trigonometric ratios for angle A.
18. A
C B
20
21
29
19.
C A
B
48
55
73
20. The perimeter of a square P with area A can be found using the formula P = 4 √ � A . If a square has a perimeter of 29.3 inches, find the area to the nearest tenth of a square foot.
21. Find the height of the lamp post to the nearest tenth of a foot.
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 centimeters and the diagonal is 16 centimeters?
24. Solve m ∠ J for the right triangle
23
10to 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)10
18.
19.
20.
21.
22.
23.
24.
25.
B:
18 ft30°
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Chapter 10 57 Glencoe Algebra 1
SCORE
1. Graph y = √ ��� x + 1 - 3. State the domain and range.
2. State the domain and range of y = -2 √ ��� x + 2 - 1.
Simplify each expression.
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 )
Solve each equation. Check your solution.
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 = ?
Determine whether the following side measures form right triangles.
13. 15, 20, 25
14. 16, 20, 30
Use a calculator to find the value of each trigonometric ratio to the nearest ten-thousandth.
15. sin 73°
16. cos 62°
17. tan 12°
Chapter 10 Test, Form 2D10
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
y
x
1.
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Chapter 10 58 Glencoe Algebra 1
For Questions 18 and 19, find the values of the three trigonometric ratios for angle A.
18.
C B
A
6
8
10
19.
C A
B
72
154
170
20. The perimeter of a square P with area A can be found using the formula P = 4 √ � A . If a square has a perimeter of 36.8 inches, find the area to the nearest tenth of a square foot.
21. Find the height of the tree to
25 ft60°
the nearest tenth of a foot.
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 centimeters and the diagonal is 20 centimeters?
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
18.
19.
20.
21.
22.
23.
24.
25.
B:
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Chapter 10 59 Glencoe Algebra 1
SCORE
1. Graph y = -2 √ ��� x - 2 - 2.State the domain and range.
2. State the domain and range of y = -8 √ ��� x + 4 - 12.
Simplify each expression.
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 )
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.
Determine whether the following side measures form right triangles.
13. 16, 49, 65
14. 5, 9, √ �� 106
Use a calculator to find the value of each trigonometric ratio to the nearest ten-thousandth.
15. sin 16°
16. cos 31°
17. tan 88°
Chapter 10 Test, Form 310
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
y
x
1.
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Chapter 10 60 Glencoe Algebra 1
For Questions 18 and 19, find the values of the three trigonometric ratios for angle A.
18. C
B
A65
5633
19.
A
B
C
6597
72
20. The radius r of a cylinder with volume V can be
found using the formula r = √ �� V − πh
, where h is the
height of the cylinder. If a cylinder has a volume of 120 cubic inches and a radius of 5 inches, find the height of the cylinder to the nearest tenth of an inch.
21. Find the distance across the lake from point A to point B to the nearest tenth of a meter.
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) 10
18.
19.
20.
21.
22.
23.
24.
25.
B:
210 mC
B
A
40˚
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Chapter 10 61 Glencoe Algebra 1
SCORE Chapter 10 Extended-Response Test10
y
xB(0, 0) C(a, 0)
A(a, b)
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. Refer to triangles ABC and DEF.
a. Yoki claims that if sin A = sin D, then m∠B = m∠E. Do you agree or disagree? Justify your reasoning.
b. Yoki also claims that sin A ≠ sin D, then m∠B ≠ m∠E. Do you agree or disagree? Justify your reasoning.
A D
BE
C F
4. Show that the triangle is a right triangle by using the Pythagorean Theorem.
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Chapter 10 62 Glencoe Algebra 1
SCORE
1. Simplify (mt2)(m3)(m2t). (Lesson 7-1)
A m6t3 B m6t2 C m9t2 D m3t2
2. Find (x + 2y)2. (Lesson 8-4)
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 8-7)
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 8-9)
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 7-6)
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. Solve the system of equations. (Lesson 6-3)
F (5, -2) G (2, 2) H (9, 3) J (-4, -2)
11. Simplify ( 3 x 2 y −
5z y 4 )
3
.
A 27 x 6 − 125 z 3 y 9
B 27 x 5 − 125 z 3 y 3
C 3 x 6 − 5 z 3 y 9
D 27 x 6 − 5 z 3 y 9
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.
4x + 3y = 145x - 4y = 33
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Chapter 10 63 Glencoe Algebra 1
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 8-9)
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
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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Chapter 10 64 Glencoe Algebra 1
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 with a slope of -5 and ay-intercept of 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. Factor xy - 3x + 4y - 12. (Lesson 8-5)
27. Find two consecutive odd integers with a product of 255. (Lesson 8-6)
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 7-6)
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 10-5)
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)
10
Part 3: Short Response
Instructions: Write your answers in the space.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30a.
30b.
-1-2-3-4 0 1 2 3 4
b
jO
20406080
100
100 200 300
120
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An
swer
s
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ight
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Chapter 10 A1 Glencoe Algebra 1
Chapter Resources
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
3
Gle
ncoe
Alg
ebra
1
Ant
icip
atio
n G
uide
Rad
ical
Exp
ressio
ns a
nd
Tri
an
gle
s
Bef
ore
you
beg
in C
ha
pte
r 10
•
Rea
d ea
ch s
tate
men
t.
•
Dec
ide
wh
eth
er y
ou A
gree
(A
) or
Dis
agre
e (D
) w
ith
th
e st
atem
ent.
•
Wri
te A
or
D i
n t
he
firs
t co
lum
n O
R i
f yo
u a
re n
ot s
ure
wh
eth
er y
ou a
gree
or
disa
gree
, wri
te N
S (
Not
Su
re).
ST
EP
1A
, D, o
r N
SS
tate
men
tS
TE
P 2
A o
r D
1.
A
n e
xpre
ssio
n t
hat
con
tain
s a
squ
are
root
is
call
ed a
ra
dica
l ex
pres
sion
.A
2.
It
is
alw
ays
tru
e th
at √
�
xy w
ill
equ
al √
� x ·
√ �
y .A
3.
1
−
√ �
3
is
in s
impl
est
form
bec
ause
√ �
3
is
not
a w
hol
e n
um
ber.
D
4.
T
he
sum
of
3 √
�
3 a
nd
2 √
�
3 w
ill
equ
al 5
√ �
3
.A
5.
B
efor
e m
ult
iply
ing
two
radi
cal
expr
essi
ons
wit
h d
iffe
ren
t ra
dica
nds
th
e sq
uar
e ro
ots
mu
st b
e ev
alu
ated
.D
6.
W
hen
sol
vin
g ra
dica
l eq
uat
ion
s by
squ
arin
g ea
ch s
ide
of t
he
equ
atio
n, i
t is
pos
sibl
e to
obt
ain
sol
uti
ons
that
are
not
so
luti
ons
to t
he
orig
inal
equ
atio
n.
A
7.
T
he
lon
gest
sid
e of
an
y tr
ian
gle
is c
alle
d th
e h
ypot
enu
se.
D
8.
B
ecau
se 5
2 = 4
2 + 3
2 , a
tria
ngl
e w
hos
e si
des
hav
e le
ngt
hs
3, 4
, an
d 5
wil
l be
a r
igh
t tr
ian
gle.
A
9.
O
n a
coo
rdin
ate
plan
e, t
he
dist
ance
bet
wee
n a
ny
two
poin
ts
can
be
fou
nd
usi
ng
the
Pyt
hag
orea
n T
heo
rem
.A
10.
Th
e m
issi
ng
mea
sure
s of
a t
rian
gle
can
be
fou
nd
if t
he
mea
sure
of
one
of i
ts s
ides
is
know
n.
D
A
fter
you
com
ple
te C
ha
pte
r 10
•
Rer
ead
each
sta
tem
ent
and
com
plet
e th
e la
st c
olu
mn
by
ente
rin
g an
A o
r a
D.
•
Did
an
y of
you
r op
inio
ns
abou
t th
e st
atem
ents
ch
ange
fro
m t
he
firs
t co
lum
n?
•
For
th
ose
stat
emen
ts t
hat
you
mar
k w
ith
a D
, use
a p
iece
of
pape
r to
wri
te a
n
exam
ple
of w
hy
you
dis
agre
e.
10 Step
1
Step
2
001_
020_
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Lesson X-1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-1
Cha
pte
r 10
5
Gle
ncoe
Alg
ebra
1
10-1
Dila
tio
ns
of
Rad
ical
Fu
nct
ion
s A
sq
uar
e ro
ot f
un
ctio
n c
onta
ins
the
squ
are
root
of
a va
riab
le. S
quar
e ro
ot f
un
ctio
ns
are
a ty
pe o
f ra
dic
al f
un
ctio
n.
In o
rder
for
a s
quar
e ro
ot t
o be
a r
eal
nu
mbe
r, th
e ra
dic
and
, or
the
expr
essi
on u
nde
r th
e ra
dica
l si
gn, c
ann
ot b
e n
egat
ive.
Val
ues
th
at m
ake
the
radi
can
d n
egat
ive
are
not
in
clu
ded
in
the
dom
ain
.
Sq
uar
e R
oo
t Fu
nct
ion
Pa
ren
t fu
nctio
n: f(
x) =
√ �
x
Typ
e o
f gra
ph
: cu
rve
Do
ma
in:
{x|
x ≥
0}
Ra
ng
e:
{y|
y ≥
0}
G
rap
h y
=3
√�x
. Sta
te t
he
dom
ain
an
d r
ange
.
Ste
p 1
Mak
e a
tabl
e. C
hoo
se n
onn
egat
ive
S
tep
2 P
lot
poin
ts a
nd
draw
a
va
lues
for
x.
smoo
th c
urv
e.
xy
00
0.5
≈ 2
.12
13
2≈
4.2
4
46
6≈
7.3
5
y
x
y =
3 x
Th
e do
mai
n i
s {x|
x ≥
0} a
nd
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 a
nd
ran
ge.
1. y
= 3 −
2 √
� x
2. y
= 4
√ �
x
3. y
= 5 −
2 √
� x
y
x
y
x
y
x
D
ilati
on
of
y =
√ �
x ;
Dila
tio
n o
f y =
√ �
x ;
Dila
tio
n o
f 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}
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
1_A
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M_C
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10
5:17
PM
Answers (Anticipation Guide and Lesson 10-1)
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Chapter 10 A2 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
6
Gle
ncoe
Alg
ebra
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 +
k, f
ollo
w t
hes
e st
eps.
Gra
ph
s o
f S
qu
are
Ro
ot
Fun
ctio
ns
Ste
p 1
D
raw
th
e gr
aph
of
y =
a √
⎯⎯ 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
, th
e gr
aph
is
in t
he
1st
quad
ran
t. I
f a
< 0
, th
e gr
aph
is
refl
ecte
d ac
ross
th
e x-
axis
an
d is
in
th
e 4t
h q
uad
ran
t.
Ste
p 2
T
ran
slat
e th
e gr
aph
⎪k⎥
un
its
up
if k
is
posi
tive
an
d do
wn
if
k 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
≥ -
1} a
nd
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
slat
ion
of
y =
√ �
x
up
3 u
nit
s;D
= {
x |
x ≥
0};
R =
{y |
y ≥
3}
tran
slat
ion
of
y =
√ �
x
rig
ht
1 u
nit
;D
= {
x |
x ≥
1};
R =
{y |
y ≥
0}
tran
slat
ion
of
y =
√ �
x
rig
ht
1 u
nit
an
d r
efl e
cted
ac
ross
th
e x-
axis
;D
= {
x |
x ≥
1};
R =
{y |
y ≤
0}
001_
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-1
Cha
pte
r 10
7
Gle
ncoe
Alg
ebra
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
dila
tio
n o
f y =
√ �
x ;
d
ilati
on
of
y =
√ �
x ;
d
ilati
on
of
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
tran
slat
ion
of
tr
ansl
atio
n o
f t
ran
slat
ion
of
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
tran
slat
ion
of
y =
√ ⎯⎯
x ;
tran
slat
ion
of
y =
√ ⎯⎯
x
dia
lati
on
of
y =
√ ⎯⎯
x r
efl e
cted
ri
gh
t 3
un
its
refl
ecte
d a
cro
ss
rig
ht
2 u
nit
s an
d u
p
acr
oss
th
e x-a
xis
the
x-a
xis;
D =
{x |
x ≥
3},
3
un
its;
D =
{x |
x ≥
2},
tran
slat
ed r
ight
4 u
nits
up
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_
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Answers (Lesson 10-1)
A01_A14_ALG1_A_CRM_C10_AN_660284.indd A2A01_A14_ALG1_A_CRM_C10_AN_660284.indd A2 12/22/10 6:27 PM12/22/10 6:27 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
Chapter 10 A3 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
8
Gle
ncoe
Alg
ebra
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
, w
her
e I
is t
he
curr
ent
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
dila
tio
n o
f y =
√ �
x ;
D =
{x |
x ≥
0},
R =
{y |
y ≥
0}
tran
slat
ion
of
y =
√ �
x
up
2 u
nit
s;D
= {x
| x
≥ 0
},
R =
{y
| y ≥
2}
tran
slat
ion
of
y =
√ �
x
rig
ht
3 u
nit
s;D
= {x
| x
≥ 3
},R
= {
y | y
≥ 0
}
tran
slat
ion
of
y =
√ �
x
up
1 u
nit
refl
ect
ed
in t
he
x-a
xis;
D
= {
x |
x ≥
0},
R =
{y |
y ≤
1}
dila
tio
n o
f y =
√ �
x
tran
slat
ed u
p 1
un
itan
d r
igh
t 1
un
it;
D =
{x |
x ≥
1},
R =
{y |
y ≥
1}
tran
slat
ion
of
y =
√ �
x
up
2 u
nit
s an
d r
igh
t 2
un
its,
refl
ect
ed
in t
he
x-a
xis;
D
= {
x |
x ≥
2},
R =
{y |
y ≤
2}
001_
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10
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PMCopyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-1
Cha
pte
r 10
9
Gle
ncoe
Alg
ebra
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
. Th
e le
ngt
h o
f th
e pe
ndu
lum
in
fee
t is
giv
en b
y L
. Gra
ph
this
fu
nct
ion
.
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
atel
y 28
4 ft
/s
3. E
RR
OR
AN
ALY
SIS
Gre
gory
is
draw
ing
the
grap
h o
f y
= -
5 √
��
�
x +
1 . H
e de
scri
bes
the
ran
ge a
nd
dom
ain
as
{x �
x ≥
-1
},
{y �
y ≥
0}.
Exp
lain
an
d co
rrec
t th
e m
ista
ke t
hat
Gre
gory
mad
e.
T
he
ran
ge
is a
ctu
ally
{y �
y ≤
0}
bec
ause
th
e g
rap
h h
as b
een
re
fl ec
ted
acr
oss
th
e x-a
xis.
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 =
10
0 √
⎯⎯ 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
slat
ion
of
V=
10
0 √
� E
on
e u
nit
to
th
e le
ft
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
slat
ion
of
V=
10
0 √
� E
on
e u
nit
do
wn
Period (sec)
23 1 04
45
Pend
ulum
Len
gth
(ft)
812
1620
001_
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Answers (Lesson 10-1)
A01_A14_ALG1_A_CRM_C10_AN_660284.indd A3A01_A14_ALG1_A_CRM_C10_AN_660284.indd A3 12/22/10 6:27 PM12/22/10 6:27 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 10 A4 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
10
G
lenc
oe A
lgeb
ra 1
A c
ub
e ro
ot f
un
ctio
n c
onta
ins
the
cube
roo
t of
a v
aria
ble.
Th
e cu
be
root
s 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, o
r al
tern
ativ
ely,
y =
3
√ �
x .
Un
like
squ
are
root
fu
nct
ion
s, c
ube
roo
t fu
nct
ion
s re
turn
rea
l n
um
bers
wh
en t
he
radi
can
d is
neg
ativ
e.
G
rap
h y
=
3 √
�
x .
Ste
p 1
Mak
e a
tabl
e. R
oun
d to
th
e n
eare
st
Ste
p 2
Plo
t po
ints
an
d dr
aw a
hu
ndr
edth
.
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
nsl
atio
n o
f t
ran
slat
ion
of
y =
3 √
�
x u
p 1
un
it
y =
3 √
�
x l
eft
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
slat
ion
of
y =
3 √
�
x
dila
tio
n o
f r
efl e
ctio
n o
f y =
3 √
�
x
u
p 2
un
its
and
y
= 3 √
�
x t
ran
slat
ed
acr
oss
th
e x-a
xis
r
igh
t 1
un
it
rig
ht
2 u
nit
s t
ran
slat
ed u
p 3
un
its
10-1
Enri
chm
ent
Cu
be R
oo
t Fu
ncti
on
s
Exam
ple
001_
020_
ALG
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dd
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10
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PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-2
Cha
pte
r 10
11
G
lenc
oe A
lgeb
ra 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
hen
you
sim
plif
y ra
dica
l ex
pres
sion
s w
ith
var
iabl
es, u
se a
bsol
ute
val
ue
to e
nsu
re
non
neg
ativ
e re
sult
s.
S
imp
lify
√ �
�
180
.
√ �
�
180
= √
��
��
��
2
� 2
� 3
� 3
� 5
P
rim
e facto
rization o
f 180
=
√ �
22
� √
�
32 � √
� 5
Pro
duct
Pro
pert
y o
f S
quare
Roots
=
2 �
3 �
√ �
5
Sim
plif
y.
=
6 √
�
5
Sim
plif
y.
S
imp
lify
√ �
��
��
�
12
0a2
· b
5 · c
4 .
√ �
��
��
�
12
0a2
� b
5 � c
4
= √
��
��
��
��
23
� 3
� 5
� a
2 � b
5 � c
4
=
√ �
22
� √
�
2 �
√ �
3
� √
�
5 �
√
�
a2 � √
��
�
b4 � b
� √
�
c4
=
2 �
√ �
2 �
√ �
3 �
√ �
5 �
⎪a⎥
� b
2 � √
�
b �
c2
=
2 ⎪ a
⎥ b2 c
2 √ �
�
30b
Exer
cise
sS
imp
lify
eac
h e
xpre
ssio
n.
1.
√ �
�
28
2. √
��
68
3. √
��
60
4. √
��
75
2 √
�
7
2 √
��
17
2 √
��
15
5 √
�
3
5.
√ �
�
162
6.
√ �
3
· √
�
6
7. √
�
2 ·
√ �
5
8.
√ �
5
· √
��
10
9 √
�
2
3 √
�
2
√ �
�
10
5 √
�
2
9.
√ �
�
4a2
10.
√ �
�
9x4
11.
√ �
��
300a
4 12
. √
��
�
128c
6
2 a
3x
2 10
a2 √
�
3
8 ⎪c
3 ⎥ √
�
2
13. 4
√ �
�
10 �
3 √
�
6
14.
√ �
�
3x2
� 3
√ �
�
3x4
15.
√ �
��
20a2 b
4 16
. √
��
�
100x
3 y
24 √
��
15
9 ⎪x
3 ⎥
2 a
b2 √
�
5
10 x
√ �
xy
17. √
��
�
24a4 b
2 18
. √
��
�
81x4 y
2 19
. √
��
��
150a
2 b2 c
2a2 b
√ �
6
9x2 y
5 ⎪ab
⎥ √
��
6c
20. √
��
��
72a6 b
3 c2
21.
√ �
��
�
45x2 y
5 z8
22.
√ �
��
�
98x4 y
6 z2
6 ⎪a
3 bc
⎥ √
��
2b
3 x
y2 z
4 √ �
�
5y
7x
2 ⎪y
3 z⎥ √
�
2
10-2
Pro
du
ct P
rop
erty
of
Sq
uar
e R
oo
tsF
or
any n
um
be
rs a
an
d b
, w
he
re a
≥ 0
an
d b
≥ 0
, √
��
ab
= √
� a � √
�
b .
Exam
ple
1
Exam
ple
2
001_
020_
ALG
1_A
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M_C
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dd
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10
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PM
Answers (Lesson 10-1 and Lesson 10-2)
A01_A14_ALG1_A_CRM_C10_AN_660284.indd A4A01_A14_ALG1_A_CRM_C10_AN_660284.indd A4 12/22/10 6:27 PM12/22/10 6:27 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
Chapter 10 A5 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
12
G
lenc
oe A
lgeb
ra 1
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
Sim
plify
ing
Rad
ical
Exp
ressio
ns
Qu
oti
ent
Pro
per
ty o
f Sq
uar
e R
oo
ts A
fra
ctio
n c
onta
inin
g ra
dica
ls i
s in
sim
ples
t fo
rm i
f n
o ra
dica
ls a
re l
eft
in t
he
den
omin
ator
. Th
e Q
uot
ien
t P
rop
erty
of
Sq
uar
e R
oots
an
d ra
tion
aliz
ing
the
den
omin
ator
can
be
use
d to
sim
plif
y ra
dica
l ex
pres
sion
s th
at
invo
lve
divi
sion
. Wh
en y
ou r
atio
nal
ize
the
den
omin
ator
, you
mu
ltip
ly t
he
nu
mer
ator
an
d de
nom
inat
or b
y a
radi
cal
expr
essi
on t
hat
giv
es a
rat
ion
al n
um
ber
in t
he
den
omin
ator
.
S
imp
lify
√ �
�
56
−
45 .
√ �
�
56
−
45 =
√ �
��
4 � 1
4 −
9
� 5
F
acto
r 56 a
nd 4
5.
=
2 � √
��
14
−
3 � √
�
5
Sim
plif
y t
he n
um
era
tor
and d
enom
inato
r.
=
2 √
��
14
−
3 √
�
5
� √
�
5
−
√ �
5
M
ultip
ly b
y √
⎯⎯
5
−
√ ⎯
⎯
5 to
rationaliz
e t
he d
enom
inato
r.
=
2 √
��
70
−
15
P
roduct
Pro
pert
y o
f S
quare
Roots
Exer
cise
sS
imp
lify
eac
h e
xpre
ssio
n.
1.
√ �
9 −
√ �
18
√
� 2
−
2 2.
√
� 8
−
√ �
24
√
� 3
−
3
3.
√ �
�
100
−
√ �
�
121
10
−
11
4.
√ �
75
−
√ �
3 5
5.
8 √
� 2
−
2 √
� 8
2 6.
√ �
2 −
5 �
√ �
6 −
5 2
√ �
3 −
5
7.
√ �
3 −
4 �
√ �
5 −
2 √
��
30
−
4
8. √
�
5 −
7 �
√ �
2 −
5 √
�
14
−
7
9.
√ �
�
3a2
−
10b6
|a
| √ �
�
30
−
10| b
3 | 10
. √
�
x6 −
y4
|x3 |
−
y2
11. √
��
100a
4 −
14
4b8
5a
2 −
6b
4 12
. √
��
�
75b3 c
6 −
a2
5 ⎪
bc
3 ⎥ √
��
3b
−
| a
|
13.
√ �
4 −
3 -
√ �
5 3
+ √
� 5
−
2
14.
√
� 8
−
2 +
√ �
3 4
√ �
2 -
2 √
� 6
15.
√ �
5 −
5 +
√ �
5 √
� 5 -
1
−
4
16.
√
� 8
−
2 √
� 7 +
4 √
��
10
4 √
� 5 -
√ �
14
−
33
10-2
Qu
oti
ent
Pro
per
ty o
f S
qu
are
Ro
ots
Fo
r a
ny n
um
be
rs a
an
d b
, w
he
re a
≥ 0
an
d b
> 0
, √
�
a −
b =
√ �
a
−
√ �
b
.
Exam
ple
001_
020_
ALG
1_A
_CR
M_C
10_C
R_6
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-2
Cha
pte
r 10
13
G
lenc
oe A
lgeb
ra 1
Skill
s Pr
acti
ce
Sim
plify
ing
Rad
ical
Exp
ressio
ns
Sim
pli
fy e
ach
exp
ress
ion
.
1.
√ �
�
28
2 √
�
7
2. √
��
40
2 √
��
10
3.
√ �
�
72
6 √
�
2
4. √
��
99
3 √
�
11
5.
√ �
2 ·
√ �
�
10
2 √
�
5
6. √
�
5 ·
√ �
�
60
10 √
�
3
7.
3 √
�
5 ·
√ �
5
15
8.
√ �
6
· 4
√ �
�
24
48
9.
2 √
�
3 ·
3 √
��
15
18 √
�
5
10.
√ �
�
16b4
4b
2
11.
√ �
��
81a2 d
4 9
| a |d
2 12
. √
��
�
40x4 y
6 2
x2 ⎪
y3 ⎥ √
��
10
13.
√ �
��
75m
5 P2
5m
2 | P
| √ �
�
3m
14.
√ �
5 −
3
√ �
�
15
−
3
15.
√ �
1 −
6
√ �
6
−
6
16.
√ �
6 −
7 ·
√ �
1 −
3
√ �
�
14
−
7
17.
√ �
�
q −
12
√
��
3q
−
6
18.
√ �
�
4h
−
5 2
√ �
�
5h
−
5
19.
√ �
�
12
−
b2
2 √
�
3
−
| b |
20.
√ �
�
45
−
4m4
3 √
�
5
−
2m2
21.
2 −
4 +
√ �
5
8
- 2
√ �
5
−
11
22.
3
−
2 -
√ �
3
6
+ 3
√ �
3
23.
5
−
7 +
√ �
7
35
- 5
√ �
7
−
42
24.
4
−
3 -
√ �
2
12
+ 4
√ �
2
−
7
10-2
001_
020_
ALG
1_A
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Answers (Lesson 10-2)
A01_A14_ALG1_A_CRM_C10_AN_660284.indd A5A01_A14_ALG1_A_CRM_C10_AN_660284.indd A5 12/22/10 6:27 PM12/22/10 6:27 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 10 A6 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
14
G
lenc
oe A
lgeb
ra 1
Prac
tice
S
imp
lify
ing
Rad
ical
Exp
ressio
ns
Sim
pli
fy.
1.
√
�
24
2 √
� 6
2.
√ �
60
2
√ �
15
3.
√ �
�
108
6 √
� 3
4. √
� 8 � √
� 6
4 √
� 3
5.
√ �
7 �
√ �
14
7
√ �
2 6.
3 √
�
12 �
5 √
� 6
90 √
� 2
7. 4
√ �
3 � 3
√ �
18
36
√ �
6 8.
√ �
��
27tu
3
3 u √
��
3tu
9.
√ �
�
50p5
5p
2 √ �
�
2p
10.
√ �
��
�
108x
6 y4 z
5 6
⎪ x3 ⎥
y2 z2 √
��
3z
11. √
��
��
56m
2 n4 p
5 2
m n
2 p2 √
��
14p
12
. √
� 8
−
√ �
6 2
√ �
3
−
3
13.
√ �
2 −
10
√
�
5
−
5 14
. √
�
5 −
32
√
��
10
−
8
15.
√ �
3 −
4 � √
�
4 −
5 √
�
15
−
5
16.
√ �
1 −
7 �
√ �
7 −
11
√
�
11
−
11
17.
√ �
3k
−
√ �
8 √
��
6k
−
4
18.
√ �
18
−
x3
3 √
��
2x
−
x2
19. √
��
4y
−
3y2
20
. √
��
9ab
−
4ab4
3 √
�
b
−
2b2
21.
3 −
5 -
√ �
2 15
+ 3
√ �
2
−
23
22.
8 −
3 +
√ �
3
12 -
4 √
�
3
−
3
23.
5 −
√ �
7 +
√ �
3 5
√ �
7
- 5
√ �
3
−
4
24.
3 √
� 7
−
-1
- √
�
27
3 √
�
7 -
9 √
��
21
−
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
the
stor
m i
n m
iles
.
a.
A t
hu
nde
rsto
rm i
s 8
mil
es i
n d
iam
eter
. Est
imat
e h
ow l
ong
the
stor
m w
ill
last
. G
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 la
st a
bo
ut
4.4
h, o
r n
earl
y 3
tim
es a
s lo
ng
.
10-2
abo
ut
12.4
s
8 √
�
3
−
9 h
≈ 1
.5 h
2 √
��
3y
−
3 y
001_
020_
ALG
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R_6
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5:17
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-2
Cha
pte
r 10
15
G
lenc
oe A
lgeb
ra 1
Wor
d Pr
oble
m P
ract
ice
Sim
plify
ing
Rad
ical
Exp
ressio
ns
1. S
POR
TS J
asm
ine
calc
ula
ted
the
hei
ght
of
her
tea
m’s
soc
cer
goal
to
be 15 − √
⎯⎯
3 f
eet.
S
impl
ify
the
expr
essi
on.
5
√ �
3
2. N
ATU
RE
In 2
010,
an
ear
thqu
ake
belo
w
the
ocea
n f
loor
in
itia
ted
a de
vast
atin
g ts
un
ami
in S
um
atra
. Sci
enti
sts
can
ap
prox
imat
e th
e ve
loci
ty V
in
fee
t pe
r se
con
d of
a t
sun
ami
in w
ater
of
dept
h
d 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, or
th
e ti
me
it t
akes
for
a
car
to a
ccel
erat
e fr
om a
sto
p to
six
ty
mil
es p
er h
our. V
= √
��
2PT
−
M
V
is
the
velo
city
in
met
ers
per
seco
nd,
P i
s th
e ca
r’s
aver
age
pow
er i
n w
atts
,M
is
the
mas
s of
th
e ca
r in
kil
ogra
ms,
an
d T
is
the
tim
e in
sec
onds
.
F
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 6
0 m
iles
pe
r h
our,
wh
ich
is
26.8
2 m
eter
s pe
r se
con
d. R
oun
d yo
ur
answ
er t
o th
e n
eare
st t
enth
.
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=
√
��
3kT
−m
.
Sol
ve t
he
equ
atio
n f
or k
.
k
=m
V 2
−3T
5. G
EOM
ETRY
Su
ppos
e E
mer
yvil
le
Hos
pita
l w
ants
to
buil
d a
new
h
elip
ad o
n w
hic
h m
edic
res
cue
hel
icop
ters
can
lan
d. T
he
hel
ipad
wil
l be
cir
cula
r an
d m
ade
of f
ire
resi
stan
t ru
bber
.
a.
If
the
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
= 12
√ �
�
2π
−
π
c.
Usi
ng
a ca
lcu
lato
r, fi
nd
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.
57 m
r
10-2
001_
020_
ALG
1_A
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M_C
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R_6
6028
4.in
dd
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/22/
10
5:17
PM
Answers (Lesson 10-2)
A01_A14_ALG1_A_CRM_C10_AN_660284.indd A6A01_A14_ALG1_A_CRM_C10_AN_660284.indd A6 12/22/10 6:27 PM12/22/10 6:27 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
Chapter 10 A7 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
16
G
lenc
oe A
lgeb
ra 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
10:
1 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
th
e n
eare
st t
enth
.
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.
2.7
2 7
3. 0
.92
1
4. 3
.62
13
5. 4
.22
18
6. 3
.92
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
1_A
_CR
M_C
10_C
R_6
6028
4.in
dd
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/22/
10
5:17
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-3
Cha
pte
r 10
17
G
lenc
oe A
lgeb
ra 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
and D
istr
ibutive
Pro
pert
ies
= 6
√ �
6
+ √
�
3
Sim
plif
y.
S
imp
lify
3 √
��
12 +
5 √
��
75 .
3 √
��
12 +
5 √
��
75 =
3 √
��
�
22 · 3
+ 5
√ �
��
52 · 3
F
acto
r 12 a
nd 7
5.
= 3
· 2
√ �
3
+ 5
· 5
√ �
3
S
implif
y.
= 6
√ �
3
+ 2
5 √
�
3
Multip
ly.
= 3
1 √
�
3
Dis
trib
utive
Pro
pert
y
Exer
cise
s S
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 1
6. 2
√ �
3 -
4 √
�
45 +
2 √
�
1 −
3
8 √
�
3
−
3 -
12
√ �
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
1_A
_CR
M_C
10_C
R_6
6028
4.in
dd
1712
/22/
10
5:17
PM
Answers (Lesson 10-2 and Lesson 10-3)
A01_A14_ALG1_A_CRM_C10_AN_660284.indd A7A01_A14_ALG1_A_CRM_C10_AN_660284.indd A7 12/22/10 6:53 PM12/22/10 6:53 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 10 A8 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
18
G
lenc
oe A
lgeb
ra 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 · 1
0 +
3 ·
4 -
8 ·
10
- 2
√ �
��
22 · 1
0
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
- 2
1 √
�
10 -
3 √
� 6
12
- 6
√ �
15
+ 1
2 √
� 2 -
6 √
��
30
17. (
2 √
� 5 -
2 √
� 3 )( √
�
10 +
√ �
6 )
18. (
√ �
2 +
3 √
� 3 )( √
�
12 -
4 √
� 8 )
4
√ �
2 2
- 2
2 √
� 6
10-3
Exam
ple
001_
020_
ALG
1_A
_CR
M_C
10_C
R_6
6028
4.in
dd
1812
/22/
10
5:17
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-3
Cha
pte
r 10
19
G
lenc
oe A
lgeb
ra 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
+ 1
2 √
��
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
1_A
_CR
M_C
10_C
R_6
6028
4.in
dd
1912
/22/
10
5:17
PM
Answers (Lesson 10-3)
A01_A14_ALG1_A_CRM_C10_AN_660284.indd A8A01_A14_ALG1_A_CRM_C10_AN_660284.indd A8 12/22/10 6:27 PM12/22/10 6:27 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
Chapter 10 A9 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
20
G
lenc
oe A
lgeb
ra 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
- 1
0 √
��
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
, wh
ere
t is
th
e su
rfac
e te
mpe
ratu
re i
n d
egre
es C
elsi
us.
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
, 10
0 √
�
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
- 1
00
√ �
�
11 ≈
7.7
5 m
/s
22. G
EOM
ETRY
A r
ecta
ngle
is 5
√ �
7
+ 2
√ �
3
met
ers
long
and
6 √
� 7 -
3 √
�
3 m
eter
s w
ide.
a.
Fin
d th
e pe
rim
eter
of
the
rect
angl
e in
sim
ples
t fo
rm. (2
2 √
�
7 -
2 √
� 3 )
m
b
. F
ind
the
area
of
the
rect
angl
e in
sim
ples
t fo
rm. (1
92 -
3 √
��
21 )
m2
10-3
001_
020_
ALG
1_A
_CR
M_C
10_C
R_6
6028
4.in
dd
2012
/22/
10
5:17
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-3
Cha
pte
r 10
21
G
lenc
oe A
lgeb
ra 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
√�
�14
9 m
2. E
AR
TH T
he
surf
ace
area
S o
f a
sph
ere
wit
h r
adiu
s r
is g
iven
by
the
form
ula
S
= 4
πr2 .
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?
63
70 k
m
3. G
EOM
ETRY
Th
e ar
ea o
f a
trap
ezoi
d is
fo
un
d by
mu
ltip
lyin
g it
s h
eigh
t by
th
e av
erag
e le
ngt
h o
f it
s ba
ses.
Fin
d th
e ar
ea
of d
eck
atta
ched
to
Mr.
Wil
son
’s h
ouse
. G
ive
you
r an
swer
as
a si
mpl
ifie
d ra
dica
l ex
pres
sion
.
6
3
√ �
�
15 f
t2
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 A
bal
l is
dro
pped
fro
m a
bu
ildi
ng
win
dow
800
fee
t ab
ove
the
grou
nd.
A h
eavi
er b
all
is d
ropp
ed f
rom
a
low
er w
indo
w 2
88 f
eet
hig
h. B
oth
bal
ls
are
rele
ased
at
the
sam
e ti
me.
Ass
um
e ai
r re
sist
ance
is
not
a f
acto
r an
d u
se t
he
foll
owin
g fo
rmu
la t
o fi
nd
how
man
y se
con
ds t
it w
ill
take
a b
all
to f
all h
fee
t.
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 f
rom
288
fee
t la
nd
s fi
rst.
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
.
Ho
use
Dec
k
h =
7 √
�
5 f
t
12 √
�
3 f
t
6 √
�
3 f
t
A (
2 √
�
14 ,
5 √
�
7 )
23 √
��
149 m
B (-
2 √
�
14 ,
7 √
�
7 )
10-3
abo
ut
2.8
s
2 √
�
2 s
- √
� 2
−
4
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Answers (Lesson 10-3)
A01_A14_ALG1_A_CRM_C10_AN_660284.indd A9A01_A14_ALG1_A_CRM_C10_AN_660284.indd A9 12/22/10 6:27 PM12/22/10 6:27 PM
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pyrig
ht © G
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-Hill, a d
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cGraw
-Hill C
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Chapter 10 A10 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
22
G
lenc
oe A
lgeb
ra 1
Enri
chm
ent
Th
e W
he
el
of Th
eo
do
rus
Th
e G
reek
mat
hem
atic
ian
s w
ere
intr
igu
ed b
y pr
oble
ms
of
repr
esen
tin
g di
ffer
ent
nu
mbe
rs a
nd
expr
essi
ons
usi
ng
geom
etri
c co
nst
ruct
ion
s.T
heo
doru
s, a
Gre
ek p
hil
osop
her
wh
o li
ved
abou
t 42
5 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
. li
ne
segm
ent
BO
√ �
2
3. l
ine
segm
ent
CO
√ �
3
4
. li
ne
segm
ent
DO
√ �
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 at
rig
ht
ang
les
to t
he
last
hyp
ote
nu
se. T
hen
dra
w t
he
new
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 . W
rite
an
exp
ress
ion
for
th
e le
ngt
h o
f th
e h
ypot
enu
se o
f th
e n
th t
rian
gle.
√
��
�
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 o
r √
��
�
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 le
ng
th √
��
18
1
1
1
1 OAB
CD
10-3
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-4
Cha
pte
r 10
23
G
lenc
oe A
lgeb
ra 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
Ori
gin
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
Ori
gin
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
s S
olve
eac
h e
qu
atio
n. C
hec
k y
our
solu
tion
.
1.
√ �
a
= 8
64
2.
√ �
a
+ 6
= 3
2 67
6 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 10
−
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_
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Answers (Lesson 10-3 and Lesson 10-4)
A01_A14_ALG1_A_CRM_C10_AN_660284.indd A10A01_A14_ALG1_A_CRM_C10_AN_660284.indd A10 12/22/10 6:27 PM12/22/10 6:27 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
Chapter 10 A11 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
24
G
lenc
oe A
lgeb
ra 1
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
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
Ori
gin
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
s S
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 -
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
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-4
Cha
pte
r 10
25
G
lenc
oe A
lgeb
ra 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 =
10
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 �
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
27
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
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Answers (Lesson 10-4)
A01_A14_ALG1_A_CRM_C10_AN_660284.indd A11A01_A14_ALG1_A_CRM_C10_AN_660284.indd A11 12/22/10 6:27 PM12/22/10 6:27 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
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Chapter 10 A12 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
26
G
lenc
oe A
lgeb
ra 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 6
0
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
160
0 ft
9.6
oh
ms
1210
wat
ts
576
ft
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-4
Cha
pte
r 10
27
G
lenc
oe A
lgeb
ra 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
, wh
ere
h i
s th
e h
eigh
t in
fee
t ab
ove
the
surf
ace
of
the
wat
er. H
ow f
ar w
ould
a s
ubm
arin
e pe
risc
ope
hav
e to
be
abov
e th
e w
ater
to
loca
te a
sh
ip 6
mil
es a
way
? R
oun
d yo
ur
answ
er t
o th
e n
eare
st t
enth
.
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 in
boa
rd f
eet
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
boa
rd f
eet
of l
um
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
in.
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 in
gal
lon
s pe
r m
inu
te,
p is
th
e n
ozzl
e pr
essu
re i
n p
oun
ds p
er
squ
are
inch
, an
d d
is
the
diam
eter
of
the
hos
e in
in
ches
. In
ord
er t
o ef
fect
ivel
y fi
ght
a fi
re, t
he
com
bin
ed f
low
rat
e of
tw
o h
oses
nee
ds t
o be
abo
ut
2430
gal
lon
s pe
r m
inu
te. T
he
diam
eter
of
each
of
the
hos
es i
s 3
inch
es, b
ut
the
noz
zle
pres
sure
of
on
e h
ose
is 4
tim
es t
hat
of
the
seco
nd
hos
e. W
hat
are
th
e n
ozzl
e pr
essu
res
for
each
hos
e? R
oun
d yo
ur
answ
ers
to t
he
nea
rest
ten
th.
11
.2 p
si a
nd
44.
8 p
si
5. G
EOM
ETRY
Th
e la
tera
l su
rfac
e ar
ea L
of
a r
igh
t ci
rcu
lar
con
e is
th
e su
rfac
e ar
ea n
ot i
ncl
udi
ng
the
area
of
the
base
. T
he
late
ral
surf
ace
area
is
repr
esen
ted
by L
= π
r √ �
��
r2 +
h2 ,
wh
ere
r is
th
e ra
diu
s of
th
e ba
se a
nd
h i
s th
e h
eigh
t.
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 of
th
e fu
nn
el, t
hat
is,
th
e ba
se o
f th
e co
ne?
x +
1 m
12 m
10
m
10-4
11.1
cm
38.5
cm
2
24.2
ft
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Answers (Lesson 10-4)
A01_A14_ALG1_A_CRM_C10_AN_660284.indd A12A01_A14_ALG1_A_CRM_C10_AN_660284.indd A12 12/22/10 6:27 PM12/22/10 6:27 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
Chapter 10 A13 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
28
G
lenc
oe A
lgeb
ra 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
each s
ide t
o r
em
ove
the s
quare
root.
x
+ 7
= x
+ 2
√ �
x +
1
Sim
plif
y. U
se t
he F
OIL
meth
od t
o s
quare
the r
ight
sid
e.
x +
7 -
x -
1 =
2 √
� x
Isola
te t
he s
quare
root
term
again
.
6
= 2
√ �
x S
implif
y.
3
= √
� x
Div
ide e
ach s
ide b
y 2
.
9
= x
S
quare
each s
ide t
o r
em
ove
the 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
Sim
pify.
4
= 4
�
The 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_
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Lesson 10-4
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
29
G
lenc
oe A
lgeb
ra 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
E
nte
r √
��
�
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
T
RA
CE
to
fin
d th
e en
dpoi
nt
of t
he
grap
h o
f th
e ra
dica
l eq
uat
ion
. Use
C
AL
C t
o de
term
ine
the
inte
rsec
tion
of
th
e gr
aph
s. T
his
in
terv
al, -
4 to
5,
wh
ere
the
grap
h o
f 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 -
4
Gra
ph e
ach
sid
e of
th
e in
equ
alit
y. F
ind
the
inte
rsec
tion
an
d tr
ace
to t
he
endp
oin
t of
th
e ra
dica
l gr
aph
.
T
he
grap
h o
f y
= √
��
�
2x -
5 i
s ab
ove
the
grap
h o
f y
= x
- 4
fro
m 2
.5 u
p to
7.
Th
us,
th
e so
luti
on i
s 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_
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Answers (Lesson 10-4)
A01_A14_ALG1_A_CRM_C10_AN_660284.indd A13A01_A14_ALG1_A_CRM_C10_AN_660284.indd A13 12/22/10 6:27 PM12/22/10 6:27 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 10 A14 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
30
G
lenc
oe A
lgeb
ra 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.
Pyt
hag
ore
an T
heo
rem
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,
the
n c
2 =
a2 +
b2.
CB
Ab
ac
F
ind
th
e m
issi
ng
len
gth
.
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
Take
the s
quare
root
of
each s
ide.
c =
13
Sim
plif
y.
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 4
5.83
3
5.3625
25c
100
110
a
40
30c
10-5
12
5
14
84
1589
5
Exam
ple
4.
5.
6.
1
6.12
5
.57
8
021_
034_
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PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-5
Cha
pte
r 10
31
G
lenc
oe A
lgeb
ra 1
Stud
y G
uide
and
Inte
rven
tion
(co
nti
nu
ed)
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
eac
h s
et o
f m
easu
res
can
b
e si
des
of
a ri
ght
tria
ngl
e.
a. 1
0, 1
2, 1
4S
ince
th
e gr
eate
st m
easu
re i
s 14
, let
c =
14,
a =
10,
an
d b
= 1
2.
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 , se
gmen
ts w
ith
th
ese
mea
sure
s ca
nn
ot f
orm
a r
igh
t tr
ian
gle.
b.
7, 2
4, 2
5S
ince
th
e gr
eate
st m
easu
re i
s 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 , se
gmen
ts w
ith
th
ese
mea
sure
s ca
n f
orm
a r
igh
t tr
ian
gle.
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 ye
s; y
es
2. 6
, 8, 1
0 ye
s; y
es
3. 8
, 8, 1
0 n
o;
no
4. 9
0, 1
20, 1
50 y
es;
yes
5. 1
5, 2
0, 2
5 ye
s; y
es
6. 4
, 8, 4
√ �
5
yes
; n
o
7. 2
, 2,
√ �
8
yes
; n
o
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
; ye
s
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
1_A
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M_C
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Answers (Lesson 10-5)
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ight
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raw
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Chapter 10 A15 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
32
G
lenc
oe A
lgeb
ra 1
Skill
s Pr
acti
ceTh
e P
yth
ag
ore
an
Th
eo
rem
Fin
d e
ach
mis
sin
g le
ngt
h. I
f n
eces
sary
, rou
nd
to
the
nea
rest
hu
nd
red
th.
1.
2.
75
36
3.
4.
30
15.7
5
5.
6.
9.
85
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.
15,
30,
34
no
; n
o
9. 1
6, 2
8, 3
2 n
o;
no
10
. 18,
24,
30
yes;
yes
11. 1
5, 3
6, 3
9 ye
s; y
es
12. 5
, 7,
√ �
�
74 y
es;
no
13. 4
, 5, 6
no
; n
o
14. 1
0, 1
1, √
��
221
yes
; n
o
25024
0
a4
9
c
2933
b
1634
b
1539a
21
72c
10-5
021_
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Lesson 10-5
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-5
Cha
pte
r 10
33
G
lenc
oe A
lgeb
ra 1
Prac
tice
Th
e P
yth
ag
ore
an
Th
eo
rem
10-5
Fin
d e
ach
mis
sin
g le
ngt
h. I
f n
eces
sary
, rou
nd
to
the
nea
rest
hu
nd
red
th.
1.
2.
3.
68
15.
49
11.
31
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.
Th
en d
eter
min
e w
het
her
th
ey f
orm
a P
yth
agor
ean
tri
ple
.
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;
no
n
o;
no
8. 9
, 2 √
��
10 , 1
1
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
reat
est
len
gth
th
at w
ill
fi t t
hro
ug
h t
he
do
or
is √
��
45 ≈
6.7
1 ft
.
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. t
elev
isio
n
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 in
.
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 sc
reen
mea
sure
s 2
in. b
y 3
in.,
then
its
dia
go
nal
is o
nly
ab
ou
t 3.
61 in
.
124
b19
11
a
60
32c
021_
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Answers (Lesson 10-5)
A15_A21_ALG1_A_CRM_C10_AN_660284.indd A15A15_A21_ALG1_A_CRM_C10_AN_660284.indd A15 12/22/10 6:27 PM12/22/10 6:27 PM
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pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
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PDF Pass
Chapter 10 A16 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
34
G
lenc
oe A
lgeb
ra 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 cm
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
eng
ths
S
tud
ent
a
b
c
Stu
den
t a
b
c
A
my
3
4
5
Fra
n
8
14
16
B
elin
da
7
2
4
25
G
us
5
12
13
E
mo
ry
9
12
15
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
122
5-in
ch
21
143
2-in
ch
25
205
0-in
ch
40
30Fr
an’s
021_
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Lesson 10-5
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
35
G
lenc
oe A
lgeb
ra 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
= 2
5 �
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 po
siti
ve i
nte
gers
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=
25
- 4
=
20
= 2
5 +
4
841
= 8
41 �
= 2
1 =
29
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, 1
5, 1
7 1
6, 3
0, 3
4
4. m
= 6
an
d n
= 5
5.
m =
10
and
n =
7
6. m
= 8
an
d n
= 5
1
1, 6
0, 6
1 5
1, 1
40, 1
49
39,
80,
89
ACB a
c b
10-5
Exam
ple
035_
042_
ALG
1_A
_CR
M_C
10_C
R_6
6028
4.in
dd
3512
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Answers (Lesson 10-5)
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swer
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pyr
ight
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lenc
oe/
McG
raw
-Hill
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ivis
ion
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The
McG
raw
-Hill
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mp
anie
s, In
c.
PDF Pass
Chapter 10 A17 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
36
G
lenc
oe A
lgeb
ra 1
Spre
adsh
eet
Act
ivit
yP
yth
ag
ore
an
Tri
ple
s
A P
yth
agor
ean
tri
ple
is
a se
t of
th
ree
wh
ole
nu
mbe
rs t
hat
sat
isfi
es t
he
equa
tion
a2 +
b2 =
c2 ,
whe
re c
is t
he g
reat
est
num
ber.
You
can
use
a sp
read
shee
t 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
prim
itiv
e P
yth
agor
ean
tri
ple
into
cel
ls A
1, B
1, a
nd
C1.
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
amp
le a
nsw
er:
For
(6, 8
, 10)
, 62 +
82
= 3
6 +
64 o
r 10
0 =
102 .
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
ents
’ 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
amp
le a
nsw
er:
(10,
24,
26)
, (15
, 36,
39)
4.
(9, 4
0, 4
1) S
amp
le a
nsw
er:
(18,
80,
82)
, (27
, 120
, 123
)
5.
(20,
21,
29)
Sam
ple
an
swer
: (4
0, 4
2, 5
8), (
60, 6
3, 8
7)
Tri
ple
s.xl
sA
1 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
hee
t 2
SI
10-5
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Lesson 10-6
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
37
G
lenc
oe A
lgeb
ra 1
Stud
y G
uide
and
Inte
rven
tion
Trig
on
om
etr
ic R
ati
os
10-6
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
−
−
hyp
ote
nu
se
co
sin
e o
f ∠
B =
le
g a
dja
ce
nt
to ∠
B
−
−
hyp
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
=
le
g o
pp
osite
∠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
each s
ide.
a =
6
Take t
he p
ositiv
e 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 =
op
p −
h
yp =
6 −
10
= 3 −
5
co
s A
= ad
j −
h
yp =
8 −
10
= 4 −
5
tan
A =
op
p −
ad
j = 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 ,
cos
A =
8 −
17
,
si
n A
=
7 −
25
, co
s A
= 24
−
25
,
tan
A =
15
−
8
tan
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
1.43
01
Exam
ple
a10 8
0.64
28
b
a
c
sin
A =
4 −
5 ,
cos
A =
3 −
5 ,
tan
A =
4 −
3
035_
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PM
Answers (Lesson 10-5 and Lesson 10-6)
A15_A21_ALG1_A_CRM_C10_AN_660284.indd A17A15_A21_ALG1_A_CRM_C10_AN_660284.indd A17 12/22/10 6:27 PM12/22/10 6:27 PM
Co
pyrig
ht © G
lencoe/M
cGraw
-Hill, a d
ivision o
f The M
cGraw
-Hill C
om
panies, Inc.
PDF Pass
Chapter 10 A18 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
38
G
lenc
oe A
lgeb
ra 1
10-6
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
righ
t tr
ian
gle.
Rou
nd
eac
h s
ide
len
gth
to
the
nea
rest
ten
th.
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
Defi nitio
n o
f cosin
e
c co
s 38
° =
13
Multip
ly e
ach s
ide b
y c
.
c =
13
−
cos
38°
Div
ide e
ach s
ide b
y c
os 3
8°.
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.
ta
n 3
8° =
a −
13
D
efi nitio
n o
f ta
ngent
13
tan
38°
= a
M
ultip
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
= 6
0°, A
C ≈
7.8,
∠
A =
46°
, AC
≈ 7.
7,
∠B
= 3
4°, A
C ≈
19.
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)
Trig
on
om
etr
ic R
ati
os
Exam
ple
13
ac
38°
035_
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PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-6
Cha
pte
r 10
39
G
lenc
oe A
lgeb
ra 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 =
77
−
85 ,
cos
A =
36
−
85
, ta
n A
= 77
−
36
sin
A =
4 −
5 ,
cos
A =
3 −
5 ,
tan
A =
4 −
3
3.
10
24
4
. 15
8
si
n A
= 12
−
13
, co
s A
=
5 −
13 ,
tan
A =
12
−
5 s
in A
= 8 −
17
, co
s A
= 15
−
17
, ta
n 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. s
in 1
8° 0
.309
0 6.
cos
68°
0.3
746
7. t
an 2
7° 0
.509
5
8. c
os 6
0° 0
.50
00
9. t
an 7
5° 3
.732
1 10
. sin
9°
0.15
64
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°, A
B =
13.
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°
10-6
Skill
s Pr
acti
ceTr
igo
no
metr
ic R
ati
os
035_
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Answers (Lesson 10-6)
A15_A21_ALG1_A_CRM_C10_AN_660284.indd A18A15_A21_ALG1_A_CRM_C10_AN_660284.indd A18 12/22/10 6:27 PM12/22/10 6:27 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
Chapter 10 A19 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
40
G
lenc
oe A
lgeb
ra 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
= 72
−
97
, ta
n A
= 65
−
72
s
in A
= 36
−
39
, co
s A
= 15
−
39
, ta
n 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. t
an 2
6° 0
.487
7 4.
sin
53°
0.7
986
5. c
os 8
1° 0
.156
4
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°, A
B =
23.
9, A
C =
9.3
∠
A =
61°
, AB
= 1
0.3,
BC
= 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
10-6
Prac
tice
Trig
on
om
etr
ic R
ati
os
25°
54 ft
h
035_
042_
ALG
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M_C
10_C
R_6
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10
5:18
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 10-6
Cha
pte
r 10
41
G
lenc
oe A
lgeb
ra 1
10-6
Wor
d Pr
oble
m P
ract
ice
Trig
on
om
etr
ic R
ati
os
1. W
ASH
ING
TON
MO
NU
MEN
T Je
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
tru
ck b
ed i
s 3
feet
ab
ove
the
grou
nd,
wh
at i
s th
e an
gle
of
incl
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
. To
the
nea
rest
deg
ree
wh
at w
ould
th
e m
easu
re o
f an
gle
BA
C b
e on
a
stan
dard
tel
evis
ion
wit
h a
4:3
asp
ect
rati
o?
( 16
−
9 h
) 2 + h
2 =
422 ;
h =
20.
6 in
.
wid
th =
36.
6 in
., m
∠A
= 2
9°,
m∠
B =
61°
16°
45°
6869
ft
556
ft
37°
035_
042_
ALG
1_A
_CR
M_C
10_C
R_6
6028
4.in
dd
4112
/22/
10
5:18
PM
Answers (Lesson 10-6)
A15_A21_ALG1_A_CRM_C10_AN_660284.indd A19A15_A21_ALG1_A_CRM_C10_AN_660284.indd A19 12/22/10 6:27 PM12/22/10 6:27 PM
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-Hill C
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PDF 2nd
Chapter 10 A20 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pte
r 10
42
G
lenc
oe A
lgeb
ra 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
ad
jace
nt
∠A
se
ca
nt
of
∠B
=
hyp
ote
nu
se
−
leg
ad
jace
nt
∠B
se
c A
= c −
b
se
c B
= c −
a
b
a
c
co
se
ca
nt
of
∠A
=
hyp
ote
nu
se
−
leg
op
po
site
∠A
co
se
ca
nt
of
∠B
=
hyp
ote
nu
se
−
leg
op
po
site
∠B
csc A
= c −
a
csc B
= c −
b
co
tan
ge
nt
of
∠A
= le
g a
dja
ce
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
= b
−
a
co
t B
= a −
b
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 =
h
yp
−
adj =
15
−
12
= 5 −
4
cs
c A
= h
yp
−
opp
= 15
−
9
= 5 −
3
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 =
17
−
8 , c
sc A
= 17
−
15
sec
A =
25
−
24 ,
csc
A =
25
−
7 ,
cot
A =
8
−
15
co
t A
= 24
−
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
tan
gen
t?
c
sc A
=
1
−
sin
A ,
sec
A =
1 −
co
s A
, an
d c
ot
A =
1 −
ta
n 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°
1.04
57
6. c
sc 4
9° 1
.325
0 7.
cot
81°
0.1
584
10-6
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 ,
cot
A =
3 −
4
035_
042_
ALG
1_A
_CR
M_C
10_C
R_6
6028
4.in
dd
4212
/22/
10
5:18
PM
Answers (Lesson 10-6)
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Chapter 10 A22 Glencoe Algebra 1
Chapter 10 Assessment Answer KeyQuiz 1 (Lessons 10-1 and 10-2) Quiz 3 (Lesson 10-5) Mid-Chapter TestPage 45 Page 46 Page 47 (Lessons 10-1 through 10-4)
Quiz 2 (Lessons 10-3 and 10-4)
Page 45Quiz 4 (Lessons 10-6)
Page 46
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
26 √ � 5
4 √ � 6 + 8 √ �� 10
5 √ � 2 + 6 √ � 5
√ �� 15 - 2 √ � 2
D
5
7
3, 8
2
2
1.
2.
3.
4.
5.
A
6 √ � 2
y
x
D = {x|x ≥ -1}; R = {y|y ≥ 0};
1.
2.
3.
4.
5.
14.70
no
C
no
no
1.
2.
3.
4.
5.
8
A
RP = 27.5,PQ = 29.2, m∠ Q = 70˚
sin B = 4 −
5 , cos B = 3 −
5 ,
tan B = 4 −
3
sin A = 3 −
5 , cos A = 4 −
5 ,
tan A = 3 −
4
1.
2.
3
4.
5.
6.
7.
D
F
D
G
D
F
D
8.
9.
10.
11.
12.
13.
14
15.
6 √ � 5 - 20 √ � 3
23
7 √ � 5 - 2 √ �� 10
3 √ � 2 + 2 √ � 6
5
29
39 √ � 2
9.49 ft
2|x| √ �� 2y
12 - 3 √ � 2
−
14
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ight
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oe/
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-Hill
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ion
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McG
raw
-Hill
Co
mp
anie
s, In
c.
An
swer
s
PDF Pass
Chapter 10 A23 Glencoe Algebra 1
Chapter 10 Assessment Answer KeyVocabulary Test Form 1Page 48 Page 49 Page 50
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
F
B
G
H
F
C
A
F
B
D
D
12. J B: ⎪2x + 1⎥
13.
14.
15.
16.
17.
18.
19.
20.
D
F
C
H
D
A
J
H
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Sample answer: the
study of relationships
among the angles
and sides of the
triangle
solving the triangle
legs
radicand
hypotenuse
radical equation
Pythagorean Theorem
converse
conjugates
cosine
Sample answer: the
trigonometric ratio
equivalent to the leg
opposite to an angle
divided by the leg
adjacent to the angle.
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Chapter 10 A24 Glencoe Algebra 1
Chapter 10 Assessment Answer KeyForm 2A Form 2BPage 51 Page 52 Page 53 Page 54
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
D
J
D
G
C
F
B
G
B
H
D
H
13.
14.
15.
16.
17.
18.
19.
20.
F
C
G
A
G
B
A
H
B: 12 m
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12. G
B
F
C
J
C
J
C
F
B
D
F
J
A
G
A
J
13.
14.
15.
16.
17.
18.
19.
20.
D
A
H
B: 14 ft
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-Hill
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McG
raw
-Hill
Co
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anie
s, In
c.
An
swer
s
PDF Pass
Chapter 10 A25 Glencoe Algebra 1
Chapter 10 Assessment Answer KeyForm 2CPage 55 Page 56
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17. 2.9040
no
yes
7
11.66
12
11
12
6 √ � 2
5 y 2 |w| √ �� 3w
3 √ � 2 + 2 √ � 3
20 √ � 5
2 √ �� 14 + 5 √ � 3
0.7431
0.9925
D = {x | x ≥ 2}; R = {y | y ≥ 1}
D = {x | x ≥ 1}; R = {y | y ≤ 5}
y
x 18.
19.
20.
21.
22.
23.
24.
25.
B:
≈ 23.2 ft
67°
≈ 12.5 cm
sin A = 48 −
73 ; cos A = 55
−
73 ;
tan A = 48 −
55
sin A = 21 −
29 ; cos A = 20
−
29 ;
tan A = 21 −
20
53.7 ft 2
√ �� 149 mi or 12.2 mi
10.4 ft
-2
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Chapter 10 A26 Glencoe Algebra 1
Chapter 10 Assessment Answer KeyForm 2DPage 57 Page 58
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
10 √ � 2
5|xy| √ �� 2x
10 √ � 5 + 15 √ � 2
9 √ � 3 - 2 √ �� 22
4
3
8.06
8
yes
no
0.7431
0.4695
0.2126
D = {x | x ≥ -1}; R = {y | y ≥ -3}D = {x | x ≥ -2}; R = {y | y ≤ -1}
4
y
x
1.
22 √ � 6
18.
19.
20.
21.
22.
23.
24.
25.
B: -3
15.20 cm
5.39 mi
43.3 ft
sin A = 36 −
85 ; cos A = 77
−
85 ;
tan A = 36 −
77
84.6 ft 2
69°
≈ 30.0 ft
sin A = 4 −
5 ; cos A = 3 −
5 ;
tan A = 4 −
3
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swer
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Chapter 10 A27 Glencoe Algebra 1
Chapter 10 Assessment Answer KeyForm 3Page 59 Page 60
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18 √ � 7
x 2 √ �� 5n
−
2| n 3 |
0
40 √ � 2 - 27 √ �� 15
6
no solution
7 −
2 , 5
√ �� 41
2 √ � 11 cm
yes
no
0.2756
28.6363
0.8572
D = {x | x ≥ 2}; R = {y | y ≤ -2}
D = {x | x ≥ -4}; R = {y | y ≤ -12}
y
x
2 ( √ �� 10 - √ � 3 )
−
7
1.
18.
19.
20.
21.
22.
23.
24.
25.
B:
8.06 miles
10.20 cm
1.5 in.
sin A = 33 −
65 ; cos A = 56
−
65 ;
tan A = 33 −
56
sin A = 72 −
97 ; cos A = 65
−
97 ;
tan A = 72 −
65
176.2 m
7 √ �� 30 - 27
−
39
52°
87.0 ft
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Chapter 10 A28 Glencoe Algebra 1
Chapter 10 Assessment Answer KeyPage 61, Extended-Response Test
Scoring Rubric
Score General Description Specifi c 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 fl aws
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.
• Satisfi es 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 simplifying radical expressions, solving radical equations, the Pythagorean Theorem, right triangles, similar triangles, and the distance formula.
• May not use appropriate strategies to solve problems.
• Computations are mostly correct.
• Written explanations are satisfactory.
• Graphs are mostly accurate.
• Satisfi es the requirements of most of the problems.
1 Nearly Unsatisfactory
A correct solution with no
supporting evidence or
explanation
• Final computation is correct.
• No written explanations or work is shown to substantiate the fi nal
computation.
• Graphs may be accurate but lack detail or explanation.
• Satisfi es 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
simplifying radical expressions, solving radical equations, the Pythagorean Theorem, right triangles, similar triangles, and the distance formula.
• 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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PDF 2nd
Chapter 10 A29 Glencoe Algebra 1
Chapter 10 Assessment Answer Key Page 61, 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. Agree; Sample answer: When sin A = sin D, m∠A = m∠ D. The triangles are right triangles, so m∠C = m∠F and, thus, m∠B = m∠E.
3b. Agree; Sample answer: When sin A ≠
sin D, m∠A ≠ m∠D. The triangles are right triangles, so m∠C = m∠F and, thus, m∠B ≠ m∠E.
4. Sample answer: a = 9, b = 8. ( √ �� 145 )
2 � 92 + 82
145 = 81 + 64
In addition to the scoring rubric found on page A28, the following sample answers may be used as guidance in evaluating extended-response assessment items.
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Chapter 10 A30 Glencoe Algebra 1
Chapter 10 Assessment Answer KeyStandardized Test PracticePage 62 Page 63
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. 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 19.
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
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Chapter 10 A31 Glencoe Algebra 1
Chapter 10 Assessment Answer KeyStandardized Test PracticePage 64
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30a.
30b. 7.2 mi
14.2 mi
$3648.08
{-4.4, -11.6}
-15 and -17 or
15 and 17
(x + 4)(y - 3)
no solution
y = -5x + 14
$10,800 at 14%,
$1200 at 10%
-1-2-3-4 0 1 2 3 4
3h6k2
−
- 2j10
{ w |w ≤ -1 or w > 2};
b
jO
20406080
100
100 200 300
120
2j + 5b = 500
j + b = 120
A22_A35_ALG1_A_CRM_C10_AN_660284.indd A31A22_A35_ALG1_A_CRM_C10_AN_660284.indd A31 12/22/10 6:27 PM12/22/10 6:27 PM