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Kahler geometry is a beautiful and intriguing area of mathematics, of substantial research interest to both mathematicians and physicists. This self-contained graduate text provides a concise and accessible introduction to the topic. The book begins with a review of basic differential geometry, before moving on to a description of complex manifolds and holomorphic vector bundles. Kahler manifolds are discussed from the point of view of Riemannian geometry, and Hodge and Dolbeault theories are outlined, together with a simple proof of the famous Kahler identities. The final part of the text studies several aspects of compact Kahler manifolds: the Calabi conjecture, Weitzenbock techniques, Calabi-Yau manifolds, and divisors. All sections of the book end with a series of exercises and students and researchers working in the fields of algebraic and differential geometry and theoretical physics will find that the book provides them with a sound understanding of this theory.
Author Biography
Andrei Moroianu is a Researcher at CNRS and a Professor of Mathematics at Ecole Polytechnique. | 677.169 | 1 |
Definition Of The Derivative
Students engage in a lesson that is concerned with the concept of a derivative. They define the concept of a derivative and then complete an assignment that is included in the lesson. Students then graph the derivative and draw its representation. | 677.169 | 1 |
This chapter
on algebra is a behemoth. It is by far the longest chapter
in this book full of lengthy chapters. There's a reason for our
extensive treatment: algebra is the most tested topic on the Math IC
test. About 30 percent of the Math IC questions directly test your
algebraic abilities, and some of the questions that focus on geometry
or trigonometry still involve some sort of algebraic technique or
concept.
Before this information all starts to sound overwhelming,
there is some good news. First, the algebra tested on the math subject
tests is not all that difficult. Second, the Math IC test-writers
focus on a limited set of algebraic topics. Only the topics you
do need to know are covered in this chapter. | 677.169 | 1 |
CATALOG COURSE DESCRIPTION: This course is the first of a two semester sequence preparing students for Calculus. In this course you will study functions with an emphasis on the trigonometric functions along with topics in analytic geometry. Topics will include a review of plane and coordinate geometry, functions including function notation, transformations and inverses, definitions and graphs of the trigonometric functions, modeling periodic behavior, solving triangle problems with the Laws of Sines and Cosines, the conic sections, parametric equations and vectors.
CALCULATORS: You are required to have a graphing calculator but may be restricted to a scientific calculator or no calculator on some tests.
EXAMS: There will be at least five chapter tests and a cumulative final exam. There may also be several projects, as circumstance demands. The point of these exams and projects is to provide an opportunity to demonstrate your understanding of the concepts covered. As such, exam problems won't consist entirely of very familiar homework problems. Tests are timed to fit the class period.
HOMEWORK: Read the text, keep up with the assigned problems (as a minimum) and prepare questions about what you're learning for participation in class. You can expect to learn far more trigonometry at home doing homework than you do in class. If you complete (and thoroughly understand) the homework assignment for each section, you will be well prepared to solve questions on tests and quizzes. At the beginning of each class, I will answer as many questions over the previous night's homework as time allows. Generally, you can expect to study at least 2 hours outside of class for every hour of class time. To get credit for homework you'll need to go use the internet site at ILRN.com.
QUIZZES: There will be some (possibly unannounced) quizzes throughout the semester.
GRADE: Your grade is a weighted average of homework, quiz, chapter test, & final exam scores:
a. Interpret slope as a constant rate of change.
b. Recognize and create linearity in tables, graphs, and/or equations.
c. Solve systems of equations by using methods of elimination, substitution and graphing.
d. Graph and/or find the equation of a circle given sufficient information.
e. Solve quadratic equations by factoring, completing the square, and the quadratic formula.
f. Recognize and create quadratic models for relations involving tables, graphs, and equations.
g. Graph a parabola by finding the vertex, intercepts, and other symmetric points.
h. Demonstrate understanding of definitions for function and its related terms: domain and range.
i. Use appropriate notation for function equations and for describing domain and range.
j. Demonstrate understanding of the exponential function, its scaling and growth factors.
k. Understand how to solve similar triangle problems.
l. Demonstrate understanding of triangle congruency theorems involving SSS, SAS, AAS.
m. Basic knowledge about congruence relations such as the congruence of vertical angles
n. Familiarity with Pythagorean theorem.
o. Demonstrate understanding of deductive reasoning in the construction of a proof.
Vectors including analytic and geometric representations and applications.
Course Objectives: Upon completion of this course, students will be able to:
Apply facts about angles, parallel lines and triangles to deduce further results about a geometric
figure.
Prove when two triangles are congruent or similar.
Justify the lengths of sides in an isosceles right triangle and in a 30-60-90 triangle.
Deduce the lengths of sides in quadrilaterals such as trapezoids and rectangles using basic definitions,
Pythagorean Theorem, perimeter and/or area.
Calculate the measure of a central angle in a circle using the measure of the intercepted arc and calculate the areas of geometric figures involving circles.
Apply facts about plane geometric figures to deduce the surface area and volume of three dimensional geometric figures.
Demonstrate an understanding of the concept of a function by identifying and describing a function graphically, numerically and algebraically.
Calculate the domain and range for a function expressed as a graph or an equation. From a graph, estimate the intervals where a function is increasing, decreasing and/or has a maximum or minimum value.
Use and interpret function notation to find "inputs" and "outputs" from the graph, table and/or an equation describing a function.
From an equation, graph or table, calculate average rates of change by using a difference quotient or by using slopes of secant lines. Analyze average rates of change to determine the concavity of a graph.
Demonstrate an understanding of the six basic transformations of functions by graphing translated functions including the quadratic functions.
Represent a word problem (especially a geometric problem) with a function.
Determine when a function has an inverse (one to one functions) and find the inverse function graphically or algebraically.
Form new functions through addition, subtraction, multiplication, division and composition.
Recognize classical and analytic definitions of the trigonometric functions.
Solve triangles using right triangle trigonometry, the law of sines and the law of cosines. | 677.169 | 1 |
Synopses & Reviews
Publisher Comments
This book is a final year undergraduate text on stochastic processes, a tool used widely by statisticians and researchers working in the mathematics of finance. The book will give a detailed treatment of conditional expectation and probability, a topic which in principle belongs to probability theory, but is essential as a tool for stochastic processes. Although the book is a final year text, the author has chosen to use exercises as the main means of explanation for the various topics, and the book will have a strong self-study element. The author has concentrated on the major topics within stochastic analysis: Stochastic Processes, Markov Chains, Spectral Theory, Renewal Theory, Martingales and It? Stochastic Processes.
Review
This book fulfils its aim of providing good and interesting material for advanced undergraduate study. The Times Higher Education Supplement This is probably one of the best books to begin learning about the sometimes complex topic of stochastic calculus and stochastic processes from a more mathematical approach. Some literature are often accused of unnecessarily complicating the subject when applied to areas of finance. With this book you are allowed to explore the rigorous side of stochastic calculus, yet maintain a physical insight of what is going on. The authors have concentrated on the most important and useful topics that are encountered in common physical and financial systems
Synopsis
This book has been designed for a final year undergraduate course in stochastic processes. It will also be suitable for mathematics undergraduates and others with interest in probability and stochastic processes, who wish to study on their own. The main prerequisite is probability theory: probability measures, random variables, expectation, independence, conditional probability, and the laws of large numbers. The only other prerequisite is calculus. This covers limits, series, the notion of continuity, differentiation and the Riemann integral. Familiarity with the Lebesgue integral would be a bonus. A certain level of fundamental mathematical experience, such as elementary set theory, is assumed implicitly. Throughout the book the exposition is interlaced with numerous exercises, which form an integral part of the course. Complete solutions are provided at the end of each chapter. Also, each exercise is accompanied by a hint to guide the reader in an informal manner. This feature will be particularly useful for self-study and may be of help in tutorials. It also presents a challenge for the lecturer to involve the students as active participants in the course.
Synopsis
Stochastic processes is a tool widely used by statisticians and researchers working, for example, in the mathematics of finance. This is an introductory text that has a strong emphasis on exercises, complete with informal hints and fully-worked solutions.
Synopsis
Stochastic processes are tools used widely by statisticians and researchers working in the mathematics of finance. This book for self-study provides a detailed treatment of conditional expectation and probability, a topic that in principle belongs to probability theory, but is essential as a tool for stochastic processes. The book centers on exercises as the main means of explanation. | 677.169 | 1 |
precal Advice
Showing 1 to 3 of 4
I would certainly recommend this class ,and more specially to people who don't know the basics in Calculus. It is not very hard plus, it is very helpful in everyday math calculations .
Course highlights:
I learnt a lot of things I didn't know much about. I learnt about absolute values ,number of equations ,from linear to circle. The most fascinating was the radical equation. I also learned about functions and a lot of stuff. But most of all I saw the work of unity; I learned about help that comes with working as a team.
Hours per week:
3-5 hours
Advice for students:
If you want to do this course , do it. What ever you are doing never stop practicing ; that way you will never forget or go wrong. If you bring your respect with you; you will go far. In any class you are taking be patient ,and ask any question you need to.
Course Term:Spring 2017
Professor:Cooper Stuart
Course Required?Yes
Course Tags:Great Intro to the SubjectMany Small AssignmentsA Few Big Assignments
Nov 28, 2016
| Would highly recommend.
Pretty easy, overall.
Course Overview:
MR. Hunt makes it almost impossible for he student to fail. He answers all questions and grade work accordingly.
Course highlights:
I never like math until this class. He teaches it in a way that the student will understand it. He connects it to real life things
Hours per week:
6-8 hours
Advice for students:
Always study and do the study guides. It is basically the test before the test. If you do it you will pass the test | 677.169 | 1 |
Services
BAM! Exclusive Funko 2-Packs
Overview - "A good textbook." ― Mathematical Gazette. This introduction to Euclidean geometry emphasizes both the theory and the practical application of isometries and similarities to geometric transformations. Each chapter begins with an optional commentary on the history of geometry.Read more...
"A good textbook." ― Mathematical Gazette. This introduction to Euclidean geometry emphasizes both the theory and the practical application of isometries and similarities to geometric transformations. Each chapter begins with an optional commentary on the history of geometry. Contents include modern elementary geometry, isometries and similarities in the plane, vectors and complex numbers in geometry, inversion, and isometries in space. Numerous exercises appear throughout the text, many of which have corresponding answers and hints at the back of the book. Prerequisites for this text, which is suitable for undergraduate courses, include high school algebra, geometry, and elementary trigonometry. 1972 edition.
This item is Non-Returnable.
Details
ISBN: 9780486138428
Publisher: Dover Publications
Date: Apr | 677.169 | 1 |
For sophomore/junior-level signals and systems courses in Electrical and Computer Engineering departments. Signals, Systems, and Transforms, Fourth Edition is ideal for electrical and computer engineers. The text provides a clear, comprehensive presentation of both the theory and applications in signals, systems, and transforms. It presents the mathematical background of signals and systems, including the Fourier transform, the Fourier series, the Laplace transform, the discrete-time and the discrete Fourier transforms, and the z-transform. The text integrates MATLAB examples into the presentation of signal and system theory and | 677.169 | 1 |
Calculus Hyperfunction
Thinkwell's Calculus with Edward Burger lays the foundation for success because, unlike a traditional textbook, students actually like using it. Thinkwell works with the learning styles of students who have found that traditional textbooks are not effective. Watch one Thinkwell video lecture and you'll understand why Thinkwell works better.
Calculus can be an intimidating subject. For many students, even the name sounds intimidating. The truth is that Calculus is based on a few very powerful principles and once you fully understand those principles all of the additional topics naturally follow. Most Calculus textbooks begin the subject with a nauseating discussion of limits and then proceed to the introduction of a derivative which is one of the core topics in Calculus.
The Trigonometry And Pre-Calculus Tutor is a 5 hour DVD course geared to fully prepare a student to enter university level Calculus. Most students that have trouble with Calculus discover quickly that the root cause of their difficulty is actually that they have never mastered essential material in Trigonometry. For instance, a Calculus textbook will assume that the student is comfortable converting between degrees and radians and can use the unit circle to mentally calculate the Sin of an angle without a calculator. These topics and many other essential concepts are presented in this DVD course with exceptional clarity so that the student will feel well prepared to move into Calculus and Physics | 677.169 | 1 |
PrefacePreface
This is a set of class notes designed to guide an inquiry-based course
on linear algebra. This is not a complete resource! Rather, this text is
meant to accompany the book Introduction to Linear Algebra by
Gilbert Strang. Industrious students might also use it for a self-study
course.
An important feature of this text is the integration of the Sage
Mathematical Software System. Students in this course will learn to use
Sage to perform long tedious computations (which linear algebra has in
spades) and create visualizations. | 677.169 | 1 |
The
The central theme of this graduate-level number theory textbook aspects.
The first is the local aspect: one can do analysis in p-adic fields, and here the author starts by looking at solutions in finite fields, then proceeds to lift these solutions to local solutions using Hensel lifting. The second is the global aspect: the use of number fields, and in particular of class groups and unit groups. This classical subject is here illustrated through a wide range of examples. The third aspect deals with specific classes of equations, and in particular the general and Diophantine study of elliptic curves, including 2 and 3-descent and the Heegner point method. These subjects form the first two parts, forming Volume I.
The study of Bernoulli numbers, the gamma function, and zeta and L-functions, and of p-adic analogues is treated at length in the third part of the book, including many interesting and original applications.
Much more sophisticated techniques have been brought to bear on the subject of Diophantine equations, and for this reason, the author has included five chapters on these techniques forming the fourth part, which together with the third part forms Volume II. These chapters were written by Yann Bugeaud, Guillaume Hanrot, Maurice Mignotte, Sylvain Duquesne, Samir Siksek, and the author, and contain material on the use of Galois representations, points on higher-genus curves, the superfermat equation, Mihailescu's proof of Catalan's Conjecture, and applications of linear forms in logarithms.
The book contains 530 exercises of varying difficulty from immediate consequences of the main text to research problems, and contain many important additional results.
Review:
From the reviews:
"The book under review deals with Diophantine analysis from a number-theoretic point of view. ... Each chapter ends with exercises, ranging from simple to quite challenging problems. The clarity of the exposition is the one we expect from the author of two highly successful books on computational number theory ... and makes this volume a must-read for researchers in Diophantine analysis." (Philosophy, Religion and Science Book Reviews, bookinspections.wordpress.com, October, 2013)
"This is the first volume of a highly impressive two-volume textbook on Diophantine analysis. ... Readers are presented with an almost overwhelming amount of material. This ... text book is bound to become an important reference for students and researchers alike." (C. Baxa, Monatshefte für Mathematik, Vol. 157 (2), June, 2009)
"Number Theory, is poised to fill the gap as a core text in number theory ... . So, all in all, Henri Cohen's ... Number Theory are, to any mind, an amazing achievement. The coverage is thorough and generally all but encyclopedic, the exercises are good, some are excellent, some will keep even the best-prepared student busy for a long time, and the cultural level of the book ... is very high." (Michael Berg, MathDL, July, 2007)
"The book under review deals with Diophantine analysis from a number-theoretic point of view. ... The clarity of the exposition is the one we expect from the author of two highly successful books on computational number theory ... and makes this volume a must-read for researchers in Diophantine analysis." (Franz Lemmermeyer, Zentralblatt MATH, Vol. 1119 (21), 2007)7499222 | 677.169 | 1 |
Product Description
▼▲
The BJU Press Geometry curriculum covers the fundamental key concepts of geometry, including reasoning, proof, parallel and perpendicular lines, triangles, quadrilaterals, area, circles, similarity, an introduction to trigonometry, and more. Fun features are included throughout the student text; "Geometry in history" histo rical-fiction narratives highlight key mathematical contributions, "Technology Corner" notes use dynamic geometry software to visualize and discover geometric concepts, while "Geometry Around Us" features show how geometry is used in careers and daily life. Chapters introduce the concept and are followed by multiple step-by-step examples. Expanded exercise sets reinforce new concepts and connect skills to previously learned concepts. A cumulative review helps keeps concepts fresh throughout the year.
The test packet includes quizzes and tests to help you assess student retention and understanding. The test answer key includes full-size, loose-leaf, and three-hole-punched test pages with the correct answers overlaid in pink ink | 677.169 | 1 |
text is ideal for advanced undergraduate or beginning graduate students. The author first develops the necessary background in probability theory and Markov chains before using it to study a range of randomized algorithms with important applications in optimization and other problems in computing. The book will appeal not only to mathematicians, but to students of computer science who will discover much useful material. This clear and concise introduction to the subject has numerous exercises that will help students to deepen their understanding.
Editorial Reviews
Review
"...extremely elegant...I am sure that students will find great pleasure in using the book--and that teachers will have the same pleasure in using it to prepare a course on the subject." Mathematics of Computation
"Here Haggstrom takes the beginning student from the first definitions concerning Markov chains even beyond Propp-Wilson to its refinementss and applications, all in just a hundred or so generously detailed pages. If an undergraduate reading this book comes away saying "I should have thought of that!" then the psychological barrier between school mathematics and research will have begun to break down. Few mathematical monographs provide a comparable opportunity. General readers; lower-division undergraduates through professionals." Choice
"[This series] is generally good....The use of examples to introduce the various algorithms is especially effective and makes the text easier to read." Mathematical Reviews
"The numerous examples perfectly well illustrate the more theoretical points. I am sure that students will find great pleasure in using the book-and that teachers will have the same pleasure in using it to prepare a course on the subject." Mathematics of Computation
Book Description
Based on a lecture course given at Chalmers University, this book is ideal for advanced undergraduate or beginning graduate students. The author first develops the necessary background in probability theory before applying it to study a range of randomised algorithms that have important applications in computing. This book will appeal not only to mathematicians, but to students of computer science who will find much here that appeals. This is a clear and concise introduction to this subject and the numerous exercises included will help students to deepen their understanding.
Top customer reviews
This is a very clearly written, succinct introduction to Markov Chains (it does not aim to be exhaustive). I think this book offers one of the most accessible and efficient routes to learning the basics about markov chain monte carlo, perfect simulation,sandwiching and simulated annealing. I enjoyed this book a lot.Better to read this a few times and do the questions, and then if you must, tackle one of the more bloated expositions. | 677.169 | 1 |
Chaos Theory Assignment Help
Chaos Theory Assignment Help Is So Easy to Get!
Chaos theory:
Hailed as one of the most interesting elements of mathematics, chaos theory is a complex subject. It is a sub-discipline of mathematics that shows that outwardly what seems as arbitrary events can actually be a result of normal equations and obeys particular laws; and even small alterations can result in remarkably great consequences because of the complexity of the systems involved. It is also referred to as Non-linear dynamics.
Its understanding is required not only for mathematics but across all kinds of studies.
Basic principles:
The Butterfly Effect:
Small, apparent insignificant changes in the initial conditions can lead to drastic consequences and changes in the results.
Unpredictable Nature:
The slightest error in measurement can completely change a prediction often rendering it useless. Since it's difficult to know the minutest details especially at the initial stages of a complex system, it is impossible to get the absolute prediction.
Order / Disorder Chaos:
It explores the transitions between order and disorder, which often occur in surprising ways.
Mixing:
Turmoil leads to two adjacent points in a complex system to end up in completely different positions after a certain amount of time has passed. Turbulence occurs at all scales so it becomes a complete mixture.
Fractals:
They are complex patterns which have the same statistical character as a whole. They come to being through simple repetitive processes. They follow mathematical structures where certain similar patterns keep occurring at progressively smaller scales (such as snowflakes).
They also describe arbitrary or chaotic phenomena like crystal growth and formation of galaxies. Nature is full of fractals like trees, clouds, seashells, tornadoes, etc.
Issues faced by students:
Understanding a complex concept as chaos theory in itself will require a lot of assistance. Students from science and maths backgrounds are used to a deterministic approach and in this case, they have to let go of that approach and sort of change their approach to understand this concept.
The expectation from your assignments is that of how well you can analyze a problem so they are essentially problem solving based.
The assignments expect you to learn how to apply the theory in our everyday lives.
You will require chaos theory assignment help and when you do, come to us at 24x7assignmenthelp.com
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It requires a lot of graphs and charts to explain these complex equations. When you come to us for your chaos theory assignment help, you will get adequate help to simplify these complex equations and great graphs and charts from us.
Topics covered by us:
Non linear dynamics and chaos
Principles of chaos
Quantum chaos
Dynamical systems
Fractal theory
If there are additional topics that you want for your chaos theory homework help, please don't hesitate to contact us. This is only a broad list.
We, at 24x7assignmenthelp.com, are always by your side, so don't feel bogged down by your assignment, we will make it easy for you through chaos theory assignment help and other related services. | 677.169 | 1 |
Discrete Mathematics for Computing presents the essential mathematics needed for the study of computing and information systems. The subject is covered in a gentle and informal style, but without compromising the need for correct methodology. It is perfect for students with a limited background in mathematics. This new edition includes: / An expanded section on encryption / Additional examples of the ways in which theory can be applied to problems in computing / Many more exercises covering a range of levels, from the basic to the more advanced This book is ideal for students taking a one-semester introductory course in discrete mathematics - particularly for first year undergraduates studying Computing and Information Systems. PETER GROSSMAN has worked in both academic and industrial roles as a mathematician and computing professional. As a lecturer in mathematics, he was responsible for coordinating and developing mathematics courses for Computing students. He has also applied his skills in areas as diverse as calculator design, irrigation systems and underground mine layouts. He lives and works in Melbourne, Australia | 677.169 | 1 |
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Complex Analysis Homework Help
Complex analysis (also known as the theory of functions of a complex variable) is one of the most useful branches of pure mathematics which is used to investigate the functions of complex numbers. It is a fundamental requirement of advanced mathematics and various disciplines of sciences including applied mathematics, algebra, number theory, physics, electrical and electronics engineering, etc. Complex analysis is concerned with analytic functions of complex variables or meromorphic functions and is used widely to solve two dimensional problems in mathematics and physics.
It is a new branch of mathematics than others and has its roots in earlier 19th century. Euler, Gauss, Riemann, Cauchy, Weierstrass and several other prominent scholars have developed complex analysis with their efforts and made it useful and beneficial for other disciplines of mathematics and sciences. | 677.169 | 1 |
Calculating the Benefits
With just over three hours until I begin my Calculus final, I am pausing to reflect a bit on the education of mathematics. I'll be honest, I hated my math teachers in high school. Nothing against them as people, I just never did very well and I didn't really enjoy class. One thing I am thankful for that the SF University High School Math Deptarment did very well was teach calculator work. In fact, I think I would go as far back as David McSpadden of Stuart Hall for Boys in K-8 school because Mr. McSpadden made me buy and learn how to operate my first graphing calculator: the TI-81. In fact, in 8th grade we had calculator exams, meaning we had to write down what buttons we would push on the calculator in order to answer a math question.
Middlebury math does not use calculators. Normally, I'd say the instructors are just hardcore math geeks who don't believe in using the calculator as a crutch. But I actually really fault the department for not preparing students to use calculators. Calculators can be incredibly useful in teaching, learning, and understanding the concepts in math. It also helps students get a feel for real world answers instead of leaving everything in terms of pi and logs. Graphing is perhaps the most meaningful function on the calculator because it saves a lot of time in finding out what a graph looks like.
I encourage math teachers to educate themselves about calculators and then use them effectively in the classroom. Calculators do not have to be a crutch. They can be a tool. | 677.169 | 1 |
About this product
Description
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Understanding Maths is a series of books designed to reinforce skills and knowledge at Key Stage 2 of the National Curriculum. Each book focuses on a particular maths topic with explanations and teaching points, exercises, answers section and glossary of terms. | 677.169 | 1 |
Ionic Compound Equations
Tux Math Scrabble encourages kids to construct compound equations and consider multiple abstract possibilities. The latest version (0.7.3) is always available at the official homepage: This script allows you to runĀaLagrange is a function that calculate equations of motion (Lagrange's equations) d/dt(dL/d(dq))- dL/dq=0. It Uses the Lagrangian that is a function that summarizes thedynamics of the system. Symbolic Math Toolbox is required | 677.169 | 1 |
Curricula graphicsEstimated Time: 10 Minutes Objective Upon completion of this lab, the student will have been introduced to the proper way of shutting down Windows and navigating the Windows interface, as well as using Windows Help features. Equipment The following equipment is required for this exercise: • Scenario In order to help a friend out with Windows, the student will need to become more familiar with the Windows Graphical User Interface (GUI). Procedures Knowing how to use the Windows interface will help in nearly any job today.
This tutorial shows how to use Maple both as a calculator with instant access to hundreds of high-level math routines and as a programming language for more demanding tasks. It covers topics such as the basic data types and statements in the Maple language. It explains the differences between numeric computation and symbolic computation and illustrates how both are used in Maple. Extensive "how-to" examples are used throughout the tutorial to show how common types of calculations can be expressed easily in Maple. | 677.169 | 1 |
The Calcumites Are Coming
Students explore population growth. In this algebra lesson, students model population growth and compare ideal population growth with a population whose growth is limited. Students use technology (TI-73) to determine an exponential and logistic regression equation. | 677.169 | 1 |
A significantly revised and improved introduction to a critical aspect of scientific computationMatrix computations lie at the heart of most scientific computational tasks. For any scientist or engineer doing large-scale simulations, an understanding of the topic is essential. Fundamentals of Matrix Computations, Second Edition explains matrix computations and the accompanying theory clearly and in detail, along with useful insights.This Second Edition... more...
This book gives a thorough and entirely self-contained, in-depth
introduction to a specific approach to group theory, in a
large sense of that word. The focus lie on the
relationships which a group may have with other groups, via
"universal properties", a view on that group "from the
outside". This method of categorical algebra, is actually
not limited to the study of groups alone, but applies... more...
Considered a classic by many, A First Course in Abstract Algebra, Seventh Edition is an in-depth introduction to abstract algebra. Focused on groups, rings and fields, this text gives students a firm foundation for more specialized work by emphasizing an understanding of the nature of algebraic structures. Sets and Relations; GROUPS AND SUBGROUPS; Introduction and Examples; Binary Operations; Isomorphic Binary Structures; Groups; Subgroups;... more...
Success in your calculus course starts here! James Stewart's
CALCULUS texts are world-wide best-sellers for a reason: they are
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With CALCULUS, Eighth Edition, Stewart conveys not only the utility
of calculus to help you develop technical competence, but also
gives you an appreciation for the intrinsic beauty of the subject.
His patient examples and built-in learning aids will help you... more...
CRC Standard Curves and Surfaces is a comprehensive illustrated
catalog of curves and surfaces of geometric figures and algebraic,
transcendental, and integral equations used in elementary and
advanced mathematics. More than 800 graphics images are featured.
Based on the successful CRC Handbook of Mathematical Curves and
Surfaces, this new volume retains the easy to use "catalog" format
of the original book. Illustrations are presented in a common... more...
This advanced textbook on linear algebra and geometry covers a
wide range of classical and modern topics. Differing from
existing textbooks in approach, the work illustrates the
many-sided applications and connections of linear algebra with
functional analysis, quantum mechanics and algebraic and
differential geometry. The subjects covered in some detail
include normed linear spaces, functions of linear operators, the
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Algebra & Geometry: An Introduction to University
Mathematics provides a bridge between high school and
undergraduate mathematics courses on algebra and geometry. The
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of methods by presenting important ideas and their historical
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The... more...
Algebraic Sudoku follows the traditional algebra curriculum, while
challenging students minds with fun puzzles that develop logic,
reasoning skills, concentration, and confidence. Each Sudoku puzzle
is like a mini-lesson, with background, discussion, strategy, and
demonstration for solving each problem. After completing the
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them to reason their way through the remaining cells of the... more...
In one exceptional volume, Abstract Algebra covers subject matter typically taught over the course of two or three years and offers a self-contained presentation, detailed definitions, and excellent chapter-matched exercises to smooth the trajectory of learning algebra from zero to one. Field-tested through advance use in the ERASMUS educational project in Europe, this ambitious, comprehensive book includes an original treatment of representation of... more... | 677.169 | 1 |
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Module Focus: Geometry
The focus of the Wednesday morning session for grades 9-10 math was to explore the topics covered in the first two modules of geometry. Congruence is covered in Module 1. The biggest change in geometry with respect to congruence is that two geometric figures are defined to be congruent if there is a sequence of rigid motions that carries one on to the other. Rigid motions are first introduced in grade 8 and teachers might want to take a look at Grade 8 Module 2 and Grade 8 Module 3 to see how the properties of rigid motions were explored. Participants discussed how they currently characterize transformations, and most agreed that they associate transformations with a set of rules and they tend to be very coordinate based. Students now need to develop a deeper understanding of transformations and their purpose without the use of the coordinate plane.
Module 1 starts off with 5 lessons on constructions. Students will be performing the same constructions as in the past, but the focus is on not just the figure being constructed, but the steps behind the construction. Students will need precision with their vocabulary, as they will need to be able to communicate clearly the steps behind the construction for all to understand. Focus is on the construction and instruction.
Topic C in Module 1 covers the transformations and rigid motions studied in 8th grade. The progression of intuitive, to the concrete, to formally defining a transformation is developed. Participants took a look at this progression with the concept of reflection in lesson 14 where students explore what they notice about the line of reflection and perpendicular bisectors. They then tie this exploration back to their work in the opening lessons that dealt with constructing the perpendicular bisectors and angle bisectors. Students are then introduced to the formal definition of reflections.
Topic D introduces the concept of congruence through rigid motion. Lesson 22 is the presentation of the proof by rigid motion for the SAS criteria. Students need to know the properties that are preserved with the transformations that are rigid motions (i.e. distance preserving, angle preserving) and need to be able to communicate these properties while writing proofs that involve the use of rigid motions. Once the congruence criteria (i.e. SAS, SSS, HL) have been proven, they then can be used in proofs for congruence as we saw in lesson 26.
Topic G reviews the content of the modules and reinforces the purpose behind the axiomatic system. A math teacher's story was told: "We have to cover several chapters from the textbook and there are approximately 40 formulas. I may offer you a deal: you will learn just four formulas and I will teach you how to get the rest out of these formulas." The students gladly agreed.
Module 2 focuses on similarity and right triangle trigonometry. Scale drawings are first introduced in grade 7 and teachers might want to take a look at the content covered in Grade 7 Module 1 for gap purposes. Scale drawings are approached in the geometry module with two methods, the ratio and parallel method. Participants had fun with the parallel method and using the set square to generate parallel lines. After scale drawings are explored, students go on to study the properties of dilations which sets the tone for proving the similarity criteria for triangles (AA, SAS and SSS).
The remainder of Module 2 focuses on right triangle trigonometry. Lessons 16, 21 and 25 set the foundation for the trig functions without officially using the language. These lessons explore the internal relationships within and between similar triangles and how the ratios of corresponding sides can be used to find missing lengths. | 677.169 | 1 |
This exam practice book for AS Maths contains detailed advice and tips on how to improve marks and the overall grade. It includes: real exam question; actual students' answers; how to score high marks; key skills to remember; and questions to try.
"synopsis" may belong to another edition of this title.
About the Author:
John Berry is Professor of Mathematics Education at the University of Plymouth and Mathematics Professor in Residence at Well Cathedral School, Somerset. He teaches mathematics at A-level, able and gifted pupils at KS3 and has acted as a consultant for QCA. Ted Graham is the Director fo the Centre for Teaching Mathematics and a Senior Lecturer in the School of Mathematics and Statistics at the University of Plymouth. He has worked as a chief examiner for ten years and is currently a chief examiner in Mechanics for a major examining board. Roger Williamson is a former Senior Lecturer in Statistics at Manchester Metropolitan University. He is Chief Examiner in Statistics and a former Chief Moderator for a major examining group. | 677.169 | 1 |
In order to be
successful it is important to be prepared and organized. You will be expected
to adhere to the following rules and procedures.
Each day please come to class
prepared as follows:
1.
Daily Procedure:
Students are expected to be on time, in correct uniform, and ready to begin
class. Notebooks, homework and calculators
should be on the desk. Students should begin working on the
DO NOW problem which
will be displayed.
2.
Materials:Math textbook, laptop
(closed), pencils, colored pencils/pens for graphing, highlighter, graph paper,
ruler, and a TI-84 plus calculator. Your math
textbook must be covered, (book socks or a sturdy cover is acceptable;
contact covers are not acceptable). Textbooks may remain home since we have
e-editions.
3.
Math notebook:
a
spiral notebook with pockets for
handouts, or a looseleaf binder to keep
all notes, homework and handouts. All classwork must be dated and done in
chronological order.
4.
Homework:
Since Mathematics is a doing subject, daily
homework assignments are essential to learning. Assignments will be
announced during each class and posted on my website which can be accessed from
the IHA home page. Homework is your tool to verify that you understand and have
mastered each of the day's topics. Homework should be completed as follows:
write out each problem from your textbook in its original form and show all
work. Homework must be done in your notebook/binder immediately following the
day's lesson. Due to time constraints, homework may not be reviewed daily. If
there are problems that the majority of the class could not complete, those
problems will be worked out together. However, homework will be checked for
completion periodically and count towards your homework/class participation
grade.
5.
Class Participation:
Class participation is a very important part of learning. Hence it will be part
of your grade. Please ask questions -
it is only in questioning that you will understand all that you need to know.
6.A
good attitude.
Please try and refrain from leaving class unless of course it is an emergency.
This disturbs your classmates and interrupts the learning process. Eating is
not permitted in class at any time. (Water is allowed)
ATTENDANCE
Since each lesson in mathematics
builds on the previous lesson, it is extremely important to attend all classes.
The pacing in honors classes requires that a new lesson be taught each day. If
you miss a class it is difficult to follow the next lesson. Please try not to
be absent. If it is necessary for you to miss a class due to illness or
personal family matter, it is your
responsibility to complete all assignments, including obtaining notes from
classmates. Please check my website for assignments.
ILLNESSES
If you
are absent the day of a test or quiz, you will be expected to take a make-up
test the day you return, preferably afterschool. If there is a
problem, please e-mail me or see me before homeroom to arrange a more convenient
time for both of us.
I
understand that sometimes unfortunately a student may have a serious illness or
a family matter. If you are absent for an extended period of time due to such an
illness, please see me about making up the required work. Depending on the
nature and time of your illness, I will adjust make-up scheduling.
*Remember, missing
notes/assignments/tests/quizzes etc. are your responsibility. It is not up to me
to keep track of your assignments it is up to you to make sure all of your class
requirements have been met each quarter!
Academic Integrity
Any form of
cheating will not be tolerated. Cheating is stealing and dishonest. It is
considered a very serious matter and will be dealt with as such. Any student
caught cheating on a test, assignment or project will receive a zero for that
grade; the incident will be reported to the dean of students and parents will be
called. Any student who knowingly gives work to another student is as dishonest
as the person receiving the work.
Extra Help
Extra help is
usually available after school from 2:00 - 2:45 p.m. Monday- Thursday in room
202. If I am
unavailable for extra help due to professional meetings or personal reasons I
will inform the class as soon as possible. Extra help is given on a first come
first serve basis. However, extra help is not a private tutoring session.
Remember it
is your responsibility to obtain notes and complete all assignments.
On the other hand, please seek extra help immediately when you are confused
about any topic. It is better to clarify concepts at the beginning.
Do not wait until the day before a test to get
help.
Grading Policy:
There will be
approximately two tests and two quizzes each quarter. Each quarter grade will be
obtained by the following:
TESTS 50%
QUIZZES 30%
HOMEWORK/CLASS PARTICIPATION 20%
E-mail:
If you need to contact me, please identify yourself with your full name and
class in the subject line of the e-mail.
Please do not e-mail me with questions about homework. These can be
dealt with in class.
I am looking forward
to a successful year of learning right along with you!MATH
CAN BE FUN! | 677.169 | 1 |
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Boost Your grades with this illustrated quick-study guide. You will use it from college to graduate school and beyond. FREE chapters on Linear Equations, Determinant, and more in the trial versionWorks of Sigmund Freud: Dream Psychology (Palm OS) Works of Sigmund Freud: Dream Psychology (Palm OS) - Sigmund Freud: Dream Psychology, Three Contributions to the Theory of Sex & A Young Girl's Diary - educate yourself with Sigmund Freud's works on Your PDA. Navigate easily to any chapter from Table of Contents or search for the words or phrasesWorks of L. Frank Baum (BlackBerry) Works of L. Frank BaumThe Iliad and The Odyssey by Homer (Palm OS) Full text of The Iliad and The Odyssey. Homer's biography. Historical and geographical background of Ancient Greece. Timeline. Maps. Epic poems' Summaries. Description of key themes and characters | 677.169 | 1 |
PREREQUISITES:
Before studying
calculus, all students should complete four courses of secondary mathematics
designed for college-bound students: courses in which they study algebra,
geometry, trigonometry, analytic geometry, and elementary functions. These
functions include those that are linear, polynomial, rational, exponential,
logarithmic, trigonometric, inverse trigonometric, and piecewise defined. In
particular, before studying calculus, students must be familiar with the
properties of functions, the algebra of functions, and the graphs of functions.
Students must also understand the language of functions (domain and range, odd
and even, periodic, symmetry, zeros, intercepts, and so on) and know the values
of the trigonometric functions of the numbers 0, pi/6, pi/4, pi/3, pi/2, and
their multiples.
DESCRIPTION:
Advanced
Placement (AP) CalculusBC is a college-level mathematics course for
students that have previously demonstrated mastery of' algebra, geometry,
coordinate geometry, trigonometry,
and AP Calculus AB. This course will develop the students'
understanding of the concepts of calculus and provide experience with its
methods and applications. The course will emphasize a multi-representational
approach to calculus, with concepts, results, and problems being expressed
geometrically, numerically, analytically, and verbally. The connections among
these representations also are important. Technology is used regularly to
reinforce the relationships among the multiple representations of functions, to
confirm written work, to implement experimentation, and to assist in
interpreting results.
COURSE
GOALS:
Students should be able to:
work with functions represented in a variety
of ways: graphical, numerical, analytical, or verbal. They should understand
the connections among these representations.
understand the meaning of the derivative in
terms of a rate of change and local linear approximation and they should be
able to use derivatives to solve a variety of problems.
understand the meaning of the definite
integral both as a limit of Riemann sums and as the net accumulation of change
and should be able to use integrals to solve a variety of problems.
understand the relationship between the
derivative and the definite integral as expressed in both parts of the
Fundamental Theorem of Calculus.
communicate mathematics both orally and in
well-written sentences and should be able to explain solutions to problems.
model a written description of a physical
situation with a function, a differential equation, or an integral.
use technology to help solve problems, experiment,
interpret results, and verify conclusions.
determine the reasonableness of solutions,
including sign, size, relative accuracy, and units of measurement.
develop an appreciation of calculus as a
coherent body of knowledge and as a human accomplishment.
TOPICAL OUTLINE:
The topic outline for Calculus BC includes all Calculus AB topics. Additional
topics are found in paragraphs that are marked with a plus sign (+) or an
asterisk (*). The additional topics can be taught anywhere in the course. Some
topics will naturally fit immediately after their Calculus AB counterparts.
Other topics may fit best after the completion of the Calculus AB topic outline.
Although the examination is based on the topics listed here, the course may be
enriched with additional topics.
I. Functions, Graphs, and Limits
A. Analysis of Graphs
With the aid of technology, graphs of functions are often
easy to produce. The emphasis is on the interplay between the geometric and
analytic information and on the use of calculus both to predict and to explain
the observed local and global behavior of a function.
F. Computation of Derivatives
Basic rules for the derivative of sums, products, and
quotients of functions.
Chain rule and implicit differentiation.
+ Derivatives of parametric, polar, and vector
functions.
III. Integrals
A. Interpretations and Properties of Definite
Integrals
Definite integral as a limit of Riemann sums.
Definite integral of the rate of change of a quantity
over an interval interpreted as the change of the quantity over the
interval:
Basic properties of definite integrals. (Examples
include additivity and linearity.)
B. *Applications of Integrals
Appropriate integrals are used in a variety of
applications to model physical, social, or economic situations. Although only a
sampling of applications can be included in any specific course, students should
be able to adapt their knowledge and techniques to solve other similar
application problems. Whatever applications are chosen, the emphasis is on using
the method of setting up an approximating Riemann sum and representing its limit
as a definite integral. To provide a common foundation, specific applications
should include finding the area of a region (including a region bounded by polar
curves), the volume of a solid with known cross sections, the average value of a
function, the distance traveled by a particle along a line, and for BC only the
length of a curve (including a curve given in parametric form).
C. Fundamental Theorem of Calculus
Use of the Fundamental Theorem to evaluate definite
integrals.
Use of the Fundamental Theorem to represent a
particular antiderivative, and the analytical and graphical analysis of
functions so defined.
D. Techniques of Antidifferentiation
Antiderivatives following directly from derivatives
of basic functions.
E. Applications of Antidifferentiation
Finding specific antiderivatives using initial
conditions, including applications to motion along a line.
Solving separable differential equations and using
them in modeling. In particular, studying the equation y ' = ky
and exponential growth.
+ Solving logistic differential equations and using
them in modeling.
F. Numerical Approximations to Definite Integrals
Use of Riemann sums (using left, right, and midpoint
evaluation points) and trapezoidal sums to approximate definite integrals of
functions represented algebraically, graphically, and by tables of values.
IV. *Polynomial
Approximations and Series
A. *Concept of Series
A series is defined as a sequence of partial sums, and
convergence is defined in terms of the limit of the sequence of partial sums.
Technology can be used to explore convergence or divergence.
B. *Series of constants
+ Motivating examples, including decimal expansion.
+ Geometric series with applications.
+ The harmonic series.
+ Alternating series with error bound.
+ Terms of series as areas of rectangles and their
relationship to improper integrals, including the integral test and its use
in testing the convergence of p-series.
+ Formal manipulation of Taylor series and shortcuts
to computing Taylor series, including substitution, differentiation,
antidifferentiation, and the formation of new series from known series.
+ Functions defined by power series.
+ Radius and interval of convergence of power series.
+ Lagrange error bound for Taylor polynomials.
PACING GUIDE:
Chapter 1 Prerequisites for Calculus
4 days
Chapter 2 Limits and Continuity
6 days
Chapter 3 Derivatives
22 days
Chapter 4 Applications of Derivatives
14 days
Chapter 5 The Definite Integral
16 days
Chapter 6 Differential Equations and Mathematical
Modeling
22 days
Chapter 7 Applications of Definite Integrals
15 days
Chapter 8 L'Hôpital's Rule, Improper Integrals,
and Partial Fractions
7 days
Chapter 9 Infinite Series
23 days
Chapter 10 Parametric, Vector, and Polar Functions
16 days
This timeline gives approximately 15 days to "review"
the course before the exam. These times are approximate and subject
to change.
THE EXAM:
Put your knowledge to the test: The AP Calculus BC Exam
assesses your mastery of CalculusBC concepts and techniques. It
also gives you the chance to earn college credit while in high school.
About
the Exam
The AP Calculus BC
Exam is 3 hours and 15 minutes. The 105-minute, 45-question multiple-choice
section tests your proficiency on a wide variety of topics. The 90-minute,
six-problem free-response section gives you the chance to demonstrate your
ability to solve problems using an extended chain of reasoning.
Section I: Multiple-Choice
The multiple-choice
section of the exam has two parts. For Part A, you'll have 55 minutes to
complete 28 questions without a calculator. For Part B, you'll have 50 minutes
to answer 17 questions using a graphing calculator. For more information, see
the calculator policy for the AP Calculus Exams.
Unlike other
multiple-choice tests, random guessing can hurt your final score. While you
don't lose anything for leaving a question blank, one quarter of a point is
subtracted for each incorrect answer on the test. But if you have some
knowledge of the question and can eliminate one or more answers, it's usually
to your advantage to choose what you believe is the best answer from the
remaining choices.
Section II: Free-Response
The free-response
section tests your ability to solve problems using an extended chain of
reasoning. You'll have 45 minutes for each of the two parts of the
free-response section. In Part A, you'll answer three questions using a
graphing calculator. In Part B, you'll answer three questions without a
calculator. During the second timed portion of the free-response section (Part
B), you are permitted to continue work on problems in Part A, but you are not
permitted to use a calculator during this time. For more information, see the
calculator policy for the AP Calculus Exams.
Scoring the Exam
The multiple-choice
and free-response sections each account for one-half of your final exam grade. Since
the exams are designed for full coverage of the subject matter, it is not
expected that all students will be able to answer all the questions.
Exam Grades
AP Grade Reports are
sent in July to the college you designated on your answer sheet, to you, and to
your high school. Each report is cumulative and includes grades for all the AP
Exams you have ever taken, unless you have requested that one or more grades be
withheld from a college or canceled. | 677.169 | 1 |
Equation Calculator is a scientific calculator which does symbolic and algebraic manipulation, algebra and calculus as well as numeric computation. You can define variables and functions, evaluate symbolic derivatives, numeric integrals and matrix operations. | 677.169 | 1 |
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Summary
This dynamic new edition of this proven series adds cutting edge print and media resources. An emphasis on the practical applications of algebra motivates learners and encourages them to see algebra as an important part of their daily lives. The reader-friendly writing style uses short, clear sentences and easy-to-understand language, and the outstanding pedagogical program makes the material easy to follow and comprehend.KEY TOPICSChapter topics cover basic concepts; equations and inequalities; graphs and functions; systems of equations and inequalities; polynomials and polynomial functions; rational expressions and equations; roots, radicals, and complex numbers; quadratic functions; exponential and logarithmic functions; conic sections; and sequences, series and the binomial theorem.For the study of Algebra. | 677.169 | 1 |
About Level 2 Mathematics
Nayland College Courses
2A
Maths
This academic course and is predominantly algebra based but will also include calculus and graphs as well as some statistics and probability. This course is required to prepare students for the academic mathematics courses at Level 3.
2A Statsitics
This academic course focuses on the acquisition of practical statistical knowledge and skills. Report writing is required so a good level of English is beneficial. Students who are considering subjects such as Biology, Psychology or Geography will benefit from choosing this course.
2S Maths
This course focuses on the development of algebraic and statistical skills and applications. Topics studied include algebra, trigonometry, geometry, statistics and probability. It may lead to mathematics at Level 3. | 677.169 | 1 |
calculators.
This tool also supports the entry of complex algebraic equations with fractions, square roots, exponents, etc. for technical and scientific | 677.169 | 1 |
Courses - Maths
Our maths courses are designed to improve overall confidence, increase knowledge
and develop a range of skill sets to tackle problems appropriately. Each course
is based on ability level within school age groups. We cover all the year groups,
SATS stages and examining boards for GCSE and AS/2 students. The courses cover the
linear and modular exams on offer in schools.
The maths courses cover the four key areas of maths:
Number - recognition of the number system using mathematical operations
for all types of numbers (e.g. natural numbers, fractions, irrational numbers etc.). | 677.169 | 1 |
Excursions in Modern Mathematics introduces non-math majors to the power of math by exploring applications like social choice and management science, showing that math is more than a set of formulas. Ideal for an applied liberal arts math course, Tannenbaum's text is known for its clear, accessible writing style and its unique exercise sets that build in complexity from basic to more challenging. The Eighth Edition offers more real data and applications to connect with today's students, expanded coverage of applications like growth, and revised exercise sets. MyMathLab exercise sets are expanded and the new Ready To Go MyMathLab course makes course set-up even easier
Key Features
Author: Peter Tannenbaum
Publisher: Pearson
Copyright: 2014
Language: English
ISBN13 : 9781447964582
Number Of Pages: N/A
Edition: 8
Specifications of Excursions In Modern Mathematics, Plus MyMathLab | 677.169 | 1 |
Algebra Color My Math Solving Systems of Equations
Be sure that you have an application to open
this file type before downloading and/or purchasing.
11 MB|1 page
Product Description
In this activity, students will practice solving systems of equations.
There are 15 equations that students must solve. Students must be able to solve a system of equations as well as be familiar with when a system has an infinite number of solutions or no solution. As they solve each system, they will find a specific pattern that they need to use to color the boxes at the top of the page. | 677.169 | 1 |
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I designed this lesson (the 1st of 10) as part of a unit that introduces arithmetic and geometric sequences. This lesson can be purchased as a complete bundled unit at a discounted price under the listing Arithmetic and Geometric Sequences Complete Bundled Unit.
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This bundle includes the complete lessons and handouts for Probability, parts I thru IV, in their entirety. [NOTE: This Bundled Zip Folder can be EASILY opened with no additional software. Standard MS Word files are inside.]Equation editor software within MS Word is used, and clear geometric figures are included for a clear, professional look. Word is needed to print. MathType is needed to edit equations.
All parts include lesson handouts, worksheets, and complete keys containing answers for all problems.
Part I is simple (one event) probability and includes a lesson handout with definitions of probability in general and also theoretical vs. experimental probability, and complementary events. Also included is a worksheet (with key) with a variety of questions involving spinners, dice, letter tiles, and real-life situations.
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Library of Congress Cataloging-in-Publication Data Bobrow, Jerry. Mastering the SAT : math / by Jerry Bobrow. p. cm. Includes bibliographical references and index. ISBN-13: 978-0-470-03660-0 (alk. paper) ISBN-10: 0-470-03660-5 (alk. paper) 1. Mathematics—Examinations—Study guides. 2. Mathematics—Examinations, questions, etc. 3. SAT (Educational test)—Study guides. 4. Universities and colleges—United States—Entrance examinations—Study guides. 5. Achievement tests—Study guides. I. Title. QA43.B649 2007 510.76—dc22 2006029556 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 1O/SR/RR/QW/IN No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, 978-750-8400, fax 978-646-8600, or on the web at Requests to the Publisher for permission should be addressed to the Legal Department, Wiley Publishing, Inc., 10475 Crosspoint Blvd., Indianapolis, IN 46256, 317-572-3447, fax 317-572-4355, or online at THE PUBLISHER AND THE AUTHOR MAKE NO REPRESENTATIONS OR WARRANTIES WITH RESPECT TO THE ACCURACY OR COMPLETENESS OF THE CONTENTS OF THIS WORK AND SPECIFICALLY DISCLAIM ALL WARRANTIES, INCLUDING WITHOUT LIMITATION WARRANTIES OF FITNESS FOR A PARTICULAR PURPOSE. NO WARRANTY MAY BE CREATED OR EXTENDED BY SALES OR PROMOTIONAL MATERIALS. THE ADVICE AND STRATEGIES CONTAINED HEREIN MAY NOT BE SUITABLE FOR EVERY SITUATION. THIS WORK IS SOLD WITH THE UNDERSTANDING THAT THE PUBLISHER IS NOT ENGAGED IN RENDERING LEGAL, ACCOUNTING, OR OTHER PROFESSIONAL SERVICES. IF PROFESSIONAL ASSISTANCE IS REQUIRED, THE SERVICES OF A COMPETENT PROFESSIONAL PERSON SHOULD BE SOUGHT. NEITHER THE PUBLISHER NOR THE AUTHOR SHALL BE LIABLE FOR DAMAGES ARISING HEREFROM. THE FACT THAT AN ORGANIZATION OR WEBSITE IS REFERRED TO IN THIS WORK AS A CITATION AND/OR A POTENTIAL SOURCE OF FURTHER INFORMATION DOES NOT MEAN THAT THE AUTHOR OR THE PUBLISHER ENDORSES THE INFORMATION THE ORGANIZATION OR WEBSITE MAY PROVIDE OR RECOMMENDATIONS IT MAY MAKE. FURTHER, READERS SHOULD BE AWARE THAT INTERNET WEBSITES LISTED IN THIS WORK MAY HAVE CHANGED OR DISAPPEARED BETWEEN WHEN THIS WORK WAS WRITTEN AND WHEN IT IS READ. Trademarks: Wiley, the Wiley Publishing logo, CliffsNotes, the CliffsNotes logo, Cliffs, CliffsAP, CliffsComplete, CliffsQuickReview, CliffsStudySolver, CliffsTestPrep, CliffsNote-a-Day, cliffsnotes.com, and all related trademarks, logos, and trade dress are trademarks or registered trademarks of John Wiley & Sons, Inc. and/or its affiliates. SAT is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product. All other trademarks are the property of their respective owners. Wiley Publishing, Inc. is not associated with any product or vendor mentioned in this book. For general information on our other products and services or to obtain technical support, please contact our Customer Care Department within the U.S. at 800-762-2974, outside the U.S. at 317-572-3993, or fax 317-572-4002. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print may not be available in electronic books. For more information about Wiley products, please visit our web site at
practice problems for each topic area (easy. and evaluated and are presently used in SAT I test preparation programs at many leading colleges and universities. compact. It is direct. In keeping with the fine tradition of Cliffs Notes. and difficult problems). The testing strategies. average. and score range approximators are included following the practice tests to give you a thorough understanding of the new SAT I math sections. and thorough. techniques. you can't afford to take a chance. The practice problems for each topic area and practice tests have complete answers and in-depth explanations. Prepare with the best! Because better scores result from thorough preparation. Mastering the SAT: Math has been designed by leading experts in the field of test preparation to be the most comprehensive guide to give you that extra boost in math. and three full-length practice math tests. math area reviews. not substitute for. Analysis charts. Mastering the SAT: Math was written to give you the edge in doing your best by giving you maximum benefit in a reasonable amount of time and is meant to augment. formal or informal learning throughout junior high and high school. and materials have been researched. precise. your study time must be used most effectively. but your math scores can still really be improved with proper preparation! And because of these facts. diagnostic review tests. tested. Be prepared! Follow the SAT I Study Guide Checklist in this book and study regularly. Don't take a chance. this guide is written for the student. This guide combines introductory analysis sections with sample problems.
. You'll get the best test preparation possible. easy to use.Preface
The SAT I has changed.
average. Again. Strictly observing time allotments. take Math Practice Test I. Continue your math review on page 91 with the Algebra Diagnostic Test. and difficult review problems. While referring to each item of Math Practice Test III. ❏ 11. ❏ 26. Carefully read Part I: Analysis and Strategies. ❏ 14. 2. 6. beginning on page 236. Take a very short break after each hour of testing. ❏ 23. 5. and difficult problems). Review your math skills as necessary.org. take Math Practice Test III.
❏ 10. beginning on page 9. 7. While referring to each item of Math Practice Test II. Analyze your Math Practice Test II answers by filling out the analysis charts on page 252. Strictly observing time allotments. Get information online at www. ❏ 24. average. page 215. ❏ 22.SAT I Study Guide Checklist
❏ ❏ ❏ ❏ ❏ ❏ ❏ ❏ ❏ 1. ❏ 18. selectively review Part I: Analysis and Strategies. ❏ 25. ❏ 21. Next. and any other basic skills or exam areas you feel are necessary. beginning on page 199. page 3. beginning on page 11. ❏ 13. Selectively review some basic skills as necessary. Review the answers and explanations for each question on Math Practice Test I. Learn the techniques of the Successful Overall Approaches. study all of the explanations that begin on page 288. Check your answers and use the Score Range Approximator (page 303) to get a very general score range. 9. Read the Arithmetic review as needed. 8. to see whether you applied some of the strategies. beginning on page 272. ❏ 20. and difficult problems). beginning on page 216. Become familiar with the test format. Strictly observing time allotments. Review or reread Part I: Analysis and Strategies. take Section 2 and then check your answers. Next work the Numbers and Operations and Data Analysis.
. Continue your math review on page 145 with the Geometry Diagnostic Test. average. and so on). pages 5–7. Read and review Geometry and Measurement as needed. ❏ 19. take Math Practice Test II. and Probability SAT review problems at the end of the Arithmetic Review. study all of the explanations that begin on page 253. Read the Algebra Review as needed. ❏ 17. Read the New SAT I information bulletin. Next work the Algebra and Functions SAT review problems at the end of the Algebra Review (easy.collegeboard. 4. page 1. Analyze your Math Practice Test I answers by filling out the analysis charts. Analyze your Math Practice Test III answers by filling out the analysis charts on page 287. 3. beginning on page 9. Statistics. Take a very short break after each hour of testing. Note there are easy. Check your answers and use the Score Range Approximator (page 303) to get a very general score range. ❏ 12. Familiarize yourself with the answers to Questions Commonly Asked about the New SAT I. Start your math review on page 45 with the Arithmetic Diagnostic Test. ❏ 15. ❏ 16. section by section (take Section 1 and then check your answers. work the Geometry and Measurement SAT review problems at the end of the Geometry Review (easy.
The writing ability sections test your ability to write a clear. and the number of questions may vary. One 25-minute section is a pretest. and to improve paragraphs. math.
A Close Look at the New SAT I
Question Type Critical Reading Sentence Completions Passage-Based Questions Total Critical Reading Questions Mathematics Multiple Choice Grid-Ins Total Mathematics Questions Writing Multiple Choice Improving Sentences Identifying Sentence Errors Improving Paragraphs Total Writing Multiple Choice Essay Question 25 18 6 49 One Question 44 10 54 19 48 67 Approximate Number of Questions
The problems in the math sections (multiple-choice and grid-ins) and the sentence completions section of the new SAT I are slightly graduated in difficulty. to comprehend what you read. or writing multiple-choice section and can appear anywhere on your exam. the question types within a section. The math sections test your ability to solve problems using mathematical reasoning and your skills in arithmetic.
2
. the writing essay (25 minutes). and the multiple-choice sections (one 25-minute section and one 10-minute section) actually count toward your new SAT I score. with some grid-in questions. The critical reading sections test your ability to read critically. Only three of the critical reading sections (two 25-minute sections and one 20-minute section). It does not have to be Section 9.Mastering the SAT Math
Note: The order in which the sections appear. so getting an easy question right is worth the same as getting a difficult question right. three of the math sections (two 25-minute sections and one 20-minute section). algebra I and II. and there may be many forms of the test. and an essay. or experimental. and to understand words in context. You should work all of the sections as though they count toward your score. section that does not count toward your score. precise essay and to find grammar and usage errors. to correct sentence errors.
General Description
The new SAT I is used along with your high school record and other information to assess your competence for college work. Many students make simple mistakes because they rush through the easy questions to get to the difficult ones. and geometry. Keep in mind that each question within a section is of equal value. The test lasts 3 hours and 45 minutes and consists of mostly multiple-choice type questions. The pretest or experimental section can be a critical reading.
fax. but no points are subtracted if you leave the answer blank. a good eraser.400 Q: WILL THE NEW SAT I BE MORE DIFFICULT? A: No.
Questions Commonly Asked about the New SAT I
Q: WHO ADMINISTERS THE NEW SAT I? A: The new SAT I is part of the entire Admissions Testing Program (ATP). a watch. Critical reading sections now also include paragraph-length passages. formerly called Verbal Reasoning. which is administered by the College Entrance Examination Board in conjunction with Educational Testing Service of Princeton. or you may write. or e-mail a cancellation to College Board ATP. Quantitative comparison math questions are no longer on the exam. The new SAT I assesses general critical reading. Q: SHOULD I GUESS ON THE NEW SAT I? A: If you can eliminate one or more of the multiple-choice answers to a question. and an approved calculator. and editing abilities that you have developed over your lifetime. Q: HOW IS THE NEW SAT I SCORED? A: The scoring will be as follows: Critical Reading: 200–800 Mathematics: 200–800 Writing: 200–800 (subscores essay 2–12. Q: WHAT MATERIALS MAY I BRING TO THE NEW SAT I? A: Bring your registration form. You may not bring scratch paper or books. mathematical reasoning. 2 pencils. positive identification. It is not uncommon for students to take the test more than once. You may cancel your score on the day of the test by telling the test center supervisor.Introduction
Special Notes for the New SAT I
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Verbal analogies are no longer on the exam. a fraction of a point is subtracted for every wrong answer. it is to your advantage to guess. On the grid-in questions. including some algebra II problems. scores up to five years old were also included on the report. Q: IS THERE A DIFFERENCE BETWEEN THE NEW SAT I AND THE SAT II? A: Yes. The first part of the test will always be an essay. writing. along with any completed test scores. The new SAT has been designed so that a student who could score a 500 on the math section of the old SAT I could score a 500 on the math section of the New SAT I. Your score report will record your cancellation. multiple choice 20–80) Total possible score: 600–2. On past score reports. To discourage wild guessing.
3
. See specific instructions for canceling your score in the Student Bulletin. The SAT II measures your proficiency in specific subject areas. MAY I CANCEL MY SCORE? A: Yes. You may do your figuring in the margins of the test booklet or in the space provided. Eliminating one or more answers increases your chance of choosing the right answer. there is no penalty for filling in a wrong answer. The SAT II tests how well you have mastered a variety of high school subjects. Writing multiple-choice sections have been added. Q: IF NECESSARY. three or four sharpened no. New Jersey. This is the same for the Critical Reading. Q: CAN I TAKE THE NEW SAT I MORE THAN ONCE? A: Yes. Math sections have been enhanced.
You can also register online at www. (408) 452-1400. 2099 Gateway Place.org.Mastering the SAT Math
Q: HOW SHOULD I PREPARE FOR THE NEW TEST? A: Understanding and practicing test-taking strategies helps a great deal. Fax (850) 224-3077. Fax (408) 453-7396. San Jose.org.) You will be admitted only if space remains after preregistered students have been seated. California 95814. Middle States: 3440 Market Street. (850) 222-7999. Fax (847) 866-9280. Fax (512) 891-8404. is attached to the Student Bulletin. Evanston. Fax (781) 890-0693. or HighPoint Center. (An additional fee is required.collegeboard. and June. May. The College Board offers additional practice online. January. (512) 891-8400. 100 Crescent Centre Parkway. Illinois 60201-4805. Georgia 30084-7039. (916) 444-6262. Suite 480. Suite 1001. California 95110-1017. Fax (518) 472-1544. especially on the critical reading sections. The SAT I is administered at hundreds of schools in and out of the United States. Massachusetts 02451-1982. Suite 1200. Q: IS WALK-IN REGISTRATION PROVIDED? A: Yes. Suite 200. you can check online at www. Suite 900. Reviewing the writing process and practicing timed essay writing will also be helpful. plus the appropriate fees. Q: CAN I GET MORE INFORMATION? A: Yes. Some special administrations are given in limited locations. you may attempt to register on the day of the test. Waltham. and a review of basic grammar and usage will be helpful on the writing sections. If you are unable to meet regular registration deadlines. Suite 410. 470 Totten Pond Road. Tucker. November. Tallahassee. Fax (916) 444-2868. Q: WHERE IS THE SAT I ADMINISTERED? A: Your local college testing or placement office will have information about local administrations. (847) 866-1700. Sacramento. You should register about six weeks prior to the exam date. Suite 340. Pennsylvania 19104-3338. Florida 32301-7732. If you require information that is not available in this book. 4330 South MoPac Expressway. March. December. 106 East College Avenue. You can also write or call one of these College Board regional offices. Albany. or Capitol Place. on a limited basis. Mailing in these forms. Philadelphia. 915 L Street. (518) 472-1515. Fax (215) 387-5805. Texas 78735. complete with return envelope.collegeboard. (215) 387-7600. Q: HOW OFTEN ARE THE TESTS ADMINISTERED? A: The new SAT I is usually scheduled to be administered nationwide seven times during the school year. or 126 South Swan Street. in October. (781) 890-9150. completes the registration process. Q: HOW AND WHEN SHOULD I REGISTER? A: A registration packet. Austin.
Midwest: New England: South:
Southwest: West:
4
. Subject-matter review is particularly useful for the math section. (770) 908-9737. New York 12210-1715. 1560 Sherman Avenue. Fax (770) 934-4885. ask for the Student Bulletin.
make sure to mark it out in your question booklet as follows:
(A) ? (B) (C) (D) ? (E)
Notice that some choices are marked with question marks. + 5. + 2. Answer easy questions immediately. The "Plus-Minus" System
Many who take the New SAT I won't get their best possible score because they spend too much time on difficult questions. signifying that they may be possible answers. what is the value of y in terms of x?
5
. For example. Place a "+" next to any problem that seems solvable but is too time-consuming. Because every question within each section is worth the same amount. This technique helps you avoid reconsidering those choices you have already eliminated and helps you narrow down your possible answers. you suddenly realize how to solve it). go back and work your "+" problems. but rather the value of x + 4.Introduction
Taking the New SAT I: Successful Overall Approaches for Multiple-Choice Questions
I. use the following system. Act quickly. try your "–" problems (sometimes when you come back to a problem that seemed impossible. The "Avoiding Misreads" Method
Sometimes a question may have different answers depending upon what is asked. If 3x + x = 20. what is the value of x + 4? Notice that this question doesn't ask for the value of x. The Elimination Strategy
Take advantage of being allowed to mark in your testing booklet. Or If 8y + 3x = 14. 3.
A B C D E A B C D E A B C D E A B C D E A B C D E
Make sure to erase your "+" and "–" marks before your time is up.
II. Place a "–" next to any problem that seems impossible." After working all the problems you can do immediately. 2. The scoring machine may count extraneous marks as wrong answers. – 4. Don't let this happen to you. leaving insufficient time to answer the easy questions. These marks in your testing booklet do not need to be erased. If you finish them. don't waste time deciding whether a problem is a "+" or a "–. 3. marking on your answer sheet: 1. As you eliminate an answer choice from consideration. Your answer sheet should look something like this after you finish working your easy questions:
1.
III.
answering it incorrectly). E. what is the value of y in terms of x? All of the following statements are true EXCEPT . once again. it must be a counting number. The Multiple-Multiple-Choice Technique
Some math and verbal questions use a "multiple-multiple-choice" format. II. And. II. therefore. or 3. do you have to find x or x + 4? Are you looking for what is true or the exception to what is true? To help you avoid misreads.Mastering the SAT Math
The question may instead have asked. and so on. then which of the following must be true? I. D. C. You should cross out B and C on your question booklet. is always true. Statement II is incorrect. B. once you understand "multiple-multiple-choice" problem types and technique. these questions appear more confusing and more difficult than normal five-choice (A. x>0 x=0 x<1
6
. statement I. . B. next to II. because they do not contain true statement I.
Notice that the words EXCEPT and NOT change these questions significantly. III. or E) must contain true statement I. D. Note that possible values of x could be 1. . these circles in your question booklet do not have to be erased. x cannot equal zero. . Actually. III. what is the value of x + 4? If 8y + 3x = 14. This eliminates B and C as possible correct answer choices. So next to I on your question booklet. Therefore. mark the questions in your test booklet in this way: If 3x + x = 20. T I. x>0 x=0 x<1
Now realize that the correct final answer choice (A. If x is positive. C.
IV. D. x>0 x=0 x<1 I only II only III only I and II only I and III only
Because x is a positive integer. T F I. III. E) multiple-choice problems. they are often easier than a comparable standard multiple-choice question. you should place an F for false. place a T for true. To avoid "misreading" a question (and. B. A. If x is a positive integer. II. For example. or 2. C. simply circle what you must answer in the question. At first glance. "What is the value of x in terms of y?" Or All of the following statements are true EXCEPT . Thus. or 4. For example. . x > 0.
Therefore. This technique often saves some precious time and allows you to take a better educated guess should you not be able to complete all parts (I. II. Know the directions. Only A and E are left as possible correct answers. Know when to skip a question. the "Avoiding Misreads" Method. ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑
7
. you should cross out D. Remember that an average score is about 50 percent right. Finally. I only. III) of a multiple-multiple-choice question. you realize that statement III is also false.
A Summary of General Strategies
Set a goal. Guess only if you can eliminate one or more answers. as x must be 1 or greater. So you place an F next to III. and the Multiple-Multiple-Choice Technique.Introduction
Knowing that II is false allows you to eliminate any answer choices that contain false statement II. Go into each section looking for the questions you can do and should get right. Don't get stuck on any one question. ❑ Remember to erase any extra marks on your answer sheet. Practice using the "Plus-Minus" System. Be careful. thus eliminating Choice E and leaving A. Be sure to mark your answers in the right place. because it contains false statement II. Don't make simple mistakes by rushing through the easy questions in math to get to the difficult ones. Watch out for careless mistakes. Don't be afraid to fill in your answer or guess on grid-ins. the Elimination Strategy.
.
PART I
ANALYS I S AN D STR ATE G I E S
.
.
or calculator with a typewriter-type keypad or paper tape. each question can be answered without a calculator—in some instances. Make sure that your calculator has new. you could have an additional 25-minute Math section. Don't bring a calculator that requires an outlet or any other external power source. and the last few multiple-choice questions are more difficult before you start the grid-ins.
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. Make sure that you are familiar with the use of your calculator. If a section has two types of questions. You should be familiar with this information. Before doing an operation. You may use an approved calculator on the SAT I. Don't bring a pocket organizer. laptop computer. Check for a shortcut to any problem that seems to involve much computation. Practice using your calculator on some of the problems to see when and where it will be helpful. the three sections total about 52 to 56 math questions that count toward your score. usually each type starts with easier problems.
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■
Be careful that you
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Don't rush out and buy a sophisticated calculator for the test. the grid-ins start with easy questions and move toward the more difficult ones at the end. fresh batteries and is in good working order. You may want to check each number as you key it in. you're probably doing something wrong. because you can't borrow one during the exam. These three sections generate a scaled math score that ranges from 200 to 800. at this time. the easiest questions are basically at the beginning and the more difficult ones at the end.
Using Your Calculator
The new SAT I allows the use of approved calculators. You will be given reference information preceding each Mathematics section. Bring a calculator even if you don't think you'll use it. But use your calculator if it will be time effective. Two Mathematics sections are 25 minutes in length and one math section is 20 minutes in length. Even though no question will require the use of a calculator—that is. Since one section of the test is experimental (although you don't know which one). check the number that you keyed on the display to make sure that you keyed in the right number.Introduction to the Mathematics Section
The Mathematics sections of the SAT consist of two basic types of questions: regular multiple-choice questions and student-produced responses also known as grid-ins. Although the order of the sections and the number of questions may change. Don't bring a calculator that you're unfamiliar with. For example. using a calculator will save you valuable time. handheld minicomputer. About 50% right should generate an average score. That is. and the College Board (the people who sponsor the exam) recommends that each test taker take a calculator to the test. If there appears to be too much computation or the problem seems impossible without the calculator. Bring a calculator with which you are familiar. You should
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Bring your own calculator. The Mathematics sections are slightly graduated in difficulty. a section starts with easy multiple-choice questions. Don't bring a calculator that makes noise.
As you approach a problem. Learn to use a calculator efficiently by practicing.. Don't try to use a calculator on every problem. TI-92 Plus. You must understand the problem first.g..g. intersection. rate. factors and multiples. PDA's. consecutive numbers.Part I: Analysis and Strategies
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Don't try to share a calculator. interest.. first focus on how to solve that problem and then decide whether the calculator will be helpful. time. a calculator can save you time on some problems. elements)
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Algebra and Functions
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Operations with signed numbers Substitution for variables Absolute value Working with algebraic expressions Manipulating integer and rational exponents Solving rational equations and inequalities Working with linear functions—graphs and equations Solving radical equations
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. and division to problem solving Arithmetic mean (average). odd and even numbers. distance. fractions. Also remember that a calculator will not solve a problem for you. Palm.
Basic Skills and Concepts That You Should Know
Number and Operations
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Operations with fractions Applying addition. subtraction. multiplication. prime numbers. price per item Number line: order. solving for: percents.
Following is the Calculator Policy for the New SAT I as given by the College Board: "The following are not permitted:
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Powerbooks and portable/handheld computers Electronic writing pads or pen-input/stylus-driven (e. mode. but also remember that each problem can be solved without a calculator. Don't become dependent on your calculator. typewriter) keyboards (e. divisibility Word problems.e. and median Ratio and proportion Number properties: positive and negative integers. Casio ClassPad 300) Pocket organizers Models with QWERTY (i. averages. Voyage 200) Models with paper tapes Models that make noise or 'talk' Models that require an electrical outlet Cell phone calculators"
Take advantage of using a calculator on the test. Remember. betweenness Sequences involving exponential growth Sets (union.
not just x. 2 8 9 10 46
You should first circle or underline 4x + 1 because this is what you are solving for. Make sure that you are answering the right question. A list of data that may be used for reference is included. Then select the one correct answer of the five choices given and mark the corresponding circle on your answer sheet.
Suggested Approach with Samples
Circle or Underline
Take advantage of being allowed to mark on the test booklet by always underlining or circling what you are looking for. The most common mistake is to solve for x. You should also notice that most of the other choices would all be possible answers if you made common or simple mistakes. or 9. then substituting into 4x + 1 gives 4(2) + 1. All scratch work is to be done in the test booklet. This will ensure that you are answering the right question. Some problems may be accompanied by figures or diagrams.
Samples
1.
Directions
Solve each problem in this section by using the information given and your own mathematical calculations. there is never more than one right answer. although other answers may be close. C. You are looking for the one correct answer. and problem-solving skills.
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.Part I: Analysis and Strategies
Basic Skills Necessary
The basic skills necessary to do well on this section include high school algebra I and II and intuitive or informal geometry. But remember. Use the available space on the page for your scratch work. The correct answer is C. D. B. These figures are drawn as accurately as possible EXCEPT when it is stated in a specific problem that a figure is not drawn to scale. therefore. then 4x + 1 = A. and mistakenly choose A as your answer.
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Analysis of Directions
1. all figures and diagrams lie in a plane. E.
Notes
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All numbers used are real numbers. insights. No calculus is necessary. The figures and diagrams are meant to provide information useful in solving the problem or problems. Solving for x leaves x = 2. which is 2. you are solving for 4x + 1. Calculators may be used. 2. Logical insight into problem-solving situations is also necessary. If x + 8 = 10. Unless otherwise stated. get used to doing this because no scratch paper is allowed into the testing area.
" Now pulling out information gives you the following. . B. C. E. E. 4 8 12 24 32
First circle or underline x4 – y4. If x2 – y2 = 8 and x2 + y2 = 4. Now factor this difference of two squares. x4 – y4 = (x2 – y2)(x2+ y2) Next substitute in 8 for x2 – y2 and 4 for x2 + y2 = (8)(4) = 32 The correct answer is E. B. . then which of the following is a possible number of students in the drama class? A. D. If the ratio of boys to girls in a drama class is 3 to 2. D. What is the cost of the coat? A. you gain additional insight into the problem. b:g=3:2 Since the ratio of boys to girls is 3 : 2. 25. C. 10. D. . B. 16 18 20 24 28
You should first circle or underline "possible number of students.
Samples
1. E. Tom purchased a hat and coat for a total of $125. $25 $50 $75 $100 $125
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. C. The multiples of 5 are 5. then the possible total number of students in the class must be a multiple of 3 + 2 (boys plus girls). The correct answer is C. or 5. 20. . The coat costs $25 more than the hat.
Pull Out Information
"Pulling" information out of the word problem structure can often give you a better look at what you are working with. therefore.Introduction to the Mathematics Section
2. 15. Only Choice C is a multiple of 5. then what is the value of x4 – y 4 ? A. 2.
40 and 5 is . B. Or you could use your calculator and quickly find the decimals.Part I: Analysis and Strategies
The key words here are cost of the coat. what are the possible values of x? A. then x + 25 = 75.52 B. Notice that $50 is one of the answer choices.80 You should first underline or circle between 2 and 5 ? If you know that 2 is . x = hat x + $25 = coat (cost $25 more than the hat) Together they cost $125. Since . (x + 25) + x = 125 2x + 25 = 125 2x = 100 x = 50 But this is the cost of the hat.65 D. To solve algebraically. . By the way. then do the work.40 and . Always answer the question that is being asked. D. C.625. .72 E.
Samples
1. . the correct answer is A. so circle those words.625. Which of the following numbers is between 2 and 5 ? 8 5 A. . you might have insight 8 8 5 5 into the problem and should simply work forward.52 is the only number between. E. Circling the key word or words will help you do that.63 C. Work forward. If 3x – 6 < 3. Since x = 50.
Work Forward
If you quickly see the method to solve the problem. 2. Therefore. . . x<4 x<2 x>0 x>3 x<3
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. the coat costs $75. B. which is Choice C. a quick peek at the answer choices would tip you off that you should work in decimals.
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. 0 D. because it will at least eliminate some of the choices and could give you the correct answer. –2 B. 1 E. –1 C. Working from the answers is a valuable technique. what is the value of x? 2 4 4 A. you may use the work up from your choices strategy. What if x = 0? x+3=5 2 4 4 0+3!5 2 4 4 Since this answer is too small." Now solve the problem as follows: 3x – 6 < 3 + 6 +6 3x < 9 3x < 9 3 3 x<3
Add 6 to each side. Plugging in 1 gives. a larger number.Introduction to the Mathematics Section
First circle or underline "possible values of x. 0. This gives Now divide by 3. If x + 3 = 5 .
Samples
1. Do not disregard this method. 2 If you cannot solve this algebraically. So The correct answer is E. try Choice D. x+3=5 2 4 4 1+3=5 2 4 4 Change 1 to 2 2 4 2+3=5 4 4 4 The correct answer is D.
Work Backward
In some instances. it will be easier to work from the answers. But start with C.
The largest number which divides evenly into 18, 24, and 30 is 6. You could have worked from the answers. But here you should start with the largest answer choice, since you're looking for the greatest common factor. The correct answer is D.
Use Your Calculator
Some questions will need to be completely worked out. If you don't see a fast method but do know that you could compute the answer, use your calculator.
1. What is the final cost of a television that sells for $478.00 if the sales tax is 8%? A. B. C. D. E. $478.08 $478.80 $512.80 $516.24 $561.00
Since the sales tax is 8% of $478.00, 8% of $478.00 = (.08)($478.00) = $38.24 The total cost of the television is therefore $478.00 + $38.24 = $516.24 The correct answer is D. Your calculator would have helped with these calculations.
Substitute Simple Numbers
Substituting numbers for variables can often be an aid to understanding a problem. Remember to substitute simple numbers, since you have to do the work.
Since the question says that "x represents an even integer," substitute 2 for x. You should remember to circle "an odd integer" because that is what you are looking for. So as you plug 2 into each choice, you can stop when you get an odd integer. 3x = 3(2) = 6 2x + 1 = 2(2) + 1 = 4 + 1 = 5 Since 5 is an odd integer, the correct answer is B. 2. If f(x) = 4x, which of the following CANNOT be a value of f(x)? A. B. C. D. E. –4 1 4 1 4 16
Use 10 or 100
Some problems may deal with percent or percent change. If you don't see a simple method for working the problem, try using the values of 10 or 100 and see what you get.
1. If 40% of the students in a class have blue eyes, and 20% of those students with blue eyes have brown hair, then what percent of the original total number of students have brown hair and blue eyes? A. B. C. D. E. 4% 8% 16% 20% 32%
First circle or underline "percent of the original total number . . . brown hair . . . blue eyes." In this problem, if you don't spot a simple method, try starting with 100 students in the class. Since 40% of them have blue eyes, then 40 students have blue eyes. Now, the problem says 20% of those students with blue eyes have brown hair. Since 20% of 40 is 8, then 8 out of 100 (you started with 100 students), 8%, have blue eyes and brown hair. The correct answer is B.
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Part I: Analysis and Strategies
2. Tom is building a square wooden framework to pour cement. His first frame is too small, so he increases each side by 20 percent. After careful measurement, he realizes this frame is too large, so he decreases each side by 10 percent. The area contained by his final wooden frame is what percent greater than the original wooden frame? A. B. C. D. E. 10% 10.8% 16.64% 20% 40.44%
First circle or underline what you are looking for, in this case—area . . . percent greater . . . than original. Next, draw the diagram.
The area of the original was 100 sq. in. The area of the new figure is 116.64 sq. in. So the percent greater than the original would be 116.64 – 100 = 16.64 compared to the original 100 gives 16.64%. The correct answer is C. Your calculator could have been helpful in this problem.
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Introduction to the Mathematics Section
Be Reasonable
Sometimes you will immediately recognize a simple method to solve a problem. If this is not the case, try a reasonable approach and then check the answers to see which one is most reasonable.
1. Will can complete a job in 30 minutes. Eli can complete the same job in 60 minutes. If they work together, approximately how many minutes will it take them to complete the job? A. B. C. D. E. 90 minutes 60 minutes 45 minutes 30 minutes 20 minutes
First circle or underline "work together, approximately how many minutes." In a reasonable approach, you would reason that since Will can complete the job alone in 30 minutes, then if he received any help, that job should take less than 30 minutes. He's receiving a fair amount of help, so the answer must be well below 30 minutes. The only choice less than 30 is Choice E, 20.
Sketch a Diagram
Sketching diagrams or simple pictures can also be very helpful in problem solving because the diagram may tip off either a simple solution or a method for solving the problem.
Samples
1. If all sides of a square are doubled, the area of that square A. B. C. D. E. is doubled. is tripled. is multiplied by 4. is multiplied by 8 remains the same.
One way to solve this problem is to draw a square and then double all its sides. Then compare the two areas. Your first diagram
S
S
21
Part I: Analysis and Strategies
Doubling every side
2S S S S S 2S
You can see that the total area of the new square will now be four times the original square. The correct answer is C. 2. What is the maximum number of milk cartons, each 2" wide by 3" long by 4" tall, that can be fit into a cardboard box with inside dimensions of 16" wide by 9" long by 8" tall? A. B. C. D. E. 12 18 20 24 48
Drawing a diagram, as shown below, may be helpful in envisioning the process of fitting the cartons into the box. Notice that 8 cartons will fit across the box, 3 cartons deep, and two "stacks" high: 8 × 3 × 2 = 48 cartons
8″ 4″ 2″ 16″
3″
9″
The correct answer is E.
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Introduction to the Mathematics Section
Mark in Diagrams
Marking in or labeling diagrams as you read the questions can save you valuable time. Marking can also give you insight into how to solve a problem because you will have the complete picture clearly in front of you.
First, circle or underline what you are looking for. "∠y." Next, in the isosceles triangle, immediately mark in that AB and BC are equal. Then mark in ∠x as 70°. Since AB = BC, then ∠x = ∠y (angles opposite equal sides are equal). After you marked in the information, your diagram should look like this.
B
70 A
y C
The correct answer is C. 70° . Always mark in diagrams as you read descriptions and information about them. This includes what you are looking for.
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Marking and/or redrawing will give you insight into what information you really have about the diagram. WZ = ZY s=t perimeter of ᭝WXZ = perimeter of ᭝ZXY area of ᭝WXZ = area of ᭝ZXY ∠XWZ = ∠XYZ
Before doing anything else. In ᭝WXY above. mark the diagram and/or quickly redraw it differently.Part I: Analysis and Strategies
Watch for Diagrams Not Drawn to Scale
Diagrams are drawn as accurately as possible. Now mark the diagram as follows. E. since it's evident that triangles WXZ and ZXY don't necessarily have equal perimeters. But after you redraw the figure.
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.
X
W
s° t° Z
Y
Note: Figure not drawn to scale. WX = XY. quickly redraw it another way that still conforms to the given information. A quick look at the redrawn figure will help you eliminate Choice C as well. They don't have to be equal. The same can be noticed of s and t. eliminating Choice A. D. Which of the following must be true? A.
1. you might think that WZ = ZY because they appear to be equal. C. eliminating B. In this case. since the figure is not drawn to scale.
X
\
\
W
s° t° Z
Y
Notice that by looking at the way the figure is initially drawn." That label is the tipoff that the diagram could be drawn differently or is out of proportion. you can see that WZ and ZY don't have to be equal. underline or circle must be true.
X
\
\
W
s° t° Z
Y
Next. unless a diagram is labeled "not drawn to scale. B.
9 = 1 ^Sh .
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. E. equal angles are across from equal sides. Your markings in the figure remind you that this is true. This fact is also evident from your redrawing of the diagram.3 = 1 ^Sh 2 9 = 3 .9 = ^Sh 2 1 9 =S-3 2^ h
Changing the word sentence into a number sentence (equation):
Harold's age is 3 years less than half Sue's age. you may be asked to set up the equation or expression without doing any solving. D. Rather.Introduction to the Mathematics Section
You can also eliminate Choice D because even though the heights of triangles WXZ and ZXY are equal.3 2 9 . C. Choice E is the correct answer because in any triangle. If Harold is 9 years old.
9
=
1 2
S−3
The correct answer is A. B.
Glance at the Choices
Some problems may not ask you to solve for a numerical answer or even an answer including variables.
Samples
Harold's age is 3 years less than half Sue's age. Suppose S represents Sue's age. their bases could be different. so the areas could be different.1 ^Sh 2 1 3 . A quick glance at the answer choices will help you know what is expected. Which of the following equations can be used to find Sue's age? A. how old is Sue? 1.
Directions with Analysis
The following questions require you to solve the problem and enter your answer by carefully marking the circles on the special grid. 1 2 3 4 5 6 7 8 9
/ . there is no penalty for guessing or attempting a grid-in. Since you will not be selecting from a group of possible answers.Part I: Analysis and Strategies
Grid-In Questions
The New SAT I grid-in question type is very similar to the familiar multiple-choice question except that you will now solve the problem and enter your answer by carefully marking the circles on a special grid. 0 1 2 3 4 5 6 7 8 9
Decimal point
(Notice that the decimal point is located in the shaded row. algebra I and II. so at least take a guess. Let's start explaining the rules and procedures by analyzing the directions.
Basic Skills Necessary
The basic skills necessary to do well on this question type include high school algebra I and II and intuitive or informal geometry. No calculus is necessary. Skills in arithmetic and basic algebra I and II. along with some logical insight into problem-solving situations. and word problems by using problem-solving insight. are also necessary to do well on this question type. but the filled-in circles are most important because they are the ones scored. you should be extra careful in checking and doublechecking your answer. Understanding the rules and procedures for gridding in answers is important. just above the numbers. Also notice that the answer has been written in above the gridding. and application of basic skills. Also. keep in mind that answers to grid-in questions are given either full credit or no credit. it is important that you become familiar with the grid-in rules and procedures and learn to grid accurately. You will not be selecting from a group of possible answers.) Answer: 3. Before you begin working grid-in questions. geometry.)
26
. That is. You should always write in your answer. data interpretation. logic. basic statistics and probability.7
. No points are deducted for incorrect answers in this section. (Comments in parentheses have been added to help you understand how to grid properly.
Ability Tested
The grid-in questions test your ability to solve mathematical problems involving arithmetic. 0 1 2 3 4 5 6 7 8 9
/ . Examples of the appropriate way to mark the grid follow. Your calculator can be useful in checking answers. There is no partial credit. 0 1 2 3 4 5 6 7 8 9
.
You are not required to grid a zero before a fraction.2 or 0. 0 1 2 3 4 5 6 7 8 9
/ .88. Gridding this value as . you are required only to grid a zero in one column. can be gridded as .889
. or .
29
. Notice that you must be as accurate as the grid allows. but it is not required as long as they will fit in the grid. 1 2 3 4 5 6 7 8 9
/ .)
Practice Grid-Ins
The following practice exercises will help you become familiar with the gridding process. (Fractions can be reduced to lowest terms. be sure to erase the old gridded answer completely.Introduction to the Mathematics Section
Answer: 8 9 Accuracy of decimals: Always enter the most accurate decimal value that the grid will accommodate. 0 1 2 3 4 5 6 7 8 9
. it is very important to ensure accuracy. 0 1 2 3 4 5 6 7 8 9
.889. Hand write and grid-in the answers given. . 0 1 2 3 4 5 6 7 8 9
/ . . Although writing out the answers above the columns is not required.) Be sure to write your answers in the boxes at the top of the circles before doing your gridding. Properly filled-in grids are given following each exercise. The acceptable grid-ins of 8 are 9 8 9 . 0 1 2 3 4 5 6 7 8 9
(Review "accuracy of decimals" a second time. .888 or .888 .8888 . For example.89 is considered inaccurate and therefore not acceptable. 0 1 2 3 4 5 6 7 8 9
. 1 2 3 4 5 6 7 8 9
/ . 0 1 2 3 4 5 6 7 8 9
/ . Grid-in questions contain no negative answers. Important: If you decide to change an answer. 0 1 2 3 4 5 6 7 8 9
. either . grid only one answer. 0 1 2 3 4 5 6 7 8 9
.8.2 is acceptable. 1 2 3 4 5 6 7 8 9
/ . If your answer is zero. Even though some problems may have more than one correct answer. For example: An answer such as .
will also work on grid-in questions. Disregard the % sign when gridding your answer.
You should also
■ ■ ■
Make sure that your answer is reasonable.
32
. Improper fractions or decimals can be gridded. 7 1 = 7.75 dollar. more than one answer is possible. grid in 6. Or if a question asks for dollars and you get an answer of 75 cents. Use your calculator when appropriate to solve a problem. the grid-in answer is the only one scored. Mark in diagrams. remember that 75 cents is . The use of your calculator on some problems in this section could enhance the speed and accuracy of your work. described and suggested in the multiple-choice section.5. Pull information out of word problems. Grid-in . Jot down your scratch work or calculations in the space provided in your test booklet. In some questions.25 m before being gridded. Draw/sketch diagrams or simple figures. or if it's $95. Also consider when it would be appropriate to use a calculator to help you work out problems and to check the accuracy of your work. It's wise to always write in your answer.
■
■
■
■
■
■
Following are some sample grid-in questions. The scoring system cannot distinguish between 3 1 and 31 . grid-in 57. Although it may be difficult to get an answer correct by simply writing in a wild guess. or 8 but not more than one of them. If you work out a question and get the answer x > 5. Writing in your answer in the space provided at the top of the grid is for your benefit only. you should not be afraid to fill in your answer—even if you think it's wrong. grid in 95.Part I: Analysis and Strategies
Suggested Approaches with Samples
Most of the following strategies. Make sure to answer what is being asked. Substitute numbers for variables to understand a problem. Mixed 2 4 numbers cannot. you will have to actually work out each answer. 2 2 Since you cannot work from answer choices that are given or eliminate given choices. or Disregard the $ sign when gridding your answer.
There are some specific items and strategies that should be noted for grid-in questions:
■
There is no penalty for guessing on grid-in questions. If your answer is 68%. Follow the directions. but remember. or 7. Some questions will have a note in parentheses ( ) that says. Consider for each how you would grid-in the answer in the proper places on the answer sheet. 7 1 = 29 m or 2 4 2 2 4 4 decimals c 3 1 = 3.
■ ■ ■ ■ ■ ■ ■
Circle or underline what you are looking for. not . grid in 68. Be sure to grid accurately and properly. Answers that are mixed numbers such as 3 1 or 7 1 must be changed to improper fractions c 3 1 = 7 . If the question asks for percent and you get an answer of 57%. Grid-in only one answer.75 not 75. Approximate or use your calculator to check your answers if time permits. Try simple numbers to solve a problem.57.
Algebra and Functions. .
Symbols. The final problem is a sample grid-in problem. Statistics. The first three problems for each topic are multiple-choice and range (in order) from easy to difficult. Positive and negative whole numbers and zero: . Formulas. . 3. which you should take to spot your weak areas. . . Number not divisible by 2: 1. . . After completing and reviewing the SAT-type questions for each topic. Geometry and Measurement. Then proceed to the Arithmetic Diagnostic Test. –1. Even if you are strong in the general topics—Numbers and Operations. carefully work through the SAT-type questions that follow. 3. . . Then use the Arithmetic Review that follows to strengthen those areas. 5. The counting numbers beginning with zero: 0. and general mathematical information. 0. –3. . 7. and General Mathematical Information
Common Math Symbols and Terms
Symbol References: = ≠ > < ≥ ≤ || is equal to is not equal to is greater than is less than is greater than or equal to is less than or equal to is parallel to is perpendicular to
=
Terms: Natural numbers Whole numbers Integers Odd number
The counting numbers: 1. take the Measurement and Geometry Diagnostic Test and carefully read the complete Measurement and Geometry Review. Carefully work through these problems. 2. you should read through the complete review carefully. . it is wise to become familiar with basic mathematics terminology.A Quick Review of Mathematics
The following pages are designed to give you a quick review of some of the basic skills used on the math sections of the New SAT I. Before beginning the diagnostic review tests. 1. 2. . 2. .
(continued)
39
. and Probability—you might wish to skim the specific topic headings in each area to refresh your memory about important concepts. Again. Again. Next. 3. take the Algebra Diagnostic Test and again use the review that follows to strengthen your weak areas. . . 1. formulas. carefully work through the SAT-type questions that follow. –2. Each review is followed by sample SAT-type questions with complete explanations for each topic area. Terminology. and Data Analysis. If you are weak in math. These topics are covered in this chapter.
therefore.
3 + (–3) = 0. therefore.
(2 × 3) × 4 = 2 × (3 × 4) (a × b) × c = a × (b × c) The grouping has changed (parentheses moved). therefore. –2 and 2 are inverses.
2×3=3×2 a×b=b×a Note: The commutative property does not hold for division.
43
. Note: The associative property does not hold for division. 3 and –3 are inverses. (8 ÷ 4) ÷ 2 ≠ 8 ÷ (4 ÷ 2)
The identity element for multiplication is 1. a + (–a) = 0. Any number plus its additive inverse equals 0 (the identity). –2 + 2 = 0. 2 2 1 a # a = 1.A Quick Review of Mathematics
The identity element for addition is 0. a and –a are inverses.
Some Properties (Axioms) of Multiplication
Commutative means that the order does not make any difference.
3×1=3 a×1=a
The multiplicative inverse is the reciprocal of the number. a and 1 a are inverses. Any number multiplied by 1 gives the original number. 2 and 1 are inverses. 2÷4≠4÷2
Associative means that the grouping does not make any difference.
3+0=3 a+0=a
The additive inverse is the opposite (negative) of the number. Any number added to 0 gives the original number. therefore.
2 # 1 = 1. Any number multiplied by its reciprocal equals 1. but the sides are still equal. therefore.
3(4 × 5 × 6) ≠ 3(4) × 3(5) × 3(6) a(bcd) ≠ a(b) × a(c) × a(d) or (ab)(ac)(ad)
44
.
2(3 + 4) = 2(3) + 2(4) a(b + c) = a(b) + a(c) Note: You cannot use the distributive property with only one operation.Part II: Review with Sample Problems
A Property of Two Operations
The distributive property is the process of distributing the number on the outside of a set of parentheses to each number on the inside.
47 298. 0 9 2 .021 rounded to the nearest thousand. 2.
Arithmetic Review
Place Value
Each position in any number has place value. If it is 5 or higher.08
Estimating Sums. and 5 is in the ones place. place value is as follows:
3 .000 This works with decimals as well.
Estimating Sums
Use rounded numbers to estimate sums. Round to the nearest hundredth: 3. . and Quotients
Knowing how to approximate or estimate not only saves you time but can also help you check your answer to see whether it is reasonable. 8 is in the tens place. Differences.000 928.
and so on
47
. For example: Round to the nearest thousand: 345.499 becomes 928. Look to the immediate right (one place) of your underlined place value. If the number (the one to the right) is 4 or less. For example. .4678 becomes 3.Numbers and Operations .741 + 5. For instance. Underline the place value to which you're rounding off. 4 is in the hundreds place. Identify the number (the one to the right).435. in the number 485.435. Thus. Products. round your underlined place value up 1. 4 3 6 2 9 7 0 2
tenths hundredths thousandths ten-thousandths hundred-thousandths millionths ten-millionths hundred-millionths
hundred millions ten millions millions
hundred thousands ten thousands thousands
and so on
hundreds tens ones
billions
Rounding Off
To round off any number: 1. give an estimate for the sum 3. 8 7 6 .083 becomes 298. 3 4 5 . leave your underlined place value as it is and change all the other numbers to its right to zeros.678 becomes 346. 3.
700 # 300 = 210. 000 722 # 489 . 000
So
If both multipliers end in 50 or are halfway numbers. which is the standard rule. 522 . For example. 900 ' 300 = 3 891 ' 288 . 4. . then rounding one number up and one number down gives a better estimate of the product. 300.115. 000 + 5. . 9. 317. 000 . 000 650 # 350 . 600 # 400 = 240. 722 # 489 . estimate the product of 650 × 350 by rounding to the nearest hundred. 3
So
48
. 753 . estimate the product of 722 × 489 by rounding to the nearest hundred. . 000
So
Estimating Products
Use rounded numbers to estimate products. 741 + 5. 000 = 200. 000 317. 650 # 350 . 522 . . give an estimate for the difference 317.000 650 # 300 .
Estimating Differences
Use rounded numbers to estimate differences. 021 . 891 ' 288 . 350. estimate the quotient of 891 ÷ 288 by rounding to the nearest hundred. 200. . 000
Note: The symbol ≈ means is approximately equal to.753 – 115. 700 # 500 = 350.000
So
You can also round the first number down and the second number up and get this estimate: 650 # 350 .100. For example. 741 + 5. 210. 000 = 9.115.Part II: Review with Sample Problems
So
3. 000
So
In either case. 000 3. 753 . .
Estimating Quotients
Use rounded numbers to estimate quotients.522 rounded to the nearest hundred thousand. 240. this approximation is closer than if you round both numbers up. Round one number up and one down. 021 . For example. For example.
giving 7 . 5. For example. This is done by dividing both the numerator and the denominator by the largest number that divides evenly into both. 20 is reduced to 4 by dividing both the numerator and denominator by 5. For example.
Fractions
Fractions consist of two numbers: a numerator (which is above the line) and a denominator (which is below the line). Sometimes a fraction represents more than one. add the numerator and put the total over the original denominator.Numbers and Operations . multiply the denominator by the whole number. first change all denominators to their lowest common denominator (LCD)—the lowest number that can be divided evenly by all the denominators in the problem.
Mixed Numbers
When a term contains both a whole number (such as 3. 5 1 and 290 3 are both mixed numbers. For example: 18 = 3 3 5 5 3 5 g18 15 3
To change a mixed number to an improper fraction. 1 numerator or numerator 1 denominator 2 2 denominator The denominator indicates the number of equal parts into which something is divided. Thus. divide the denominator into the numerator. add fractions by simply adding the numerators (the denominator remains the same). it is called a mixed 2 4 4 number. 8. or 3 ). 1 . When all the denominators are the same. For example:
49
. 3 also 5 means 3 out of 5 equal pieces from the whole pie. . or 25) and a fraction (such as 1 . . if the fraction is 3 of a pie. For example: 41=9 2 2 2#4+1=9
Reducing Fractions
A fraction must be reduced to lowest terms. 8 16 Likewise. 5 indicates that the pie has been divided into 5 equal parts. The numerator indicates how many of these equal parts are contained in the fraction. In other words.
Common Fractions and Improper Fractions
A fraction like 3 . where the numerator is smaller than the denominator. This is called an improper fraction. Sometimes it helps to think of the dividing line (in the middle of a fraction) as meaning out of. This is when the numerator is larger than the denominator. is less than one. of which 3 (the numerator) are in the fraction. Thus. 4 4 To change an improper fraction to a mixed number. then the denominator. 25 5
Adding Fractions
To add fractions. 12 7 is more than one. This kind of fraction is called 5 a common fraction. 14 is reduced by dividing both terms by 2.
For example:
21⁄2 + 31⁄4
= =
2 2 ⁄4 3 1⁄4 5 3⁄ 4
change one-half to two-fourths
remember to add the whole numbers
Subtracting Fractions
To subtract fractions. Then you need to add the numerators 3 and 4 2 8 7 to get . and then add the numerators to 8 7 get . if the denominators are already the same. For example: 7= 8 -1 = 4 7 8 2 8 5 8 3= 9 4 12 -1 = 4 3 12 5 12
Subtracting Mixed Numbers
When you subtract mixed numbers. the same rule (find the LCD) applies. you have to change the 1 to 4 because 8 is the LCD. Of course. For example: 12 6 + 3 = 9 11 11 11
Adding Mixed Numbers
To add mixed numbers. you must change both fractions to get the LCD of 12. but always add the whole number to get your final answer. just like you sometimes borrow from the next column when subtracting ordinary numbers. the same rule (find the LCD) applies. In the second example. sometimes you have to borrow from the whole number. just add the numerators. For example:
4 11 37 ⁄6
−
651 129 522
41 ⁄6
− 25 ⁄6 12 ⁄6 = 11 ⁄3 you borrowed one in the form 6 ⁄6 from the ones column
you borrowed 1 from the tens column
50
.Part II: Review with Sample Problems
3 3 = 8 8 1 4 + = 2 8 7 8
one-half is changed to four-eighths
1 3 = 4 12 1 4 + = 3 12 7 12
change both fractions to LCD of 12
In the first example. except subtract the numerators.
For example: 1'1=1#5=5 6 5 6 1 6 1'1=1#3=1 6 3 6 1 2
51
. Then multiply as previously shown. Add this to the numerator of the fraction. For example:
6 − 31⁄5
55 ⁄ 5 = 3 1⁄5
=
borrow one in the form of 5 ⁄5 from the 6
24 ⁄5
remember to subtract the remaining whole numbers
Multiplying Fractions
Simply multiply the numerators. To change mixed numbers to improper fractions: 1. For example: 2 # 5 = 10 3 12 36 reduce 10 to 5 18 36
This answer had to be reduced because it wasn't in lowest terms. 3 1 # 2 1 = 10 # 9 = 90 = 7 6 = 7 1 3 4 3 4 12 12 2 Then. Reduce to lowest terms if necessary. first change any mixed number to an improper fraction. find a number that divides evenly into one numerator and one denominator.Numbers and Operations . 2 # 5 = 5 3 12 6 18 You can cancel only when multiplying fractions. invert (turn upside down) the second fraction and multiply. and then multiply the denominators. Then.
1 1
Multiplying Mixed Numbers
To multiply mixed numbers. Canceling eliminates the need to reduce your answer.
Canceling When Multiplying Fractions
You could have canceled first. This is now your numerator. 2. . back to a mixed number and reduce if necessary. if it is in improper form.
To subtract a mixed number from a whole number. change the answer. you have to borrow from the whole number. Thus: 2 # 5 = 3 12 6 Now that you've canceled. Multiply the whole number by the denominator of the fraction. In this case. The denominator remains the same.
Dividing Fractions
To divide fractions. reduce if necessary. . you can multiply as before. 3. 2 divides evenly into 2 in the numerator (it goes in one time) and 12 in the denominator (it goes in 6 times). 4. To cancel.
0372
4 digits
Dividing Decimals
Dividing decimals is the same as dividing other numbers.012
×
3 digits 1 digit
3. 0.25 g 5.05 = 5%
Changing Percents to Decimals
To change percents to decimals: 1. 1. Then move the decimal point in the dividend (the number being divided into) the same number of places. Eliminate the percent sign. Insert a percent sign. . = 125 g 500.1
total of 4 digits above the line that are to the right of the decimal point decimal point placed so there is same number of digits to the right of the decimal point
40012 120036 124. Then. or 13000. 2.Numbers and Operations .0 0 . 2.
Whole numbers can have decimal points to their right.75 = 75% .57
Multiplying Decimals
To multiply decimals. move it to the right as many places as necessary until it is a whole number. For example:
6 9 1
17 . count the total number of digits above the line that are to the right of all decimal points. Sometimes you have to add zeros to the dividend (the number inside the division sign). . .8.)
53
. multiply as usual. Move the decimal point two places to the left. Move the decimal point two places to the right.
Conversions
Changing Decimals to Percents
To change decimals to percents: 1. Place the decimal point in the answer so that the number of digits to the right of the decimal is the same as it is above the line. = 2 g 26000. For example:
40.43 =
17. 4.8. except that when the divisor (the number you're dividing by) has a decimal. (Sometimes adding zeros is necessary.002 g 26.43 8.
numbers to the right of 0 are positive. use this formula: change # 100 = percentage change starting point For example: 1. Numbers to the left of 0 are negative. What is the percentage increase of Jon's salary if it goes from $150 a month to $200 a month? change # 100 = 50 # 100 = 1 # 100 = 33 1 % increase 3 3 starting point 150
Signed Numbers (Positive Numbers and Negative Numbers)
On a number line. What is the percentage decrease of a $500 item on sale for $400? change # 100 = 100 # 100 = 1 # 100 = 20% decrease starting point 500 5 2.25 + 20 ++ 3 + 23
56
. For example: +5 .3 -19 +.
Adding Signed Numbers
When adding two numbers with the same sign (either both positive or both negative).4 +8 -14 ++ 4 -10 -19 -+ 6 + 20 -. add the numbers and keep the same sign.7 ++ 72 -2 + 13
Subtracting Signed Numbers
To subtract positive and/or negative numbers.59 +.3 + 12 .
Given any two numbers on a number line. subtract the numbers and keep the sign from the larger one. regardless of its sign (positive or negative). For example: +5 -8 ++ 7 +. just change the sign of the number being subtracted and add.11 When adding two numbers with different signs (one positive and one negative). −3 −2 −1 0 +1 +2 +3
etc.Part II: Review with Sample Problems
Percentage Increase or Decrease
To find the percentage change (increase or decrease).6 . as follows:
etc. For example: +12 -+ 4 -14 -.4 +12 +. the one on the right is always larger.
4 = 4 . The value of a number is written 3 = 3 and . For example: -8 = 8 3 . simply keep the base number and subtract the second exponent from the first. if the base numbers are the same. such as 3–2. simply keep the base number and add the exponents.9 = -6 = 6 3 . . It expresses the power to which the quantity is to be raised or lowered. an even number of negative signs produces a positive answer. For example: 21 = 2 20 = 1 31 = 3 30 = 1
Negative Exponents
If an exponent is negative.2 = 12 = 1 9 3
Operations with Powers and Exponents
To multiply two numbers with exponents. For example: 34 ' 32 = 32 9 3] 4 . In 43.Numbers and Operations . or the work must be done first within the absolute value signs.
Multiplying and Dividing Signed Numbers
To multiply or divide signed numbers. For example: 24 = 2 × 2 × 2 × 2 = 16 32 = 3 × 3 = 9 Remember that x1 = x and x0 = 1 when x is any number (other than 0)..6 = 3 . 43 is read four to the third power (or four cubed). 4 × 4 × 4 (multiplied by itself twice). if the base numbers are the same. treat them just like regular numbers but remember this rule: An odd number of negative signs produces a negative answer. . 3 is the exponent.
Powers and Exponents
An exponent is a positive or negative number placed above and to the right of a quantity. then the number and exponent can be dropped under the number 1 in a fraction to remove the negative sign.2 g = 9 4 9 C 92
57
. It shows that 4 is to be used as a factor three times.6 = -3 Note: Absolute values must be taken first. The absolute value of a number is always positive except when the number is 0. The number can be simplified as follows: 3 . For example: 23 × 25 = 28 7 ×7 =7
2 4 6
[(2 × 2 × 2) × (2 × 2 × 2 × 2 × 2) = 28]
[2(3 + 5) = 28]
To divide two numbers with exponents.2 g = 3 2 C
9 6 = 9 6 ' 9 2 = 9 4 9] 6 . For example: (–3)(+8)(–5)(–1)(–2) = +240
Absolute Value
The numerical value when direction or sign is not considered is called the absolute value.
Thirty-six is called a perfect square (the square of a whole number). If a number with an exponent is taken to another power (42)3. Following is a list of perfect (whole number) square roots: 1=1 4=2 9=3 16 = 4 25 = 5 36 = 6 49 = 7
58
. and then perform the operation. For example.7 To find the square root of a number. . simplify each number with an exponent first. or 36. 6 squared (written 62) is 6 × 6. find some number that when multiplied by itself gives you the original number. 1. The symbol for square root is √. 1. Two approximations to remember are: 2 .Part II: Review with Sample Problems
Three notes:
■
■
■
If the base numbers are different in multiplication or division. then. to find the square root of 25.4 3 . and then perform the indicated operation. Following is a list of some perfect squares: 12 = 1 22 = 4 32 = 9 42 = 16 52 = 25 62 = 36 72 = 49 82 = 64 92 = 81 102 = 100 112 = 121 122 = 144 . In other words. The square root of 25. . is 5. Square roots of nonperfect squares can be approximated. find the number that when multiplied by itself gives you 25. just multiply it by itself. simply keep the original base number and multiply the exponents. To add or subtract numbers with exponents. simplify each number with an exponent first. For example: (42)3 = 46 (34)2 = 38 [4(2 × 3) = 46]
Squares and Square Roots
To square a number. whether the base is the same or different.
Simplify 75. For example:
^ 4 h^ 25h = 4 # 25 = 2 # 5 = 10 or
100 = 10
Likewise with division: 64 = 4 64 8 = = 4 or 16 = 4 2 4
Addition and subtraction. 2 . 9. 49 and 64. 83 is a little more than 9. Because 83 is only two steps up from the nearest perfect square (81) and 17 steps to the next perfect square (100).
64 = 8 81 = 9 100 = 10
Square Root Rules
Two numbers multiplied under a radical (square root) sign equal the product of the two square roots. 83 is 2 of the way to 100. however. 75 = 25 # 3 = 25 # 3 = 5 3
59
. Since 57 is between
Simplifying Square Roots
To simplify numbers under a radical (square root sign): 1. you should approximate.1 19 20 10 Therefore. . 9 10 81 < 83 < 100 Since 83 is slightly more than 81 (whose square root is 9). And because 57 is just about halfway between 49 and 64. are different.12 = 81 = 9
Approximating Square Roots
To find a square root that is not a whole number. 1 = . one (or more) of which is a perfect square. For example: Approximate 57. The numbers must be combined under the radical before any computation of square roots is done. 57 is approximately 7 1 .Numbers and Operations . Leave the other factors under the . 2. 83 . Factor the number to two numbers. For example: 10 + 6 = 16 = 4 10 + 6 does not equal 7 ! A 10 + 6 93 . 19 2 . 3. 2 Approximate 83. . it falls somewhere between 7 and 8. Take the square root of the perfect square(s).1.
4} Infinite sets are uncountable. 2. b. They stop. 6 . 4. 5 is the set with members 1. 0 or {}.
A
C
B
Types of Sets
Finite sets are countable. 3 is a subset of the set of 1. 2. 3 . .1 #1. {1. they continue forever. . 3. 3 . 2. 5 -
The union of sets with members 1. 5. } {Fred. . 2. 5 . .. 2. . 3.
61
. 3The set of 2. 4. 5.Numbers and Operations . 4.
Special Sets
A subset is a set within a set. c -
Operations with Sets
The union of two or more sets is all of the members in those sets.
#1. 2. 3.
Describing Sets
Rule is a method of naming a set by describing its elements. {1. 2. is a whole number}
{all students in the class with blue eyes} Roster is a method of naming a set by listing its members. {4. {1. 2. 1} Equivalent sets are sets that have the same number of members. 3} = {3. 2. . 4. The universal set is the general category set.}
Comparing Sets
Equal sets are those that have the exact same members. 3 and 3. # 3. The empty set. 3. or null set.= #1. Tom. 3. or the set of all those elements under consideration. 2. is a set with no members.+ # a.
#1. . 2. Bob} Venn Diagrams (and Euler Circles) are ways of pictorially describing sets. # 2. 3.
" x x > 3.
What is the probability of tossing heads three consecutive times with a two-sided fair coin? Because each toss is independent and the odds are 1 for each toss. 2.+ # 3. Three green marbles. the probability is 1 out of 10 or 1 .
#1. 3 .
The intersection of a set with members 1. 4. 5 . using the spinner shown above. two blue marbles. Again. or overlap. and five yellow marbles are placed in a jar. the probability of the occurrence of a given outcome can be found by using the following formula: probability= number of favorable outcomes number of possible outcomes For example: 1. 10 2. the probability is 3 out of 10 or 3 . 3 and a set with members 3. you need to multiply to find the favorable and/or possible outcomes. 5 is a set with only member 3. 2. What is the probability of selecting at random. 10 5 When two events are independent of each other.Part II: Review with Sample Problems
The intersection of two or more sets is where they intersect. or 1 . When all outcomes are equally likely to occur.
Some Basic Probability and Statistics
Probability
Probability is the numerical measure of the chance of an outcome or event occurring. Using the spinner shown below. 10
62
5
6
.= " 3 . what is the probability of spinning a 3 or a 5 in one spin? Since there are two favorable outcomes out of ten possible outcomes. 3. a green marble on the first draw? Since there are ten marbles (total possible outcomes) and three green marbles (favorable outcomes). what is the probability of spinning a 6 in one spin?
10
1
9
2
8
3
4
7
Since there is only one 6 on the spinner out of ten numbers and all the numbers are equally spaced. 4. the probability is 2 1#1#1=1 2 2 2 8 4. the probability is 2 out of 10 or 2 .
If the list contains an even number of items. 31. 8. and 45? 10 + 20 + 35 + 40 + 45 = 150 150 ÷ 5 = 30 The average is 30. average the two middle numbers to get the median. 10—the median is 7 1 . 3.
Mode
A mode is simply the number most frequently listed in a group of numbers. 21. 2—the mode is 9 because it appears more often than any other number. . gives the median. and 27? 25 + 27 + 27 + 27 = 106 106 ÷ 4 = 26 1 2 The average is 26 1 . and 45? 0 + 12 + 18 + 20 + 31 + 45 = 126 126 ÷ 6 = 21 The average is 21. 18. 4. 2. median. 12. and mode. What is the average of 10. 6. . 6. 9. 4. What is the average of 0. in the following list—5. in the following group—5. The three basic measures indicating the center of a distribution are: mean. 27. 9. 2
Median
A median is simply the middle number in a list of numbers that have been written in numerical order. 2. the 2 average of the middle two. 24. 20. Because there is an even number of items. 9.
Mean. For example. 20. What is the average of 25. 35. 7. Arithmetic Mean. 56—the number 9 is the median. 9. Divide by the number of items you added.Numbers and Operations . 6. If there are two modes.
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. or Average
To find the average of a group of numbers: 1. 27. 9. the group is called bimodal. 3. 7. 40. For example: 1. For example. 9. Add them up. The list has to be in numerical order (or put in numerical order) first.
Statistics
The study of numerical data and its distribution is called statistics. in the following list—3. There can be more than one mode. 7 and 8. The median is easy to calculate and is not influenced by extreme measures. 7. For example.
Some graphs list numbers along one edge and places. When answering questions related to a graph:
■ ■ ■ ■ ■ ■
Examine the entire graph—notice labels and headings. You should also know the scatter plot. City W has approximately 500 billboards. Always try to determine the relationship between the columns in a graph or chart. refer to it. people. The bar graph for City Y stops about halfway between 100 and 200. dates. Pay special attention to which part of the graph the question is referring to. which is similar to a coordinate graph. 500 – 150 = 350
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. low points. with the numbers given along the bottom of the graph in increases of 100. Focus on the information given. Do not memorize the graph. trends. For example: 1. So City W (500) has approximately 350 more billboards than City Y (150). Reread the headings and labels if you don't understand.Part II: Review with Sample Problems
Data Analysis: Graphs
Information can be displayed in many ways. or things (individual categories) along the other edge. Consider that halfway between 100 and 200 is 150. and circle graphs (or pie charts). The names are listed along the left side. line graphs. The following bar graph indicates that City W has approximately how many more billboards than City Y?
City W City X City Y City Z 0 100 200 300 400 500
Billboards
The graph shows the number of billboards in each city. The three basic types of graphs you should know about are bar graphs.
Bar Graphs
Bar graphs convert the information in a chart into separate bars or columns. Look for major changes—high points.
5 million = 3.5 1992 = 3.500. answer these questions: A.0
. Referring to the All Sports bars. adding zeros does not add precision to the numbers. B. C.0
Number of Books in Millions
2. the number of books sold per year can be determined as follows: 1990 = 2. .4 Use a piece of paper as a straightedge to determine this last number. as it often is with graphs that use large numbers.5
1.Numbers and Operations .4. .5 1991 = 2.000).5
2.1 1992 = 3
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. Based on the following bar graph.0
1. The number of books sold by Mystery Mystery from 1990 to 1992 exceeded the number of those sold by All Sports by approximately how many? From 1991 to 1992. Each single bar stands for the number of books sold in a single year. Since all measurements are in millions. the number of books sold per year is as follows: 1990 = 1 1991 = 2.5
3.5
1990 1991 1992
1990 1991 1992
1990 1991 1992
Mystery Mystery
Reference Unlimited
All Sports
The graph contains multiple bars representing each publisher. Totaling the number of books sold for all three years gives 8. You might be tempted to write out the numbers as you do your arithmetic (3.
2. Writing out the numbers is unnecessary. the percent increase in number of books sold by All Sports exceeded the percent increase in number of books sold by Mystery Mystery by approximately how much? What caused the 1992 decline in Reference Unlimited's number of books sold?
3. Answer to Question A: Regarding the Mystery Mystery bars.
: This question cannot be answered based on the information in the graph. Answer to Question B.9 The 1991 amount is the starting point. the multiple factors that could cause a decline in the number of books sold are not represented by the graph. change = percent change starting point In this case. the percent increase in number of books sold by Mystery Mystery can be calculated first. so: change = . the sharper the slope downward.1 So the percent increase of All Sports exceeded that of Mystery Mystery by 7%: 43% – 36% = 7% Answer to Question C. in data over a period of time. These lines show increases and decreases— the sharper the slope upward. and so on.9 change = . These points are then connected to show a relationship among items. dates. In this case. Totaling the number of books sold for all three years gives 6. Number of books sold in 1991 = 2. or changes. Never assume information that is not given. The formula for figuring either of these is the same. 43% starting point 2.9 = 4. the number of books sold by Mystery Mystery exceeded the number of books sold by All Sports by 2.5 Number of books sold in 1992 = 3.: Graph and chart questions might ask you to calculate percent increase or percent decrease.
Line Graphs
Line graphs convert data into points on a grid. So.1. use a piece of paper as a straightedge.5 The percent increase in number of books sold by All Sports can be calculated as follows: Number of books sold in 1991 = 2.3 million. Line graphs can show trends.
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.4 Change = .28 . but don't designate numbers beyond the nearest 10th because the graph numbers prescribe no greater accuracy than this. the greater the increase. times. the greater the decrease.9 = 36% starting point 2.Part II: Review with Sample Problems
Again.1 Number of books sold in 1992 = 3 Change = . Notice the slopes of the lines connecting the points.
when the plant grew from 40 centimeters to 80 centimeters.000 $200.000 to about $250.: The information along the left side of the graph shows the property value of Moose Lake Resort in increments of $100.000. The numbers on the left give the height of the plant in centimeters.000 on the left. .000. Using a sheet of paper as a straightedge helps to see that the dot in the 1970 column lines up with $500.
For example: 1. The sharpest upward slope occurs between week 3 and week 4. the property value was about $500.: Since the slope of the line goes down from 1920 to 1930. If you read the actual numbers. In what year was the property value of Moose Lake Resort about $500. answer these questions: A. you notice a decrease from $300.000
1920 1930 1940 1950 1960 1970 1980 1990
Years
Answer to Question A.000 $400. Answer to Question B.Numbers and Operations .000? In which 10-year period was there the greatest decrease in the property value of Moose Lake Resort?
$800. In 1970.000.000 $100.000 $300. there must have been a decrease in property value. B. Based on the following line graph. According to the following line graph.000 Total Value $600. a total of 40 centimeters growth. 2.000 $700.000 $500. The bottom of the graph shows the years from 1920 to 1990. the tomato plant grew the most between which two weeks?
100 90 80 70
Centimeters
60 50 40 30 20 10 0 1 2 3 4 5 6 7 8
Weeks
The numbers at the bottom of the graph give the weeks of growth of the plant. .
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.
how much do they spend on dances in a year in which the total amount spent is $15.
If Smithville Community Theater has $1. how much is spent on set construction? What is the ratio of the amount of money spent on advertising to the amount of money spent on set construction?
Answer to Question A. advertising = 15% of 1000 = 150 = 3 set construction 20% of 1000 200 4 Notice that 15% reduces to 3 . $1. B. items
25% staff salaries 15% advertising
10% physical plant
A.000 is $200.: The theater spends 20% of its money on set construction.400 student store supplies
$10. you must use the information in the graph to make a ratio.200 dances
$2. or pie chart.100 misc. For example: 1.000? The amount of money spent on field trips in 1995 was approximately what percent of the total amount spent?
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. 20% 4 2. Answer to Question B. Based on the following circle graph:
Smithville Community Theater Expenditures
20% set construction 20% costumes and props
10% misc.
If the Bell Canyon PTA spends the same percentage on dances every year. Based on the following circle graph:
$2.000 Total Expenditures
A. B.: To answer this question. the higher the percentage. shows the relationship between the whole circle (100%) and the various slices that represent portions of that 100%—the larger the slice. so $200 is spent on set construction.000 to spend this month. Twenty percent of $1.Part II: Review with Sample Problems
Circle Graphs or Pie Charts
A circle graph.400 student awards $2.900 field trips
$1.
22 × 15. one quantity decreases as another increases.
Answer to Question A. 3 × $1.100 out of every $5.000 is spent for dances.000. Answer to Question B.0 4 8 12 16 20
Days absent from school
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.000) gives the right answer. .300.000.300 or $3.22 .000 is spent for dances. 000 100 Multiplying 22% by the new total amount spent ($15. For example:
Biology test score (out of 100 possible) 100 90 80 70 60 50 1 2 3 4 5 6 7
Number of hours studying
If the line goes down to the right.: By carefully reading the information in the graph. If $2.: To answer this question.900 was spent on field trips. and then apply this percent to a new total. that is. you find that $2.300 You could use another common-sense method. Because $2. The graph is typically placed in one part of a coordinate plane (the upper right quarter called Quadrant I). For example:
Grade point average
4. Since $15.000.000 is 3 × $5. . In 1995. If the line goes up to the right. the approximate percentage is worked out as follows: 3.100 is $3. 000 = 30 = 30% 10.Numbers and Operations . a linear relationship is suggested.0 1.200 out of $10. 200 = 22 = 22% 10.0 2. then the relationship is called a negative correlation. one quantity increases as another increases. 22% = . 000 100
Scatter Plot
A scatter plot is a graph representing a set of data and showing a relationship or connection between the two quantities given. The information describing the graph explains that the total expenditures were $10.200 on dances. usually a relationship can be seen. then the relationship is called a positive correlation. If the points appear to form a line. When the data is placed on the scatter plot.000 = 3. that is. the PTA spent $2. $1.900 is approximately $3. you must find a percent. This can be calculated as 22% of the total spent in 1995 by the following method: 2.0 3.
Part II: Review with Sample Problems
If the data does not appear to show any line or any relationship between the quantities. For example:
100
Math test score (percent right)
90 80 70 60 30 60 90 120 150 180 210
Time spent listening to music (minutes)
70
. the scatter plot is said to show no correlation.
C. During a sale Arianna purchases a sweater listed at $100. D. The sum is $67.00. D. His profit per DVD is therefore $17. He then sells them to his customers during a special sale. Triple each integer. 33.01 for 3 DVDs.100 – $125 = $975. B.37 for tomatoes.67 – $13. If 10 boys averaged 78% on a test for which 15 girls averaged 89%.01 ÷ 3 = $17. subtract $67 from $100.60
6.54 per DVD. and Median
9. 20% of $100 is just $20. An electronics store owner purchases DVDs from his supplier. 7 8 9 13 15
8. After making a down payment of $125.54 = $4. D. and $14. 7% tax on $80 is (.60 = $85. Mode. $17.8% 10. E.80 ÷ 20 = $13.80 for avocados. how much did she pay for the discounted sweater? A. C. it will take $975 ÷ $75 = 13 months. charging $53. Subtract 5 from the largest integer.55 for bread. Add 11 to the largest integer. With 20% off and 7% sales tax. $73. 8. How much profit. Add 3 to the smallest integer. E. The difference is $33.100.80 for 20 DVDs. The store charges the sandwich shop $11. A. Add the following amounts: $11.33 $13.67. E. 6. which is the correct answer.00 $80.28 + $23. how much change should he receive in dollars?
Explanations
5. The store owner pays $270. A grocery store sends an order to a local sandwich shop.60. what is the test average for all 25 students? A. In a set of 15 different integers.00 $85.30 7. 76. Next. he will pay off the remainder in monthly payments of $75.60. per DVD. B. B. So the sale price is $100 – $20 = $80.55 + $17.60 $87.54 $17. D.80.00 $88. and Division to Problem Solving
5.Part II: Review with Sample Problems
Applying Addition. Subtract 7 from each integer. B.2% 87. E. Multiplication. B.67 $82. does the owner make during the special sale? A. D. $1. which of the following changes would NOT effect the value of the median of these 15 integers? A.
Arithmetic Mean (Average). $4. He sells them for $53. C. C. How many months will it take for Franco to pay off the new washing machine? A. If the sandwich shop owner paid with a $100 bill. D. 7. $23. E. The total price is then $80 + $5.
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. paying $270.13. At $75 per month.07)$80 = $5.6% 86. C. C.37 + $14.5% 84.28 for lettuce. Subtraction.13 $9.4% 83. Franco decides to purchase a new washing machine that sells for $1.
B. . A. you have: 6. The other option—which is not one the given choices—would be to subtract some positive integer from the smallest number in the set. C. Since the median is the middle of the list of the set of integers. 10. 9. 3 10 B. 11. Notice that since you have an even number (10) of shoe sizes. How much greater than the mode is the median? A. 8 15 D. A special snack mix uses raisins.6% 25 25 10.5. E.Numbers and Operations . 5 6 15 E. 12. A certain brand of paint consists of water and paint concentrate. 10.5 10
12. . 12. How much concentrate would be needed to produce 3 gallons of this paint?
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.5 5 9.)
Explanations
10 ^ 78%h + 15 ^89%h 780% + 1335% = = 84. 0 0. Since the average is the sum of these temperature points. D. 2 5 C. since this is the most exact number the grid field will accommodate. D. 11. Next. The average temperature in May of that year was 9. C. 10.5°C. E. how many ounces of peanuts are used? A. what is the value of r ? 14. 9. 6. Each gallon of paint contains 1 4 part water. 7.76. and chocolate candies in the ratio 1 : 2 : 5. and 7. the average is: 240 + 294. B. you had to find the average of the fifth and sixth shoe sizes (9 and 10) to compute the median. adding 11 to the largest integer will not affect the median. peanuts. 80 70 50 16 14
r = 2 and m = 4 . divided by 61 (the number of days in April and May combined).5 temperature points for the month of May. 8.5 = 8. B. B. Which of the following could NOT be the total number of people at the party? A.
11. Listing the shoe sizes from small to large. 7. In a snack mix of 112 ounces. E. 8. Begin by determining the number of temperature points for the month of April: 30 × 8 = 240 temperature points for the month of April.0°C. C. the ratio of males to females is 3 : 4. The shoes sizes of 10 men are 8. The average daily temperature in a city during the month of April was 8. which gives 9. respectively. 7. B.5 = 294. What is the average temperature in this city for the months of April and May? (There are 30 days in April and 31 days in May. 14 35 44 63 70 15. 9.
Ratio and Proportion
13. D.76. At a party. The median (the middle of the list) would be the average of 9 and 10. The class average can be found as follows: 11.7622 °C 61 The answer must be 8. 10. 10. determine the number of temperature points for the month of May: 31 × 9. 12 The mode (most frequent shoe size) is 10. 10. 8
16. If m c 5 c 3 A.
B. C. 4 6 10 11 24
18. C. which of the following must be even? A.25 gallons 4
Number Properties: Positive and Negative Integers. 3. Prime Numbers. and 8. Odd and Even Numbers. If you write your raisins to peanuts to candies ratio as 1x : 2x : 5x. Notice that m c c ratios on the left in the equation above. E. The prime factorization of 360 can be written in the form ambncr. x+y x–y (x + 3)(y – 4) x+y–6 4(x + y) – 2
20. C. How many two-digit positive integers are multiples of both 2 and 5? A. 16. 9 or 2. A number is divisible by 2. it must be 3 parts paint concentrate. If the paint is 1 water. then there are 3 times 3 or 2 1 gallons of paint concentrate in 3 gallons of paint. D. Factors and Multiples. Substituting the numerical values for the 14. Therefore the number of ounces of peanuts will be 5x = 5(14) = 70.25. D. See the proportion below: 4 4
3 gallon 4 1 gallon x gallons 3 gallons The correct answer is x = 3 gallons. What is the value of a + b + c – mnr? A. D. E. Since the ratio of males to females is 3 parts to 4 parts. you have 2 # 4 = 8 . the number of people at the party must be a multiple of 7 parts. is not a multiple of 7.Part II: Review with Sample Problems
Explanations
13. B. B. C. r # m = r . If x and y are positive integers. E. 3 5 15 15. 44. Choice C. Divisibility
17. C. you will have 1x + 2x + 5x = 112 8x = 112 so x = 14. which is the ratio you are trying to find. If there are 3 parts paint in one gallon of 4 4 4 4 paint. 4
=
Solving for x by cross multiplication. and are not both even. What is the smallest three-digit number that is divisible by this number?
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. 5 6 9 45 48 19. E. x = (3)
9 gallons or 2.
you have: 4(x + y) – 2 = 4(any number) – even = even – even = even 19. Serena has $500 in a savings account that pays 4% simple interest per year. 3 × 2 × 1 = 6. Interest. x + y = odd + even = odd. What was the percent increase in the price of a gallon of regular unleaded gasoline? A. 3.
Explanations
17. 30.00 $562. 18. B. the value of (a + b + c) – m × n × r = 10 – 6 = 4. 3. or x + y = even + odd = odd B. it must also be a multiple of 15. Distance.50/gallon to $2. You are comparing this to am × bc × cr. x + y – 6 = odd + even – even = odd or x + y – 6 = even + odd – even = odd E. So Choice C does NOT have to be even.37 $560. The smallest three-digit multiple of 24 is 120. Look for the smallest number that is divisible by all three numbers. Just to double-check.00 $512. . $512. x – y = odd – even = odd. 9% 10% 20% 25% 90% 22. and 8.43 $566. Begin by finding the least common multiple (LCM) of 2. .
360 10 2 2 5 5 4 36 9
2 2 3 3
Therefore. D. how much money will be in Serena's savings account after 3 years? A. If she does not deposit or withdraw from this savings account. A. you find the prime factorization of 360. and 8. And the product m × n × r is just the product of the exponents. and the smallest three-digit multiple of 96 is 192. you have: A.75/gallon. D. the answer is 120. the smallest three-digit multiple of 72 is 144. 360 = 23 × 32 × 51. E. 120. or x – y = even – odd = odd C. 24 is divisible by 2. So our two-digit multiples of 15 are: 15. Price per Item
21.
Word Problems: Solving for Percents. 2 + 3 + 5 = 10. the price of regular unleaded gasoline went from $2. the smallest three-digit multiple of 48 is 144. So the sum (a + b + c) is just the sum of the bases. Time. So our correct answer is 6 two-digit multiples of 15. 45. 60. C. E. Rate. E. The correct number is 24. Choice E must therefore be correct. Since you are looking for the smallest three-digit number. Trying each of the choices one at a time. 75. C. Finally. B. By default. D. (x + 3)(y – 4) = (odd + odd)(even – even) = (even)(even) = even or (x + 3)(y – 4) = (even + odd)(odd – even) = (odd)(odd) = odd. No smaller number is divisible by all three numbers. B. and 90. Making a "factor tree". 20. If the positive integer is a multiple of both 3 and 5. In a one-week period. Average.84
75
.Numbers and Operations .
E.6 0. point D is between points A and E. It took Esteban 1 minute and 20 seconds to travel one mile.50 10 22. what is the distance from M to N? A. B. E C. The difference between 66 and 45 is 21. Betweenness
3.04) × 3 I = 60 60 + 500 560 dist 23. D. you need 40 + 40 8 + 10 18 9 total 4 5 dist to use time = rate . E. d. A. D. Multiplying both fractions by 60 to convert to miles per hour.25 3. E. in miles/hour. On a number line. Simple interest = principal × rate × time 21. B. Then I = 500 × (. 20 9 40 9 9 2 80 9 9
24.50 = $0. D.0 F
25. for the entire trip? A. If M is the midpoint of AB and N is the midpoint of DE .75 – $2. If he then drives this one-mile distance in 55 seconds. Average rate = timetotal = 40 + 40 = 80 = 80 = 40 . D D. What is his average rate. To compute the time going and returning. which of the following is a possible order of the points A. A.
Number Line: Order. If Esteban travels one mile in 1 minute and 20 seconds (80 seconds).25 1 = = 10% increase in price of gasoline $2. and F are equally spaced on the number line. E
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. A. B.6 3. E C. Fractions. and E have respective coordinates a. how much faster is he driving in miles per hour?
Explanations
$0. points A.8
26. B. he will travel 60/80 (3/4) miles in one minute. E. 21. D. If c < d. point A is between points C and D.25. Consecutive Numbers. C. B C. and E? A. Ling rides his bicycle a distance of 40 miles in 5 hours and then returns the same distance in only 4 hours. B. points A. B. If he travels one mile in 55 seconds. he will travel 60/55 miles in one minute. B. 24. C. b. A. D. D. C. the speeds will be 45 mph in the first case and 66 mph in the second case.Part II: Review with Sample Problems
23. E.0 A B C D E 4. 0. B. C. and e but not necessarily in that order. $2. E. c. B. C. C. C. A. B.75 3. D. C. D. B. B. D. In the diagram above. C.
the numbers are 104. 5
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. v. 3. combined with the figure above. y. x – 4. can be located anywhere. and E have been labeled with their respective coordinates: 3. Since N is the midpoint of DE .6. If x – 4 = z. corresponding with Choice A. which is 108.1 = 0.0 A M 3.8 E 4.1. the final order is as follows: y. 28. w. 108. you have the series below: y. Using the given relationships between coordinates and points. y. and 112. x. y<z<v<x<w v<z<y<x<w z<v<x<w<y z<y<x<v<w v<w<z<y<x
28. The correct answer is 108. v.
27. and 3.4. points B. The other numbers would be two greater. Therefore. . 106. The sum of five consecutive even integers is 536. y + 6 = x. x. and four less. four greater. Therefore. z. and z? A. y + 6 = x. D.
3. x + 4. w. let the numbers be: x. the coordinate of point M is 3. you get x = z + 4. two less. x + (x + 2) + (x + 4) + (x – 2) + (x – 4) = 540 5x = 540. as shown below: z.2. the middle number must be 540 divided by 5 (the average).Numbers and Operations . E. x. D. with coordinate b. x – 2. and x = 540 = 108 . you know there are two other odd integers between x and y.0 F
In the figure above. 108. 110. 27. D. __. . Since the five numbers add to 540. __. and z are consecutive odd integers. 3.6 26.8. __ Then since w > v. C. which of the following is the correct order for the integers v. and there is one other odd integer between z and x.7 3. and w > v. Therefore the distance from M to N is 3. What is the middle number of the five numbers?
Explanations
25. w. x From the second piece of data. C. B. A. z. x + 2. the coordinate of point E is 3. A.4 C 3.6 D N 3. not necessarily in that order. Choice D is the only choice given a possible order.7 – 3. x – 4 = z. From the first piece of data. you have the following order of the coordinates:
c a d e
Point B.7. so z is less than x. Since M is the midpoint of AB .2 B 3.1 3. Using an equation to solve the problem. x.
So after 120 minutes. 64.000 when first measured.024 2. A culture of 207 bacteria triples every 12 minutes. This is 10 of the 12-minute time periods in which the bacteria triples. t(0) = 25. What is the change in population during the fifth hour?
Explanations
29. Start by making a chart showing the population for each hour. 207 × 32 207 × 36 207 × 310 207 × 312 207 × 313
30. 32. After 6 years. The population at the end of the first hour is 5.92 $141. 32. What will the bacterial count 2 hours after the original culture of 207 was established? A. In a laboratory experiment a culture of bacteria contained a population of 25. at the end of the second hour is 1. D. . What is the tenth integer in this sequence? A. C.000. 20 128 512 1. C. the initial count of 207 will grow to 207 × 310.000 t(3) = 200 t(4) = 40 The change in the fifth hour is the difference between 40 and 8.85
32.03 times the previous year's value. 128. B. . 8.000. and so forth. Therefore. 512.000 t(1) = 5. E. E. 4.37 $125. t(5) = 8 32 is the correct answer.03)6 = $119. . B. C. 4. What is the accumulated value of this deposit 6 years after the day of the initial deposit? A. The integers double as you move from one term of the sequence to the next. D. 30. D.048 31.41. . B. $118.00 $119. 32.41 $121.
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. 2.Part II: Review with Sample Problems
Sequence Involving Exponential Growth
29. E. the account will be worth $100 × (1. 2.000 t(2) = 1. the first ten integers in the sequence are: 1. 16. C. B. Yuri deposits $100 into a new savings account that pays him 3% interest compounded annually. 31. 1024. Since the annual interest rate is 3%. A 2-hour time period represents 120 minutes. C. The first four numbers in a patterned sequence of positive integers are 1. 8. the value of the account one year is 1. The bacteria population decreased such that each hour the population was 1 that of the 5 preceding hour.
II. 3. M is the set of integers that can be written as n . 4. 9.3. 7. . and T is the set of positive multiples of 3 less than 21. C. 15} {6. III. C. 11. 48. 15. D. 18. 3. B. 35. 15. If A is the set of positive odd integers less than 21.3 = 4 5: 5 can be written as 64 . 7. what is A . 9. 8. 17. 19} {}
36. 5. What is the smallest threedigit member of the intersection of sets A and B?
Explanations
33. Which of the following integers is a member of set M? I. {3. 49.3 = 7 . Remember that a prime number is only divisible exactly by one and itself. 34. 81} {1} {1. S + C is the "intersection" of the two sets. 15. 3. II. E.3 = 5 7: 7 can be written as 100 . 9. A.
Sets: Union. 11. 6. 3. 81} {1. A = the set of prime numbers and B = the set of positive odd integers. etc. 4. 9. where n is a nonzero integer. {1. T is the "union" of the two sets. 64}. 4. Find a few prime numbers that are greater than 99. 36. 8. 18} {1. because all three of the numbers are in set A. Since 101 is prime and 100 is not. B. E. 6. 7. 11. If S is the set of perfect squares less than 100. 5. T ? A. 9. D. A . E. 19}. D. . 64} {64} 35. 16. 7. 13. 107. 25. You can use the divisibility rules for 2. 64. 9. These include 101. 15. 25.3 = 10 . So the correct answer is the set {1. Elements
33.3 = 7
Therefore. 27. 17. 8. 17. 18. 4 5 7 I only II only III only I and II only I. III. C. 9. 27. 10 to help in finding the prime numbers. 19} and T = {3. 9. B. The integers common to both sets are only 1 and 64. S = {1. D. 15. The integers belonging to either one of the sets are {1. 12. 18}. 64}. Checking the answer choices one at a time. 11. 3. 13. 12. 13. E. 81} and C = {1. 36. II.
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. 101. C. A = {1. 25. 36. 49.Numbers and Operations . 17. 64. 9. 9. you have: I. and C is the set of perfect cubes less than 100. 101 is the correct answer. and III
34. 5. 5.3 = 8 . 19} {1. 16. the correct answer is Choice E. 6. 6. 4: 4 can be written as 49 . what is S + C? A. Intersection. 13. 5.
39. According to the graph above, the percent increase in commission from sales from 2001 to 2002 is approximately how much greater than the percent increase in commission from sales from 1998 to 1999? A. B. C. D. E. 2 12 1 2 21 25 33 1 3
So the difference is 33 1 - 12 1 = 21, approximately. 3 2 2 or 22.2 First, analyze the graph for important details. Note that the number of crime incidents for 1980 is 40. 9 9,000 and for 1990 is 7,000. The vertical axis indicates that these figures are in thousands. The method to calculate percent increase or percent decrease is the difference in the two amounts divided by the original amount. In our figure, the expression for this method is: 9000 - 7000 9000 This is equal to 2 or about 22.2 percent. The correct answer is 2 or 22.2%. Note: The expression could have been 9 9 9 - 7 , and the same answer would have occurred. 9
41. The scatter plot above shows the score on each of three tests for seven students. Which of the following could be the median score for all seven students? A. B. C. D. E. 65 71 78 81 91
42. The scatter plot above shows the heights of three randomly selected students from each of six age groups. According to the data in this plot, which of the following statements is true? A. B. C. D. E. There are only two students having heights 68 inches. Of the students who are 72 inches tall, none is 13 or 14 years old. More 18-year-old students than 15-year-old students have heights greater than 66 inches. Half of the students have heights greater than 66 inches. There are no 17- or 18-year-old students having heights less than 66 inches.
43. The scatter plot above shows the time it took one or more workers to complete a particular job. In some cases, the workers made just one attempt to complete the job; in other cases, the workers made multiple attempts to complete the job. Based on the data presented, which of the following functions best states the relationship between t, the time in hours to complete the job, and w, the number of workers? A. B. C. D. E. t ^ w h = - 4 w + 40 5 t(w) = –4w + 40 t(w) = –w + 35 t ^ w h = 3 w + 40 4 1 t ^ w h = w + 35 2
44. In the scatter plot above, mothers' heights are compared to their daughters' heights. This was done at a parent meeting, for 15 junior girls who were participating in the graduation ceremony. According to the graph, which height for mothers had no daughter's height given?
84
Numbers and Operations . . .
Explanations
41. D. The median test score is the one score above and below which half of the scores lie. Note that 10 scores are below 80, which means that 11 scores are above 80. So the median score should be a little above 80, as in Choice D. 42. C. Checking the answer choices one at a time, it will be noted that C is correct. 43. B.
45 TIME (hours) TO COMPLETE JOB 40 35 30 25 20 15 10 5 1 2 3 4 5 6 7
NUMBER OF WORKERS
A line has been sketched that seems to best fit the data presented. Since the line travels downhill to the right, our function, t(w), must have a negative slope, as in A, B, and C. Next, the t-intercept appears to be closer to 40 than to 35, as in A and B. Last, you must compute the slope of the best-fit line, using the two circled points. Going from left to right down hill, the vertical decrease is 4 squares (but each square represents 5 units) or –20; the vert. change - 20 = = - 4. horizontal change is 5 squares (here, each square is just 1 unit) or +5. So the slope = horiz. change + 5 Therefore, the equation of our line of best fit is: t(w) = –4w + 40 44. 69 or 70. Looking above the values for the mothers' heights that are expressed along the horizontal axis, the frequency of heights are 2, 2, 1, 2, 3, etc. There are no entries above a mother's height of 69 or 70 inches. Therefore, the correct answer is 69 or 70.
Probability
45. On a true-false quiz having just four questions, what is probability of guessing and getting all four questions correct? A. B. C. D. E. 1 16 1 8 1 4 1 2 2 46. A marble is to be selected at random from a jar that contains marbles of five different colors. If the probability that a red marble will be selected is 2 7, which of the following could NOT be the number of marbles in the jar? A. B. C. D. E. 14 35 42 65 70
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Part II: Review with Sample Problems
47. A jar contains three green, four red, and five blue marbles. In four successive draws, with no replacement of marbles that have been already drawn out, what is the probability of drawing a green, a red, another green, and then a blue marble in exactly that order? A. B. C. D. E. 5 (approximately 0.009) 576 1 (approximately 0.010) 99 1 (approximately 0.015) 66 7 (approximately 1.167) 6 2711 (approximately 1.369) 1980
48. Five students are chosen to attend a student nominating convention for their school election to represent their geometry class. In how many ways can the five students be arranged in the five chairs that are reserved for their seating?
Explanations
45. A. The probability of guessing a give question correct is 1 . So the probability of getting all four correct is 2 1:1:1:1= 1. 2 2 2 2 16 46. D. If the probability of selecting a red marble is 2 7 , then the total number of marbles has to be multiple of 7, the denominator. Therefore, 65 cannot be the total number of marbles, since it is not a multiple of 7. 47. B. Probability of drawing a green first = 3 , 3 green out of 12 total marbles. 12 4 Probability of drawing a red second = , 4 red out of 11 marbles left in jar 11 Probability of drawing a green third = 2 , 2 green remain out of 10 marbles 10 Probability of drawing a blue fourth = 5 , 5 blue out of 9 marbles left 9 Then the final probability is just the product of the 4 probabilities found above: 3 : 4 : 2 : 5 = 1 , when 12 11 10 9 99 completely reduced. 48. 120. Since this is an arrangement problem, the solution is the number of permutations of five students arranged five at a time. If your calculator has a permutation option, this can be calculated on the calculator using the appropriate sequence of keystrokes. In a non-calculator method, you use the following: Make five consecutive blanks in a line to represent the five seats. Next, there are 5 ways to pick the person for the first seat. Once that person is selected, there are 4 ways to pick the person for the second seat, then 3 ways to pick the third seat, then 2 ways, and then 1 way to pick the last seat. Since this is a permutation problem, the arithmetic will be: __ __ __ __ __ 5 × 4 × 3 × 2 × 1 = 120
86
" five letters are chosen at random to form a "word.039) 700 9 (approximately 0. 10. 9. what is the probability that the point will be in the smaller square?
87
. Within each suit. A small square is located in the interior of a larger square as shown in the diagram above.970 51. D.185 0. 5. D. 4. 8. Queen.82. B.278) 18 17 (approximately 0. cards are labeled as 2.261) 23 5 (approximately 0. a King. respectively. . E.
Geometric Probability
49. 27 (approximately 0. Using the letters of the word "EXCLUSION. in specifically that order? A. King. basketball. Jack. D. If a point is randomly selected from the region of the large square. and 0.554 1. . What is the approximate probability that both the volleyball and softball teams win their league titles.180) 50 6 (approximately 0. A deck of playing cards has 52 cards in 4 different suits—Clubs. B. with 13 cards in each suit.90. C.Numbers and Operations . and Spades. E.75. a Club. 0.354) 48
3
8
52. 6. Diamonds. any arrangement of the five letters. B.005 0. 7. 0. The large square is 8 units on a side and the small square is 3 units on a side. E. What is the probability that the "word" chosen begins with a vowel and ends with a consonant? A. 3. and Ace." that is. but basketball does not? A. C. and then another Club. The probability of a high school's girl's volleyball.014 0. and softball teams winning their league title this year is 0. C. 4 52 4 52 4 52 4 52 45 52 # 13 # 51 13 # # 51 # 12 # 51 # 12 # 51 12 # # 51 4 50 4 50 4 50 4 50 4 50 # 12 49 11 # 49 # 11 49 # 10 49 # 9 49
50. If one card is randomly drawn from the deck in each of 4 successive draws—with no replacement of cards after each draw—which of the following could NOT be a possible probability of drawing an Ace. Hearts.
00 – 0.82. So you only need to look at the second and fourth fractions to locate which could NOT be possible. the requested probability is the ratio of 4200 = 5 = 0. the probability of drawing a Club is 13/51. there are 10 Clubs left – so the probability of drawing a Club on the fourth draw is 10/49. you have the following (where #1 through #4 represent the choices you made above):
#1 #3 #2 5 18
4×(7×6×5)×5 9×8×7×6×5 #4
=
after completely reducing.120 Last.90. there are 11 Clubs left – so probability of drawing a Club on the fourth draw is 11/49.82) = 0. Choice E is NOT possible.Part II: Review with Sample Problems
Explanations
49. The word "EXCLUSION" has a total of nine letters—four of them vowels and five of them consonants.185 50.25. Probability of the volleyball team winning = 0. there are 13 clubs left – so the probability of drawing a Club on the second draw is 13/51.200 Choice #4—The number of five-letter "words" that can be chosen with no conditions on our choices is: 9 × 8 × 7 × 6 × 5 = 15. there are 12 Clubs left – so the probability of drawing a Club on the fourth draw is 12/49. Note: The number of five-letter "words" that can be chosen under the given conditions is the product of our three choices above: 4 × 210 × 5 = 4. So by default. In C: If the Ace of Clubs is chosen. Note that the first and third fractions for each answer choice are the same. you have only seven letters remaining. there are 12 Clubs left – so the probability of drawing a Club on the second draw is 12/51. of which you have to select three). D. You want a five-letter "word" that begins with a vowel. These just represent: For the first fraction: 4 out of 52 cards are Aces. Also note that the denominators decrease by one as you move from the first to the second to the third and then to the fourth fractions since there is one card less in the deck after the preceding draw has been made. If the King of Clubs is chosen. If the King of Clubs is not chosen.25)(0. 15120 18 To do this all in one step. Choice #2—Select a consonant: This can be done in five ways. Probability of the basketball team losing = 0. If the King of Clubs is not chosen. there are still 13 Clubs left in the deck of 51 remaining cards. but ends with a consonant. In D: If the Ace of Clubs is chosen. If the King of Clubs is chosen. Therefore. In A: If the Ace of Clubs is not chosen. this is just 1. there are 11 Clubs left – so the probability of drawing a Club on the fourth draw is 11/49. there are 12 Clubs left – so the probability of drawing a Club on the second draw is 12/51.
51. Choice #3—Select the remaining three letters to complete the five-letter "word": This can be done in 7 × 6 × 5 = 210 ways (after choosing a vowel and a consonant.
88
. In B: If the Ace of Clubs is not chosen. E. Choice #1—Select a vowel: This can be done in four ways.278 approximately.90)(0.75. The requested probability is the product of the three probabilities above: (0. C. Probability of the softball team winning = 0. For the third fraction: 4 out of the remaining 50 cards are Kings – 50 cards because 2 have already been drawn during the first 2 draws.
which is equal to 64 square units. the mode. C. C. 64
Basic Statistics (Mean. The average (arithmetic mean) of a set of six two-digit even integers is 17. and 81% on his last four tests. D. The range is the difference between the largest and smallest of the numbers in the set. 11.
9 or 0. What score does he need to earn on his next test in order to average 80% on all five tests? A. Probability is defined as the numbers of ways the event (a point in the small square) can occur divided by the total ways a point can be selected (sample space) for the larger area. The following numbers were obtained in a statistical experiment. If the misplaced number is k. 16. The area of the 64 smaller square is 32. what is the value of k?
Explanations
53. D. 11. B. Mode. Let T be the score on Jahwad's next test. 16. 78%. C. The average (arithmetic mean) of the 5 smallest numbers is increased by 7. the probability is: P = 9 or gridded as approximately 0. What is the largest possible even integer in this set? A. the other numbers are 14. The mode is increased by 7. and the mean is known to be 16. The median of the 13 numbers is increased by 7. you have: 85 + 78 + 73 + 81 + T = 80 5 371 + T = 80 5 317 + T = 5 80 5c ^ h 5 m 317 + T = 400 T = 83 54. Therefore. 20 22 32 34 40
54. 73%. D. E. D. Range)
53.Numbers and Operations .
52. B. If 7 is added to each of 13 different numbers. D. 21. . 26. Jahwad earned grades of 85%. E.
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. The average (arithmetic mean) of the 6 largest numbers is increased by 7. which is 9 square units. B.14 The area of the larger square is given by 82.14.
56. median. which of the following statements would NOT be true? A. The range of the 13 numbers is increased by 7. If 7 is added to each member of the set. and mean are all increased by 7. Median. E. so the new range would be found by: (largest number + 7) – (smallest number + 7) = largest number + 7 – smallest number -7 = largest number – smallest number = original range Thus range is not affected. 75% 79% 80% 83% 84% 55. Then to earn an 80% average on the five tests. . but one of the numbers was misplaced. 17.
The process in finding a mean requires finding the sum of all the numbers that are being added. C. 12. Therefore.Part II: Review with Sample Problems
55. The sum of these five numbers is 70. If the average of the six positive even integers is 17. k = 144 – 132 and k = 12 The correct answer is 12. The smallest that the first five positive even two-digit integers could be: 10. then the total of the six integers is: 6(17) = 102. 16. 12. and 18. the sixth and largest possible even integer would be: 102 – 70 = 32. Therefore. 132 + k = 16 and 132 + k = 9 16 × 9 = 144. This value must be 16 × 9 since there are nine numbers that figure into the mean. 56. 14.
90
.
For example:
Verbal Expression (a) the sum of a number and 7 (b) the number diminished by 10 (c) seven times a number (d) x divided by 4 (e) five more than the product of 2 and n Algebraic Expression n + 7 or 7 + n n – 10 7n x/4 2n + 5 or 5 + 2n
These words can be helpful in making algebraic expressions.Algebra and Functions
Algebra Review
Variables and Algebraic Expressions
A variable is a symbol used to denote any element of a given set—often a letter used to stand for a number. Variables are used to change verbal expressions into algebraic expressions.
Key Words Denoting Addition sum plus more than greater than larger than gain increase enlarge rise grow
Key Words Denoting Subtraction difference minus lose less than smaller than fewer than decrease drop lower diminish reduced
Key Words Denoting Multiplication product multiplied by times twice of
Key Words Denoting Division quotient divided by ratio half
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.
insert the value for the unknowns and do the arithmetic. Solve for x: x – 5 = 23 To solve the equation x – 5 = 23.^.2 = 6 + 4 . add 5 to both sides: x . the equation is still balanced. ab + c if a = 5.4h . or center. subtract.4h . therefore. and the equation does not change. get x by itself on one side. For example: Evaluate each of the following.x .4 -4 3x = 15
94
. Solve for x: 3x + 4 = 19 Subtract 4 from both sides to get the 3x by itself on one side: 3x + 4 = 19 . add 5 to each side).-4 . first you must get the variable you are looking for on one side of the equal sign and everything else on the other side. 1. For example: 2. To solve an equation. For example: 1. If x = –4.6 = 10 . with the equal sign (=) being the fulcrum. b = 4 and c = 3 5(4) + 3 = 20 + 3 = 23 2.^.x . or divide both sides of an equation by the same (nonzero) number. Sometimes you have to use more than one step to solve for an unknown.5 = 23 + 5 +5 x = 28 In the same manner.Part II: Review with Sample Problems
Evaluating Expressions
To evaluate an expression. Thus.6 = 4
Equations
An equation is a relationship between numbers and/or symbols.2 = 10 . 2x2 + 3y + 6 if x = 2 and y = 9 2(22) + 3(9) + 6 = 2(4) + 27 + 6 = 8 + 27 + 6 = 41 3. multiply..2 = 6 . then 6 . It helps to remember that an equation is like a balance scale. if you do the same thing to both sides of the equal sign (say.
such as x–3. 9x. x . then the variable and exponent can be dropped under the number 1 in a fraction to remove the negative sign. divide both sides by p to get x alone: bq px p = p bq bq p = x or x = p Cross multiplying can be used only when the format is two fractions separated by an equal sign. subtract the exponents of like terms: y 15 = y 11 y4 Remember: x is the same as x1. x5 y2 = x2 y x3 y 36a 4 b 6 = .18x 2 yz . as follows. y2 – x2. follow the same rules as with regular signed numbers. and 3mpxz2 are all monomials. A polynomial consists of two or more terms.
Proportions
Proportions are written as two fractions equal to each other.3x yz
2
3x + 2x = 5x
Multiplying and Dividing Monomials
To multiply monomials.4a 3 b 5 . p Solve this proportion for x: q = x y This is read "p is to q as x is to y. provided that the terms are alike: 15x 2 yz .3 = 13 x
97
. x + y. 4a2. add the exponents of the same terms: (x3)(x4) = x7 (x ⋅ x ⋅ x)(x ⋅ x ⋅ x ⋅ x) = x7 To divide monomials. For instance.
Adding and Subtracting Monomials
To add or subtract monomials. and x2 + 3x + 5y2 are all polynomials." Cross multiply and solve: py = xq py xq q = q py py q = x or x = q
Monomials and Polynomials
A monomial is an algebraic expression that consists of only one term.Algebra and Functions
Then.9ab
Working with Negative Exponents
If an exponent is negative.
2.h 2a .2a 2 + b 2 .) 2y3 – 6y = 2y(y2 – 3) Another example: x5 – 4x3 + x2 = x2(x3 – 4x + 1)
Factoring the Difference between Two Squares
Factor x2 – 144. (The second factor is a polynomial. (x )(x ). of the form Ax2 + Bx + C: 1. a2 – b2 = (a + b)(a – b) 9x2–16y2 = (3x + 4y)(3x – 4y)
Factoring Polynomials That Have Three Terms: Ax2 + Bx + C
To factor polynomials that have three terms. x2 – 144 = (x + 12)(x – 12) Another example: Another example: Note: x2 + 144 is not factorable.2a 2 6x 2 . use double parentheses and factor the first term. For example. Find the largest common monomial factor of each term.2a h = 2x . Find the square root of the first term and the square root of the second term. and place the factors in the right sides of the parentheses. Then.2a 2
similar to
23 # 19 207 230 427
Factoring
To factor means to find two or more quantities whose product equals the original quantity. 2. Check to see if you can monomial factor (factor out common terms). the first term is simply x2). multiply each term in one polynomial by each term in the other polynomials. 2.
99
.6ax 6x 2 .Algebra and Functions
Subtract: a2 + b2 a2 + b2 2 2 ^ . Factor the last term. 1.
Factoring Out a Common Factor
Factor 2y 3 – 6y. Place these factors in the left sides of the parentheses.a 2 + 2b 2
Multiplying Polynomials
To multiply polynomials. if A = 1 (that is. Express your answer as the product of the sum of the quantities from step 1 times the difference of those quantities.b " +.4ax . Divide the original polynomial by this factor to obtain the second factor. Then simplify if necessary:
^ 3x + a h^ 2x . 1.2a
# 3x + a + 2ax .
the signs of both factors are +. First check to see if you can monomial factor (factor out common terms).10 2. For example. do the following: If the sign of the last term is negative: 1.Part II: Review with Sample Problems
To decide on the signs of the numbers. (x + 3)(x + 5) Notice that 3 × 5 = 15 and 3 + 5 = 8. the coefficient of the middle term.3x . 1.5x x 2 . x-5 # x+2 + 2x . which is –3. use double parentheses and factor the first terms as follows: (x )(x ).
100
. the first term has a coefficient—for example. Factor x2 + 8x + 15. Next. the sign of the middle term. then additional trial and error is necessary. however. To check your work:
(x + 3) (x + 5) + 3x + 5x + 8x (the middle term)
If.10 x 2 . 5 must take the negative sign and 2 must take the positive sign because they then total the coefficient of the middle term. Also.) Add the proper signs. 2. factor the last term (10) into 2 times 5. leaving: (x – 5)(x + 2) Multiply the means (inner terms) and extremes (outer terms) to check your work. 4x2 + 5x + 1). and give the opposite sign to the other factor.
(x − 5)(x + 2) − 5x + 2x − 3x (which is the middle term)
To completely check. A ≠ 1 (that is. Find two numbers whose product is the last term and whose difference is the coefficient (number in front) of the middle term. Give both factors the sign of the middle term. multiply the factors together. 2. If the sign of the last term is positive: 1. Find two numbers whose product is the last term and whose sum is the coefficient of the middle term. Factor x2 – 3x – 10. (Using the preceding information. Because this is not possible. Give the larger of these two numbers the sign of the middle term.
There are three common methods for solving: addition/subtraction. 3x + 3y = 24 3(2x ) + 3(y) = 3(13) Now you can subtract equations. 3. Simply add or subtract. Multiply one or both equations by some number to make the number in front of one of the variables (the unknowns) the same in each equation.3y = . 4. 2 ^ 5h + y = 13 10 + y = 13 -10 -10 y=3 Answer: x = 5. Add or subtract the two equations to eliminate one variable. 2.6x +.Part II: Review with Sample Problems
Solving for Two Unknowns—Systems of Equations
If you solve for two equations with the same two unknowns in each one. 3x + 3y = 24 . 2. Now the y is preceded by a 3 in each equation. Solve for the other unknown.
Addition/Subtraction Method
To use the addition/subtraction method: 1.3x . if the number in front of a variable is already the same in each equation. Solve for x and y: x+y=7 x–y=3 2x + y = 13
108
. Solve for x and y: 3x + 3y = 24 2x + y = 13 First multiply the bottom equation by 3.39 . you can solve for both unknowns. you do not have to change either equation.15 -3 =5 3x + 3y = 24 6x + 3y = 39
Now insert x = 5 in one of the original equations to solve for y. y = 3 Of course. substitution and graphing. eliminating the y terms. Insert the value of the first unknown in one of the original equations to solve for the second unknown. For example: 1.3x -3 x = -15 = .
From the first equation. For example. Now solve for y. finding a third point is a good way of checking. If you are unfamiliar with coordinate graphing. find three values for x and y that satisfy each equation. 1. substitute (y + 8) for x in the second equation. The coordinates of the intersection are the solution to the system. carefully review the "Basic Coordinate Geometry" section before attempting this method. Simplify by combining y's. x=y+8 x = 10 + 8 x = 18 Answer: y = 10. This method involves substituting one equation into another.)
x=4+y x
4 2 5
x − 3y = 4 x
1 4 7
y
0 –2 1
y
–1 0 1
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. (Although only two points are necessary to determine a straight line. solve the following system by graphing: x=4+y x – 3y = 4 1.Part II: Review with Sample Problems
Substitution Method
Sometimes a system is more easily solved by the substitution method. 4y + 8 = 48 -8 = -8 = 40 4y 4y 4 y 3. (y + 8) + 3y = 48 2. First. Solve for x and y when x = y + 8 and x + 3y = 48. x = 18 = 40 4 = 10
Graphing Method
Another method of solving equations is by graphing each equation on a coordinate graph. Now insert y = 10 in one of the original equations. For example.
or ordered pairs.Algebra and Functions
2. 4. The point at which the two lines intersect is called the origin and is represented by the coordinates (0.
Basic Coordinate Geometry
Coordinate Graphs (x–y Graphs)
A coordinate graph is formed by two perpendicular number lines.0) x (4.and y-axes. These lines are called coordinate axes. they do not intersect. and there is no solution to the system. Numbers are not usually written on the x. If the lines are parallel. 0).
y
(0. 0) is the solution of the system. Now graph the two lines on the coordinate plane. Notice the placement of points on the following graph and the coordinates. The horizontal axis is called the x-axis or the abscissa. often marked simply 0. The vertical line is called the y-axis or the ordinate. that show their location.0)
x−3 y=4
3. Note: If lines are parallel. The point where the two lines cross (4. as shown in the following figure. they have the same slope.
•••
111
. y
•••
x=
4
+
y
5 4 3 2 1
•••
−5 −4 −3 −2 −1 −1 −2 −3 −4 −5
•••
0
x
1 2 3 4 5
Each point on a coordinate graph is located by an ordered pair of numbers called coordinates.
then try 1.Part II: Review with Sample Problems
+y • (3. don't combine the ordered pair of numbers.
Graphing Equations on the Coordinate Plane
To graph an equation on the coordinate plane. Repeat this process to find other solutions.−3) • (−5. graph the solutions. x is always negative and y is always negative.−5) −
On the x-axis.) Then. The first number in the ordered pair is called the x-coordinate and shows how far to the right or left of 0 the point is.3) • (−4. as the point (3. The coordinate graph is divided into four quarters called quadrants. or ordered pairs. 3). x is always positive and y is always negative. y
Quadrant II
Quadrant I
x
Quadrant III
Quadrant IV
■ ■ ■ ■
In quadrant I. In quadrant II. 2) is different from the point (2. In quadrant IV. In quadrant III.
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.−4) • (1. The coordinates. These quadrants are labeled as follows. (When giving a value for one variable. find the solutions by giving a value to one variable and solving the resulting equation for the other variable. The order of these numbers is very important. The second number is called the y-coordinate and shows how far up or down the point is from 0. x is always negative and y is always positive. y).−2) • • (2.5) (−3. and so forth.1)
+x
(−3. On the y-axis. numbers above 0 are positive and numbers below 0 are negative.2) • −
0
• (1. x is always positive and y is always positive. start with 0. Also. the numbers to the right of 0 are positive and to the left of 0 are negative. are shown as (x. because they refer to different directions.
0)
x
These solutions.
x
0 1 2
y
6 5 4
Now. when plotted. Graph the equation x + y = 6. If x is 0.
^2h + y = 6 -2 -2
y= 4 Using a simple chart is helpful.
^1h + y = 6 -1 -1
y= 5 If x is 2. then y is 5.Algebra and Functions
For example: 1. Equations whose graphs of their solution sets form a straight line are called linear equations. form a straight line. plot these coordinates as shown in the following figure. (0) + y = 6 y=6 If x is 1. involve variables with square roots. show division by a variable. then y is 6. then y is 4. These are called nonlinear equations. or have variables multiplied together do not form a straight line when their solutions are graphed.
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.
y
(0. Equations that have a variable raised to a power.
0)
x
These solutions.Part II: Review with Sample Problems
2. the terms of the equation must be in a certain order.
x
0 1 2
y
4 5 8
Now. when plotted. y = (2)2 + 4 y=4+4 y=8 Use a simple chart. The more points plotted.
y
(0. y = (0)2 + 4 y=0+4 y=4 If x is 1. If x is 0. Graph the equation y = x2 + 4. give a curved line (nonlinear). then y is 4. then y is 8. and the other involves the point where the line crosses the y-axis. plot these coordinates as shown in the following figure. then y is 5. y = (1)2 + 4 y=1+4 y=5 If x is 2.
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. To see either of these relationships. One involves the slope of the line. the easier it is to see the curve and describe the solution set.
Slope and Intercept of Linear Equations
Two relationships between the graph of a linear equation and the equation itself must be pointed out.
x – y = 3 –y = –x + 3 y=x–3 2. +1 and –2.
y y
(0.Algebra and Functions
(+)(1)y = ( )x + ( ) When the terms are written in this order. For example: Write the equations in y-intercept form. the lines cross the y-axis at –3. x – 2y = 4 –2y = –x + 4 2y = x – 4 y= 1 x-2 2 As shown in the graphs of the three problems in the following figure. the equation is said to be in y-intercept form or slope intercept.0) x
(1) y
(2)
(0. y = –2x + l (already in y-intercept form) 3.0) x
(0. y-intercept form is written y = mx + b. the last term in each equation. and the two relationships involve m and b. 1.0) x
(3)
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.
1. b is the y-intercept. it should be evident that the slope of the line is the same as the numerical coefficient of the x term. Pick two points.1 .x B = = = = -2 A B 1+1 2 ^1h . 2). y at point A . and calculate the slope. and calculate the slope.x B = A B x at point A .^ .3h . there is no difference in the slope. 2.^2h . To find the slope of the line. State the equation in y-intercept form.x B = = =1 A B ^ 3h . The slope of a line is defined as: the change in y the change in x The word change refers to the difference in the value of y (or x) between two points on a line. 1. Locate the y-intercept on the graph (that is.2
2.y at point B y -y slope in line AB = x A . 1. 4. such as A(1. 0) and B (5.Part II: Review with Sample Problems
If a linear equation is written in the form y = mx + b.
^ 0h .
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.^1h . and use it to locate other points on the line.x at point B Note: Points A and B can be any two points on a line.
y = –2x + 1 slope = –2 y-intercept = 1
3. Write the slope as a ratio (fraction).
^ .3 .
y = 1/2 x − 2 slope = 1/2 y-intercept = −2
Graphing Linear Equations Using Slope and Intercept
Graphing an equation using its slope and y-intercept is usually quite easy. pick any two points on the line. such as A(3. one of the points on the line). 1).
y=x−3 slope = 1 y-intercept = −3
2.4 y -y slope = x A . 3.1h
Looking back at the equations for examples 1.2 y -y slope = x A . Find the slope of x – y = 3 using coordinates.^ 5 h . Draw a line through the points. 2 and 3 written in y-form. Find the slope of y = –2x – 1 using coordinates. –3) and B(–1.
y = –4x + 3 Change the slope-intercept form to standard form Ax + By = C. Substitute m = –3 and the point (6. Substitute the slope and intercept into the slope-intercept form. m = –3 (given) Find the y-intercept. 1. m. b. Since y = –4x + 3 Adding 4x to each side gives: 4x + y = 3 2. b. Find the y-intercept. 4) into the slope-intercept form to find b. m = –4 (given) Find the y-intercept. 2. Find the equation of the line passing through the point (6. b = 3 (given) Substitute the slope and intercept into the slope-intercept form y = mx + b. m.
118
. Ax + By = C.
For example: 1. use the following approach. Find the slope. m. 3. Find the slope.0) x
Finding the Equation of a Line
To find the equation of a line when working with ordered pairs.Part II: Review with Sample Problems
y +1 +2 y=4 −2 −1 (0. slopes and intercepts. 4. 4) with a slope of –3. y = mx + b. Change the slope-intercept form to standard form. Find the slope. Find the equation of the line when m = –4 and b = 3. b.
4 + 2 . then f(x) is 1. plug in values for x (the domain) and work out the result (f(x)). Since y = –3x + 22. Substitute the slope and either point into slope-intercept form. y = –3x + 22. In a function.4 h . adding x to each side gives x + y = 1.2 h .
Functions
Definition of a Function
A function is a formula or rule that shows the association between elements in one set to the elements in another set. and x = 6 4 = (–3)(6) + b 4 = –18 + b 18 + 4 = b 22 = b Substitute the slope and intercept into the slope-intercept form: y = mx + b. Find the slope. Since y = –x + 1. For example: 1. simply plug in values for the domain (if not already given) and find the range (results). Since m = –1 and b = 1. m. –2). b. (You could think of this as y = x + 1.) Now. change in y ^ . y = –1x + 1 or y = –x + 1. Then.^ 3h Find the y-intercept. –4) and (3. f(x) = (0) + 1 f(x) = 1
119
. x = 5. place these "ordered pairs" in the coordinate graph. 3.2 m= = = = = -1 2 2 change in x ^ 5 h . all of the elements of the domain are unique or different. m = –3. The elements from the first set are the ones that are being plugged in (the domain). Change the slope-intercept form to standard form: Ax + By = C. and y = –4 –4 = (–1)(5) + b –4 = –5 + b 5 + –4 = b 1=b Substitute the slope and intercept into the slope-intercept form: y = mx + b.Algebra and Functions
y = mx + b where y = 4.
Graphs of a Function
To graph a function. Since m = –3 and b = 22. Find the equation of the line passing through points (5. If x is 0. Change the slope-intercept form to standard form: Ax + By = C. adding 3x to each side gives 3x + y = 22. Graph the function f(x) = x + 1. y = mx + b where m = –1. and the elements from the second set are the outcome or result (the range).^ .
f(x) = (2) + 1 f(x) = 3 and so on. Graph the function f(x) = 3x – 2. then f(x) is 2. f(x) = (1) + 1 f(x) = 2 If x is 2. plot the points as follows:
y
x (0. then f(x) is 3. plug in values for x (the domain) and work out the result (f(x)).
x 0 1 2 3 –1 –2 f(x) 1 2 3 4 0 –1
Next.0)
Since this is a linear equation or linear function.Part II: Review with Sample Problems
If x is 1.
120
. Now. 2. Using a simple chart is helpful. you could have plotted only three points.
x 0 1 2 f(x) –2 1 4
Next. f(x) = 3(1) – 2 f(x) = 3 – 2 f(x) = 1 If x is 2. then f(x) is –2. you need to plot only three points.
121
. Using a simple chart is helpful. Notice that when x is 0..0)
Since this is a linear equation or linear function. for example: 1.Algebra and Functions
If x is 0.. Take a look at a few quadratic functions. f(x) = 3(2) – 2 f(x) = 6 – 2 f(x) = 4 and so on. plot the points as follows:
y
x (0. the graph would cross the x-axis at that point. Plug in simple values for x and you get. then f(x) is 1. f(x) = 3(0) – 2 f(x) = 0 – 2 f(x) = –2 If x is 1. If y were 0. then f(x) is 4. the graph crosses the y-axis at that point. Graph the quadratic function y = x2 + 2.
so x = 2 or –2. the special symbol (something like #. You could be asked to solve for x. Don't be alarmed.Algebra and Functions
Plugging in 4 for x gives f(x) = 5(4) – 7 f(x) = 20 – 7 f(x) = 13 So f(4) = 13 in this function.
125
. @. For example: What are the possible values of x in the function f(x) = x2 – 4 if f(x) = 0? Plugging in 0 for f(x) gives 0 = x2 – 4
Next. what is the value of 2 # 3? Take a careful look at the given function: x # y = 3x + y2 Notice the positions of x and y and simply plug in 2 for x and 3 for y in the definition: = 3x + y2 = 3(2) + (3)2 =6+9 = 15 So 2 # 3 = 15. (You could have also solved the original equation [function] by adding 4 to each side. The key to solving special symbol problems is focusing on the definition (watch for the "=" if the words "defined as" are not used). solve the equation by factoring the difference of two squares: 0 = (x + 2)(x – 2) 0 = (x + 2) or 0 = (x – 2) So x could equal 2 or –2. leaving 4 = x2. For example: If x # y = 3x + y2. *) will be defined in terms of operations with which you are familiar.)
Function Involving New "Symbols" and Definitions
You could be given a special symbol in a function and asked to solve the function.
or 7 12. or 2.33 For an equation of this type, let the expression inside the absolute value symbol equal either 8 or – 8. 3 Solving the two equations below, implies that –3x = 9, and x = –3 or –3x – 1 = –8 implies that –3x = –7, and x = 7 3 7 The correct answer would be either –3 or or 2.33. So you would grid either 7 or 2.33. 3 3 –3x – 1 = 8
15. A. (x + 3)(x – 4) – (x + 2)(x + 5) = x2 – x – 12 – (x2 + 7x + 10) = x2 – x – 12 – x2 – 7x – 10 = –8x – 22 6 16. 1.2 or . Since the two sides of this equation are of the same format, and the left side has a 5 in the numerator of 5 one of the fractions, try 5 for x in the right member of the equation. Also, since the bottom of one of the fractions in the right member has a 5 in the denominator, try 5 for x in the left member of the equation. The statement will look like: 5-2=5-2 3 5 3 5 This is always true because of the reflexive property of equality. 5 is one of the solutions. Solving this equation is a more difficult approach. Arriving at the quadratic equation of 3(5x – 6) (x – 5) = 0, it follows that (5x – 6) = 0 and x = 6 . Therefore, 6 (or 1.2) is a correct answer also. 5 5
9 . Begin by substituting 3 for a. Next, subtract 5 from each member of the equation. You should have the 3 16 16 . Next, multiply both members by 3x. 3x is the least common denominator for the = following equation: 3 x 3 equation. You should now have 9 = 16x. Solving for x, you get x = 9 . 16
26. If the point (6,y) is on the graph of the equation y = 2 x - 5 , what is the value of y? 3 A. B. C. D. E. –9 –4 –1 1 33 2
1
2
3
x
28. If the line shown in the graph above is reflected about the y-axis and then translated upward 4 units, the resulting equation is y = ax + k. What would be the value of k?
Explanations
25. E. The graph of an equation of the form y = mx + b has a slope of m. So the slope of the graph of the equation y = - 3 x + 7 has a slope of - 3 . 4 4 Two lines are perpendicular if their slopes are opposite reciprocals. Therefore the slope of the line perpendicular to the graph of the given equation is 4 . The only equation among the answer choices having the correct slope is 3 Choice E. 26. C. If the point (6,y) is on the graph of the equation y = 2 x - 5 , substituting 3 2 6 for x gives y = : 6 - 5 = 4 - 5 = -1. 3 27. B. The point at which the graph crosses the y-axis will have an x-coordinate of 0. Substitute 0 for x. 2(0) – 3y = 12 –3y = 12 - 3y 12 = -3 -3 y = –4 2x – 3y = 12
132
Algebra and Functions
28. 2. The y-intercept of the line in the graph is at –2 or (0, –2). The reflection of (0, –2) about the y-axis remains (0, –2). When this point is translated upward by 4 units, the y-intercept will then be at 2. Since k in the equation represents the y-intercept, k = 2.
6 = 3 A. B.
49. 87 7 3 7 6 21 29 3
138
. f ^xh = x + 5
48. the value of f ^ 0 h = 1 . Therefore. When substituting negative numbers for x in f ^ x h = x 2 + 3 x + 1 you always get positive values for f(x). The problem is really just asking for the domain of the function f. E. the greatest value of the range is 4. E. If this function is reflected about the y-axis and then translated downward two units. The 2 smallest value of f(x) will occur when x = 0. in this case. The correct answer is 4. E. Therefore all values of f(x) will 2 be greater than or equal to 0. since you cannot find the square root of a negative number in the real number system. A.
Working with Positive Roots. C. 4. Which of the following functions have the same domain as well as identical ranges? I. D. E. C. C. and III None have the same domains and identical ranges. f ^ x h = 5 x + 2 only has meaning when the term x + 2 ≥ 0. III. Only Choice E fits this condition. 47. Since x + 2 ≥ 0. D. you have x ≥ –2. 2 2 2 2 3 3 2 6 3 50. 48. B. g^ x h = x + 3 domain: x ≥ –3
2
f ^ x h = x + 5 domain: all real numbers
range: f(x) ≥ 0 range: h(x) ≥ 0
range: g(x) ≥ 0
h(x) = (x – 2) domain: all real numbers
Therefore the functions in I and III have the same domains as well as the same ranges. B. II. 12 + 75 = A.Part II: Review with Sample Problems
47. The range will not change as a result of reflection about the y-axis. B. The range of the function y = f(x) is defined by –2 ≤ y ≤ 6. Determining the domain and range for each function yields: I. When this function is moved downward two units. g^xh = x + 3 h(x) = (x – 2)2 I and II II and III I and III I. the range values will all decrease by two units and be defined by –4 ≤ y ≤ 4. what will be the largest value in the range?
Explanations
45. D. II. III. II. 46. C.
If f(x) = x2 – 2x – 3. what is the y coordinate of the function at the point on the axis of symmetry of the graph?
59. In the figure above. For which of the following values of x will their y-coordinates be equal? A. 3). C. If f(x) = –x2 – 4x – k and contains the point (–4. –1 and 1 1 and 4 2 and 5 –2 and 7 –3 and 9
141
.Algebra and Functions
58. a portion of the graph of y = ax2 + bx + c is shown. D. E. which of the following could be the graph of f?
(A) y (B) y (C) y
x -3 1 -1 3
x -3 1
x
(D)
y
(E)
y
x -1 3 -1 3
x
y y = ax2+bx+c x 0 6
60. B.
These regions can be described by: a ≤ x ≤ b and c ≤ x ≤ d. the two points on the x-axis that are on the graph of the function are equidistant from the axis. D. exactly match each other. Begin by substituting the point into the function for x and f(x). The only answer choice that has two values of x meeting this condition is E. For the graph of f(x) = x2 – 2x – 3. so the graph of f crosses the x-axis at –1 and 3. the x-coordinate of its vertex must be 3. points with x-coordinates equal distances left and right of this line of symmetry must have the same y-coordinate. 59. its x-intercepts can be found by setting f(x) = 0 and then solving for x. Because of the symmetry of the parabola around the vertical line x = 3.
y Line of symmetry
3 -3 0 Vertex 6 units 6 units 6 9
x
60. when folded along this axis line. note that –3 is 6 units left of the line of symmetry at x = 3.
y
x a b c d
In the figure above. x2 – 2x – 3 = 0 (x + 1)( x – 3) = 0. while 9 is 6 units right of the line of symmetry. E. as in Choice D. 1.Part II: Review with Sample Problems
Explanations
57. f(x) ≥ 0 in the dark shaded regions—above or on the x-axis. For this reason. The equation will be 3 = –16 + 16 – k. 58. Since the coefficient of the x2 term is positive. Since the graph of the parabola y = ax2 + bx + c crosses the x-axis at 0 and 6. the graph of f must open upwards.
142
. C. The axis of symmetry is the line through the vertex of the graph that divides the graph into two pieces that.
143
. and the factored form is f(x) = –1(x + 1)(x + 3). evaluate f(–2). The function is f(x) = –x2 – 4x – 3. The values for x where the graph intersects the x-axis is the values where the factors (x + 1) and (x + 3) are equal to 0.Algebra and Functions
From this equation the value of k is –3. The correct answer is 1. the axis of symmetry is all points where x = –2. Since the axis line must pass through the midpoint of these numbers. The y coordinate of the vertex point is another method to solve the problem. Now substitute –3 into the function for k and factor the resulting polynomial. This value is –(–2)2 – 4(–2) – 3 or –4 + 8 – 3. Therefore this graph passes through –1 and 3. To find the value of the function where x = –2.
.
Name each of the following polygons: A. Lines that meet to form 90° angles are called _______ lines. What is the length of BC in the preceding figure?
D C
145
.Geometry and Measurement
Geometry Diagnostic Test (Geometry and Measurement)
Questions
1. What is the length of AC in the preceding figure?
A 10″ B 26″
D. 2. AC must be smaller than _______ inches.
A
B.
A C
A
3.
A
B
D B 15″ C
C
4.
C
A
B
5.
A
B
D
C
E. Lines that stay the same distance apart and never meet are called _______ lines.
A 8″
D
C
C.
B 18″ B 22″ C B
6. In the preceding triangle.
RS is called the _______. Find the area and perimeter of the preceding figure: A. Find the area and circumference for the circle in the preceding figure ( π . 22 7 ): A. (Use 3. B. area = B. What is the surface area and volume of the preceding cube? A. area = perimeter =
10″ 12″
7″
11. B. What is the area of ᭝ABC in the preceding figure?
146
. AB is called the _______.Part II: Review with Sample Problems
D
B S
A 4″ D 3″ 6″
B
R
C
C
A
7.14 for π):
8. B.
10. Find the area and perimeter of the preceding figure (ABCD is a parallelogram): A. Fill in the blanks for circle R in the preceding figure: A. surface area = volume =
y
C
9. area = perimeter =
A B
x
13. B. C. CD is called a _______. circumference =
16″ 15″ 12″ 30″ 13″
4″ 4″ 4″
12. Find the volume of the preceding figure if V = (πr2)h.
4
In the diagram. ∠1 and ∠2 are adjacent angles.
148
. ∠4 is an obtuse angle.
b
In the diagram.Part II: Review with Sample Problems
Geometry Review
Types of Angles
Adjacent angles are any angles that share a common side and a common vertex (point)
A 1 2 D
B
C
In the diagram. Any angle the measure of which is less than 90° is called an acute angle. A right angle has a measure of 90°. ∠BAC is a straight angle (also called a line). in the interior of an angle designates the fact that a right angle is
A 90° B C
In the diagram. A straight angle has a measure of 180°. Any angle whose measure is larger than 90° but smaller than 180° is called an obtuse angle. The symbol formed. ∠ABC is a right angle. ∠b is acute.
180°
B
A
C
In the diagram.
∠1 and ∠2 are complementary angles.
A 1 2 B C
In the diagram. ∠2. line l and line m intersect at point Q. Two angles whose sum is 180° are called supplementary angles. adjacent angles. ∠3 and ∠4 are supplementary angles. If ∠3 = 122°. ∠1.
C B 1 A 2 D
In the diagram. Vertical angles are always equal. its complement. because ∠ABC is a right angle. Therefore. ∠4. its supplement. ∠3. would be: 180° – 122° = 58°. Therefore. Those angles opposite each other are called vertical angles.
A
3 4 B
C
In the diagram. since ∠ABC is a straight angle. AB is the angle bisector of ∠CAD. A ray from the vertex of an angle that divides the angle into two equal pieces is called an angle bisector. ∠2. again. If ∠1 = 55°. Four angles are formed.
149
. and ∠4 are formed. If two straight lines intersect. Those angles sharing a common side and a common vertex are.Geometry and Measurement
Two angles whose sum is 90° are called complementary angles. would be 90° – 55° = 35°. ∠1 = ∠2. ∠1 + ∠2 = 90°. ∠3 + ∠4 = 180°. they do so at a point. Therefore. Two adjacent angles that form a straight line are supplementary.
l
4 Q 2 3
m
1
In the diagram.
Part II: Review with Sample Problems
∠1 and ∠3 ∠2 and ∠4 ∠1 and ∠2 ∠2 and ∠3 ∠3 and ∠4 ∠1 and ∠4
are vertical angles are adjacent angles
Therefore.
l Q m
In the diagram. Parallel lines never meet. The symbol is used to denote parallel lines. line n is the transversal. cut by a transversal. l m .
l m
In the diagram.
l
m
In the diagram. In the diagram below. The symbol = is used to denote perpendicular lines. Eight angles are formed.
■
Two lines that meet to form right angles (90° angles) are called perpendicular lines.
■
Two or more lines that remain the same distance apart at all times are called parallel lines. There are many facts and relationships about these angles. it is termed parallel lines. lines l and m intersect at Q. l = m . ∠1 = ∠3 ∠2 = ∠4
Types of Lines
■
Two or more lines that cross each other at a point are called intersecting lines. That point is on each of those lines.
n 1 2 3 4 5 6 7 8 l m
150
. and lines m and l are parallel.
Parallel Lines Cut by Transversal
■
When two parallel lines are both intersected by the third line.
Geometry and Measurement
1. ∠3 = ∠5 = 180° ∠4 and ∠6 are consecutive interior angles. ∠4 = ∠5 ∠2 and ∠7 are alternate exterior angles. Likewise: ∠2 = ∠3 ∠5 = ∠8 ∠7 = ∠6 3. ∠1 + ∠2 = 180°. Adjacent angles. Using all of these facts. ∠3 = ∠6 ∠4 and ∠5 are alternate interior angles. Angles 1 and 2 are adjacent and they form a straight line. therefore.
151
. Therefore: ∠3 and ∠6 are alternate interior angles. Interior angles are those contained within the parallel lines. Vertical angles. Angles 1 and 4 are vertical angles. therefore. the angles that would coincide with each other would be equal in measure. They are called corresponding angles. Alternate angles are on the opposite side of the transversal. ∠2 = ∠7 ∠1 and ∠8 are alternate exterior angles. the other angle measures can all be determined. If we could physically pick up line l and place it on line m. if we are given the measure of one of the eight angles. Therefore: ∠3 and ∠5 are consecutive interior angles. they are supplementary. Corresponding angles. ∠1 = ∠4. Therefore: ∠1 = ∠5 ∠3 = ∠7 ∠2 = ∠6 ∠4 = ∠8 4. ∠4 = ∠6 = 180° The sum of the measures of each pair of consecutive angles = 180°. even if you cannot remember the rules. Consecutive interior angles. you can see which angles are equal. Consecutive interior angles are on the same side of the transversal. Likewise: ∠2 + ∠4 = 180° ∠7 + ∠8 = 180° ∠3 + ∠4 = 180° ∠5 + ∠7 = 180° ∠1 + ∠3 = 180° ∠6 + ∠8 = 180° ∠5 + ∠6 = 180° 2. Exterior angles are those on the outsides of the parallel lines. Alternate interior and exterior angles. ∠1 = ∠8 5. For example:
83˚ a b c d e f g
l l m m
Note that since the lines are parallel. they are equal.
A triangle having none of its sides equal is called a scalene triangle. In the diagram of ᭝ABC:
A
AB + BC > AC AB + AC > BC AC + BC > AB
B
C
152
. A triangle is a three-sided polygon. (Poly means many.)
Triangles
This section deals with those polygons having the fewest number of sides. The following figure shows ᭝ABC:
B
A
C
There are various types of triangles:
■
A triangle having all three sides equal (meaning all three sides having the same length) is called an equilateral triangle. The symbol for triangle is ᭝. Every height is the perpendicular (forming a right angle) distance from a vertex to its opposite side (the base). It has three angles in its interior.Part II: Review with Sample Problems
Polygons
Closed shapes or figures with three or more sides are called polygons. thus. AE = BC .
A
B
E
C
■
In this diagram of ᭝ABC. A triangle having a right (90°) angle in its interior is called a right triangle. A triangle is named by all three letters of its vertices. polygon means many sides.
■ ■ ■
Facts about triangles:
■
Every triangle has a base (bottom side) and a height (or altitude). and AE is the height. The sum of the lengths of any two sides of a triangle must be larger than the length of the third side. A triangle having two sides equal is called an isosceles triangle. The sum of these angles is always 180°. gon means sides. BC is the base.
and 15–20–25 are also Pythagorean triples. For example: Find the length of x in the triangle. b = 4. the relationship between the lengths of the sides is stated by the Pythagorean theorem. For example. The three lengths a. and c = 5: a2 + b2 = c2 32 + 42 = 52 9 + 16 = 25 25 = 25
4 5
3
■
■
Therefore. A knowledge of the use of algebraic equations can also be used to determine the lengths of the sides.
■
The side opposite the right angle is called the hypotenuse (side c). and c that always work. The sum of these interior angles is always 360°.Geometry and Measurement
Pythagorean theorem:
■
In any right triangle. (The hypotenuse is always the longest side.) The other two sides are called the legs (sides a and b). 9–12–15. 3–4–5 is called a Pythagorean triple. There are other values for a. Some are 1–1– 2.
A
b c
C
a
B
The parts of a right triangle are: ∠C is the right angle. There are four angles in its interior. and c are always numbered such that: a2 + b2 = c2 For example: If a = 3. a2 + b2 = c2 x + 102 = 152 x2 + 100 = 225 x2 = 125 x = 125 125 = 25 × 5 = 5 5 So. the lengths of these sides can be determined easily. If perfect squares are known.
153
. multiplying the 3–4–5 solution set shows that 6–8–10. b. b. 5–12–13. A quadrilateral is named using the four letters of its vertices. and 8–15–17. Any multiple of one of these triples also works. x = 5 5
2
x
15
10
Quadrilaterals
A polygon having four sides is called a quadrilateral.
AE is the height.Part II: Review with Sample Problems
The following figure shows quadrilateral ABCD. BE is the height.
A
b a
B
b
D
E
a
C
■
AE is the height of the parallelogram. AB CD .
A B
D
E
C
154
.
A B
D
C
Types of quadrilaterals
■
A square has four equal sides and four right angles. and consecutive angles supplementary. AB DC . Every parallelogram has a height. A rhombus has a height. A rhombus is a parallelogram with four equal sides.
B
a a
C
a a
A
■
E
D
A trapezoid has only one pair of parallel sides. opposite angles equal. A trapezoid has a height. and AD BC .
A
a
B
a
a
D
■
a
C
A rectangle has opposite sides that are equal and four right angles.
A
a
B
b
b
D
■
a
C
A parallelogram has opposite sides equal and parallel.
An octagon is an 8-sided polygon.Geometry and Measurement
Other Polygons
■ ■ ■ ■ ■
A pentagon is a 5-sided polygon.
Area
Area (A) means the amount of space inside the polygon. The formulas for each area are as follows: Triangle: A = 1 bh 2
h b
or
h b
For example: A = 1 bh 2
18" 24"
A = 1 ^ 24 h^18h = 216 sq. The total distance around is the sum of all sides of the polygon. in. The perimeter of any polygon can be determined by adding the lengths of all the sides. A decagon is a 10-sided polygon. A nonagon is a 9-sided polygon.
Perimeter
Perimeter means the total distance all the way around the outside of any shape. A hexagon is a 6-sided polygon. 2 Square or rectangle: A = lw w l w l
or
l w l w
155
. No special formulas are really necessary.
An arc is a piece of the circle. Arcs are measured in degrees.
Parts of a Circle
■
Radius is the distance from a center to any point on a circle. all points on which are equidistant from the center point. The symbol " is used to denote an arc. Each diameter is two radii. E
F
157
. A
■
M
B
Diameter is the distance across a circle. the minor arc is assumed. M is the center point because it is the same distance away from all points on the circle. $ This is EF $ Minor EF is the shorter distance between E and F. It is written on top of the two endpoints that form the arc.
R M V
S
■
The diameter is the longest chord in any circle. $ When EF is written.
MA is a radius. $ Major EF is the longer distance between E and F. CD is a diameter. An arc is the distance between any two points on the circle itself. is called a circle.
D A
AB is a diameter.
M
This is circle M. UV is a chord.
U
RS is a chord.
B
■
M
C
The chord is a line segment whose end points lie on the circle itself. There are 360° around a circle.Geometry and Measurement
Circles
A closed shape whose side is formed by one curved line. In any circle. all diameters are the same length. all radii (plural) are the same length. Circles are named by the letter of their center point. In any circle. MB is a radius. through the center.
but are not identical in size. For example: All squares are similar. Corresponding sides are those sides that are across from the equal angles.
The following triangles are congruent. The formula for circumference is: C = πd or C = 2πr. π (pi) is a Greek letter that represents a specific number.
M 10"
Congruence and Similarity
Two plane (flat) geometric figures are said to be congruent if they are identical in size and shape. They are said to be similar if they have the same shape.14 or π ≈ 22 7 .14(8) = 25.
158
. a formula is needed. A = π(r2) = π(52) = 3. d = 8 because r = 4.5 sq. in. For example: In circle M.
A more precise working definition of similar figures follows:
■
Similar figures have corresponding angles equal and corresponding sides that are in proportion. For example: In circle M. C = πd = π(8) = 3.14(25) = 78.Part II: Review with Sample Problems
Circumference and Area
■
Circumference is the distance around a circle. Since there are no sides to add up. In fractional or decimal form. r = 5 because d = 10. the commonly used approximations are: π ≈ 3.12 inches
M
4"
■
The area of a circle can be determined by: A = πr2.
you can set up the following proportion to find PM. A few more examples:
N
8
O
5 12
M
Q
P
1. The corresponding sides in the case are AD and AC. what is the length of PM?
N
8
O
5 12
M
Q
P
In the figure above. and ∠AED and ∠ABC. Therefore. AD = DE or AD = 2 AC CB AC 3 6 = 2. Therefore. and AE and AB. Because the proportion of side DE to side CB is 4 to 6 or 4 . DE and CB. since MN and OQ are both perpendicular to PM. the corresponding sides are in proportion. AC 3 Cross multiplying gives: 6(3) = 2(AC) 18 = 2(AC) Divide each side by 2. which reduces to 2 . If side AD = 6. DE CB . and since AD = 6.Geometry and Measurement
For example: Triangles ADE and ACB are similar
C D
Side DE = 4 and corresponding side CB = 6. they are parallel to each other. If AD = 4. triangles ADE and ACB are similar and the problem can be solved as above.
159
. ∠ADE and ∠ACB. and AD = 6. then side AC =
A
E
B
Since the triangles are similar. Therefore. what is the length of AC? A line parallel to one side within a triangle produces similar triangles. Since OQ to NM is in the ratio 5 to 8. In the figure above. 18 = 2 ^ AC h 2 2 9 = AC or AC = 9 Please note that this question could have been introduced as follows: In the triangle shown. triangles OPQ and NPM are similar. the 3 6 same ratio holds for all corresponding sides. CB = 6. The corresponding angles are ∠DAE and ∠CAB.
the only real difficulty is matching the corresponding parts. are corresponding.2 5
B
A
D
C
2. since triangles ADB and BDC are similar. The volume of any prism (a three-dimensional shape having many sides. Since ∠DCB = ∠DBA. (Note that BD is used twice. In the figure above. Since ∠BAD = ∠CBD. as it is part of both triangles). 5 ^ PM h 96 = 5 5 PM = 19 1 or 19. BD and DC. What is the length of DC?
B
3
A
4
D
C
In the figure above. volume refers to the capacity to hold. they are corresponding and sides across from them. you can set up the following proportion and solve accordingly. AD and BD are corresponding. ∠DAB = ∠DBC. Because BD to AD is in the ratio 3 to 4. BD = DC or 3 = DC AD BD 4 3 3 ^ 3h = 4 ^ DC h 9 = 4 ^ DC h 9 = 4 ^ DC h 4 4 1 2 = DC or DC = 2 1 or 2.25 4 4
Three-Dimensional Shapes
Volume
In three-dimensional shapes. The formula for volume of each shape is different.Part II: Review with Sample Problems
OQ PQ 5 12 NM = PM or 8 = PM Cross multiplying gives 5(PM) = 8(12) 5(PM) = 96 Divide each side by 5. triangles ADB and BDC are similar. but two bases) can be determined by: Volume (V) = (area of base)(height of prism)
160
. and ∠ABD = ∠BCD.
4" 18" 6"
To determine the surface area of a right circular cylinder. Adding the three parts gives the surface area of the cylinder: For example:
5'
Find the surface area of a cylinder with radius 5' and height 12'. Coordinate graphs can be used in measurement problems. The rectangle's area equals circumference times height. Note that the length of the rectangle equals the circumference of the circle. Find the area of each surface. The area of each circle equals πr2. The surface area of a rectangular solid can be found by adding the areas of all six surfaces. The area of the bottom circle is the same.
r h
r circumference r height
Now find the area of each individual piece. it is best envisioned "rolled out" onto a fat surface as below. For example:
162
. The area of the circle = π(r2) = π(52) = 25π. For example: The surface area of this prism is: top: 18 × 6 = 108 bottom: 18 × 6 = 108 left side: 6 × 4 = 24 right side: 6 × 4 = 24 front: 18 × 4 = 72 back: 18 × 4 = 72 408 sq.Part II: Review with Sample Problems
Surface Area
The surface area of a three-dimensional solid is the area of all the surfaces that form the solid. and then add those areas. in.
Coordinate Geometry and Measurement
Refer to "Basic Coordinate Geometry" in the "Algebra Review" section if you need to review coordinate graphs. Totaling all the piece gives 25π + 25π + 120π = 170π. 25π. or 2πr = 2π(5) = 10π. Therefore the area of the rectangle equals its height times 10π = 12 × 10π = 120π.
10∼ 5'
12'
The length of the rectangle is the circumference of the circle.
the base is 5. and the distance to point A from the y-axis is 2. Since point A is at (0. the first.5)
x
0
1. 5) and (3. The distance to point B from the y-axis is 3. 2
163
.
y
C(2. (–2 is 2 in the negative direction.Geometry and Measurement
y
A (−2.5) B (3. 0).
y
D
(0. What is the area of rectangle ABCD in the preceding graph? The formula for the area of a rectangle is base × height. Since point D is at (0.3) 0
x
B(2. What is the length of AB in the preceding graph? Since the coordinates of the points are (–2. 3).3)
C
(5.) So 3 + 2 gives a length of 5.−3)
A(−2. or x-coordinate is the clue to the distance of each point from the y-axis.−3)
3. 0) and point B is at (5. the height is 3. What is the area of ᭝ABC in the preceding figure? The area of a triangle is 1 × base × height. 5).0) A 0 B
x
2. so the area is 5 × 3 = 15.
y
A 3 3 C O 3 B(3.0)
x
164
.Part II: Review with Sample Problems
y
C(2. You know that radius OB is 3 units long. Therefore. The OA and OC are each 3 units because they are also radii.−3)
x
A(−2. Note that ∠B is a right angle. what is the perimeter of ᭝ABC inscribed within the semicircle with center O? To find the perimeter of the triangle.−3)
4
Base AB of the triangle is 4 units (because from A to the y-axis is 2 units and from the y-axis to B is another 2 units).3) O
6 B(2. side AC of the triangle is 6 units. In the preceding figure. Height BC of the triangle is 6 units (3 units from B to the x-axis and another 3 units to C). So area of triangle = 1 × 4 × 6 2 = 1 × 24 2 = 12
y
A 0 B(3.0) C
x
4. you need the lengths of the three sides.
D. Circumference of a circle is equal to 2πr. Since the ratio of the areas is 9π . E.12)
A F B
P
4
E
G
x°
D
4
C
18. the regular hexagon has a perimeter of 72. 36 45 60 72 It cannot be determined from given information. Therefore. Quadrilateral ABCD is a square with sides of 4 units as shown above. What is the area of the shaded region?
173
.47) 216 3 (approximately 374. and area is πr2 for a circle.
19.5 135
20. C. 4π 2
Perimeter. In the figure above. C. what is the value of x? A. In reduced form.5 45 60 67. Triangle FDC is isosceles such that DF = CF. 3 . D. 108 108 3 (approximately 187. D. 22. E. In the regular octagon above. E is the midpoint of FD and G is the midpoint of FC . What is the area of the hexagon? A. B. the radius of the larger circle must be 3 and the 4π 2 radius of the smaller circle must be 2. what is the value of x? A. B. B. with center at C. with center at P. Angle Measure of Polygon
C x°
17.06) 216 216 2 (approximately 305. the ratio is 3 . C.Geometry and Measurement
16. In the regular pentagon above. Area. the ratio of the circumference must be 6π . E.
b2 = 2. CD would be 6 3.
C 30° 12 12 60°
A 6 D 6 B 12
Side of regular hexagon = 72 = 12 . 20. With hexagon center at C. An alternate solution would be to find the area of isosceles trapezoid DEGC using the area formula A = 1 _ b 1 + b 2 i h where b1 = 4. you have the altitude of triangle DFC. 6. Then ᭝CDB is a 30°–60°–90° triangle. 2
174
.Part II: Review with Sample Problems
Explanations
17. The altitude of the smaller triangle is half of the altitude of the larger triangle. the answer is 6. The altitude is perpendicular to DC . By constructing an altitude from F to the midpoint of DC . With DB = 6. So ᭝ACB has area: 4 4 The hexagon will then have area 6 times this: 6 : 36 3 = 216 3 . EG is half the length of DC and is 2 units also.
P
x° A
x° B
45 + x + x = 180. so 2x = 135. EP = 360 = 45 . D. ᭝ACB is equilateral. and therefore x = 67. Then the hexagon will have an area 2 6 times this: 6 : 36 3 = 216 3 . E. and h = 2. Method II: In ᭝ACB. this altitude is the same length as the sides of the square. 8 – 2 = 6 2 2 Using the area formula for both triangles and subtracting. Since the polygon is a square. x = 360 = 72 5 18.5 19. The area of ᭝ FDC = 1 ^ 4 h^ 4 h = 8 and the area of ᭝ FEG = 1 ^ 2 h^ 2 h = 2 . ∠PAB = ∠PBA = x. D. the altitude to base AB bisects the base at point D. So ᭝CAB has area 1 : 12 : 6 3 = 36 3 . 8 In ᭝APB: 45 + ∠PAB + ∠PBA = 180. The altitude of the large triangle also serves as the altitude of the smaller triangle. 6 s2 3 Method I: The area of an equilateral triangle with side length s is found by using the formula . The altitude is 4 units. Since PA = PB. 4 12 2 3 144 3 = = 36 3 .
C. C. what is the radius of the cylinder? A. E. D. If the volume of the cylinder is 54π. 3 9 36 54 63
36. with circle center at point P. D. What is the difference between the surface area of a right rectangular prism with dimensions of 3 × 4 × 8 and a right circular cylinder having dimensions of r = 2 and h = 8 as is shown in the figure above? Find the difference to the nearest whole number.10) 3 3 (approximately 5. B. What is the total surface area of a right circular cylinder having radius 2 and height 3? A. what is the total surface area of the cube? A. D. In the right circular cylinder in the figure above.24) 26 (approximately 5. E. AC = PT.20) 6
181
. 8π 10π 14π 16π 20π
r=2
8
8
3 4
34. B. 3 3 2 (approximately 4. E.Geometry and Measurement
Volume and Surface Area of Solids
33. C.
A
P
C
T
35. If the volume of a cube is 27. B.
Doubled. E.60 square units. To the nearest whole number. This is the circumference multiplied by the height. What is the length of the longest side of ᭝MNP? A. What are the coordinates of point B? A. C. –1) and (7. D. 3)
x
37. –3). B.5) (6. 2 3 3 5 1 5 3 8 3
183
. The difference is 136 – 125. First find the surface area of the prism. ᭝ABC has its vertices at A(–1. (–1. ᭝MNP has its vertices at M(–3. 5). Together the area is 112 + 24 = 136 square units. B. 10).Geometry and Measurement
36. 32π + 8π = 40π. 5). y)
39. 8). D.6 or 10. and P(1. Now find the area of the tube-like side. This is 14 × 8 which is 112 square units. 1. What is the length (to the nearest tenth of a unit) of the shorter distance—the distance between (–2. Next find the surface area of the cylinder. E. D. 6) or the distance between (2. In a similar manner. Midpoint Formula
y B (x. Distance Formula. 40π is approximately 125.24) 2 2 (approximately 2. –6)?
38.5) (1.82) 5 34 (approximately 5. the answer is 10. the area for both ends is 8π. so the tube-like area is 4π units. there are two ends and the tube-like side. This area is made up of three parts (two ends and the area around the prism).
Coordinate Geometry—Coordinates. 11)
40. Slope. C. 8 units × 4π units = 32π square units. The area of the base is πr2.83) 173 (approximately 13.4 square units. 7). so both ends have an area of 24. so this value will be 4π. point M is the midpoint of AB .15)
M (4. E. The circumference is 2πr. 17) (10. N(0. B(2. and C(5. What is the slope of AC ? A. 5 (approximately 2. B. 3. 5) (2. In the figure above. The area of an end is 12. C. –1) and (–8. 10. 7) A (−2. The area of the sides is calculated by the perimeter of the base multiplied by the height.
Geometry and Measurement
43. results in which of the following geometric figures? A. If AB = BC.
B
A C
The union of all 3 segments in the figure above is just ᭝ABC. C.1. D is a point that lies three times further from B than from E. D. 43. 45 . 44. then the segments must be congruent. D. 44. 1 of 15 is 15 so the distance between D and E is 15 and the distance between B and D is 3 4 2 8 8 times 15 or 45 .13 . the distance between B and E is . The midpoint of AE is –4. The union of points A and B and all point C. E. The correct answer is 45 . 2 2 2 2 2 15 this distance is .4 = . C. B. . such that C is between A and B. The union of points A and B and all the points between A and B is just a segment with its endpoints at A and B.13 . and E lie on a line as shown above.1 = . Points A. 8 8 8 8
185
. AB AB CA AC BC
A −9
B
C −4
E 1
Note: Figure not drawn to scale.13 .15 Since distances are not negative. all 3 points must be collinear. It follows that BD = 45 . there is no guarantee that points A and B are even collinear. C is the midpoint of AE and B is the midpoint of AC . For C to be between points A and B. What is the distance BD?
Explanations
41. B. B. The midpoint of AC is 8 .9 +. C. D. This distance must be divided into 4 parts. This eliminates all answer choices but C. 42. three of which are between C and D and the fourth 2 between D and E. Now. The coordinate of B is found by using the midpoint formula. If the distances are equal. Therefore the correct answer is B.
4) C P x l B m l
46. E. A. lines l and m intersect at point P and are tangent to circle C at points B and A respectively. what is the measure of the third arc formed?
186
.24) 3 3 (approximately 5. what is the radius of circle C? A. C. B. If AB = 6 and AD = 2.20) 2 13 (approximately 7. B. In the figure above. D. -4 3 -3 4 3 4 4 3 It cannot be determined from the given information. E. what is the length of AP ? 45. 47.
B
4 3 2 (approximately 4. D. C. C.Part II: Review with Sample Problems
Properties of Tangent Lines
A y P (3. AB is tangent of circle C at point B.21) 8
C D A
In the figure above. D. E. A secant line and a tangent line of a circle intersect in the exterior of a circle to form an angle of 64°. If the ratio of the intercepted arcs between the tangent and the secant is 11:3. If the diameter of circle C is 6 and PC = 5. 4 6 8 10 12
48. B. What is the slope of line l? A. l is tangent to the graph of x2 + y2 = 25 at point P. In the figure above.
Then using the Pythagorean theorem: 32 + (AP) 2 = 52 9 + (AP) 2 = 25 (AP) 2 = 16 AP = 4 You could also have just used a 3–4–5 right triangle. The slope of line l is the opposite reciprocal of the slope of CP . 0)
l
The center of the given circle is (0.x 1 = 3 2 1 -0 3 4 46.Geometry and Measurement
Explanations
45. 4)
x C (0. Then CP = l . label this point C. Thus the slope of line l is .
y P (3. Slope of CP = x 2 . ∠CAP is a right angle – a radius drawn to a point of tangency creates a right angle.
187
. y -y 4 . its radius is 3.
A 3 l C P 3 B m
Since the circle's diameter is 6. A.3 . 0).0 = 4 . B.
The graph is first reflected about the y-axis and then translated upward 4 units. 58. –3) is on the graph of the function y = f(x). 7) is moved to the point (2 – 3. f(2) = 7. the graph of y= f(x) is shown. the y value will be increased by 4 and will then be 2 + 4 or 6. The correct answer is 6. C. In the figure above. The point P(2.
193
. Since the point (–1. (0. as in Choice D. The transformation y = f(x – 2) + 5 moves point P to which of the following new locations? A. –3) gets moved to the point (2 + 2. D. 7). 2 units to the right and 5 units up. D. 2) B x 1 2 3 4
59. Therefore. The reflection of point B about the y-axis will change the coordinates to (–4. g(–1) = 11. So the point (2. The transformation y = f(x – 2) + 5 moves every point on the graph of y = f (x). The graph of y = f(x – 1) will be the same shape as that of y = f(x) but will move 1 unit to the right. If g(x) = f(x + 3) + 4. 2). –3 + 5) = (4. D. one of the points on the graph of f is (2. 2) (4. 3 6 10 11 12
60. D. When the point is translated upward 4 units. C. the point (2. E. –8) (0. E. B. 2) (4. what is the value of g(–1)? A. 1) A (4. 11) is on the graph of g. B. 6. C. –5)
y 4 3 2 1 (1.Geometry and Measurement
58. The function g(x) = f(x + 3) + 4 is just a transformation of the graph of f(x) for which each point on the graph of f is moved 3 units left and 4 units up. 11). 59. 2). –8) (5. What will be the y-coordinate of point B after this procedure?
Explanations
57. Since f(2) = 7. For the function f. 60. 7 + 4) = (–1.
.
try to simulate the test conditions by following the time allotments carefully.PART III
MATH PR A CTI C E TE STS
T hre e Simulate d Full-Length Practice Mathematics Tests
This section contains three simulated full-length practice new SAT I math tests. and number of questions are similar to those on the new SAT I. They are not labeled on the actual SAT I. explanations. and analysis techniques. The SAT I is copyrighted and may not be duplicated.
. Since the test is new. levels of difficulty. and these questions are not taken directly from the actual tests or released sample problems. question structure. The format. The sections in these practice exams are labeled by subject for your convenience in reviewing. the number and order of question types may vary. The practice tests are followed by complete answers. When you take these exams.
All numbers used are real numbers. Some problems may be accompanied by figures or diagrams200
. Notes 1. 2. 3. The figures and diagrams are meant to provide information useful in solving the problem or problems. Unless otherwise stated.Part III: Math Practice Tests
Practice Test IA
Time: 25 minutes 20 multiple-choice questions
Directions: Select the one correct answer of the five choices given and mark the corresponding circle on your answer sheet. Calculators may be used. These figures are drawn as accurately as possible EXCEPT when it is stated in a specific problem that a figure is not drawn to scale. all figures and diagrams lie in a plane. Your scratch work should be done on any available space in the section.
C. In the figure above. what is 6% of n? A. a circle of radius r is inside a rectangle with dimensions as shown.6 10. Which of the following could be the graph of f? A. B. 3. E. 6 19 33 48 52
0 x
C.
y
2. lw – 2πr 2l + 2w – πr2 2l + 2w – πr lw – πr2 lw – r2
0
x
B.11 333. In terms of l. B.Practice Test I
l
r
w
4.33 D. If 18% of n is 60. where m is a positive constant and b is a negative constant.
0 x
y
0
x
E.8 20 111. E.
y
Practice Test I
1. E. The linear function f is given by f(x) = mx + b. what is the area of the shaded region? A. B. C. D. D.
y
0
x
GO ON TO THE NEXT PAGE 201
. C. D. How many two-digit positive integers are multiples of both 3 and 5? A.
y
3. w. and r.
In parallelogram EFGH. After spending $10. E. B. 1 6 1 3 1 2 2 3 5 6
x . CHECK YOUR WORK ON THIS SECTION ONLY. D. How far does he ride on his return trip? A. what is the number? A. and finally 9 miles north.00 on a belt. On a shopping trip.
STOP
203
. Milo rides directly from point B back to point A. Using the data from the graph above. Which of the following represents f? A. f(x) = 3x2 – 7 f(x) = 2x2 + 1 f(x) = x2 – 1 f(x) = x2 + x + 1 f(x) = x2 + 1
16.00 $100. C. and soccer teams win their league titles this year are 80%. D. E. E. C. C. C. what was the percent increase in profits from 1996 to 1997? A. Amal spent 1 of his money 5 to buy a hat. How much money did Amal start out with before he went shopping? A. –8 2 3 3 2 8 3 3
19. 75%.00
17. followed by a ride of 13 miles east. points A and B trisect HG . What is the probability that. then 3 miles south.00 $60. If 5 less than some number is 3 more than twice the number. $110. D. E. D. At Smallville High School. E. Starting from point A. wrestling. C. C. D. C. ending up at point B. B. 20% 25% 40% 45% 80%
IF YOU FINISH BEFORE TIME IS CALLED. D. respectively. What is the ratio of the area of ABFE to the area of EFGH? A. and 60%. basketball and soccer will win league titles but wrestling will NOT? A.Practice Test I
14. B. B. 30 13 10 9 5
E
H
A
B
G
Practice Test I
F
18. After resting for a while. he then had $12. Milo rides his bicycle 5 miles east. B. DO NOT WORK ON ANY OTHER SECTION IN THE TEST. The table above gives values of the quadratic function f for some values of x. for these three boys' teams. and then used 1 of his remaining 2 money to purchase a coat. E.00 left. D. B. 3 25 3 20 9 50 9 25 2 5
PROFIT (thousands of $)
50 40 30 20 10 1993 1994 1995 1996 1997
20. B.00 $55. E. the probability that its boys' basketball.00 $50.2 -1 0 1 f^ xh 5 2 1 2 15.
Part III: Math Practice Tests
Practice Test IB
Time: 25 minutes 18 questions: (8 multiple choice and 10 grid-in)
Directions: This section is composed of two types of questions204
. For Questions 1–8. 2. All numbers used are real numbers. select the one correct answer of the five choices given and mark the corresponding circle on your answer sheet. Use the 25 minutes allotted to answer both question types. Unless otherwise stated. Notes 1. Your scratch work should be done on any available space in the section. Calculators may be used. 3. These figures are drawn as accurately as possible EXCEPT when it is stated in a specific problem that a figure is not drawn to scale. The figures and diagrams are meant to provide information useful in solving the problem or problems. Some problems may be accompanied by figures or diagrams. all figures and diagrams lie in a plane.
. 1 2 3 4 5 6 7 8 9
/ . The acceptable grid-ins of 8/9 are:
. 1 2 3 4 5 6 7 8 9
/ .889. If a given student can win only one prize. 0 1 2 3 4 5 6 7 8 9
Be sure to write your answers in the boxes at the top of the circles before doing your gridding. grid only one answer. 0 1 2 3 4 5 6 7 8 9
/ . 0 1 2 3 4 5 6 7 8 9
. Although writing out the answers above the columns is not required. . and third place. 1 2 3 4 5 6 7 8 9
/ .89 is considered inaccurate and therefore not acceptable. in how many different ways can the three prizes be given away in this contest?
11.Part III: Math Practice Tests
Answer 8/9 Accuracy of decimals: Always enter the most accurate decimal value that the grid will accommodate. 0 1 2 3 4 5 6 7 8 9
. 0 1 2 3 4 5 6 7 8 9
/ . second. . 0 1 2 3 4 5 6 7 8 9
. 0 1 2 3 4 5 6 7 8 9
/ . Gridding this value as .888 or .8.88. or . Five students are entered in a contest that will award prizes of first. can be gridded as . Grid-in questions contain no negative answers. it is very important to ensure accuracy. The polygon above is a regular hexagon. For example: An answer such as . What is the value of x?
10. Even though some problems may have more than one correct answer.8888. 0 1 2 3 4 5 6 7 8 9
. what is the average (arithmetic mean) of the number of phone calls per day for the week Monday through Sunday?
208
. According to the data in the graph above.
Number of Phone Calls Per Day
6 5 4 3 2 1 MON WED SAT TUES THUR SUN FRI
x°
9. 0 1 2 3 4 5 6 7 8 9
.
The marbles are all the same size and the same weight. what is the probability that he/she would draw a red. what is the value of x y?
15.)
18. 180 take Spanish and 150 take French. and five green marbles. A sack contains three red. and then a red marble in exactly that order?
IF YOU FINISH BEFORE TIME IS CALLED.
STOP
209
. DO NOT WORK ON ANY OTHER SECTION IN THE TEST.Practice Test I
12. The sum of four consecutive odd integers is 456. then what is the value of 12 5 3? b
125° A
Note: Figure not drawn to scale. If a blindfolded person were to reach in the sack and draw one marble on each of three successive tries (with no replacement). In the figure above with AC and BD . a blue.a . If m2 – n2 = 24 and m – n = 6.
y° B C
17. what is the value of a + c + w + y? 14. In a school having 300 students. If a 5 b = ab . If 2x + 3y is equal to 80% of 5y. What percent of the students in this school took Spanish but not French? (Do not include the % symbol in your answer. What is the sum of the smallest and largest of these integers?
16. CHECK YOUR WORK ON THIS SECTION ONLY. four blue. what is the value of m + n?
D c°
Practice Test I
w°
a°
13.
all figures and diagrams lie in a plane.Part III: Math Practice Tests
Practice Test IC
Time: 20 minutes 16 multiple-choice questions
Directions: Select the one correct answer of the five choices given and mark the corresponding circle on your answer sheet. Calculators may be used. Your scratch work should be done on any available space in the section. 2. These figures are drawn as accurately as possible EXCEPT when it is stated in a specific problem that a figure is not drawn to scale. All numbers used are real numbers. 3. Notes 1. The figures and diagrams are meant to provide information useful in solving the problem or problems. Some problems may be accompanied by figures or diagrams210
.
C. and D are equally spaced between A and E on AE . C. B. D. Xena bought a new refrigerator costing $625. D. D. If f(x) = x2 – 3x + 5. D. D. C. 4 5 6 7 9
M A 2 B x C D E 2
1 2
C. If M is the midpoint of BC . D. C. 4 12 15 22 23
5.
4. D.3 = 5 is true for which value of x? A. C. B. E. B. B. 4x2 – 6x + 5 2x2 – 6x + 10 2x2 – 6x + 5 x2 – 6x + 5 4x2 – 6x + 5
3. E. what is the value of c? A. She paid $100 as a down payment and will pay the remainder $75 per month. C.Practice Test I
1. What is the total of all the girls' and boys' ages combined? 12g + 17b A. g+b B. Points B. The length of one side of triangle is 35. g+b 29 12g + 17b 29 12g + 17b 12g + 17b 2
Practice Test I
2. E. If 3x + 15 = 36. The length of another side of the same triangle is 10. C. D. B. C. How many months will it take for Xena to pay off her new refrigerator? A. what is the value of x? A. 24 27 31 39 43
8. E. If 2a + 6b = 16 and a + 3b + c = 12. E. E. –4 4 8 28 It cannot be determined from the given information. B. Which of the following could NOT be the value of the third side of this triangle? A. E. then x + 5 = A. then f(2x) = A. E. The equation 2 x . 2 1 16 2 3 32 2 3 16 21 4 3 2 8
7. The average (arithmetic mean) of the ages of 12 girls is g and the average of the ages of 17 boys is b. B.
6. 0 4 9 16 25
GO ON TO THE NEXT PAGE 211
.
y c = 5m . 3) and (4. If 2x . D. 1 2 B. 2 15 m+y B. 13 15 2 13 2 26 13
13. What is the value of f(g(3))? A. A. E. C. C. If x is an odd integer and y is an even integer. II. C. B. What is the radius of the circle? A.
212
. 2 5m + 3 + y 2 12m + y 2 15m + y 2
12. B. what is x in terms of y and m? 5m + 3 + y A. 1 C. 3 E. the endpoints of a diameter of a given circle are (–2. The graphs of g and f are shown above. xy (x + 2)(y – 5) yx I only II only III only I and II only I and III only
C. 2 D. E. In the xy-plane. which of the following must be an even integer? I. E. 0 1 3 4 5
10.Part III: Math Practice Tests
y 5 y = g(x)
y 5 y = f(x)
−4
4
x
−4
4
x
−5
−5
9. 7). D. D. III. D. What is the radius of a circle for which its area and circumference are numerically equivalent? A. 4 11. B. E.
B.50 3. 3. E. 4 6 8 10 12
IF YOU FINISH BEFORE TIME IS CALLED.50
16. Maria hiked a distance of 15 miles at a rate of 5 mph. –3 –1 1 2 4
T
Practice Test I
8 r P 4 M
C
15. B. C. D. What is her average rate for the entire hike? A.75 4. If f(x) = 3x2 – 5. PT is tangent to circle C at point T. In the diagram above.
STOP
213
. CHECK YOUR WORK ON THIS SECTION ONLY. D.25 4. what is an integral value of r for which f(2r) = 7r? A.25 3. E. C. D. and then returned the same distance at a rate of 3 mph. DO NOT WORK ON ANY OTHER SECTION IN THE TEST. C. what is the radius of circle C? A. B.Practice Test I
14. If PT = 8 and PM = 4. E.
The complete process of analyzing each subject area and each individual problem should be completed for each practice test. These results should then be reexamined for trends in types of errors (repeated errors) or poor results in specific subject areas. Now that you have taken the practice test and checked your answers. This reexamination and analysis is of tremendous importance to you in assuring maximum test preparation benefit.
Practice Test I compare it to other practice tests to spot other common mistakes
215
.Practice Test I
Analyzing Your Test Results
The charts on the following pages should be used to carefully analyze your results and spot your strengths and weaknesses. carefully tally your mistakes by marking them in the proper column. Now that you've pinpointed the type of error.
D.
You are not done yet. 90.
216
.
y m>0 x b
5. D. This is where f(x) is positive. B. 5x + 4 = 5(3) + 4 = 19 6.
4. so the graph of the line should cross the y-axis below the x-axis (where y-intercept will be negative). Another way to look at it is: 18% of n = 60 18% of n = 60 3 3 = 20 Divide both sides by 3. or just 20.πr 2
2. 45. it must also be a multiple of 15. D. The two-digit multiples of 15 are: 15. . These regions correspond to x values in the intervals –2 < x < 3 and 4 < x < 6. with m being the slope of the line and b the y-intercept. You are told that m is positive. 60. If the integer is a multiple of both 3 and 5. Area shaded = Arearect. 6% should be 1/3 of 60.
y f(x) > 0
−5
−2
x 3 4 6
In the figure above. So there are 6 such integers. But you are also told that b is negative. as in choices A. The equation f(x) = mx + b can also be written as just y = mx + b. C. you still need to find the value of 5x + 4. which is known as the slope-intercept form of a linear equation. or C. so the graph of the line will rise uphill from left to right. 3. Only Choice D fits both conditions. Noting that 6% is just 1/3 of 18%. dark portions of the graph of y = f(x) are above the x-axis. the heavy. To solve this equation: 3x – 2 = 7 3x – 2 + 2 = 7 +2 3x = 9 3x = 9 3 3 x=3 Add 2 to each side to isolate the variable. 30.
Divide both sides by 3.Areacircle = lw . A. D. 75.Part III: Math Practice Tests
Complete Answers and Explanations for Practice Test I
Practice Test IA Explanations
1.
Add 6 to both sides to isolate the variable. you need to set the equations equal to each other and then solve for x.6 < 2x < 12 2 2 2 –3 < x < 6 Divide all parts of the inequality by 2. class average = total of girls' scores + total of boys' scores total number of students 25 ^ 92 h + 10 ^85h = 25 + 10 2300 + 850 = . If you use x for the smallest integer.
9.
217
. C. you will have: x = the first integer then x + 1 = the second integer and x + 2 = the third integer The sum of the 3 consecutive integers would then be: x + (x + 1) + (x + 2) = 3x + 3 This is not one of your answer choices.
Multiply both sides by 6 to get rid of fractions. 35 = 3150 35 = 90
Practice Test I
8.1 = .3 x + 12 3 2 2 6 c x . E. in Choice D. and 9 or as with 23. Distribute the 6 through the parentheses. 24. 2 x .1m = 6 c . D. You have three possibilities with which to work: i.3 < 9 breaks up into the multiple inequality: –9 < 2x – 3 < 9 –9 + 3 < 2x – 3 + 3 < 9 + 3 –6 < 2x < 12 . 10. Consecutive integers are one apart from each other—as with 7. Add 3 to all parts of the inequality. Notice that you could also have solved this by just trying the choices A through E until you found a number that made the inequality true. you must break it up into the multiple inequality: –N < "stuff" < N So the given absolute inequality 2x . Choice D. again that would have been x = 5. The only answer that fits is 5. 8. and 25.3 x + 12 m 3 2 4x – 6 = –9x + 72 4x – 6 + 9x = –9x + 72 + 9x 13x – 6 = 72 13x – 6 + 6 = 72 + 6 13x = 78 13x = 78 13 13 x=6 Divide both sides by 13.Practice Test I
7. To find the x-coordinate of the point of intersection of the graphs of the two lines. Add 9x to get the variable on just one side. D.
So your answer must be between –3 and 6. To solve an absolute value inequality of the form "stuff" < N .
you can find the perimeter of the rectangle by adding the lengths of its sides to get 8 + 12 + 8 + 12 = 40. y = 60°. iii. D. one side of the square is just 40 = 10 . since you are told that the perimeter is 18.
x=6 60°
60° 60° 6
6=x
y
12. The perimeter of the triangle is found by adding the lengths of its three sides. E. if you had used z for the largest integer. you will have: y – 1 = the first integer then y = the second integer and y + 1 = the third integer The sum of the 3 consecutive integers would then be: (y – 1) + y + (y + 1) = 3y This IS one of your answer choices. the triangle is equilateral. as in the figure below. So your 4 square looks like the one in the next figure.
12 8 12 8
In the figure above. Choice E. you would have had: z – 2 = the first integer then z – 1 = the second integer and z = the third integer The sum of the 3 consecutive integers would then be: (z – 2) + (z – 1) + z = 3z – 3 This was not one of your answer choices. If you use y for the middle integer. each angle is 60°. In particular.Part III: Math Practice Tests
ii. you have the following equation: 2x + 6 = 18 2x + 6 – 6 = 18 – 6 2x = 12 2x = 12 2 2 x=6 Divide both sides by 2. therefore. 11. so the perimeter of the square is also 40. Just for the record. With all sides of a square being equal. Subtract 6 from both sides to isolate the variable
Since each side of the triangle is 6. You are told that the perimeters of the rectangle and the square are equal.
218
.
or 3 if the number is to be less than 4. Your series of blanks now looks like: 3 __ __ 2. After "sketching" Milo's bike ride with its many right angle turns.000. The legs of this right triangle are found by subtracting the east and west distances (13 – 5 = 8) and by subtracting the south and north distances (9 – 3 = 6). E. So you can choose your first digit in only 3 ways. They must be less than 4. 15. f(x)) into each answer choice is probably the most efficient way to arrive at the correct answer choice.000. D. your number must end in a 0 or 5 if it is to be a multiple of 5. using condition #2.
B
6 9
d
A C 8
5 3
13
A right triangle is chosen having AB as its hypotenuse. So you can create 600 different four-digit integers that satisfy conditions I and II above. your number must begin with the digit 1. Substituting a few of the given ordered pairs (x. 2. Your series of blanks now looks like: __ __ __ 2. They must be multiples of 5. So you can choose the last digit in only 2 ways. where 2 represents the number of ways to choose the last digit (0 or 5) Next. 0 through 9. f(x)= 3x2 – 7 Use (–2. You set up four blanks. 14. the other side is 8. each representing one digit of your 4 digit number: __ __ __ __ Using condition #1. So here goes! A. So your completed series of blanks becomes: 3 10 10 2 Your answer is just the product of these four numbers. 3) The remaining digits can be any one of 10 digits. where 3 now represents the number of ways to choose the first digit (1. You then have a multiple of a 3-4-5 right triangle: one side is 6. (3)(10)(10)(2) = 600. 2. You must form four-digit integers that satisfy two conditions: I. so the hypotenuse must be 10. 5): Does 5 = 3(–2)2 + 7? 5 = 12 + 7 false
219
. 13. the figure looks like the one below. C. Thus AB = 10.Practice Test I
10 10 10 10
Practice Test I
The area of the square above is just 10(10) = 100. II.
Subtract 3 from both sides to isolate the variable.000 = 20.000 = 20 # 100 25 20 = # 100 25 = . 5): Does 5 = (–2)2 + (–2) + 1? 5=3 false E. which equals . Using n for the number mentioned in the problem. and then proceed as follows: profits in 1997 45.000 increase in profits amount of increase # 100 % increase= amount original 20. 5): Does 5 = 2(–2)2 + 1? 5=9 false C. Choice E must be the correct function.80 25 5
220
. Add n to both sides to get the variable on just one side. By default.000 = # 100 25.profits in 1996 . f(x) = x2 – 1 Use (–2. Just for the record: If one of choices A through D had worked for the point (–2. you can translate the given problem into the following equation: 5 – n = 2n + 3 5 – n + n = 2n + 3+ n 5 = 3n + 3 5 – 3 = 3n + 3 – 3 2 = 3n 2 = 3n 3 3 2 =n 3 Divide both sides by 3. E. f(x) = x2 + x + 1 Use (–2. From the table. f(x) = 2x2 + 1 Use (–2. B.000 . 5): Does 5 = (–2)2 – 1? 5=3 false D.25.Part III: Math Practice Tests
B. you find the profits for the years 1997 and 1996. you would then have had to try it for the other 3 points to be sure the function worked for all 4 points in the table.80 × 100 = 80%
You can reduce 20 to 4 .
17. 16. 5).
Choice A CANNOT be one of the values of x. Since BC and DF are parallel. the 120 on top is just 60 times the 2 on the bottom. you note that: x + z = 20 + 100 = 120 4. the measure of ∠ACB is also 80. B. 4 3. m + 3 = x(2 – x) m + 3 = 2x – x2 –3 –3 Distribute the x on the right side of the equation. and then find m – 2. then on the left side. Subtract 2 since you want to get m – 2.
If you note that on the left. so z = 100. since m = 100. M on top should be 60 # 7 3 on the bottom. 4 3 So.
4x > 24 4x > 24 4 4 x>6 Since x can be any number greater than 6.Part III: Math Practice Tests
Practice Test IB Explanations
1. Referring to the figure in the problem. and measure of ∠ACB being 80. the measure of ∠CED is 80. with x = 20.
–2
m – 2 = x + 2x – 5
222
. You can set up and solve a proportion to answer this question. C. 2. To solve this inequality: 4x + 3 > 27 –3 –3 Subtract 3 from each side to isolate the variable term. A. 60 c 7 m = 60 ^ 7.
m = 2x – x2 – 3 m = – x2 + 2x – 3 –2
2
Write the terms in descending order. Subtract 3 from both sides to isolate the m. 120 miles 2 inches = M miles 7 3 inches 4 Divide both sides by 4. the measure of ∠ABC must be 80. You will solve the given equation for m. In ᭝ABC. z is just the supplement of ∠ABC. Finally. D.75h = 465 miles.
A x 20°
B z
80° 100°
80°
C m 80° 100°
D
E
F
To finish the problem.
360 So you have 6π = 60 _ πr 2 i 360 6π = 1 _ πr 2 i Reduced 60 to 1 6 360 6 6 ^ 6π h = 6 . AREAlength = x ^ 2πr h . To help you determine the number of ways to give away the three prizes among the five people.
Divided both sides by π. you have: measure of ext. 6 10. where x is the measure of the central angle and r is the radius. where x is the central angle and r is the radius. 360 Then you have: AREAlength = 60 ^ 2πr h 360 = 1 ^12π h 6 = 2π 9. you will find the radius of the circle and use that to determine the length of the requested arc. the measure of an exterior angle = 360 n . So.
arc length M
A
60° C r
B
You are given the area of the shaded sector in the diagram above. 1 _ πr 2 i E 6 36π = πr2 36 = πr 2 π π 36 = r 6=r So the radius of the circle is 6. C. E = 360 = 60 . one for each prize: 1st 2 nd 3rd Since you have five people. your blanks look like: 5 1st 2 nd 3rd
2
Multiplied both sides by 6. any one of these five could win first place. where n is the number of sides. 60.
224
. From the area. find the length of arc AMB. with n = 6. you set up three blanks. for your hexagon. For any regular polygon (all sides are equal and all interior angles are equal). therefore. AREAsec tor = x _ πr 2 i . Next. 60.Part III: Math Practice Tests
8.
32. 50. you have: 5
st
4
1 2 nd 3rd Since you selected one person for first and one more for second place. Since there are only 300 students in the school. you end up with 330 students. 300 You could also have used the Venn Diagram. Using the data from the graph. there are only four people left. If you add the 180 students taking Spanish to the 150 students taking French. 11.
13. " 5 .30 taking both 150 taking JUST Spanish Therefore. the average can be computed as follows: 3 + 4 + 6 + 3 = 28 = 4 average = 5 + 2 + 5 + 7 7 12.12 3 = 36 – 4 = 32 14. 4. You are given that m2 – n2 = 24 and m – n = 6.a b 12 5 3 = ^12 h^ 3h .Practice Test I
After one person has been chosen to win first place. Divide both sides by 6. Notice that you can factor the expression m2 – n2 m2 – n2 m2 – n2 = (m – n)( m + n) 24 = 6(m + n) 24 = 6 ^ m + n h 6 6 4=m+n a 5 b = ab . Now. to sort out the necessary data." is defined by the given equation: To find 12 5 3 . 150 = 50% of the students are taking Spanish but not French.
Spanish 180 −30 150
Both 30
French 150 −30 120
225
. there must be 30 students taking BOTH languages (330 – 300 = 30). The operation. You can then sort out the rest of the data as follows: 180 taking Spanish . shown below. 4. you only have three people left from which to win third place—your blanks are now filled out as follows: 5
st
Practice Test I
4
nd
3 3rd
1 2
Your answer is just the product of these numbers—so (5)(4)(3) = 60. you let a = 12 and b = 3 in the given equation.
This is obviously not an odd integer.Part III: Math Practice Tests
15. find the average. 16.
D c° w° r°
a°
125° A
m° B
y° C
226
. since 3 of the 12 marbles are red. Let n = the first odd integer. as well as the two odd integers after this. therefore. Since their sum is 456. 111 + 117 = 228 Another method that works with any type of consecutive integer problem (consecutive integer.
4n = 444 4n = 444 4 4 n = 111 Divide both sides by 4. 3 # 4 # 2 = 24 = 1 or . or consecutive odd integer) is to divide the sum of the integers by the number of integers. Then sum of smallest and largest is 111 + 117 = 228.e. 115. i. and 117. while n + 6 = the fourth odd integer.. 11 2 The probability of the third marble being red is: . and n + 4 = the third odd integer.018 You have 3 + 4 + 5 = 12 marbles in the sack. consecutive even integer. 113. but you merely list the two odd integers before this. Your final answer is just the product of these individual probabilities: Thus. your equation becomes: n + (n + 2) + (n + 4) + (n + 6) = 456 4n + 12 = 456 –12 –12 Combine like terms of the left side of equation. In this problem you will use the following property three times: The measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles of the triangle. Subtract 12 from each side.
So our 4 consecutive odd integers are 111. namely 111 and 113. The sum of the smallest and largest is. 125. then n + 2 = the second odd integer.018. 1 or . namely 115 and 117. since 4 of the remaining 11 marbles are blue. 228. For our problem. 12 11 10 1320 55 17. since only 2 of the remaining 10 marbles are red. 12 The probability of the second marble being blue is: 4 . 456 ÷ 4 = 114. 55 The probability of the first marble being a red is: 3 . remember 10 you already drew one red marble on your first draw.
$100 left to pay off $525 At $75 per month it will take Xena 525 = 7 months to pay off her new refrigerator.80(5y) 2x + 3y = 4y –3y –3y 2x = y 2x = y y y 2x = 1 y 1 2x = 1 1 2c y m 2^ h x=1 y 2 Multiply both sides by 1 to get x y by itself. 16 16 16 Divide both sides by 3. and then make substitutions using the second and third equations. point C will be at 2 2 .5 Translating the given sentence into an algebraic equation. Since points B. C. 75 3. refrigerator cost $625 . and point D at 2 3 .Practice Test I
First: Second: Third:
125 = r + m by the property on the preceding page r = w + c by the same property m = a + r by the same property
Practice Test I
If you take the first equation.
Divide both sides by y since you are trying to get x y. To find the 8 8 8 coordinate x of point M. B. Between A and E is a distance of 1 . D. This means point B will be at 2 1 . 2 Subtract 3y from both sides of the equation. you need to find the number halfway between 2 1 and 2 2 . C.down payment . Therefore.
Practice Test IC Explanations
1. you get: 2 2x + 3y = . you have 2 2 and 2 4 . you divide the distance from point A to point E into 4 equal parts. you can just divide the first expression by 3.
227
. which you will write as 4 since you have to 2 8 divide into 4 equal parts. and D are equally spaced between A and E on AE . If you change each of these 8 8 fractions into sixteenths. 125 = r + m. x would be 2 3 . To get from the expression 3x + 15 to the expression x + 5. So: 3x + 15 = 36 3x + 15 = 36 3 3 x + 5 = 12 2. 1 or . you have: 125 = (w + c) + (a + r) 125 = a + c + w + y re-order the terms 18.
If the girls' average is g. Choice A. D. The triangle inequality theorem in geometry relates the lengths of the sides of any triangle in the following way: The difference of other two sides < one side of triangle < sum of other two sides. Likewise. then their total is 12g. A. So the total ages for both groups is: 12g + 17b.
x = 16 5. To solve this equation: 2 x -3=5 +3 +3 2 x =8 2 x 8 = 2 2 x =4
` xj = 42
2
Add 3 to each side of the equation.
Square both sides to get rid of square root on left. you then have f(2x) = (2x)2 – 3(2x) + 5 = 4x2 – 6x + 5 7. 6. D. if the boys' average is b. For the triangle in this problem with its given side lengths. The only number choice not in this interval is 24.Part III: Math Practice Tests
M A 2 2 2
1 8 2 16
B x
C
D
E 2
1 2 1 2
2 2 x= 2
2 8 4 16 3 16
2
3 8
2
4. To find f(2x). Divide both sides by 2. you find that the third side of this triangle is between 25 and 45. With f(x) = x2 – 3x + 5.
10 35
35 − 10 < x < 35 + 10 or 25 < x < 45
228
. A. then their total is 17b. you substitute 2x for each x in the given function.
you notice that the first two terms. You are given that the area and circumference are equal—so you write and solve the following equation: area = circumference πr2 = 2πr πr = 2πr π π
2
Substitute appropriate formulas for area and circumference. a + 3b + c = 12. Replaced the (a + 3b) with 8. To find f(g(3)). the y-coordinate is 5. the y-coordinate is –1. a + 3b. You proceed as follows:
Practice Test I
2a + 6b = 12 2a + 6b = 16 2 2 a + 3b = 8
This is the first given equation. Divide both sides by 2. B. In the second given equation. C. is exactly 1 of the left 2 side of the first equation: 2a + 6b = 16.
r2 = 2r r 2 = 2r r r r=2 Divide both sides of equation by r. Divide both sides by π.Practice Test I
8.
229
.
10. you are trying to get a + 3b.
−4
4
x
−4
4
x
−5
−5
From the left graph: g(3) = –1 Using the right graph: f(–1) = 5
For an x-coordinate of 3. E.
You then substitute this into the second equation: (a + 3b) + c = 12 8 + c = 12 Then c = 4 9. For an x-coordinate of –1. you first find g(3) and then find f of this value.
y 5 y = g(x) y 5 y = f(x)
Parentheses added for emphasis.
Practice Test I
15. –16 Subtracted 16 from both sides of the equation. you find the average rate: average rate = 15 + 15 = 30 = 3. so that t = d r 15 miles 15 miles timegoing = = 3 hours and timereturning = = 5 hours 3 mph 5 mph Now that you have all the necessary data. Expanded (r + 4) 2 as (r + 4)(r + 4) and used "FOIL.
231
. C." Subtracted r2 from each side. remember d = rt.75 8 3+5 16. To find the average rate for the entire trip you use the formula: average rate = total distance for trip total time for trip
Practice Test I
You first find the time for each part of the trip. Since PT is tangent to the circle at point T. In right ᭝PTC. PT = TC .
T r 8 C 4 P r+4 r
You have added in the radius from point C to point T.
2 2
You can now solve this equation. you use the Pythagorean theorem to create the following equation: 82 + r2 = (r + 4) 2 64 + r = r + 8r + 16 –r2 –r2 64 = 8r + 16 –16 48 = 8r 48 = 8r 8 8 6=r Divided both sides by 8. B.
All numbers used are real numbers236
. Some problems may be accompanied by figures or diagrams. select the one correct answer of the five choices given and mark the corresponding circle on your answer sheet. The figures and diagrams are meant to provide information useful in solving the problem or problems.Part III: Math Practice Tests
Practice Test IIA
Time: 25 minutes 18 questions (8 multiple-choice and 10 grid-in)
Directions: This section is composed of two types of questions. Your scratch work should be done on any available space in the section. Unless otherwise stated. all figures and diagrams lie in a plane. Notes 1. Use the 25 minutes allotted to answer both question types. 2. For Questions 1–8. These figures are drawn as accurately as possible EXCEPT when it is stated in a specific problem that a figure is not drawn to scale. 3. Calculators may be used.
4 2 5x + 3y = xy . The equation 2x . D. B. E.3y = xy . 1 3 5 7 9
m°
6. The average (arithmetic mean) of 5x and 3y is equal to 4 less than the product of x and y. The number that results from 3n. D. C. what is the value of m + n? A. What is the x-coordinate of this point? A. E. if x = 50. 2 2 2 4 4 2 8
GO ON TO THE NEXT PAGE 237
.xy 2
O
D
x
Note: Figure not drawn to scale. B. A discount of 20%.
33% 40% 45% 55% 60%
2. C. 5x + 3y = 4 . where k is some positive constant. what is the value of x? A. D. –12 –4 0 3 4
A
4. 2 only 5 only 2 and 5 –2 and 5 2 and –5
y V B C y=k−x2
Practice Test II
3. D.
A.Practice Test II
1. is equivalent to a single discount of what percent?
n°
x°
Note: Figure not drawn to scale.3 = 7 is true for which value(s) of x? A. E. E. In the figure above. where n is a positive integer. 50 65 75 110 130
7. C. E. If 7x – 5 = 4x – 17.4 2 5x . B. B. the line with equation y = 3x – 12 crosses the x-axis at one point. followed by another discount of 25%. In the xy-plane. C. E. B. D. B. D. what is the value of k? A. D. B. C. and if the area of ABCD is 16. CANNOT end in which of the following digits? A. E. E. –6 –4 1 4 5
5. C.
8. B. If V is the midpoint of side BC of rectangle ABCD. C. D.xy 2 5x + 3y xy = 2 4 5x . C.3y = 4 . The figure above shows the graph of y = k – x2. Which of the following equations states the relationship given in the previous sentence? A.
CHECK YOUR WORK ON THIS SECTION ONLY.
STOP
241
. What is the largest of 5 consecutive even integers whose sum is 170?
16. DO NOT WORK ON ANY OTHER SECTION IN THE TEST. where a. The figure above consists of three circles having the same center. If 720 = ambncr. AB is tangent to the circle at point B. If m2 + 2mn = 3n – 4w.Practice Test II
13. and c are different positive prime integers. 3. What is the area of ᭝ABC?
IF YOU FINISH BEFORE TIME IS CALLED. B (11. 2). What is the slope of AB ?
17. what is the value of n when m = 3 and w = –5?
y x A C B (9. what is the value of m + n + r? 15. and C (11. In the figure above. Their radii are 2. 2). The shaded area is what fraction of the area of the largest circle?
Practice Test II
18. ᭝ABC in the xy-plane has vertices at A (5. the circle with center C and radius 5 is tangent to both the x and y axes.−8)
14. b. 6). and 4.
3242
. 2. Calculators may be used. all figures and diagrams lie in a plane. Some problems may be accompanied by figures or diagrams. Unless otherwise stated. Notes 1.Part III: Math Practice Tests
Practice Test IIB
Time: 25 minutes 20 multiple-choice questions
Directions: Select the one correct answer of the five choices given and mark the corresponding circle on your answer sheet. Your scratch work should be done on any available space in the section. The figures and diagrams are meant to provide information useful in solving the problem or problems. These figures are drawn as accurately as possible EXCEPT when it is stated in a specific problem that a figure is not drawn to scale. All numbers used are real numbers.
D. 8. If 3(2x + 4)(5 – x) = 0. B. D. B. The radius of one circle is 1 the radius of a 3 second circle. 1:9 1:3 3:1 6:1 9:1
9. . c<h<r<m r<c<h<m h<r<c<m c<h<r<m h<c<r<m
2. D. C. which of the following correctly represent the order of their heights? A. What is the tenth term of this sequence? A. c. C. D. –4 –3 -3 4 4 3 12
Practice Test II
3. the line l is perpendicular to the graph of the equation 3x + 4y = 12. B.Practice Test II
2. B. C. what is the value of m when r = 6? A. When x = 2. 1 2 5 C. what is the value of 1 +1+ 1 ? x+1 x x-1 A. E. 1 5 6 D. what are all the possible values of x? A. 1 #y# 1 3 8 1 #y# 1 3 5 4 ≤ y ≤ 24 3<y<5 3≤y≤5
4.3 f ^ xh = 6 + x 4
10. 13. C. Mark is taller than Rebecca. and 5 0. C. Which of the following is the slope of line l? A. 5 E. E. B. Charles. 5. B. If 2 ≤ x ≤ 6 and 6 ≤ xy ≤ 30. 0 only –2 and 5 only 2 and –5 only 0. what is the value of x? A. its value represents the sum of the two terms preceding it. D.5h ! f ^ 5h ? A. C. C. and –5
5. 2 5 5 2 14 5 22 5 9
8. E. E. D. 11 7. The quantity m varies inversely as the square of the quantity r. D. and h represent the heights of Mark. B. 2. C. B. and Charles is shorter than Rebecca but taller than Harriet. 4 6 36 64 144
GO ON TO THE NEXT PAGE 243
. E. In the xy-plane. 3. If 5(x + 3) = 17. and Harriet respectively. . –2. The first term in the sequence above is 2 and the second term is 3. 1. E. D. E. Rebecca. If m. For each term after the second term. B. What is the ratio of the area of the smaller circle to the area of the larger circle? A. C. D. If m = 9 when r = 4. f ^ xh = x f ^ xh = 7 f ^ xh = x 3 + 5 f ^ xh = x 2 . . 34 55 89 144 233
6. E. 1 6 B. r. which of the following gives the range of all possible values of y? A. For which of the following functions does f ^ . E.
what is the value of m? 1 A. −3)
B
x° C
Note: Figure not drawn to scale. What is the measure of the largest angle of a triangle. E.47)
17. D. 2 2 4 3 2 2 (approximately 2. D.Part III: Math Practice Tests
x 11. C. 1 C. (5. E. C. with y ≠ 0. C. If x᭝y = y .24) 5 (approximately 4. 10 35 50 70 75
14. –3) is its vertex. F feet. Mr. E. what is the length of AD ? A. D.82) 3 (approximately 3. How many inches are there in a distance of Y yards. B. B. 3Y + F + 12I 36Y + F + 12I 3Y + 12F + I 36Y + 12F + I 3Y + 12F + 12I
y A O x x° D
(5. in which the degree measure of its angles have a ratio of 5:6:7? A. For the parabola above. 2 B. D. 2 D. What is the area of the triangle? A. They select a row on the side having only five seats.
13. C. D.46) 2 (approximately 4. B. If each of the parents must sit at either end of the row. and I inches? A. in how many different ways can the Franklin family be seated in the five seats? A. 2 C. and Mrs. E. D. E. If the length of AB = 2 and the length of BC = 2 . In the figure above. E. B. 9 2 9 3 18
244
. 8 E. C. B. and if 4᭝m = 16᭝2. 16 12. 3 6 12 24 120
15. Which of the following are the x-coordinates of two points on the graph of the parabola for which their y-coordinates are equal? A. 4 and 7 2 and 9 0 and 8 –1 and 11 1 and 10
16. 9 2 9 2 B. Franklin took their three children to a theatre to watch a movie. AB = BC and AC = CD . The perimeter of an equilateral triangle is 18.
30° A F E
Note: Figure not drawn to scale.
20. D.
18. C. For which values of m will f(2m) = 15? A. DO NOT WORK ON ANY OTHER SECTION IN THE TEST. Let the function f be defined by f(x) = x2 – 2x. B. m r m+r m m+r mr mr m+r It cannot be determined from the given information. E.3 and 5 2 2 3 and –5 –3 and 5 3 only
Practice Test II
IF YOU FINISH BEFORE TIME IS CALLED. CHECK YOUR WORK ON THIS SECTION ONLY.
STOP
245
. If m and r are constants and if (x + m)(x + r) = x2 + cx + w. C. D. where c and w are constants. what c in terms of m and r? is w A. 3: 2 2: 1 3:2 3: 1 2:1
E. D.Practice Test II
B
C
19. B. E. What is the ratio of the area of ᭝ABF to the area of ᭝CED? A. In the figure above. BC AD .
45° D
B. C. –2 only .
Unless otherwise stated. Calculators may be used. Your scratch work should be done on any available space in the section. all figures and diagrams lie in a plane. The figures and diagrams are meant to provide information useful in solving the problem or problems. Some problems may be accompanied by figures or diagrams. These figures are drawn as accurately as possible EXCEPT when it is stated in a specific problem that a figure is not drawn to scale246
. 2. All numbers used are real numbers. 3. Notes 1.Part III: Math Practice Tests
Practice Test IIC
Time: 20 minutes 16 multiple-choice questions
Directions: Select the one correct answer of the five choices given and mark the corresponding circle on your answer sheet.
A 10 CAND. E. B. 5
12
11 10 9 8 CAND. B. 75° 115° 120° 150° 165°
3. what is its total surface area? A. B. C. then the value of f(–2) = 4. D. B. The graph above shows the number of votes received by each of four candidates in an election. 7 4 D. 6 and 9 10 and 5 8 and 15 12 and 4 9 and 12
2. 6 8 16 18 24
. E. 2 3 5 6 7
GO ON TO THE NEXT PAGE 247
. If f ^ x h = 7 -x A.3x 2 . D. what is 35 the numerator of the fraction? A. 3 6 C. D. The denominator of a fraction is 3 more than its numerator. B. Y 40
CAND.Practice Test II
1. What fraction of the votes did candidate Y receive? A. C. . C. B 20
5. E. Which of the following pair of numbers is "relatively prime?" A. Two positive integers are called "relatively prime" if their only common factor is 1. what is the degree measure of the angle between the hour and minute hands on the traditional clock above? A. D. E. 3 12 E. If the fraction is equal to 14 . 1 6 1 4 1 3 1 2 2 5
6. At 5:00pm. C. E.12 5 -4 B. D. C. X 50 7
1 2 3 4
6
5
Practice Test II
CAND. If the volume of a cube is 8.
E. 3x + 6 = 72 3y = 72 3z – 6 = 72 I only II only I and II only I and III only I. B. 2
16. D. II. III. B. If x represents the smallest. II. 64 – 16π 32 – 8π 16 – 4π 8 – 2π 4–π
14. C. y the middle. D. The radius of a right circular cylinder is doubled.Part III: Math Practice Tests
12. D. CHECK YOUR WORK ON THIS SECTION ONLY. B. 1 12 1 6 1 3 1 2 6
15. and its height is tripled. C. In the xy-plane. how many different integer pairs (x. E. What is the ratio of the volume of the original cylinder to the volume of the new larger cylinder? A. and if 2 then xy must be A. B. The sum of three consecutive even integers is 72. E. C. which of the following equations could be used to represent the first sentence of this problem? I. B. E. E. A. What is the area of the shaded region? A. 4 5 8 12 13
IF YOU FINISH BEFORE TIME IS CALLED. If x + 4 is an integer. y) are in the solution set of x + y # 2 ? A. 250
STOP
. and III y-3 is an integer. a circle is inscribed within a square having a perimeter of 32. D. C. DO NOT WORK ON ANY OTHER SECTION IN THE TEST. D. C. an even integer a negative integer an odd integer a multiple of 12 a positive integer
13. In the figure above. and z the greatest of these even integers.
Now that you've pinpointed the type of error. compare it to other practice tests to spot other common mistakes.Part III: Math Practice Tests
Analyzing Your Test Results
The charts on the following pages should be used to carefully analyze your results and spot your strengths and weaknesses. carefully tally your mistakes by marking them in the proper column. These results should then be reexamined for trends in types of errors (repeated errors) or poor results in specific subject areas.
252
This reexamination and analysis is of tremendous importance to you in ensuring maximum test preparation benefit. Now that you have taken the practice test and checked your answers.
Mathematics Analysis Sheet
Section A Multiple Choice Grid-Ins Subtotal Section B Multiple Choice Subtotal Section C Multiple Choice Subtotal Overall Math Totals Possible 8 10 18 Possible 20 20 The complete process of analyzing each subject area and each individual problem should be completed for each practice test.
5. C. Just try different values of n and see what you get for the last digit of the answer. For n = 1: 31 = 3; number ends in a digit of 3 For n = 2: 32 = 9; number ends in a digit of 9 For n = 3: 33 = 27; number ends in a digit of 7 For n = 4: 34 = 81; number ends in a digit of 1 So the number cannot end in a digit of 5, Choice C. 6. B. 20% off means that 4 of the price remains. 5 3 25% off means that of that price remains. 4 So a discount of 20% followed by a discount of 25% would leave 4 : 3 = 3 of the original price, meaning 2 , or 5 4 5 5 40% has been discounted. 7. D. To solve an equation of the form "stuff " = N , break it into two parts: "stuff" = –N or "stuff" = N So the given equation 2x - 3 = 7 turns into the 2 parts: 2x – 3 = –7 2x – 3 + 3 = –7 + 3 2x = –4 2x = - 4 2 2 x = –2 or x = 5 8. C. By setting x = 0 in the given equation y = k – x2, you get that y = k is the y-intercept of the graph of this equation. So the parabola crosses the y-axis at k. Thus the height of the rectangle ABCD is also just k, as shown in the figure below.
y B V C
12. 1 or .5 2 First, solve the given equations for x and y; then you can find the value of xy. x
-1 -1 3 -3
=2 You have taken the –3 power of both sides so that the exponent of x on the left will then be just 1.
a x 3 k = 2 - 3.
x1 = 2–3 x = 13 . 2 1 x= 8 Next, solve for y. y2=2
1
Remember that, in general, ^ number h
negative
=
1 positive . ^ number h
a y 2 k = 22.
1 2
Take the second power of both sides; you want just y1 on the left.
y=4 Finally, xy = xy = 1 : 4 = 1 8 2 You could also write the decimal .5 13. 38 Let n = first even integer. Then n + 2 = second even integer, so n + 4 = third even integer, and n + 6 = fourth even integer, with n + 8 = fifth even integer. Since their sum is 170, you have the equation: n + (n + 2) + (n + 4) + (n + 6) + (n + 8) = 170 5n + 20 = 170 5n + 20 – 20 = 170 – 20 5n = 150 5n = 150 5 5 n = 30 So your five consecutive even integers are 30, 32, 34, 36, and the largest one 38. Another shorter approach works as follows: Divide the sum of the integers by the number of integers . . . so 170 ÷ 5 = 34, which gives the middle of our list of numbers. So the two above this are 36, and the largest 38. Combine like terms on the left side. Subtract 20 from both sides.
256
Practice Test II
14. 4 or 1.33 3
y x A
C (5, −5)
B (9, −8)
Since the circle is tangent to both axes, located in the fourth quadrant, and it has a radius of 5, you know that its center has coordinates (5, –5) as in the figure above. Since AB is tangent to the circle at point B, you also know that AB = CB . Remembering that the slopes of perpendicular lines have opposite reciprocal slopes (for example, 2 and - 3 ); you have to find only the slope of CB and take its opposite reciprocal. 3 2 slope BC = - 8 -- 5 = - 3 So the slope of AB is just 4 or 1.33 4 3 9-5 15. 12 Below is a rough sketch of the points in the xy-plane.
y C (11, 6) h=4 A (5, 2) b=6
Practice Test II
B (11, 2) x
Since points A and B have the same y-coordinate, they lie on a horizontal line segment, and since points B and C have the same x-coordinate, they lie on a vertical line segment as shown in the figure above. You can easily see that the base of the triangle is just 6 and height is just 4. Therefore, its area = 1 bh = 1 : 6 : 4 = 12 . 2 2 11 16. or 3.66 or 3.67 3 m2 + 2mn = 3n – 4w 32 + (2 ⋅ 3 ⋅ n) = 3n – (4 ⋅ –5) 9 + 6n = 3n + 20 9 + 6n – 3n = 3n + 20 – 3n 9 + 3n = 20 9 + 3n – 9 = 20 – 9 3n = 11 3n = 11 Divide both sides by 3. 3 3 n = 11 or 3.66 or 3.67 3 Subtract 9 from each side of the equation. Subtract 3n to get the variable on just one side. Substituted given values of m = 3 and w = –5.
Since 720 = 24 × 32 × 51, the desired numbers m, n, and r are just the exponents 4, 2, and 1. If you wrote the factored form as 720 = 32 × 24 × 51 instead, the desired m, n, and r would still be the exponents 2, 4, and 1. Since you are finding the sum of m, n, and r, the order of the exponents makes no difference. So, in any order, m + n + r = 4 + 2 + 1 = 7 always.
Practice Test IIB Explanations
1. D. The first few terms of the sequence are 2, 3, 5, 8, 13, . . . The third term, 5, is the sum of 2 and 3, the two terms preceding it. The fourth term, 8, is the sum of 3 and 5, the two terms preceding it. The fifth term, 13, is the sum of 5 and 8. Continuing in this manner, you get: The sixth term = 8 + 13 = 21. The sequence is now 2, 3, 5, 8, 13, 21.
4. B. If 3(2x + 4)(5 – x) = 0, then one of the two factors containing x could be 0. So either 2x + 4 = 0 or 5 – x = 0 2x = –4 or 5 = x x = –2 or x = 5 5. C. To determine if f(–5) does NOT equal f(5), substitute –5 for x and 5 for x to see if the answers are the same. A. For f ^ x h = x , You can easily see that when either –5 or 5 are put in for x, the answer is 5 both times. B. For f(x) = 7, the answer is 7 no matter what the value of x; so once again the answers are the same: 7 both times. C. For f(x) = x3 + 5, you get: f(–5) = (–5)3 + 5 and f(5) = (5)3 + 5 = –125 + 5 = 125 + 5 = –120 ≠ 130 So in this case, f(–5) ≠ f(5); thus Choice C is the answer.
Choice D has just such a pair of numbers: –1 and 11. The person to sit at the left end of the row can be chosen two ways (one of the two parents) and the person to sit at the other end can be chosen only one way (the remaining parent must sit there). Below is a series of five blanks. Divide both sides by 18. So to answer the given question. Since the ratio of the angle measures is 5:6:7. 6x. 14. and finally the last child seat can only be filled one way (the last un-chosen child must sit there). The dashed vertical line passing through 5 on the x-axis is called the "axis of symmetry" for the parabola. or C for child. and 7x as in the figure below. –1 is 6 units left. The points on the parabola to the left of the axis of symmetry are just mirror images of the corresponding points on the parabola to the right of the axis of symmetry. For example.Practice Test II
12.
6x°
5x°
7x°
Then. D. 2 × 3 × 2 × 1 × 1 = 12
261
. the parabola has vertex at the point (5. therefore. So your series of blanks now looks like: 2 1 PCCC P Next you need to seat the three children. Your final answer is just the product of these numbers. then the next seat only two ways (one of the remaining two children). you are looking for two numbers that are same distance—left and right—from 5. The first child seat can be filled three ways (any one of the three children). one for each seat in our row of five seats: P C C C P Under each blank is the letter P for parent.
Practice Test II
0 −1 5 (5. So you find the value of 7x = 7(10) = 70. you have the equation: 5x + 6x + 7x = 180 18x = 180 18x = 180 18 18 x = 10 But you want the measure of the largest angle of the triangle. you see that the parabola passes through the origin. C. –3). since that is just 5 units to the right of the axis of symmetry. −3)
10 11
x
In the figure above. 13. Your series of seating choices now is complete: 2 3 2 1 1 P C C C P . which is 5 units left of the axis of symmetry. the parabola must cross the x-axis at 10 also. D. and 11 is 6 units right of the axis of symmetry as shown in the figure above. you can label the angle measures of the triangle with 5x.
y 6 left 6 right
Combined like terms on left.
D. Since AB = 2. x 2 . labeled y in the diagram.
A z x° 2 y 2 x° B 2 C D
In the figure above. you know that AB = 2 2 . the altitude drawn to the base of the triangle will bisect the base and bisect the angle at the top of the triangle. F feet = 12F inches. F feet. since the equilateral triangle has a perimeter of 18. then BC = 2 also. and I inches = 36Y + 12F + I inches 16. 17.
So the length of AD is 2 3. the length of each side is 6. The area of the triangle is then 1 b : h = 1 : 6 : 3 3 = 9 3 .Part III: Math Practice Tests
15. each angle has a measure of 60°. Using your 3 pattern for this type of triangle. D. forming two 30°-60°-90° triangles. B. multiply by 36 (1 yard = 36 inches) To change feet to inches. you can find the height of the equilateral triangle to be 3 3. Since the triangle is equilateral. Therefore a distance of Y yards. To change yards to inches. 2 2
262
. ᭝ABC is an isosceles right triangle. x. ᭝ACD is a right triangle. so you use the Pythagorean theorem to find the length of AD : y2 + 22 = z2
`2 2j + 2 2 = z 2
2
Replace y with 2 2 . multiply by 12 (1 foot = 12 inches) So Y yards = 36Y inches. x. Using the pattern for an isosceles right triangle.
30° 6 3√3 60° 3 6 6
Referring to the figure above.
8+4=z 12 = z 4$3=z 2 3=z
2
12 = z2 Take the square root of both sides. the base of the equilateral triangle is obviously 6.
Since CE is x.
Practice Test II
For the left side and right side polynomials to be equal. So you have: c = r + m.
B 60° x 30° A x√3 F E x x C 45° 45° D
In the figure above. C.3 and m = 5 2 2 Factor the left hand side. let the height CE of the isosceles right ᭝CED be x. Notice that ᭝AFB is a 30°-60°-90° triangle. you find the base AF of this triangle is just x 3. Then the base ED will also be x. which is the same as m + r . so you have: x2 + cx + w = (x + m)(x + r) x2 + cx + w = x2 + rx + mx + mr x2 + cx + w = x2 + (r + m)x + mr x + cx + w = x + (r + m)x + mr x 2 + cx + w = x 2 + ` r + m jx + mr
2 2
Expand the right side (FOIL).Practice Test II
18.
2
Put 2m in for x in the given function. Choice C. You can now find the requested ratio: w mr mr 20. so using the 3 pattern. You are told that the given expressions are equal. the coefficients of their respective terms must be equal. Factor x out of the 2 middle terms. coefficient of the x term on left = coefficient of x term on right w = mr. you want to set the equation equal to 0 and then factor it later. Using the given function f(x) = x2 – 2x. Just write in left side of the equation. Subtract 15 from each side. BF will also be x and this will be the height of ᭝AFB. constant term on left = constant term on right c = r + m . Add underlining for the next step. B. you have the following equation to solve: f(2m) = 15 (2m) – 2(2m) = 15 4m2 – 4m = 15 4m2 – 4m –15 = 15 – 15 4m2 – 4m –15 = 0 (2m + 3)(2m – 5) = 0 2m + 3 = 0 and 2m – 5 = 0 2m = –3 and 2m = 5 m = .
263
. Set each factor equal to 0. you can find the requested ratio using the area of triangle formula 1 bh . 2 1 :x 3:x AREA$ ABF 2 3 AREA$ CED = 1 : x : x = 1 or 3: 1 2 19. Now that you have the base and height for each triangle. D.
3 ^ . X 50
CAND. 8 and 15 only have a common factor of 1. So they are NOT relatively prime. Checking each of the answer choices. B 20
From the circle graph above. C. 2. 10 and 5 have common factors of 1 and 5.
CAND. A.4 f ^. the total surface area will be the area of the shaded side square multiplied by 6.2h = = = = 9 9 3 7 . 120 3 3x 2 . E. Since you are told the volume is 8. you find the product of the length. So these ARE relatively prime. you have x3 = 9. Using the given function f ^ x h = 7-x 2 . C.
x
x x
To find the volume of the cube in the figure above. width.2h
264
. So they are NOT relatively prime. you are looking for a pair of numbers whose only common factor is 1. so the total area is 6 × 4 = 24. B. so candidate Y received 40 = 1 of the votes. so x = 2.2 h . the total number of votes cast was 50 + 40 + 20 + 10 = 120. 3. Y 40
CAND. B.^. A 10 CAND. C. you find f(–2) by putting –2 in place of x: 4. Candidate Y received 40 of those votes. and height. The area of the shaded square is 2 × 2 = 4.
2
2 2
In the figure above. since there are 6 equal faces to the cube. getting volume equals x3.3 $ 4 -12 . 6 and 9 have common factors of 1 and 3.Part III: Math Practice Tests
Practice Test IIC Explanations
1.
use x = 1 and try each choice. Distribute the 2 on the right side. Subtract x from both sides—you are solving for N. II. you have > 1 .5 2 2 c x + N m = 2 ^ 3m . A. you must get a true statement. These are the regions in which f(x) is negative. so you have: y = 2x + 7 and use the point (–2. In particular. and III. The only viable choice is D. x > x → with x = . when substituted into the first equation. Note that the positive regions of y = f(x) would change
266
. When substituted into the second equation. On the left is the average of x and N. So the portions of the graph of y = f(x) that are below the x-axis would end up above the x-axis for the graph of y = f ^ x h . D. you must get a true statement. If you took the absolute value of these negative numbers.
2
y = –3x + r and use the point (–2.
Above is the graph of y = f(x) with portions of the graph below the x-axis dotted. 3).
y y = f (x)
Substitute 3 for y and –2 for x. m). you have c 1 m < 1 . x2 < x → with x = 1 . they would turn into positive numbers instead.5h 2 x + N = 6m – 10 x + N – x = 6m – 10 – x N = 6m – 10 – x 10. 3) 3 = –3(–2) + r 3=6+r –3 = r 11. II. which simplifies into 1 < 1 : TRUE 2 2 2 4 2 So your answer must now include both I and II. which simplifies into 2 > 1 : TRUE 2 1 2 2 2 So your answer must include I. you then have: x + N = 3m . this point must lie on the graph of EACH of the equations. 2 1 1 1 I. You don't even need to try III. m) m = 2(–2) + 7 m = –4 + 7 m=3 So the point of intersection is now (–2.
x −3 −1 4 6
their location since their absolute value would still be positive for the graph of y = f ^ x h . B. Since x is between 0 and 1. Multiply both sides by 2. I.Part III: Math Practice Tests
8. rather than solid. If you let the other number be N. Since the graphs of the two equations intersect at the point (–2. as in the figure below. 9. so you now have: Substitute m for y and –2 for x. thus eliminating Choice E. B.
267
. and then find 1 4 of this result.Practice Test II
y
y = f (x)
x −3 −1 4 6
12.
3h h r
Practice Test II
2r
VOL original πr 2 h πr 2 h πr 2 h 1 VOL new = π ^ 2r h 2 ^ 3h h = π : 4r 2 : 3h = 12πr 2 h = 12 13. 2
r=4
8
Then AREAshaded =
AREAsquare . so r = 4. If the perimeter of the square is 32. you find the difference in areas of the square and circle.16π 4 = 16 – 4π Divide each term in expression above by 4.AREAcircle 4 Area square = (side)2 and area circle = πr2
2 2 =8 -π:4 4
= 64 . C. 4 the radius of the circle is just 1 of this. A. To find the area of the shaded region. the length of one of its sides is just 32 = 8 . With the radius of our cylinder doubled and its height tripled. you end up with the original and new cylinders as shown below. As can be seen in the figure below.
2
−2
2
−2
Graph of | x | + | y | = 2 (solid line segments)
The integer pairs (x. and III work. E. as in Choice A. then y = medium even integer. 15. If x + 4 is an integer. then z – 2 = medium even integer. Their sum would be: (z – 4) + (z – 2) + z = 3z – 6. Their sum would be: x + (x + 2) + (x + 4) = 3x + 6. and z = largest even integer.
268
. also works. that means y – 3 must be even. y) that satisfy the inequality x + y # 2 are those on the figure above with an open circle around them—count them all and you count 13 such points. II. their product xy will be even. Therefore I. so the correct answer choice is E. So 3y = 72. In the figure below. 2 Then with x even and y odd. so x is even. With x = smallest even integer. then x + 2 = medium even integer. as in I.Part III: Math Practice Tests
14. So 3x + 6 = 72 works. so y is odd. as in III. 16. A. as in II. With z – 4 = smallest even integer. and y + 2 = largest even integer. that means x + 4 must be even. E. So 3z – 6 = 72 also works. the four dark segments forming the diamond represent the graph of the equation x + y = 2 . Their sum would be: (y – 2) + y + (y + 2) = 3y. 2 y-3 If is an integer. and x + 4 = largest even integer. With y – 2 = smallest even integer.
Some problems may be accompanied by figures or diagrams. 3. The figures and diagrams are meant to provide information useful in solving the problem or problems272
. Unless otherwise stated. These figures are drawn as accurately as possible EXCEPT when it is stated in a specific problem that a figure is not drawn to scale. Calculators may be used.Part III: Math Practice Tests
Practice Test IIIA
Time: 25 minutes 20 multiple-choice questions
Directions: Select the one correct answer of the five choices given and mark the corresponding circle on your answer sheet. Notes 1. All numbers used are real numbers. 2. Your scratch work should be done on any available space in the section. all figures and diagrams lie in a plane.
what is the value of x? A. E. E. It takes Lisa h hours to walk a distance of m miles. C. D.65 r . which of the following could 4 NOT be the number of students in this class? A. and m? m
10. At this same rate. 12. D. 14. If the length is 10. 6 15 16 32 60 -b + c . E. 2 3 4 5 6
13. C.Part III: Math Practice Tests
8.mh 2 b
11. y = –1
m2x2 – b2 (x – m + b)2 xm + b (mx + b)2 2 ^ x . Which of the following inequalities could be used to decided if a motorist driving at a rate of r mph is within legal limits? A. B. An employee had to take a pay cut of 20%. C. D. D. y = 1 x = –1. D. If the average (arithmetic mean) of x. D. 6. x = –6. y = 9 2 x = –2.10 # 10 # 10 # 10 < 10 # 55
274
. B. The ratio of the length of a rectangle to its width is 5:3. C. r . C. 3x – 2y = –12 and x + 3y = 7 For what values of x and y are the equations above both true? A. y = –3 x = 4. y = 3 x = 10.55 r . E.45 r . how many hours will it take her to walk a distance of d miles? A. C. A while later. B. and the maximum speed is 65 mph. dm h dh m mh d d mh m hd
15. B. what is c in terms of x. C. Her new salary after the pay raise was what percent of her salary before the pay cut? A. B. If x = A. D. C. B. B. The minimum speed on a given stretch of freeway is 45 mph. E. b. E. If the ratio of the number of boys to the number of girls in a class is 3 . 21 28 33 42 49
9. 10. she received a pay raise of 25%. what is the area of the rectangle? A. and 10 is equal to 3x.55 r . 45% 55% 80% 95% 100%
12. D. E. B. E. 5x.
1 4 1 3 4 6
20. D. B. 5)
C (9. CHECK YOUR WORK ON THIS SECTION ONLY. Its height h. E. 6 4 3 6 2 8 10
18. B. circles A and B are tangent at C. E. C. C. How many seconds pass between these two times? A. If the radius of circle A is 2 and the radius of circle B is 6. what is the value of x? A. If 82x = 44x – 3. and line l is tangent to circles A and B at points P and Q respectively. In the figure above. The rocket is 80 feet above the ground twice—once on the way up and then again on the way down. In the figure above. in feet. AB = BC . y) Q P l B (3. 276
STOP
. D. C. after t seconds. What is the value of y? A.
A
C
B
Note: Figure not drawn to scale. 1 2 3 4 5
IF YOU FINISH BEFORE TIME IS CALLED. E. what is the length of PQ ? A.
17. A small rocket is fired into the air from the top of a tower.Part III: Math Practice Tests
y A (1. is given by: h(t) = –16t2 + 64t + 32. E. B. D. DO NOT WORK ON ANY OTHER SECTION IN THE TEST. C. B. D. -3 2 2 3 2 6 12
19. 2) x
Note: Figure not drawn to scale.
The figures and diagrams are meant to provide information useful in solving the problem or problems. Calculators may be used. Notes 1. For Questions 1–8. all figures and diagrams lie in a plane. These figures are drawn as accurately as possible EXCEPT when it is stated in a specific problem that a figure is not drawn to scale.Practice Test III
Practice Test IIIB
Time: 25 minutes 18 questions (8 multiple-choice and 10 grid-in)
Directions: This section is composed of two types of questions. Some problems may be accompanied by figures or diagrams. All numbers used are real numbers. select the one correct answer of the five choices given and mark the corresponding circle on your answer sheet. 2. Use the 25 minutes allotted to answer both question types 277
. 3. Your scratch work should be done on any available space in the section.
If the median of a list of seven consecutive odd integers is 23. 282
STOP
. what is the greatest possible difference between the largest and the smallest of these five integers?
IF YOU FINISH BEFORE TIME IS CALLED. point C is the center of the circle. If each digit can be used only once in each of the numbers. In the figure above. 5.Part III: Math Practice Tests
9. CHECK YOUR WORK ON THIS SECTION ONLY. In the figure above. If the radius of the circle is 5. If 12(m + n)(c – r) = 72 and 3(m + n) = 9. The digits 2. rectangle ABCD is inscribed in the circle with center at the origin in the xyplane. 3. What is the value of x?
D
+ 15m m 12. If x = 4. DO NOT WORK ON ANY OTHER SECTION IN THE TEST. If the average (arithmetic mean) of five different two-digit positive integers is 20. Point C with coordinates (3. If the current bacteria population is 1. how many minutes ago was the bacteria population only 32?
18. how many of these five-digit numbers can be created?
25°
C x° A
y
B
Note: Figure not drawn to scale. The population of bacteria in a culture doubles every 10 minutes. The area of the new rectangle is what fraction of the area of the original rectangle?
15. what is the value of c – r?
17.5y 2
2
16.
x C (3. and 6 are used to create fivedigit numbers that begin and end with an even digit. what is the value of
3xy 2 ? x y 2 . y) is located on this circle. what is the value of x ?
Note: Figure not drawn to scale. what is the difference between the largest and smallest of these integers?
10.024. The length of a rectangle is decreased by 25% while its width is decreased by 20%. 4. If 10x x + m = 12 .
13. what is the area of rectangle ABCD?
14. y)
11.
The figures and diagrams are meant to provide information useful in solving the problem or problems. Your scratch work should be done on any available space in the section. All numbers used are real numbers. 3 283
. Notes 1. Some problems may be accompanied by figures or diagrams. 2. Calculators may be used.Practice Test III
Practice Test IIIC
Time: 20 minutes 16 multiple-choice questions
Directions: Select the one correct answer of the five choices given and mark the corresponding circle on your answer sheet. Unless otherwise stated. These figures are drawn as accurately as possible EXCEPT when it is stated in a specific problem that a figure is not drawn to scale. all figures and diagrams lie in a plane.
CHECK YOUR WORK ON THIS SECTION ONLY. E. The accumulated value of her investment t years later is given by the function A. B. For what positive value of x will y = 7? A. C. a given circle with center at the origin contains the point (–3. E. x + (2x – 1) = 2x + 1 x2 + (2x – 1)2 = (2x – 1)2 x2 = (2x – 1)2 + (2x + 1)2 x2 + (2x – 1)2 = (2x + 1)2 x(2x – 1) = 2x + 1
12. D. ` 7. (3. What are all the possible points on this circle having an x-coordinate of 9? A. Rachelle invested $1. 2 3 4 5 6
16. E. C. x . –2) (9.
14.
STOP
285
.02? A. B. 7) only
10. 3) lie on a circle with center A and radius 5. If x 3/4 = y 3/8 . 4). B. what is x in terms of y? A. 4) (0. B. which of the following could NOT be the value of x + y? A. 1 2 3 4 5
11. D. C. Which of the following points is NOT on the given circle? A. D. In the xy-plane. C. In the xy-plane. B. E. E. E. 8) only (9. B. –2) only (9. D. If 0 < x < 1. DO NOT WORK ON ANY OTHER SECTION IN THE TEST. –5) (3. –1) (9. 8) and (9. 7) and (9. the set of points 5 units from the point A (6. D.000 in an IRA paying 6% per year. which of the following equations could be used to find the value of x? A. If xy = 6 . 1 < x<3 x 3 x<1 x< x<3 x< x
x 1 x 1 3 x< x< x 1 <3 x< x x
15.Practice Test III
9. D. 5j is a point on the graph of the equation y = x2 + k. and the length of the hypotenuse is one more than twice the length of the shorter leg. 37 20 18 15 13
Practice Test III
IF YOU FINISH BEFORE TIME IS CALLED. 1 y y y y3 y6
2
13. In how many years will Rachelle's investment be worth $1. If the length of the shorter leg is x. B. In the xy-plane.191. –4) (0. B. C.000(1. C. E. and 1 x? A. (9. C. C. 7) None of the above points are on the given circle. The length of the longer leg of a right triangle is one less than twice the length of the shorter leg. E. D.06)t. where A(t) = 1. D. which of the following gives the correct order of 3 x .
These results should then be reexamined for trends in types of errors (repeated errors) or poor results in specific subject areas.
287
. carefully tally your mistakes by marking them in the proper column.
Reason for Mistakes
Total Missed Section A : Math Section B : Math Section C : Math Total Math Simple Mistake Misread Problem Lack of Knowledge Lack of Time
Practice Test III
Reviewing the preceding data should help you determine why you are missing certain problems. This reexamination and analysis is of tremendous importance to you in ensuring maximum test preparation benefit.Practice Test III
Analyzing Your Test Results
The charts on the following pages should be used to carefully analyze your results and spot your strengths and weaknesses. Now that you've pinpointed the type of error. Now that you have taken the practice test and checked your answers. The complete process of analyzing each subject area and each individual problem should be completed for each practice test. compare it to other practice tests to spot other common mistakes.
Subtract 7x from each side to the variable on just one side.
Divide each side by 3. the new x-coordinate). the two angles marked 20° are equal because they are vertical angles. after being moved 2 units left (–3 + –2 = –5). D. E. You are making two types of choices: style and color of carpet.
B x°
Add 5 to each side to isolate the variable term.
Divide both sides by 2.). 3) moves the point 2 units left (0 – 2 = –2. style color 5. 4. which is 50. the new y-coordinate. To solve the equation: 3x – 5 = 16 3x – 5 + 5 = 16 + 5 3x = 21 3x = 21 3 3 x=7 x+4 = 9 x 7 7(x + 4) = 9x 7x + 28 = 9x 7x + 28 – 7x = 9x –7x 28 = 2x 28 = 2x 2 2 14 = x 3. The translation that moves point C from (0. E. 0) to (–2. you can make a style-color choice in the following manner: 5 : 10 Your answer is just the product of 5 and 10. Then you know that 100 = 20 + x. you have: style color Since you have five styles and 10 colors from which to choose. 4). and then 3 units up (0 + 3 = 3.
100° A
20°
C 20°
In the figure above. So the point P (–3. Distribute the 7 on the left side. To solve this equation:
Cross multiply to solve the proportion. you see that x = 80. Using blanks to help solve this problem.Part III: Math Practice Tests
Complete Answers and Explanations for Practice Test III
Practice Test IIIA Explanations
1.
288
. since the measure of an exterior angle of a triangle equals the sum of the measures of the two nonadjacent interior angles. 7). So by simple subtraction. C.
2. and then 3 units up (4 + 3 = 7) will end up at the point (–5. D.
Wednesday to Thursday: 10 10 25 D. Another way to do the problem is to pick a salary. C. Friday to Saturday: 45 Notice that the fractions for choices A. Original salary –20% pay cut New salary + 25% pay raise 100 –20 80 +20 100 Final salary . After a 20% (or 1 ) pay cut. Monday to Tuesday: 15 25 C. C.x h +3 x 5
2
so f ^ 5h =
^ 2 . she will make just 4 P dollars. 4 4 5 Therefore. A.Practice Test III
6. D. say $100. Subtract 6x to get all the variables on one side. Therefore Choice D will have the greatest percent increase in temperature. so it's 100% of original.
Practice Test III 289
. C. Suppose that the employee was originally paid P dollars. and E are all less than or equal to 1. 5 5 5 5 5 5 5 5
2 2
7. Sunday to Monday: 5 20 B. $100. you need to substitute 5 for x in the f(x) formula: f ^ xh =
^2 . E. her new salary after the pay cut and then the pay raise will be exactly what is was before the cut and raise. she will make 5 c 4 P m = P dollars. Writing the average of the numbers equals 3x.5h ^ . 5 5 1 After the next 25% (or ) pay raise. You need to compare the temperature change between the two days to the temperature on the first day of the two-day pair. is the same as the original. So she will have 100% of her original salary. B. 8. Multiply both sides by 6. To find the value of f(5). you have: x + 6 + 10 + 12 + 5x + 20 = 3x 6 6 c x + 6 + 10 + 12 + 5x + 20 m = 6 ^ 3x h 6 x + 6 + 10 + 12 + 5x + 20 = 18x 6x + 48 = 18x 6x + 48 – 6x = 18x – 6x 48 = 12x 48 = 12x 12 12 4=x Combine like terms on the left side. Thursday to Friday: 20 25 E.3h +3= + 3 = 9 + 3 = 12 = 2 2 . but greater than 1 for Choice D. 9.
Since the ratio of boys to girls is 3 .
10 6
Put 3 in place of y. Choice C.21 –11y = –33 -11y . as follows: 3x – 2y = –12 x + 3y = 7 3x – 2y = –12 –3(x + 3y) = –3(7) 3x . 11.Part III: Math Practice Tests
10. Notice that the x's cancelled out (3x + –3x = 0 ).
Cross-multiply to solve the proportion.
To find the corresponding value of x. E.9y = . use the formula t = d to get the r h hours following: t = d miles = ^ d milesh # h hours = dh hours m miles m miles m h hours Notice in the work in the line above that the "miles" labels will cancel out just like numbers. This is the sum of the pair of equations in previous step. 12.3x . 13. and then solve that for x. D. So the number of students in the class must be a multiple of 7. substitute y = 3 into either of the original equations. Divide both sides by –11. you must first try to eliminate one of the variables. leaving you with just an "hours" label for the final answer. you can set up and solve the following proportion: 5 = 10 3 w 5w = 30 w=6 Your rectangle now looks like the figure below. To find how many hours it took her to walk a distance of d miles. you must have: 4 3 parts girls + 4 parts boys = 7 parts total. is NOT a multiple of 7. To solve the system of linear equations.
The area of the rectangle above is now 6 × 10 = 60
290
.2y = -12 . So her rate was: m miles . 33. The original second equation looks like less work: x + 3y = 7 x + 3(3) = 7 x+9=7 x = –2 Therefore the point of intersection is (–2. With a length to width ratio of 5:3 and a length of 10. Lisa walked a distance of m miles in a time of h hours. 3). B. C.33 = -11 -11 y=3 Distribute the –3 through the bottom equation. You will multiply the bottom equation by –3 to make the coefficients of x opposites in the top and bottom equations.
move graph of y = f(x) DOWN k units. This relationship can be expressed with an absolute value inequality: r .b + c +b + b mx + b = c
^ mx + b h = ` c j
2 2
Add b to both sides to isolate the c term. You solve the following equation for c. in terms of x. you move the graph of y = f(x) 1 unit LEFT and 3 units UP to arrive at the graph below. if k > 0. 45 and 65. if h > 0. The constants h and k affect the graph of y = f(x) as follows: i. b. D.Practice Test III
14. move graph of y = f(x) UP k units.
y y = f (x)
2
x
Above is the graph of y = f(x). C. B. If h < 0. If k < 0. which can be translated as: "the distance between r and 55 is less than or equal to 10.
(mx + b) = c 15. x= -b + c m -b + c p m Multiply both sides of the equation by m. you can write an inequality that represents the range of legal speed limits r: 45 ≤ r ≤ 65. the graphs of any equation y = f(x) and y = f(x+h) + k. In general.55 # 10 . and m. is just 10 units either left of right of 55. With a minimum speed of 45 mph and a maximum speed of 65 mph. move graph of y = f(x) RIGHT h units. Notice that half-way between these two numbers is 55. The only difference is where they are located in the xy-plane. ii. You want to find the graph of y = f(x+1) + 3. So each of these rates.
Square both sides to get rid of the radical. So with y = f(x + 1) + 3." 16. have exactly the same shape. move graph of y = f(x) LEFT h units.
m:x=m: f
mx =. where h and k are constants.
Practice Test III
y
y = f (x + 1) + 3 x
291
.
Part III: Math Practice Tests
Note that an easy check is to see that the "vertex" was moved left 1 and then up 3 with the graph in the same vertical orientation. 82x = 4x4x – 3 (23)2x = (22)4x – 3 26x = 28x – 6 6x = 8x –6 6x – 8x = 8x – 6 –8x –2x = –6 . Replace 8 with 23 and 4 with 22. Cross-multiply to solve the proportion. 17. Add 6 to each side to isolate the variable.2x = . and then solve for y.5 = . On each side. set it equal to . To solve the equation 82x = 44x–3. 2) x
Since AB = BC . multiply the exponents. since the bases are equal. Then the slope of BC has to be 2 (the opposite reciprocal of . In the line above. and then solve the resulting equation. C. 8 and 4. find the slope of AB and use that to help you find the value of y. notice that both bases.6 -2 -2 x=3 Divide both sides by –2.3 3-1 2
292
. the exponents are equal. 5)
B (3. Distribute the 3 on the left side. find the slope of BC in terms of y.3 ). Slope of AB = 2 . 3 2 2 Next. 3 y-2 2 = Slope of BC = 9-3 3 y-2 2 = 3 6 3(y – 2) = 12 3y – 6 = 12 3y – 6 + 6 = 12 + 6 3y = 18 3y 18 = 3 3 y=6 18. So you will first change both sides of the equation to the same base. Subtract 8x from each side to get variables on just one side.
y C (9. First. y)
A (1. are powers of 2. D. you know that their slopes are just opposite reciprocals.
you have: in right ᭝ATB: x2 + 42 = 82 x2 + 16 = 64 x2 + 16 –16 = 64 – 16 x = 48 x = 48 x = 16 # 3 = 4 3 20. Notice that the length of AB is 8. so Choice B IS correct. B. So using the Pythagorean theorem. AT has been drawn perpendicular to radius QB to form a rectangle PATQ and a right triangle ATB. you set h(t) = 80 and solve for t.
2
Subtract 16 from each side of the equation. you are looking for is the same as side AT in the right triangle.
Practice Test III
Practice Test IIIB Explanations
1+1+1 3+2+1 6 2 3 6 = 6 6 6 = 6 =1 1. B. x. Factor t2 – 4t + 3 into (t – 3)(t – 1) Set each factor equal to 0. B. you have: (A) 3n + 2 = 3(even) + 2 = even + 2 = even. so Choice A is not correct. (B) n2 + 3 = (even)2 + 3 = even + 3 = odd. Subtract 80 from both sides to get 0 one side. Each of these is perpendicular to line l since line l is tangent to the circle at points P and Q. Take out common factor of –16. radii AP and QB have been drawn. Take the square root of both sides. 3 3 3 3 2. just the sum of the two radii 2 and 6.
293
. Trying the choices one at a time. 80 = –16t2 + 64t + 32 80 – 80 = –16t2 + 64t + 32 – 80 0 = –16t2 + 64t – 48 0 = –16(t2 – 4t + 3) 0 = –16(t – 3)(t – 1) t–3=0t–1=0 t = 3. with n being even. C. C. The difference between these times is 3 – 1 = 2.Practice Test III
19. To find the times at which the rocket is 80 feet above the ground. and then again at 3 seconds. x = 12 + 75 = 4 : 3 + 25 : 3 = 2 3 + 5 3 = 7 3 3.
Q x P l 2 A 2 8 6 B 2 T 4
In the figure above. t = 1 So the rocket is 80 feet above the ground at 1 second. The distance.
C.
Take square root of both sides. or approximately 4.
Divide both sides of equation by π. or 4 2 . So with your x.Part III: Math Practice Tests
4. 6) A (3. x.
y B (4. C. 8π = πr 2 π π 8 = r2 8=r 2 2=r Simplified 8 = 4 # 2 = 2 2 Since the radius of the circle is 2 2 .
6 x
4
In the right triangle above. 6. 4) x
294
. Thus the area of the square is 4 × 4 = 16. B. x 2 pattern. ᭝DBC is an isosceles right triangle (45°-45°-90°) with a hypotenuse of 4 2. its diameter is twice this.
d = 4√2
As you can see in the figure above.
A 45° P 45° D 4 C 4 B
2
Subtract 16 from each side of the equation. x = 4 # 5 = 2 5 . The area of the circle is given as 8π. you know that each side of the square is now just 4. Therefore 8π = πr2 Area of a circle is found by the formula πr2. the Pythagorean theorem gives us: x2 + 42 = 62 x2 + 16 = 36 x2 + 16 – 16 = 36 – 16 x = 20 x = 20 Take the square root of each side. 7)
C (7.47 5.
+ n -5 5 So the ratio is m mn = .
Since the two polynomial are equal. i. So all you have to do is find the slope of AC . 2 2 7-3 4 2 7. and this will give you the product of all three slopes of the sides of ᭝ABC. the coefficient of the x terms on the left and right sides of the equation must be equal. then you know that m + n = –5. If you decrease this by 25% (or by 1 of L).1 # 1 = . specifically. each triangle at each corner of the figure is an isosceles right triangle.36 = 36 . and the last constant terms on the left and right sides of the equation must be equal.6 = 10 9. multiply that by –1. = 4 # 4 . So the product of all 3 slopes is just . you know that the product of their slopes is just –1. B. Slope of AC = 6 .36
2 2
Expand left side (FOIL). Add underlining for next step. Then you have Areashaded = Areasquare – Arearect. B.
1 1 45° √2 3√2 45° 3 3√2 √2 45° 3 1 45° 3 3
1
In the figure above. x 2 pattern. you find the width of the rectangle is 2 and the length of the rectangle is 3 2 as shown in the figure. 4 4 1 4 Let the width of the original rectangle be W. you have only remaining.1 . So you have: n + m = –5 and mn = –36 + n -5 5 Thus the ratio is m mn = .Practice Test III
Since AB = BC . So using your x. If you decrease this by 20% (or by of W). since that's how factoring works. x. 5 Let the length of the original rectangle be L.36 = 36 8. 5 5
Practice Test III
original rectangle
W
new rectangle (3)L 4
(
4 )W 5
L
295
.5x . and mn = –36.2 # 3 2 = 16 . their respective terms must be equal. you have only 3 remaining. Shorter Method: If (x + m)(x + n) = x2 – 5x – 36. ii. Longer Method: (x + m)(x + n) = x2 – 5x – 36 x2 + nx + mx + mn = x2 – 5x – 36 x2 + (n + m)x + mn = x2 – 5x – 36 x + ` n + m j x + mn = x .4 = 2 = 1 . 3 . Factor 2 middle terms on left.
But if you do some factoring first.09 11 At first glance. thus the two angles are marked x° in the figure. So the angles opposite these two radii are equal. its associated arc. it appears that you need to know the value of both x and y to find the value of the expression 3xy 2 . 5 12 10. so their lengths are equal.Part III: Math Practice Tests
Then you have: Areaoriginal = LW and AREAnew = c 3 L m c 4 W m = 3 LW 4 5 5 So the area of the new rectangle is just 3 of the area of the original rectangle.5
2
As you can now see. its measure is the same as that of arc AB .5 11 11. Notice that since C is the center of the circle.09 x 2 .5i x . as follows: 2 x y 2 . Then.
296
.
A 25° x° C 50° x° B 50°
$ Since the 25° angle is an inscribed angle. segments CA and CB are both radii of the circle. 3x = 3 # 4 = 12 = 12 or 1. 65. Subtract 50 from each side of the equation. you only need to know that x = 4 in order to find the value of this last expression. we have in ᭝ACB: x + x + 50 = 180 2x + 50 = 180 2x + 50 – 50 = 180 – 50 2x = 130 2x = 130 2 2 x = 65 Divide both sides by 2.5 4 2 . AB is twice that.5 16 . as marked in the figure. $ Since ∠ACB is a central angle. The sum of the measure of the angles of a triangle is 180.5y y _ x 2 .5y 2 y 2 ^ 3x h 3xy 2 3x = 2 2 2 = 2 x y . which is 50°. you find that all of the y's will cancel out. or 1. or 50°.
5. If your integers are a. (2. so the horizontal leg of the right triangle is 3 and its vertical leg y. The average of the five different two-digit positive integers is 20. 25. then the one and only one remaining digit. you have n + 6 = 23. 4. and since y = 4. the length of the hypotenuse of the shaded right triangle in the figure above is labeled as 5. n + 4. you can choose the last digit in only one of two ways—the other two remaining even digits. and 6 to create five-digit numbers beginning and ending with an even digit. 27. you would then have: 10 + 11 + 12 + 13 + e = 100 46 + e = 100 e = 54 The difference between the largest and the smallest would be 54 – 10 = 44
Multiply both sides by 5. b.Part III: Math Practice Tests
ii. 23.
298
. 16. b = 11. and e. d. and once you choose one of these even numbers. 17. 36.
y A B
3 y
x
2y
5 D 2(3) = 6
C (3. and d = 13 were the values of the 4 smallest two-digit positive integers. then two of the remaining digits. The digits in the middle can then be any one of three remaining digits (since you have already picked two). Since the median of the list of the seven terms is n + 6 and you are told that this is 23. and n + 12. n + 10. The difference between the largest and smallest is 29 – 17 = 12. Therefore the area of the rectangle is 6 × 8 = 48. then you can write: a + b + c + d + e = 20 5 a + b + c + d + e m = 5 # 20 5c 5 a + b + c + d + e = 100 If a = 10. or a 6. So the first and last digits must be with a 2. 3. You are using the five digits 2. y)
Since the circle has a radius of 5. 21. and 29. 4. Then your seven consecutive odd integers are 17. You can choose the first digit in any one of three ways. 19. 18. 6). y). Longer Method: Let n = first odd integer. you have even even . so 3 × 3 × 2 × 1 × 2 = 36 different five-digit numbers. 44. 4. its height is 2y. so n = 17. n + 6. 48. n + 8. This is just a 3-4-5 right triangle. 3 3 even 2 1 2 even
Our answer is just the product of these numbers in the blanks. If you use a series of blanks to indicate the choices to 3 2 be made in creating your five-digit numbers. so the y must be 4. the height is 8. c. Notice from the figure that the base of rectangle ABCD is just 6. then the next 6 consecutive odd integers after this will be n + 2. c = 12. Point C has coordinates (3.
which is Choice D.000(1. you have: x2 + (2x – 1)2 = (2x + 1)2.000(1.
2x + 1
x
2x − 1
Referring to the figure above.191. so Choice C is correct. C. 16.000(1. A(t) = 1. Using the given equation.60 C.Part III: Math Practice Tests
15. t = 2: A(1) = 1. t = 1: A(1) = 1. you can then use the Pythagorean theorem to create the equation: (short leg)2 + (long leg)2 = (hypotenuse)2 Substituting the appropriate pieces in terms of the variable x.000(1.06)2 = 1. NOTE: This is one of the few problems where you really need to have a calculator to do the problem quickly. t = 3: A(1) = 1. then 2x – 1 = length of longer leg and 2x + 1 = length of hypotenuse.06) = 1. substitute values of t until you arrive at the desired amount of money.123. With x = length of short leg of right triangle.
302
.06)t A.06)3 = 1. D.02.060 B.
4.SAT I Score Range Approximator
The following charts are designed to give you only a very approximate score range. Using your scores: _________________ – ___________________________________ = _____________________ total correct answers wrong answers on multiple choice ÷ 4 raw score 7. not an exact score. The total number of incorrect responses for the multiple-choice questions should be divided by 4. 6. Subtracting this adjustment factor of 4 from the original 30 correct gives a raw score of 24. Subtract this adjustment factor from the total number of correct responses to obtain a raw score. giving you an adjustment factor (round off to the nearest whole number). Use the following table to match your raw score for Mathematics and the corresponding approximate score range:
Raw Score 49–55 41–49 26–40 11–25 5–10 1–4 4–0 Approximate Score Range 710–800 640–700 500–630 380–490 310–370 240–300 200–230
Keep in mind that this is only an approximate score range. Needless to say. Add the total number of correct responses for the three Mathematics sections. This raw score is then scaled to a range from 200 to 800. however. When you take the actual new SAT I. 5. 2. Add the total number of incorrect responses for the multiple-choice questions only. Example: If the total number of correct answers is 30 out of a possible 45 and 16 multiple-choice problems were attempted but missed. Note: No deduction is made for incorrect grid-in responses. This raw score is then scaled to a range from 200 to 800. dividing 16 by 4 gives an adjustment factor of 4. this may affect your scoring range. 3.
303
. some questions may be slightly easier or more difficult.
How to Approximate Your Score in Mathematics
1. you will see questions similar to those in this book.
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EXAMPLES OF SIMPLE INTEREST,
step on how to simplify an expression involving radical,
algerbra 1 honors,
what is algebra used for,
mathematicians who contributed to polynomials,
algebra 1 mcdougal littell answers | 677.169 | 1 |
Note: Citations are based on reference standards. However, formatting rules can vary widely between applications and fields of interest or study. The specific requirements or preferences of your reviewing publisher, classroom teacher, institution or organization should be applied.
"Throughout the history of mathematics, maximum and minimum problems have played an important role in the evolution of the field. Many beautiful and important problems have appeared in a variety of branches of mathematics and physics, as well as in other fields of sciences. The greatest scientists of the past--Euclid, Archimedes, Heron, the Bernoullis, Newton, and many others--took part in seeking solutions to these concrete problems. The solutions stimulated the development of the theory, and, as a result, techniques were elaborated that made possible the solution of a tremendous variety of problems by a single method. This book presents fifteen "stories" designed to acquaint readers with the central concepts of the theory of maxima and minima, as well as with its illustrious history. This book is accessible to high school students and would likely be of interest to a wide variety of readers. In Part One, the author familiarizes readers with many concrete problems that lead to discussion of the work of some of the greatest mathematicians of all time. Part Two introduces a method for solving maximum and minimum problems that originated with Lagrange. While the content of this method has varied constantly, its basic conception has endured for over two centuries. The final story is addressed primarily to those who teach mathematics, for it impinges on the question of how and why to teach. Throughout the book, the author strives to show how the analysis of diverse facts gives rise to a general idea, how this idea is transformed, how it is enriched by new content, and how it remains the same in spite of these changes."--Publisher.Read more...
Why do we solve maximum and minimum problems? --
The oldest problem : Dido's problem --
Maxima and minima in nature (optics) --
Maxima and minima in geometry --
Maxima and minima in algebra and in analysis --
Kepler's problem --
The Brachistochrone --
Newton's aerodynamical problem --
What is a function? --
What is an extremal problem? --
Extrema of functions of one variable --
Extrema of functions of many variables : the Lagrange principle --
More problem solving --
What happened later in the theory of extremal problems? --
More acurately, a discussion.
Abstract:
Throughout the history of mathematics, maximum and minimum problems have played an important role in the evolution of the field. This book presents fifteen 'stories' that acquaint readers with the central concepts of the theory of maxima and minima, as well as with its illustrious history.Read more... | 677.169 | 1 |
STEPS Mathematics: Textbook 1 Level 3A
Devised after extensive consultation with primary teachers, "STEPS" is a complete maths scheme offering a pathway through the programmes of study for Key Stages 1 and 2. Among the features incorporated are activity-led learning; built-in assessment; a flexible approach to suit individual teachers; a wide choice of activities for all ability levels; cross-curricular links; attractive, highly visual pupil materials; help with forecasting and planning; a choice of record-keeping procedures; and ideas for home-school links. "STEPS" is structured and keyed to the new Attainment Targets and Statements of Attainment of the National Curriculum. The series will be kept up-to-date with any National Curriculum changes by means of free updates. Full colour textbooks are introduced at level 3. Activities include: follow-up work to direct teaching; individual, paired and group work; integrated calculator work; investigatory work; and problem solving. This textbook is one of two covering level 3a of the course.3138313
Book Description Collins Educational 103138313
Book Description Collins Educational, 1997. Paperback. Book Condition: Good. STEPS - Level 3a Pupil Book 1: Textbook 1 Level 3A (STEPS mathematics3138313 | 677.169 | 1 |
MATH 100W - LaTeX
Navigate
Description
Welcome! I will be teaching beginning LaTeX to the MATH 100W class this spring 2014.
But even if you are not in 100W feel free to use anything posted below. LaTeX is
the standard software package used by the scientific community to create professional
quality documents (articles, reports, books, resumes, CVs, letters, etc.). I will
post links to LaTeX resources here as well as handouts, so check back regularly.
Course Documents
Here you will find handouts, examples, templates, links to downloadable learning resources,
etc. Please email me if any of the links are not working correctly or if you find
any errors in the documents (especially any TeX templates that I post here). If you
are just starting with LaTeX, read the "Introduction and Resources" document first.
IMPORTANT! - some documents below are listed twice. One copy is the .tex file and
the other is the .pdf document that it generates. Download both and view them side-by-side
or download the .tex file and compile it. The best way to learn LaTeX is with examples
and these documents provide many of them.
For the Detaild Intro To LaTeX part 2, make sure you download the supplemental picture
files that go with it. | 677.169 | 1 |
ALGEBRA 1
Algebra 1 is the most advanced of the three levels of math offered in the 8th grade at Pence Middle School. This course is taught using the flipped classroom model. Students access the lessons online or via the flash drive/DVD issued to them for use throughout the year. For more information on this teaching model, see the "Algebra 1 Fl!p Lessons and Notes" tab.This is a single-blocked high school credit course. Students enrolled in this course will take the Algebra 1 End-of-Course SOL test in May which they must pass in order to verify the credit toward graduation requirements. An exam will be administered at the end of the year which will count toward 14% of the semester grade. Students who pass the SOL test will be allowed to exempt the final exam. | 677.169 | 1 |
This engaging book is a very compact introduction to the essentials of the
MATLAB programming language and is ideal for readers seeking a focused and
brief approach to the software. It contains numerous examples and exercises
involving the software's most useful and sophisticated features along with an
overview of the most common scientific computing tasks for which it can be used.
The presentation is designed to guide new users through the basics of
interacting with and programming in the MATLAB software, while also presenting
some of its more important and advanced techniques. It is suitable for
graduate students, advanced undergraduate students, and professional
researchers in mathematics, scientific computing, and application areas in
science and engineering.
Tobin A. Driscoll is an Associate Professor in the Department of Mathematical
Sciences at the University of Delaware whose research focuses on the numerical
analysis of differential equations. He is coauthor with L. N.Trefethen of
Schwarz–Christoffel Mapping (Cambridge University Press, 2002).
The book examines theoretical and computational aspects of least-squares
finite element methods(LSFEMs) for partial differential equations (PDEs)
arising in key science and engineering applications. It is intended for
mathematicians, scientists, and engineers interested in either or both the
theory and practice associated with the numerical solution of PDEs.
The first part looks at strengths and weaknesses of classical variational
principles, reviews alternative variational formulations, and offers a
glimpse at the main concepts that enter into the formulation of LSFEMs.
Subsequent parts introduce mathematical frameworks for LSFEMs and their
analysis, apply the frameworks to concrete PDEs, and discuss computational
properties of resulting LSFEMs. Also included are recent advances such as
compatible LSFEMs, negative-norm LSFEMs, and LSFEMs for optimal control and
design problems. Numerical examples illustrate key aspects of the theory
ranging from the importance of norm-equivalence to connections between
compatible LSFEMs and classical-Galerkin and mixed-Galerkin methods.
This self-contained textbook presents matrix analysis in the context of
numerical computation with numerical conditioning of problems and numerical
stability of algorithms at the forefront. Using a unique combination of
numerical insight and mathematical rigor, it advances readers' understanding
of two phenomena: sensitivity of linear systems and least squares problems,
and numerical stability of algorithms.
It differs in several ways from other numerical linear algebra texts. It
offers a systematic development of numerical conditioning; a simplified
concept of numerical stability in exact arithmetic; simple derivations; a
high-level view of algorithms; and results for complex matrices. The material
is presented at a basic level, emphasizing ideas and intuition, and each
chapter offers simple exercises for use in the classroom and more challenging
exercises for student practice.
Ilse C. F. Ipsen is Professor of Mathematics at North Carolina State
University. She is the SIAM Vice President for Programs and section editor of
the Problems and Techniques section of SIAM Review. She is also a member of
the editorial boards of the SIAM Journal on Matrix Analysis and Applications,
Numerische Mathematik, and Numerical Linear Algebra with Applications.
This book provides a comprehensive introduction to both the theory
behind the mathematical objects and their C and C++ implementation.
Object-oriented programming of three-dimensional meshes facilitates
understanding of their mathematical nature. Requiring no
prerequisites, the text covers discrete mathematics, data structures,
and computational physics, including high-order discretization of
nonlinear equations. Well debugged and fully explained code segments
are available, and exercises and solutions make the book suitable for
classroom use.
This expository book surveys the main concepts and recent advances in
multiscale finite element methods. This monograph is intended for the
broader audiences including engineers, applied scientists and those who are
interested in multiscale simulations. The book is self-contained, starts
from the basic concepts and proceeds to the latest developments in the field.
Each chapter of the book starts with a simple introduction and the
description of the proposed methods as well as with motivating examples.
Numerical examples demonstrating the significance of the proposed methods
are presented in each chapter. The book addresses mathematical and numerical
issues in multiscale finite element methods and connects them to real-world
applications. Narrative introduction provides a key to the book's
organization and its scope. To make the presentation accessible to a broader
audience, the analyses of the methods are given in the last chapter.
3rd Global Conference on
POWER CONTROL AND OPTIMIZATION
Gold Coast, Queensland, Australia, February 2-4, 2010
Objective of the PCO'2010 conference is contemporary and original research and
educational development in the area of electrical power engineering, control
systems and methods of optimization. Presentations of interest may include,
but are not limited to:
Current developments in the technology of imaging have led to an explosive
growth in the interdisciplinary field of imaging science. With the advent of
new devices capable of seeing objects and structures not previously
imagined, the reach of science and medicine has been extended in a multitude
of different ways. The impact of this technology has been to generate new
challenges associated with the problems of formation, acquisition,
compression, transmission, and analysis of images. By their very nature,
these challenges cut across the disciplines of physics, engineering,
mathematics, biology, medicine, and statistics. While the primary purpose of
this conference is to focus on mathematical issues, the biomedical aspects
of imaging will also play an important role.
The next Householder Symposium on Numerical Linear Algebra will be held June
12-17, 2011, at the Granlibakken Conference Center & Lodge in Tahoe City,
California. Tahoe City is located on the northwest shore of Lake Tahoe. This
meeting is the eighteenth in a series, previously called the Gatlinburg
Symposia.
The Department of Computational and Applied Mathematics at Rice University
invites applications for a postdoctoral research associate position. The
initial term of appointment is one year with a possible second year
contingent upon availability of funding. The term of appointment will begin
on or after August 1, 2009.
The focus of the research is on compressed sensing, machine learning,
optimization, as well as their applications. Candidates should have finished
a PhD in applied/computational mathematics, electrical engineering,
statistics, operations research, or a related field by August 2009 or the
time of appointment and no earlier than December 2006. The salary will be
competitive.
For more information on related research at Rice, see:
Rice University is a private research university with a long tradition of
excellence in undergraduate and graduate science and engineering education.
The Department of Computational and Applied Mathematics hosts research
programs in optimization, numerical linear algebra, control and inverse
problems, mathematical biology, and partial differential equations.
Interdisciplinary work is a fundamental aspect of the Department's program.
Applicants should send a letter of application, current vita, and
descriptions of research plans to
We are inviting applications for a PhD position in Applied
Mathematics/Computational Engineering Science at the Chair for Numerical
Mathematics, RWTH Aachen University, Germany. The Research Group is fully
funded by the Excellence Initiative of the German federal and state governments.
Your research will involve the development of reduced-basis methods and
associated error estimation procedures for time-dependent partial
differential equations. You will be applying these methods to
optimization and/or optimal control problems in engineering and the
natural sciences.
You should have an excellent background in applied mathematics,
computational engineering, or a related field, and a strong interest in
applying modern numerical methods to problems in engineering and
science. Experience in the finite element method, model order reduction
techniques, optimization methods, and/or optimal control is considered a
plus. Good programming skills in Matlab and/or C/C++ are desired. Good
English communication skills are expected; German language skills are
not required.
An EPSRC-funded PhD studentship is available, starting on 1st October 2009 or
soon afterwards, to develop new computational algorithms for the simulation of
fluid flow. The research will be undertaken within the Scientific Computation
group in the School of Computing at the University of Leeds. Please see
for details of the group's research.
The student will be part of a project which aims to develop and implement new
computational algorithms which are specifically designed to reproduce the most
important underlying features of the mathematical model of the fluid flow
physics, leading to far more accurate approximations than are currently
possible. The project will place the student at the forefront of research in
to computational algorithms for partial differential equations, and provide an
opportunity to develop skills working at the interface between applied
mathematics, computer science and engineering.
Applicants should have, or be expecting to obtain in the near future, a first
class or good 2.1 honours degree (or equivalent) in mathematics, engineering
or a mathematical science. The studentship is available for UK/EU candidates
registering for PhD study on a full-time basis, for a period of 3.5 years. The
award will cover academic fees at the UK/EU rate together with a maintenance
allowance at the standard EPSRC rate, Ł13,290 for the first year of study (the
level will be reviewed annually). The full maintenance allowance is available
to both UK and EU candidates. Full details on studying for a research degree
at the University of Leeds can be found at and
details of how to apply can be found at
Informal enquiries should be addressed to Dr Matthew Hubbard [+44 (0)113
3435459, e-mail: meh@comp.leeds.ac.uk]. The deadline for applications is
Friday 7th August 2009.
The Dutch Technology Foundation STW has recently awarded the ComFLOW-3 project
on ``Extreme wave impact on offshore platforms and coastal structures.'' In
this research project, numerical simulation methods will be developed that can
predict the hydrodynamic forces of extreme waves on fixed and floating
constructions (dykes, offshore platforms, etc.). It continues a long-year line
of development of the ComFLOW code.
Three PhD positions are available within the project. Keywords are turbulence
modelling, wave propagation, absorbing boundary conditions and local grid
refinement. The research is carried out by the University of Groningen
(Mathematics) and the University of Delft (Maritime Engineering), in
cooperation with MARIN, Deltares and many offshore companies worldwide.
The candidates should have experience in developing numerical flow models
(CFD). Their background should be in applied mathematics, maritime engineering,
civil engineering, or comparable. More information can be obtained from
prof.dr. A.E.P. Veldman, Computational Mechanics and Numerical Mathematics,
University of Groningen, +0503633988, a.e.p.veldman@rug.nl. See also the
ComFLOW website | 677.169 | 1 |
ristide Halanay, Judita Samuel
ISBN: 0792346750;
Издательство: Kluwer Academic Publishers
This volume presents some of the most important mathematical tools for studying economic models. It contains basic topics concerning linear differential equations and linear discrete-time systems; a sketch of the general theory of nonlinear systems and the stability of equilibria; an introduction to numerical methods for differential equations, and some applications to the solution of nonlinear equations and static optimization. The second part of the book discusses stabilization problems, including optimal stabilization, linear-quadratic optimization and other problems of dynamic optimization, including a proof of the Maximum Principle for general optimal control problems. All these mathematical subjects are illustrated with detailed discussions of economic models. Audience: This text is recommended as auxiliary material for undergraduate and graduate level MBA students, while at the same time it can also be used as a reference by specialists. | 677.169 | 1 |
Algebra, Grades 7+ (eBook)
Be sure that you have an application to open
this file type before downloading and/or purchasing.
10 MB|128 pages
Product Description
The 100+ Series, Algebra, offers in-depth practice and review for challenging middle school math topics such as radicals and exponents; factoring; and solving and graphing equations, practice pages to support standards-based instruction. | 677.169 | 1 |
Transcription
1 Summer Math Packet Student Name: Say Hello to Algebra 2 For Students Entering Algebra 2 This summer math booklet was developed to provide students in middle school an opportunity to review grade level math objectives and to improve math performance. Algebra 2-1 -
2 Say Hello to Algebra 2 One goal of Argyle Magent Middle School is to promote increased math performance at all grade levels. Completing the summer math booklet allows each student and parent within the school to work together to achieve this goal. Students who complete the summer math booklet will be able to: Increase retention of math concepts, Increase the level of proficiency on the Maryland School Assessment Work toward closing the gap in student performance. Student Responsibilities Students will be able to improve their math performance by: Completing the summer math booklet, Reviewing math skills throughout the summer. Student Signature Grade Date Parent Responsibilities Parents will be able to promote student success in math by: Supporting the math goals of Argyle Magnet Middle School Monitoring student completion of the summer math booklet, Encouraging student use of math concepts in summer activities. Parent Signature Date This summer math booklet was adapted from the Sail into Summer with Math! booklets and from Introductory Algebra 6 th Edition by Keedy/Bittinger, published Addison Wesley, Algebra 2-2 -A polynomial of degree n (in one variable, with real coefficients) is an expression of the form: a n x n + a n 1 x n 1 + a n 2 x n 2 + + a 2 x 2 + a 1 x + a 0 where a n, a n 1, a n 2, a 2, a 1, a 0 are2.3. Finding polynomial functions. An Introduction: As is usually the case when learning a new concept in mathematics, the new concept is the reverse of the previous one. Remember how you first learnedOperations with Algebraic Expressions: Multiplication of Polynomials The product of a monomial x monomial To multiply a monomial times a monomial, multiply the coefficients and add the on powers with theMathematics Placement The ACT COMPASS math test is a self-adaptive test, which potentially tests students within four different levels of math including pre-algebra, algebra, college algebra, and trigonometry.
Sheets Algebra Cheat Sheets provide you with a tool for teaching your students note-taking, problem-solving, and organizational skills in the context of algebra lessons. These sheets teach the conceptsPage of Review of Radical Expressions and Equations Skills involving radicals can be divided into the following groups: Evaluate square roots or higher order roots. Simplify radical expressions. Rationalize
8. Radicals - Multiply and Divide Radicals Objective: Multiply and divide radicals using the product and quotient rules of radicals. Multiplying radicals is very simple if the index on all the radicalsRational Equations Overview of Objectives, students should be able to: 1. Solve rational equations with variables in the denominators.. Recognize identities, conditional equations, and inconsistent equations.
The verbal answers to all of the following questions should be memorized before completion of algebra. Answers that are not memorized will hinder your ability to succeed in geometry and algebra. (EverythingFactoring Methods When you are trying to factor a polynomial, there are three general steps you want to follow: 1. See if there is a Greatest Common Factor 2. See if you can Factor by Grouping 3. See if6.7 Factoring Trinomials: The ac Method 6.7 OBJECTIVES 1. Use the ac test to determine whether a trinomial is factorable over the integers 2. Use the results of the ac test to factor a trinomial 3. For Rational Exponents Objective: Convert between radical notation and exponential notation and simplify expressions with rational exponents using the properties of exponents. When we simplify
Chapter 9. Simplify Radical Expressions Any term under a radical sign is called a radical or a square root expression. The number or expression under the the radical sign is called the radicand. The radicandThis assignment will help you to prepare for Algebra 1 by reviewing some of the things you learned in Middle School. If you cannot remember how to complete a specific problem, there is an example at the
Click on the links below to jump directly to the relevant section What is algebra? Operations with algebraic terms Mathematical properties of real numbers Order of operations What is Algebra? Algebra is
To the applicant: The following information will help you review math that is included in the Paraprofessional written examination for the Conejo Valley Unified School District. The Education Code requires
1. Sit anywhere in the concentric circles. Do not move the desks. 2. Take out chapter 6, HW/notes #1-#5, a pencil, a red pen, and your calculator. 3. Work on opener #6 with the person sitting across fromFactoring Polynomials 4-1-2014 The opposite of multiplying polynomials is factoring. Why would you want to factor a polynomial? Let p(x) be a polynomial. p(c) = 0 is equivalent to x c dividing p(x). RecallUC Davis, School and University Partnerships CAHSEE on Target Mathematics Curriculum Published by The University of California, Davis, School/University Partnerships Program 006 Director Sarah R. Martinez,
Chapter R.4 Factoring Polynomials Introduction to Factoring To factor an expression means to write the expression as a product of two or more factors. Sample Problem: Factor each expression. a. 15 b. x7-6 Choosing a Factoring Model Extension: Factoring Polynomials with More Than One Variable Essential question: How can you factor polynomials with more than one variable? What is the connection between
2. Solutions of Linear Equations in One Variable 2. OBJECTIVES. Identify a linear equation 2. Combine like terms to solve an equation We begin this chapter by considering one of the most important tools
The following pages include many problems to practice factoring skills. There are also several activities with examples to help you with factoring if you feel like you are not proficient with it. ThereLESSON 6.2 POLYNOMIAL OPERATIONS I Overview In business, people use algebra everyday to find unknown quantities. For example, a manufacturer may use algebra to determine a product s selling price in orderMATH 90 CHAPTER 6 Name:. 6.1 GCF and Factoring by Groups Need To Know Definitions How to factor by GCF How to factor by groups The Greatest Common Factor Factoring means to write a number as product. a
A Systematic Approach to Factoring Step 1 Count the number of terms. (Remember****Knowing the number of terms will allow you to eliminate unnecessary tools.) Step 2 Is there a greatest common factor? Tool
Algebra 2 PreAP Name Period IMPORTANT INSTRUCTIONS FOR STUDENTS!!! We understand that students come to Algebra II with different strengths and needs. For this reason, students have options for completing
MyMathLab ecourse for Developmental Mathematics, North Shore Community College, University of New Orleans, Orange Coast College, Normandale Community College Table of Contents Module 1: Whole Numbers and
7.7 Negative Integer Exponents 7.7 OBJECTIVES. Define the zero exponent 2. Use the definition of a negative exponent to simplify an expression 3. Use the properties of exponents to simplify expressionsBrunswick High School has reinstated a summer math curriculum for students Algebra 1, Geometry, and Algebra 2 for the 2014-2015 school year. Goal The goal of the summer math program is to help students9.5 Quadratics - Build Quadratics From Roots Objective: Find a quadratic equation that has given roots using reverse factoring and reverse completing the square. Up to this point we have found the solutions | 677.169 | 1 |
Algebra, Present and Past
What is Algebra? Algebra is one of the many branches of mathematics. This branch of math utilizes mathematical equations to describe the relationships between any two things that may change over a period of time. Whenever a math equation has to represent something that fluctuates and does not stay the same, letters or symbols are often used to represent the varying quantity. This letter or symbol is called a variable because it varies.
The very start of modern algebra began with the ancient Babylonians. They were the very first people who used an advanced math system for their time. This math used an algebraic system to do calculations, and it enabled the Babylonians to utilize formulas and calculate the answer using unknown values for a problem class that would be solved in the modern day by quadratic equations, linear equations, and indeterminate linear equations. During this same time period in the first millennium BC, most of the mathematicians in India, China, Greece, and Egypt were using geometric methods to solve such equations. The word algebra comes from the Arabic word al-jabar, which means reunion of the broken parts. A Persian Muslim mathematician named Muhammad ibn Masa al-khwarizmi wrote a book in 820 whose title is translated to mean The book of Summary Concerning Calculating by Transposition and Reduction. There is debate in math circles on whether the mathematician Diophantes, who has always been known as the father of Algebra, or Al-Khwarizmi should actually be named the father of Algebra. There is support for Al-Khwarizmi instead of Diophantes because a lot of his work on reduction is still currently in use today. He also explained how to solve a quadratic very extensively.
In the fourteenth century the word algeber referred to bone setting, which is quite similar to the original meaning of the word. The mathematical form of the word that is spelled algebra had made its debut by the sixteenth century. A man by the name of Robert Recorde was the very first person to use the term algebra in its mathematical meaning. He was also the inventor of the equality symbol(=). The original meaning of the word Algebra would refer today to the sub class of elementary algebra. There are numerous different fields of algebra in the modern definition of the word.
Some of these other mathematic fields include elementary algebra, s-algebra, abstract algebra, geometric algebra, boolean algebra, linear algebra, and many more. The mastery of elementary algebra must be accomplished before calculus or any of the other advanced mathematics can be learned. Many careers today require that at least elementary algebra has been mastered. The symbolism that is first introduced in elementary algebra extends throughout all the fields of mathematics. This is because the symbolism is the letters in the language of math. Modern fields of algebra are vastly more advanced than the first equations and formulas that were developed by the Babylonians. Numerous jobs and careers in the modern work environment require that algebra is mastered as a legitimate requirement for employment. | 677.169 | 1 |
ISBN-10: 053844052X
ISBN-13: 9780538440523
Edition: 16 provides comprehensive coverage of personal and business-related mathematics. In addition to reviewing the basic operations of arithmetic, students are prepared to understand and manage their personal finances, as well as grasp the fundamentals of business finances. Basic math skills are covered in a step-by-step manner, building confidence in users before they try it alone. Spreadsheet applications are available on the Data CD and a simulation activity begins every chapter. Chapters are organized into short lessons for ease of instruction and ease of learning.
Raymond M. Kaczmarski has been involved with business education throughout his career as a high school teacher, a district-wide curriculum administrator, and a university teacher-educator. Although he has retired from the Detroit Public Schools, his interest in education continues through his writing, working with student vocational organizations, and consulting on career-technical education issues.
Dr. Robert A. Schultheis has been a high school teacher, a university teacher educator, and a univeristy management information systems faculty member specializing in networking. He is the coauthor of textbooks in both high school business education and college-level information systems. He has retired from Southern Illinois University at Edwardsville but continues to write, consult organizations on information system matters, and maintain web sites for | 677.169 | 1 |
Category: Mathematics
This ebook investigates the geometry of quaternion and octonion algebras. Following a finished ancient advent, the e-book illuminates the designated homes of three- and four-dimensional Euclidean areas utilizing quaternions, resulting in enumerations of the corresponding finite teams of symmetries. the second one 1/2 the publication discusses the fewer established octonion algebra, targeting its extraordinary "triality symmetry" after a suitable examine of Moufang loops. The authors additionally describe the arithmetics of the quaternions and octonions. The publication concludes with a brand new conception of octonion factorization. subject matters coated contain the geometry of complicated numbers, quaternions and three-dimensional teams, quaternions and four-dimensional teams, Hurwitz crucial quaternions, composition algebras, Moufang loops, octonions and 8-dimensional geometry, necessary octonions, and the octonion projective plane.
In this conscientiously revised 5th version, the Short Course has been stated thus far and displays a contemporary and sensible method of LaTeX utilization. New chapters were extra on illustrations and the way to take advantage of LaTeX on an iPad.
Key features:
An example-based, visible strategy and a mild advent with the Short Course
A distinctive exposition of multiline math formulation with a Visual Guide
A unified method of TeX, LaTeX, and the AMS enhancements
A fast creation to making shows with formulas
From past reviews:
Grätzer's ebook is a solution.
―European Mathematical Society Newsletter
There are a number of LaTeX publications, yet this one wins arms down for the attractiveness of its technique and breadth of coverage.
―Amazon.com, better of 2000, Editor's choice
A amateur reader may be capable of research the main crucial good points of LaTeX adequate to start typesetting papers inside a number of hours of time… An skilled TeX person, however, will discover a systematic and distinct dialogue of LaTeX fea
tures.
―Report on Mathematical Physics
A very precious and useful gizmo for all scientists and engineers. ―Review of Astronomical Tools
Stunning contemporary effects by way of Host–Kra, Green–Tao, and others, spotlight the timeliness of this systematic creation to classical ergodic concept utilizing the instruments of operator conception. Assuming no past publicity to ergodic conception, this ebook presents a contemporary starting place for introductory classes on ergodic conception, specifically for college students or researchers with an curiosity in practical research. whereas simple analytic notions and effects are reviewed in different appendices, extra complicated operator theoretic issues are built intimately, even past their fast reference to ergodic concept. accordingly, the e-book can be compatible for complex or special-topic classes on practical research with purposes to ergodic theory.
Topics include:
• an intuitive advent to ergodic theory
• an advent to the fundamental notions, buildings, and traditional examples of topological dynamical systems
• Markov operators, and the comparable suggestion of an element of a degree maintaining system
• compact teams and semigroups, and a strong device of their examine, the Jacobs–de Leeuw–Glicksberg decomposition
• an advent to the spectral thought of dynamical structures, the theorems of Furstenberg and Weiss on a number of recurrence, and purposes of dynamical structures to combinatorics (theorems of van der Waerden, Gallai,and Hindman, Furstenberg's Correspondence precept, theorems of Roth and Furstenberg–Sárközy)
Beyond its use within the school room, Operator Theoretic features of Ergodic thought can function a invaluable origin for doing examine on the intersection of ergodic thought and operator theory
Considered a vintage via many, a primary direction in summary Algebra is an in-depth advent to summary algebra. interested by teams, earrings and fields, this article supplies scholars an organization starting place for extra really good paintings through emphasizing an realizing of the character of algebraic structures.
* This classical method of summary algebra makes a speciality of applications.
* The textual content is aimed at high-level classes at faculties with powerful arithmetic programs.
Today complicated numbers have such frequent sensible use--from electric engineering to aeronautics--that few humans may anticipate the tale at the back of their derivation to be packed with event and enigma. In An Imaginary Tale, Paul Nahin tells the 2000-year-old historical past of 1 of arithmetic' so much elusive numbers, the sq. root of minus one, sometimes called i. He recreates the baffling mathematical difficulties that conjured it up, and the colourful characters who attempted to unravel them.
In 1878, while brothers stole a mathematical papyrus from the traditional Egyptian burial web site within the Valley of Kings, they led students to the earliest recognized prevalence of the sq. root of a adverse quantity. The papyrus provided a selected numerical instance of the way to calculate the amount of a truncated sq. pyramid, which implied the necessity for i. within the first century, the mathematician-engineer Heron of Alexandria encountered I in a separate undertaking, yet fudged the mathematics; medieval mathematicians stumbled upon the concept that whereas grappling with the that means of destructive numbers, yet brushed off their sq. roots as nonsense. by the point of Descartes, a theoretical use for those elusive sq. roots--now known as "imaginary numbers"--was suspected, yet efforts to resolve them resulted in severe, sour debates. The infamous i ultimately gained recognition and used to be positioned to exploit in advanced research and theoretical physics in Napoleonic times.
Addressing readers with either a normal and scholarly curiosity in arithmetic, Nahin weaves into this narrative pleasing ancient evidence and mathematical discussions, together with the applying of complicated numbers and services to special difficulties, comparable to Kepler's legislation of planetary movement and ac electric circuits. This publication could be learn as a fascinating background, nearly a biography, of 1 of the main evasive and pervasive "numbers" in all of mathematics.
A interesting exploration of ways insights from machine algorithms could be utilized to our daily lives, aiding to resolve universal decision-making difficulties and remove darkness from the workings of the human mind
All our lives are restricted via restricted area and time, limits that supply upward push to a selected set of difficulties. What may still we do, or depart undone, in an afternoon or an entire life? How a lot messiness may still we settle for? What stability of latest actions and well-known favorites is the main pleasurable? those could seem like uniquely human quandaries, yet they aren't: desktops, too, face a similar constraints, so laptop scientists were grappling with their model of such matters for many years. And the options they have came across have a lot to educate us.
In a dazzlingly interdisciplinary paintings, acclaimed writer Brian Christian and cognitive scientist Tom Griffiths exhibit how the algorithms utilized by desktops may also untangle very human questions. They clarify how one can have higher hunches and whilst to depart issues to probability, easy methods to take care of overwhelming offerings and the way most sensible to connect to others. From discovering a wife to discovering a parking spot, from organizing one's inbox to knowing the workings of reminiscence, Algorithms to dwell By transforms the knowledge of desktop technology into concepts for human living.
All mathematicians be aware of that math can be utilized for a good many stuff. Educators, who've to put on a generalist's hat more often than not, infrequently get the chance to illustrate that truth. As a bunch, they're principally overworked, pressured to house their sessions, mountains of bureaucracy and but one way or the other locate time to take care of of their box. With all of this occurring, it's demanding to discover possibilities for college kids to have interaction with and be focused on fixing a few of their difficulties. even though, regardless of all of those stumbling blocks, a few academics were in a position to enhance such collaborations, and this e-book is a suite of stories on their success.
I used to be very inspired via what the folk have been in a position to do in those tasks. the main fascinating ones have been initiatives the place the economic associate reaped vast financial merits. i used to be surprised to learn that during one undertaking the paintings of 1 scholar kept the sponsoring corporation $17,000,000 (this isn't really a typo). As I learn via this ebook, i started getting principles for constructing relationships among my tuition and native companies. We at present have a few relationships within the quarter of computing yet not anything within the region of mathematics.
These papers describe actual luck tales within the sector of utilized arithmetic and that good fortune is multi-faceted. in fact, the corporations achieve monetary virtue, I frequently questioned in the event that they had the decency to not less than give a contribution a refund to the college of beginning. the scholars and college achieve useful adventure and bragging rights within the mathematical global that's immense albeit immeasurable. this is often one of the main attention-grabbing math books that i've got learn lately.
Outlines idea and strategies of calculus, emphasizing robust knowing of thoughts, and the fundamental ideas of research. reports basic and intermediate calculus and lines discussions of elementary-point set concept, and homes of continuing services.
The use of Clifford algebras in mathematical physics and engineering has grown speedily lately. while different advancements have privileged a geometrical technique, this booklet makes use of an algebraic process that may be brought as a tensor made from quaternion algebras and gives a unified calculus for a lot of physics. It proposes a pedagogical creation to this new calculus, in keeping with quaternions, with functions generally in particular relativity, classical electromagnetism, and common relativity. | 677.169 | 1 |
Understanding Satellite NavigationTeaches the fundamentals of satellite navigation systems, using MATLAB as a visualization and problem solving tool
Worked out numerical problems are provided to aid practical understanding
Learn the fundamentals of satellite navigation technology, backed up by practical simulations and visualizations in MATLAB through simulationsAbout the author – Asia Pacific, where he teaches on the M.Tech course on Satellite Communications and Satellite Navigation. He serves as visiting FacultyAbout the Author: - Asia Pacific, where he teaches on the M.Tech course on Satellite Communications and Satellite Navigation. He serves as a visiting Professor EOD. Hardback. Book Condition: new. BRAND NEW, Understanding Satellite Navigation, Rajat Acharya, How to improve your efficiency when working with a satellite navigation system* How to use MATLAB for simulation, helping to visualize concepts* Various possible implementation approaches for the technologyThe most significant applications of satellite navigation systems * Teaches the fundamentals of satellite navigation systems, using MATLAB as a visualization and problem solving tool* Worked out numerical problems are provided to aid practical understanding* On-line support provides MATLAB scripts for simulation exercises and MATLAB based solutions, standard algorithms, and PowerPoint slides. Bookseller Inventory # B9780127999494 | 677.169 | 1 |
Showing 1 to 13 of 13
Linear Systems
Part 3
APPLICATIONS
Translating words into symbols
+
Add
Give
Increased
Plus
More (than)
And
Combined with
All together
In addition to
Total
Exceeds
Take away
Subtract
Decreased
Minus
Remove
Without
Lower
Less (than)
Difference
of/between
E
Math Overview
Chapter Overview
When wanting to locate something, people use coordinates. Coordinates can be very
helpful in transformations.
Above are the quadrants listed for the Cartesian Coordinate System.
Once a shape is in a different location, the c
Systems in Action
Designing Efficient Systems
-
You are efficient when you use the least amount of energy needed for a task. Also
meaning you arent wasting energy. If not used in science, the term could also
describe time and money.
-
In homes, things are
Math Overview
Chapter Overview
Scientific Notation could be used to simplify numbers. Scientific Notations include
decimals and powers.
e.g.
If you wanted to write 6,549,300,000 as a scientific notation you would find how
many zeros are in the number (in
Math Overview
Chapter Overview
Term Number
o is part of a sequence. It isnt the value, but instead is used for the order.
e.g.
SEQUENCE
1, 1, 2, 3, 5
THE TERM VALUE IS: 1, 1, 2, 3, 5
THE TERM NUMBER IS: 1, 2, 3, 4, 5
Algebraic Expression
o The algebraic e
Math Overview
Chapter Overview
The
RADUIS
is the length of the outer liner to the middle of the circle. The radius
is often used in equations to find the diameter, circumference, area of the circle, and the
volume of a cylinder.
Radius
The
DIAMETER is dou
Math Overview
Chapter Overview
Repeating Decimal
sometimes decimals are long and never ending. Those include REPEATING
DECIMALS. They are not random numbers, but instead a set of numbers that
repeats itself. Instead of writing the full number, its best i
SYSTEMS IN ACTION
GETTING TO WORK
How to calculate work . . .
F D=W
The unit used to measure work is Joule (J).
Heres an example question.
A teacher moved her desk 4 m to the other side of the room. She used the force of
150 N.
To solve this problem, well
Math Overview
Chapter Overview
Types of Angles
Acute less than 90
Right exactly 90
Obtuse more than 90, but less than 180
Straight exactly 180
Reflex more than 180, but less than 360
Full Rotation exactly 360
Types of Triangles
Triangles can be defined by
Math Overview
Chapter Overview
Graphs
Graphs are used to display and sort data. There are many graphs that are used
like
Line Graph
Pie Graph (Circle Graph)
Stem-and-leaf plot
Pictographs
Bar Graphs
Line Graphs
Scatter Plot
Double-Line Graph
Double-Bar G
Math Overview
Chapter Overview
To add and subtract fractions, it
is best if both fractions have the
same denominator. If they dont
you simply find equivalent
fractions for both of the
fractions.
e.g.
You have to add to .
First, you have to find the lowest | 677.169 | 1 |
Be sure that you have an application to open
this file type before downloading and/or purchasing.
742 KB|3 pages
Product Description
Students never get enough practice factoring, but now they can! Students will need to factor many problems just to solve the maze, The Factor Maze has 2 Mazes: (ax^2 + bx + c), and Rational Expression Maze
1) Factoring Maze: Start at the top left hand corner, and solve the maze by factoring every polynomial you come across. Write the factorization under the polynomial for easy reference. You can only move one square up, down, left, or right when it SHARES a factor with the current square. You have solved the maze when you exit at the bottom right.
2) Rational Expression Maze: Simplify the following problems. Move through the maze by locating the answer for each problem until you reach "End Here" You must show your work for full credit. | 677.169 | 1 |
MatBasic is a calculating, programming and debugging environment using special high-level programming language designed for solving mathematical problems. MatBasic programming language allows execution of difficult mathematical calculations, involving an exhaustive set of tools for the purpose ofImprove whole number computation skills from simple to complex problems. Math facts exercises are included. Randomly generated problems are structured and worked out number by number as you would on paper. Generate printed worksheetsthat reinforce the on-screen activities. Multi-sensory approach | 677.169 | 1 |
9780078092480
0078092485231.42
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Summary
Basic Mathematical Skills with Geometry, 8/e by Baratto/Bergmanis part of the latest offerings in the successful Hutchison Series in Mathematics. The eigth edition continues the hallmark approach of encouraging the learning of mathematics by focusing its coverage on mastering math through practice.This worktext seeks to provide carefully detailed explanations and accessible pedagogy to introduce basic mathematical skills and put the content in context. The authors use a three-pronged approach (I. Communication, II. Pattern Recognition, and III. Problem Solving) to present the material and stimulate critical thinking skills. Items such as Math Anxietyboxes, Check Yourselfexercises, and Activitiesrepresent this approach and the underlying philosophy of mastering math through practice. The exercise sets have been expanded, organized, and clearly labeled. Vocational and professional-technical exercises have been added throughout. Repeated exposure to this consistent structure should help advance the student's skills in relating to mathematics. The book is designed for a one-semester basic math course and is appropriate for lecture, learning center, laboratory, or self-paced courses. It is accompanied by numerous useful supplements, including McGraw-Hill's online homework management system, MathZone. | 677.169 | 1 |
GEOMETRY END OF COURSE ASSESSMENT PRACTICE TEST ANSWERS
Geometry end of course assessment practice test answers
Mathematics is fundamental to existence. CPM geometry end of course assessment practice test answers all students in learning mathematics through problem solving, reasoning, and communication. Includes student text, plagiarism free and adhere to standard referencing style Harvard, APA etc…. Students solve problems that are usually more skills oriented Students develop strong problem-solving steps to solving word problems in math. A vector is a quantity with both magnitude and direction. Is the amount of money earned a function of the number of wins. And if you live in the area, please drop in at one of my constituent coffees, held at 10am on the first Saturday of each month, at the Human Bean in Geometry end of course assessment practice test answers. Links are offsite and download and viewing options vary InterActMath offers about 20 free textbooks for free online use. Polygon Vocabulary AssignmentReview Booklet regular Review Booklet honors Chapter 9 reviewResource Page 9. 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Although glencoe mcgraw hill algebra 1 answers degree to which students should be required to master the skills of symbolic manipulation is often a topic of debate among educators, he is up 'til midnight. CPM Math isn't working for my kids at all October 2000 What is the most effective way for a parent to give the school feedback about math textbooks and teaching. The banks include questions from Regents Exams dating back to 1866. Students will effectively communicate mathematical ideas, reasoning, multiplying, and dividing complex numbers. Online Scientific Calculator that does fractions, free pre algebra tutors, free answer to math problems, solving radicals calculator, ti 84 factoring program, elementary algebra examples, algebra double variables. Then, continuing the solving focus of Chapter 3, algebra problem solver, how do you solve linear equations, math 1 9th grade, Simplifying Radicals. 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We were also told by these tutors that the majority of students receiving math tutoring outside of the schools have been from Westerville Central High School. Finish Dilations Packet Test tomorrow. Copyright 2003 Help With Fractions All Rights Reserved Privacy Policy. In your composition book, show the notes and examples used in the video and solve as many practice problems geometry end of course assessment practice test answers possible. See your tutoring offers. In this article, I will show you how you can make better use of your CASIO calculator i. Thanks for sharing a very informative blog…thanks a lot…………. Lial, John Hornsby, David The Meritnation app is the digital equivalent of a guide book for Indian students. A teacher made a pair of foam dice to use in math games. The good news is, if you need to send out that math tutoring Homework math help middle school many units still need to be assembled after you have worked 5 hours of your shirt. 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Write a note describing what you're looking for-we'll pass it to a handful of our letter for greetings tutors and they'll reach out to you ASAP. The topics covered in this course include Math Solver Softonic 8 8 User 7. You can drag to adjust the size and shape of the frame. Simplifying radical machine, free work sheet for grade9, year ten surds test notes, SA, name a pair of numbers whose greatest common factor is the same as on of the numbers, algebrator substitution method, soft math. Please either fix the camera so it's NOT HORRIBLE or just get rid of it. Our job is to write about the troubling math problems that spread across social media like wildfire. Combinations math problems, complex fractions calculator. Add up test scores and see what the. Because the equation solver uses a numerical method, it only works with equations with a single variable. Louis has a large collection of Math exams with solutions, including precalculus. 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In the academic years 2010/11 and 2011/12 the Maths, Stats and OR (MSOR) Network supported a set of 32 projects on 'Mathematical Sciences HE Curriculum Innovation' through funding of around £250,000. This work was completed as part of the Mathematical Sciences Strand of the National HE STEM Programme.
The National HE STEM Programme was an initiative aiming to enable the HE sector to engage with schools, enhance curricula, support graduates and develop the workforce, operating through a three-year grant from the Higher Education Funding Councils for England and Wales.
This funding was distributed via a series of funding calls. Around 70% of the funding was allocated to addressing the recommendations of the HE Mathematics Curriculum Summit. In order to allow for interesting innovation which could not be predicted, calls for funding always included an open call for projects fitting the National HE STEM Programme aims. Around 30% of the funding allocated was for new innovations discovered this way.
Counting everybody who was named as a project collaborator or as an author in one of the publications (but not those who, for example, spoke at one of the workshops), this represents the work of more than 120 individuals working at 41 UK higher education institutions, two professional bodies, two schools, three non-UK universities and various companies There were over 50 workshops, seminars and conference presentations associated with this work.
The projects we supported have lots to share – good practice advice, evaluated innovative approaches, problem banks and other curriculum resources that you can pick up and use right away, and much more. A Summary booklet provides details of the aims, objectives, outputs and outcomes of each project with Links to access the resources created by each project. Projects are arranged into themed sections.
Developing graduate skills: A booklet was published collecting case studies of successful methods to improve graduate skills development – skills that employers require from graduate employees and academics seek in incoming PhD students – within a mathematical context. Three mini-projects were commissioned based on these case studies and provided evidence that some of this practice is suitable for transfer elsewhere. In addition, mathematics-specific resources and teaching practice on speaking and writing skills were developed and shared.
Engaging with employers: Projects working with employers, employees or professional bodies, either in delivery of a curriculum approach or providing input to develop good practice advice or curriculum resources that you can use. This includes resources giving an idea of what it is like to work as a mathematician and a survey of graduates' views of the mathematics HE curriculum.
Industrial problems: Banks of real world problems developed in consultation with industrial partners made available for undergraduate projects in mathematics and statistics.
Problem-solving: Two projects working to share good practice and develop curriculum resources on the teaching of problem-solving. We say mathematics develops problem-solving but do we actually know how to develop problem-solving as a skill in our students?
Maths Arcade: An innovative practice involving developing mathematical thinking, providing student support (particularly at the transition to university) and building a staff and student mathematical community. A case study booklet gives details of its implementation at eight universities.
Student-centred Approaches: Projects working to accommodate student needs or taking a student-centred view on improving the undergraduate experience. Including methods for supporting students in different contexts, helping engineers better understand their mathematics and providing adjustments for students with disabilities.
Assessment: A major project conducted research to answer questions about what alternative methods of assessment can offer, evidence of validity and guidance on the process of changing your teaching to involve a new assessment type.
Audio-visual media in teaching and learning: Investigating the recording of lectures and other teaching and learning content, and the effectiveness of learning through audio-visual media.
Projects have completed research and collected good practice advice, developed innovative practice or produced and shared curriculum resources to address various issues in mathematical sciences HE curriculum development. The work includes the need to develop graduate skills and take account of employer requirements, while remembering to ground this in mathematical content and take account of the needs of the discipline. How the collected resources affect the ability of the higher education mathematical sciences community to more effectively develop graduate mathematicians depends on how well these are taken up. This substantial set of projects in curriculum development has produced outputs with the potential to be very useful. Please use them!
Posted on8 May 2012|Comments Off on Links to work on HE Curriculum Innovation in Mathematical Sciences
In May I am giving a presentation at four workshops with the National HE STEM Programme Mathematical Sciences Strand. The workshop title is 'Maths Strand Outputs in the National HE STEM Programme' and this will be repeated in Manchester, London, Cardiff and Birmingham. I will be speaking on 'Work on HE Curriculum Innovation in Mathematical Sciences'. In this talk I will name several resources that my project has produced. These are listed below so I can give an easy link to participants.
Our work until that point was all covered in the HE STEM special issue of MSOR Connections 11(3). This includes final reports from our first call projects, interim reports from our second call projects and initial plans from our third call projects.
Posted on16 April 2012|Comments Off on Using social media to engage students – a list
I am running a workshop today on using social media to engage students, particularly mathematical sciences undergraduates. I think this is an emerging area about which little is known. I've tried to think of some examples of what you might do with these technologies. What do you think of my list? I'd be pleased to hear suggestions for additions, or stories about when you've tried this and how it went, in the comments.As part of that talk I looked into the link between use of lecture recordings and achievement. One study identified as a positive behaviour as students coming to class then using the video recording to revisit points they struggled with. On the other hand, skipping lectures to watch the videos instead seemed to be a detrimental approach.
I also considered what might be the effect of lecture capture on attendance. The studies I found seemed to indicate a split here. Traditional, non-interactive lectures where the students watched, listened and copied what the lecturer wrote on the board observed a decrease in attendance. Those lectures which included an interactive component did not observe such a decrease in attendance. The implication might be that if the video recording faithfully replicates the lecture experience then students see little point in attending.
These results, taken together, seem to suggest that increasing interactivity in lectures encourages students into the positive behaviour mode. A few things are being conflated here and it's all based on small scale studies, but a question is raised about whether traditional lectures are really that effective. My talk tomorrow will draw on this theme to suggest methods to increase interactivity.
The direct inspiration for this topic being on the workshop schedule is an American RadioWorks documentary Don't Lecture Me, part of a series on 21st century 'college' (in the American sense) education.
Part of this talks about students' preconceived ideas about the physical world and the effect this can have on their understanding of physics, saying:
One reason it's hard for students to learn physics is that they come into class with a very strong set of intuitive beliefs about how the physical world works… It turns out though that many of these intuitive notions do not square with what physicists have discovered about how things actually work. Most people's intuition tells them if you drop two balls of different weights from the second story of a building, the heavier ball will reach the ground first. But it doesn't – and this is a very difficult concept for most students to understand because they already have a concept in their mind that's in conflict with this new concept.
When they looked at the test that I gave to them, some students asked me, "How should I answer these questions? According to what you taught me, or according to the way I usually think about these things?" That's when it started to dawn on me that something was really amiss.
This sort of thing isn't just happening at the applied end of the spectrum; it can happen in pure maths too. I remember reading some work by Lara Alcock and Adrian Simpson, Ideas from Mathematics Education, which discusses students' preconceived or intuitive ideas of mathematical concepts ("concept images") – using examples such as functions, limits, groups – and how these are relied on by students above formal definitions, even when the two fail to coincide significantly. Among much else of interest in that book, they say:
Pre-existing concept images might override or interfere with the use of the definition, even when the latter is known.
This brings me to a video I saw a while ago by Derek Muller on the effectiveness of science videos. The part I want to focus on is when Muller studies the responses of students who watch a video passively. In the video, when what is said differs from a participant's conceptual understanding they don't notice, their test scores before and after the learning stay the same and they actually become more confident in their misconception.
I'm not sure YouTube has a very thorough peer-review policy and I haven't read the original research but the idea is interesting. Don't Lecture Me makes a similar claim about traditional lectures:
The traditional, lecture-based physics course produces little or no change in most students' fundamental understanding of how the physical world works. Even students who can solve physics problems and pass exams leave the traditional lecture class with many of their incorrect, intuitive notions intact.
There's a question here about how anyone becomes a physicist. The answer given in the piece is that roughly 10% of students are motivated to teach themselves. David Hestenes is quoted saying: "They essentially learn it on their own". It may be that the best students (and future researchers) are learning in spite of the teaching, not because of it.
So if simply watching a teacher talk through correct material isn't helping to challenge students' misconceptions, what can be done?
Muller advocates presenting students with common misconceptions. In the video he describes an experiment in which participants are shown a video in which their misconception is presented by an actor and then challenged in a discussion with another actor. The participants reported finding the video harder to watch but their test scores increased.
In Don't Lecture Me (and in life), Mazur advocates a method called peer instruction. In this, students are asked a multiple-choice question in class and allowed to vote on the correct answer via an audience response system. They are then asked to discuss their answer with students sitting near them. If two students' answers differ then whoever is correct ought to be able to convince the other of this.
What is common about these methods is the use of discussion to challenge misconceptions. Muller uses actors while Mazur uses peers, but in neither case does an authority figure tell anyone the correct answer wholesale. I'd say using discussion to challenge misconceptions is clearly indicated as a potential strategy, with peer instruction the better for a lecture environment.
In Don't Lecture Me, Mazur says peer discussion works because the peer recently shared the conceptual difficulties. He says:
That's the irony of becoming an expert in your field. It becomes not easier to teach, it becomes harder to teach because you're unaware of the conceptual difficulties of a beginning learner.
I expect the approach works because students are evolving their intuitive concept towards the formal version, rather than trying to memorise a second, formal definition in parallel (or in conflict) with their intuitive one. Alcock and Simpson suggest mathematicians are still using concept images to think mathematically, but that they are doing so with "sophisticated images which they can rely on to closely match the [formal] definition".
A while ago Sally Barton and I did a study of a lecturer's use of audience response system (electronic voting system, clickers?) questions in class. He took fifteen minutes once a fortnight to present a quiz of five questions to students, with the aim of encouraging students to keep up to date with their lecture notes. After voting on the answers, students were told the correct answer and directed to the module webpage for worked solutions.
First, we asked students to rate on a scale their approach to answering the questions from "I think carefully about the questions asked" to "I don't think, I just choose answers at random". The students whose answer suggested they were more engaged with the quizzes reported taking remedial action much more than those who seemed less engaged. However, the 'more engaged' students reported that they were able to keep up to date with lecture notes in this module and others (where quizzes weren't used) equally well. This suggests the quizzes were not needed as an extra incentive to keep up to date for these students. The 'less engaged' students tended to take little remedial action, even when they had not known the answer and had simply guessed correctly, suggesting that the quizzes were not encouraging those less engaged students to interact with the teaching materials.
When they would take remedial action, the action taken most often by the 'less engaged' students was not to work through the problem again, check the model solution or read lecture notes, but was to discuss the problem with their friends.
If we're right, that this group of students are least likely to engage with formal teaching material but perfectly agreeable to discussion with peers, and if this result generalises, then peer instruction could have real positive consequences for these least engaged students.
A curriculum barrier for students with disabilities is the delivery of mathematical learning resources such as lecture notes, problem and solution sheets in inaccessible formats. The current practise of repeatedly re-typesetting notes in to produce particular formats is expensive in the long run. We will develop methods, instructions and examples by which a single master copy may be used to automatically produce a variety of formats. Thus all resources are updated from a master enabling departments to make proactive adjustments. The methods will be appropriate for use by individual lecturers/departments with access to a small range of mathematical/assistive technologies. | 677.169 | 1 |
1123607219
ISBN: 0023607211
Edition: 3
Publisher: Prentice Hall PTR
AUTHOR
Johnsonbaugh, Richard
SUMMARY
This best-selling book provides an accessible introduction to discrete mathematics through an algorithmic approach that focuses on problem- solving techniques. This edition has the techniques of proofs woven into the text as a running theme and each chapter has the problem-solving corner. The text provides complete coverage of: Logic and Proofs; Algorithms; Counting Methods and the Pigeonhole Principle; Recurrence Relations; Graph Theory; Trees; Network Models; Boolean Algebra and Combinatorial Circuits; Automata, Grammars, and Languages; Computational Geometry. For individuals interested in mastering introductory discrete mathematics.Johnsonbaugh, Richard is the author of 'Discrete Mathematics' with ISBN 9780023607219 and ISBN 00236072 | 677.169 | 1 |
auto calculator Our auto calculator enables you to calculate the value of complex expressions directly usingCalculator 4 M Everybody needs a calculator. "Calculator 4 M" easy to use and beautifully designed to do things better than your phone or handheld calculator ever did | 677.169 | 1 |
algebra
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Examples from texts
There is a great variety of ways in which Boolean algebra and HK can be formulated.
Back in advanced algebra the year before, Mr. Sanchez had taught them how to convert into base two (turning regular numbers into ones and zeros), all the while claiming that this knowledge was going to get them computer jobs one day. | 677.169 | 1 |
Paperback Textbook
Publisher:
McGraw-Hill
Supplemental materials are not guaranteed for used textbooks or rentals (access codes, DVDs, workbooks)....
Show More keep her or him motivated and excited to learn. They can practice the problems they find challenging, polish skills they've mastered, and stretch themselves to explore skills they have not yet attempted. This book features exercises that increase in difficulty as your child proceeds through it. This book is appropriate for a 4th grade student working above his or her grade level, or as a great review and practice for a struggling 5th or 6th grad | 677.169 | 1 |
Exponential functions: Applications of biology, finance and medicine
Be sure that you have an application to open
this file type before downloading and/or purchasing.
268 KB|11 pages
Product Description
Students will use their knowledge of exponential functions of the form y = a(b)^(x/c) to solve exponential equations. The application problems require students to create a function to model the situation, then solve for either the value of the exponent, or the result of a given value of the exponent.
The use of negative exponents is necessary (for instance, to find when 1/8 = 1/2 ^ (x/2), they could use 2^ -3 = (2^ -2)^(x/2), to get -3=-2x/2 and solve for x)
The first 2 pages involve solving equations where students can manipulate the equations to have the "same base" to solve for the exponents as above. The final 4 pages have students using logs (or a graphing calculator if students have not learned about logs) to solve for the variable as the same base method is not possible. Students who do not yet know logs can complete this exercise, but solve for the answer using the intersection of curves on the graphing calculator. Quick instructions are included, and the document is editable to use this option.
6 pages of .doc word file plus 5 pages of pdf answers are included.
See Making Math Matter MMM for other calculus, precalculus and algebra application assignments and handouts.
Can be used as homework, handout, activity, worksheet or an assessment. | 677.169 | 1 |
Numbers And Proofs: NUA01ERS & PROOFS
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'Numbers and Proofs' presents a gentle introduction to the notion of proof to give the reader an understanding of how to decipher others' proofs as well as construct their own. Useful methods of proof are illustrated in the context of studying problems concerning mainly numbers (real, rational, complex and integers). An indispensable guide to all students of mathematics. Each proof is preceded by a discussion which is intended to show the reader the kind of thoughts they might have before any attempt proof is made. Established proofs which the student is in a better position to follow then follow.
Presented in the author's entertaining and informal style, and written to reflect the changing profile of students entering universities, this book will prove essential reading for all seeking an introduction to the notion of proof as well as giving a definitive guide to the more common forms. Stressing the importance of backing up "truths" found through experimentation, with logically sound and watertight arguments, it provides an ideal bridge to more complex undergraduate maths. | 677.169 | 1 |
Introduction to Number Theory Learn the fundamentals of number theory from former MATHCOUNTS, AHSME, and AIME perfect scorer Mathew Crawford. Topics covered in the book include primes & composites, multiples & divisors, prime factorization and its uses, base numbers, modular arithmetic, divisibility rules, linear congruences, how to develop number sense, and much more. | 677.169 | 1 |
If you think a Cartesian coordinate is something from science fiction or a hyperbolic tangent is an extreme exaggeration, you need "Trigonometry DeMYSTiFieD,"Автор: Coburn Название: College algebra and trigonometry with mathzone ISBN: 0073229806 ISBN-13(EAN): 9780073229805 Издательство: McGraw-Hill Цена: 5807 college algebra text is written in a friendly and an easy to understand manner in order to help students understand the concept presented. This feature combined with ample examples, various types of exercises, and well thought out, real-world applications give the student the right tools to succeed. There are specific features and exercise problems to incorporate graphing calculator technology for those interested, however the material is presented in a way so that it may be skipped for those not utilizing technology. Купить
Автор: Young, Cynthia Y. Название: Algebra and trigonometry ISBN: 0470222735 ISBN-13(EAN): 9780470222737 Издательство: Wiley Цена: 50204 Anyone trying to learn algebra and trigonometry may think they understand a concept but then are unable to apply that understanding when they attempt to complete exercises. This book helps them overcome common barriers to learning the concepts and builds confidence in their ability to do mathematics. Купить
Автор: Young Название: Algebra and Trigonometry, 3rd Edition ISBN: 0470648031 ISBN-13(EAN): 9780470648032 Издательство: Wiley Цена: 17 third edition of Cynthia Youngs Algebra bridge the gap between classroom instruction and independent homework by overcoming common learning barriers and building confidence in students ability to do mathematics. Written in a clear, single voice that speaks to students and mirrors how instructors communicate in lecture, Youngs hallmark pedagogy enables students to become independent, successful learners. Varied exercise types and modeling projects keep the learning fresh and motivating. Young continues her tradition of fostering a love for succeeding in mathematics by introducing inquiry-based learning projects in this edition, providing learners an opportunity to master the material with more freedom while reinforcing mathematical skills and intuition. The seamless integration of Cynthia Youngs Algebra s vision of building student confidence in mathematics because it takes the guesswork out of studying by providing them with a clear roadmap: what to do, how to do it, and whether they did it right. Купить | 677.169 | 1 |
Showing 1 to 6 of 6
Activity 1.2.6 Airfoil Simulation
Introduction
Airfoils have a complex geometry designed to direct airflow. This velocity change results in
forces that affect aircraft performance. This can be simulated using a computer to provide an
estimate of expected
Activity 1.2.9 Glider Design Using AERY
Software
Introduction
The AERY Glider Design Software package provides an easy to use interface for
creating a glider design. The software package will perform the calculations required
to predict the stability of f
pre calculus Advice
Showing 1 to 3 of 4
I recommend taking these classes because they are fun! They've helped me out in many ways that I can't explain.
Course highlights:
My favorite part of this course was the teacher explaining important lessons and I learned that teaching is pretty difficult and you ned skills.
Hours per week:
3-5 hours
Advice for students:
Study Study Study, so you can pass.
Course Term:Spring 2017
Professor:Ferguson
Course Required?Yes
Course Tags:Background Knowledge Expected
Feb 17, 2017
| Would recommend.
This class was tough.
Course Overview:
It takes a lot of work and extra time out of class to understand. You have to take the time and use your peers and teacher for help because pre cal can be very hard if you mess around.
Course highlights:
Being able to have a good teacher and knowing the material after studying it.
Hours per week:
9-11 hours
Advice for students:
Take all of your notes, ask questions and do all of your homework, You have got to study extra hard and ask questions.
Course Term:Winter 2017
Professor:Mr. Johnson
Course Required?Yes
Course Tags:Math-heavyBackground Knowledge ExpectedGo to Office Hours
Dec 28, 2016
| Would recommend.
Not too easy. Not too difficult.
Course Overview:
I recommend this course because the teacher offers both individual and group work to enhance our ways of studying and understanding.
Course highlights:
I learn that working in groups or partner is faster than individual and more effective. Even if one partner doesn't understand, the other partner could help because not every student is brave enough to ask the teacher.
Hours per week:
3-5 hours
Advice for students:
I advice students to take their time actually doing problems when working in group because sometimes, students teach better than teachers. | 677.169 | 1 |
This book is for sophomore-level or junior/senior-level first courses in linear algebra and assumes calculus as a prerequisite. This thorough and accessible text, from one of the leading figures in the use of technology in linear algebra, gives students a challenging and broad understanding of the subject. The author infuses key concepts with their modern practical applications to offer students examples of how mathematics is used in the real world. Each chapter contains integrated worked examples and chapter tests. The book stresses the important roles geometry and visualization play in understanding linear algebra | 677.169 | 1 |
Algebra Standard in US High School
May 3, 2016
To know what kind of culture America has in High School Math, teaching a typical class in first year university or Community College will give you a very good idea. The image below is only funny here, never funny or brought up in Singapore.
so what other kinds of funny errors do they have a lot (and laugh it off)? Here is another off my colleague's door. There are so many of these, someone publish it into a book.
Door Joke of a US college Mathematician
So when you teach the students, they really got all confused in their algebra. Off the top of my head, here is a list of what's considered common in the US:
But like an Olympic swim clinic, my job here is to correct a future Olympian's error or bad habit. And different students have different bad habits. So this is like a Dharma Master trying to help his student with his affliction. With constant diligence, the student will hope to let go and reduce his kleshas.
Let me end with another so-called funny Math joke. If you are a Singaporean who commit this error and your parents saw it, it will hardly be funny | 677.169 | 1 |
Features
STEMReader has a range of tools and features to help you understand equations and maths notation. In particular, STEMReader allows you to explore equations using audio, text transcription or a map view. You can even find out what unfamiliar symbols mean.
There are 3 different versions of STEMReader with different tools available. These are:
STEMReader Basic: this is the free version of STEMReader. You can still have equations read aloud, but not all the features within STEMReader are available.
STEMReader Premium (coming soon): you will be able to experience all the features of STEMReader.
STEMReader Exam (coming soon): this is a locked down version of STEMReader which removes features which would not be suitable to access during an assessment, test or exam. The exam mode is available if you have a valid license for STEMReader Premium.
Purchasing for a school, college or university? Contact us for volume arrangements. | 677.169 | 1 |
The purpose of this class is to complete the development of the real number system, which was begun in Algebra I. The first semester will consist of a more intense exploration of concepts learned in Algebra I. During the second semester, the focus will be on extending the concepts learned in the first semester while introducing nonlinear equations, trigonometry, logarithms, probability and statistics, and the complex number system.
This class will focus on the study of the plane geometry system while reviewing concepts learned in Algebra I. During the last nine weeks, Algebra I concepts will be extended as the real number system is developed further.
This course includes material that focuses on arithmetic operations, problem-solving techniques, estimation of answers, measurement skills, geometry, data handling, simple statistics, and the use of algebraic formulas to solve problems. The emphasis of the course is on the ability to understand and apply functional mathematics to solve problems that students might encounter in the work place. The course will employ the use of video, print, hands-on laboratories and practical problem-solving activities.
This course is a continuation of Technical Algebra I. It will include work on basic algebra skills, geometry, and an introduction to trigonometry. The course will employ the use of video, hands-on laboratories, and practical problem-solving activities.
Trigonometry is the study of angles, triangles, circular function, and their relationships. Topics covered are: radian and degree measure, trigonometric functions of a circle and a triangle, identities, polar coordinates, and complex numbers. Pre-calculus students will study various topics such as: functions, coordinate geometry, logarithms, probability, statistics, and an introduction to the calculus.
The study of calculus has been a core subject area of science and engineering for many years and is now included as an important topic in many other disciplines. The students will study differential and integral calculus.
This course introduces the student to computer programming in the BASIC language. The student will learn about the hardware of a computer and what to expect of it. The main part of this course will be learning to write computer programs in the BASIC language. The student will also learn applications of the computer programs.
This class is designed for students who need one more year of mathematics, but do not want to take Algebra II, Trigonometry, or Calculus. The students will review arithmetic fundamentals that have been previously studied. Throughout the course of study, students use their skills to solve a variety of business problems that demonstrate how widely arithmetic is used in the business world. The range of topics covered also provides students with a broad introduction to the business content and terminology that they will study in greater detail in advanced business classes. | 677.169 | 1 |
WJEC Computer Science 'Technical Topics' Worksheets
Lovely resource - I like the writing style, very well suited to
all levels with a GCSE Computing class. C Charles,
Computing Teacher & Independent Reviewer
Truth Tables
Logic Gates & Diagrams
Denary ↔ Binary
Denary ↔ Hexadecimal
Binary ↔ Hexadecimal
Binary Codes and Characters
Digital Images
Digital Sound
HTML - Text and Formatting
HTML - Images and Hyperlinks
Algorithms
Selection Control Flow
Iteration Control Flow
Handling Data in Algorithms
Trace & Testing
Fantastic collection of worksheets designed to help students nail
the trickier areas of the specification. Provides practice and
reinforcement of 15 'technical topics' – including binary logic,
number systems and algorithms
Each worksheet:
is cross-referenced to the 2012 WJEC Computer
Sciencespecification
includes student-friendly notes and worked examples
Full solutions allow quick, easy marking – by teachers or
students
Mark sheet enables students to track their progress and
performance
Use in class or as homeworks; to support your teaching or for
revision | 677.169 | 1 |
Wednesday, July 25, 2012
Imagine for a moment that you are a student and math isn't your favorite subject. Now imagine that you were told that you could master Algebra I in about 20 minutes a day. What would your reaction be? Our first thoughts when asked to review No-Nonsense Algebra from Math Essentials were something along the lines of "that sounds too good to be true - but if it IS true, sign us up!" So here I am to tell you how it's working!
We have been reviewing a couple of the resources from Math Essentials. We received two books - from the Mastering Essential Math Skills series, Problem Solving, and the newer No-Nonsense Algebra.
Math Essentials presents a series of workbooks using a system of teaching developed by Richard W. Fisher that helps motivate students to learn and produces dramatic results. The program is easy to follow and requires only 20 minutes per day because the lessons are short, concise, and self-contained. The newest product in the series is No-Nonsense Algebra, and this was the focus for our family's review.
No-Nonsense Algebra is an easy-to-understand text and teaching system that includes built-in reviews for mastery, and eliminates distractions. Each lesson includes a clearly explained introduction to the topic, a section of helpful hints, examples with step-by-step solutions, written exercises, and a review section. The idea is to teach algebra in a logical and systematic manner.
In the back of the book is an access code to the online teaching videos that correspond with each lesson. The author teaches the lessons himself in a video segment, guiding the student through the topic and some examples. Each day Landon worked on Algebra, he watched the teaching video, and did the examples and review in the workbook. He didn't complain about it, and was done in half an hour or less each day, depending on the length of the video. And so far, he's been getting the right answers almost all the time. (The solutions are in the back of the book, but we remove them so that I do the grading, and so that it's not so easy to cheat.)
After a few lessons, Landon informed me that he would like to do the entire Algebra I course using No-Nonsense Algebra, and I had no objections!! In fact, I was doing a happy dance. On the inside, of course! He likes the teaching videos, which work well even with our internet connection being a little spotty at times. He can pause the video or replay a part that he needs to see again without any trouble. The problems that he needs to work out are enough to get familiar with the concept taught, and to prove that he understands it, without being overly time-consuming. The lesson is covered in just one or two pages, and the pages themselves are clear and uncluttered.
I knew we had a winner on our hands when I was privileged to hear the following conversation a couple weeks ago. Landon had a friend over and they were talking about school (the friend is homeschooled as well). When this young man mentioned that he was not liking algebra and having trouble with it, Landon immediately said, "You should try the one I've got - No-Nonsense Algebra. The lessons are short so they don't take all day. There's a workbook and you go online to watch the videos that teach it and then you do the lessons." Not only that, but he went to get the book to show the kid!! Once again, I was doing that happy dance on the inside, because Landon doesn't give endorsements out like candy. He generally doesn't give endorsements for anything that is school-related. When I saw the friend's mom, I made sure to give her a heads-up that Landon had been doing an unpaid and unsolicited promo for a math book and promised to bring the book along for her to see soon.
What we liked best:
concise lessons - one or two pages of teaching, exercises and review presented efficiently so that it doesn't take an intimidating amount of time.
the clean, uncluttered pages mean there's less distraction.
by working out the solutions and writing down answers in a separate notebook, the text is non-consumable. There is an emphasis on writing down each example and the steps to the solution as an important step in the learning process.
the video lessons are clear and easy to understand.
chapter tests and a final text to measure progress and understanding.
What we weren't crazy about:
I did find a couple of minor typos in the solutions. Fortunately, on problems that I could easily solve on my own so I trusted myself and Landon on those ones. I hope that there aren't mistakes when we get to the more challenging (for me!) problems.
I also need to mention the other book that we were looking at - Problem Solving from the Mastering Essential Math Skills series. This series of books is targeted to specific skill areas in math, helping students in middle school and high school grades improve their mastery and understanding of topics like integers, geometry, fractions, decimals, and problem solving. These are also designed for about 20 minutes a day of work. The Problem Solving book reviewed skills in reading and understanding graphs, basic operations with whole numbers and fractions, working with percentages and with decimals, all in the context of word problems. Kennady did some of the lessons in this book, and did okay with the section on graphs, but with this book being targeted to middle school and up, we found that the rest of it was still a bit beyond her. We will probably be using it as a supplement in future though. Again, the lessons were clear and concise, with just a page or two to teach each lesson, and even Kennady (the math-phobe!) wasn't freaked out by it.
Would this no-nonsense approach to algebra be right for your homeschool? Here's what you need to know:
No-Nonsense Algebra is available for $27.95, which includes the book and access to the online video lessons. The workbook Problem Solving in the Mastering Essential Math Skills series is $11.95. You can view all the Math Essential products and prices at the Math Essentials website.
Math Essentials is offering a special promotion in July, August, and September. Any order that includes Mastering Essential Math Skills Book 1, OR Mastering Essential Math Skills Book 2, OR No-Nonsense Algebra, will include a free copy of Geometry (a $14.95 value) and a free Homework Kit (a $4.99 value.)
Some of my Crewmates reviewed other titles in the Math Essentials series. You can visit the Schoolhouse Review Crew blog for more information about these products and to read other Crew member reviews.
Disclaimer: As part of the Schoolhouse Review Crew, we received a complimentary copy of these two books and access to the corresponding online video lessons in exchange for our honest opinions. | 677.169 | 1 |
How to Solve It: A New Aspect of Mathematical Method
Hardcover
Item is available through our marketplace sellers.
Overview anagrams. Generations of readers have relished G. Polya's deft--indeed, brilliant--instructions on stripping away irrelevancies and going straight to the heart of the problem.
Advertising
Editorial Reviews
Mathematical Monthly — E. T. Bell
Mathematical Review — Herman Weyl
Scientific Monthly
I recommend it highly to any person who is seriously interested in finding out methods of solving problems, and who does not object to being entertained while he does it.
American Journal of Psychology
Any young person seeking a career in the sciences would do well to ponder this important contribution to the teacher's art. — A. C. Schaeffer
Mathematics Magazine
Every mathematics student should experience and live this book
Mathematical Monthly - E.TMathematical Review - Herman Weyl
American Journal of Psychology - A.C. Schaeffer
Any young person seeking a career in the sciences would do well to ponder this important contribution to the teacher's art.
Mathematical Monthly - E. TAmerican Journal of Psychology - A. C. Schaeffer
Any young person seeking a career in the sciences would do well to ponder this important contribution to the teacher's art.
From the Publisher
" "—E. T. Bell, Mathematical Monthly
"[This] elementary textbook on heuristic reasoning, shows anew how keen its author is on questions of method and the formulation of methodological principles. Exposition and illustrative material are of a disarmingly elementary character, but very carefully thought out and selected."—Herman Weyl, Mathematical Review
"I recommend it highly to any person who is seriously interested in finding out methods of solving problems, and who does not object to being entertained while he does it."—Scientific Monthly
"Any young person seeking a career in the sciences would do well to ponder this important contribution to the teacher's art."—A. C. Schaeffer, American Journal of Psychology
"Every mathematics student should experience and live this book"—Mathematics Magazine
"In an age that all solutions should be provided with the least possible effort, this book brings a very important message: mathematics and problem solving in general needs a lot of practice and experience obtained by challenging creative thinking, and certainly not by copying predefined recipes provided by others. Let's hope this classic will remain a source of inspiration for several generations to come."—A. Bultheel, European Mathematical Society
Related Subjects
Meet the Author | 677.169 | 1 |
See also GEOMETRIC MECHANICS - Part II: Rotating, Translating and Rolling (2nd Edition) This textbook introduces the tools and language of modern geometric mechanics to advanced undergraduates and beginning graduate students in mathematics, physics and engineering. It treats the fundamental problems of dynamical systems from the viewpoint of Lie group symmetry in variational principles. The only prerequisites are linear algebra, calculus and some familiarity with Hamilton's principle and canical Poisson brackets in classical mechanics at the beginning undergraduate level.The ideas and concepts of geometric mechanics are explained in the context of explicit examples. Through these examples, the student develops skills in performing computational manipulations, starting from Fermat's principle, working through the theory of differential forms on manifolds and transferring these ideas to the applications of reduction by symmetry to reveal Lie-Poisson Hamiltonian formulations and momentum maps in physical applications.The many Exercises and Worked Answers in the text enable the student to grasp the essential aspects of the subject. In addition, the modern language and application of differential forms is explained in the context of geometric mechanics, so that the importance of Lie derivatives and their flows is clear. All theorems are stated and proved explicitly.The organisation of the first edition has been preserved in the second edition. However, the substance of the text has been rewritten throughout to improve the flow and to enrich the development of the material. In particular, the role of Noether's theorem about the implications of Lie group symmetries for conservation laws of dynamical systems has been emphasised throughout, with many applications. | 677.169 | 1 |
ATHEMATICS
SubGenre:Algebra / Intermediate
Language:English
Pages:312
Developing Mathematical Fluency
by Grayson H. Wheatley and George E Abshire
Overview
Mathematical fluency involves being able to reason mathematically and compute accurately, efficiently and flexibly (NCTM, 2000). The meaningful and engaging activities in this book have been carefully designed and tested in classrooms. Emphasis is placed on problem solving, fractions, decimals, percent and algebraic thinking. Fundamental to the implementation of this mathematics program is the Problem Centered Learning instructional strategy, in which students work in groups and explain their reasoning to other students. This book is also appropriate for home schooling, tutors, and interested parents. The book now ships with a CD containing digital files of the student pages. These files will be especially useful for teachers using smartboards. Furthmore, it will make copying pages for student use much easier. If you have already purchased the book and want the digital file, it can be purchased separately.
Only available in ePub for iBooks.
Description
While many persons see education as the process of acquiring knowledge that involves memorizing facts and
procedures, mathematics is not just a set of rules. An alternative way of thinking about mathematics is as the
activity of constructing patterns and relationships. In this view, mathematics is something people do. Knowledge is
not acquired but constructed by the individual as he or she solves problems. In today's fast changing society with
numerous new challenges, it is important that students give meaning to their mathematical activity and be able to solve
problems not seen previously.
Mathematics is reasoning, not just memorization. While it is useful to know certain facts and procedures, it is
essential that these facts and procedures develop with understanding. For example, a student who thinks in tens
can see immediately that 34 + 16 is 50 without going through some arbitrary set of steps by making marks on
paper. Mathematics is more like learning to find your way around a park with many trails. With experience that
involves exploration, a person can build a mental map of the park and move through the park without getting lost. In
the same way children can build networks of schemes that allow them to cope with novelty and solve problems they
have not previously considered. Just memorizing facts and procedures can actually be debilitating since it bypasses the
important activity of building inter-connected mathematical ideas.
About the author
eBook - $18.99 (ePub - This title is only available in iBooks Fixed Layout eBook format. Please use an Apple device to view this file.) | 677.169 | 1 |
This third edition of the successful outline in linear algebra--which sold more than 400,000 copies in its past two editions--has been thoroughly updated to increase its applicability to the fields in which linear algebra is now essential: computer science, engineering, mathematics, physics, and quantitative analysis. Revised coverage includes new problems relevant to computer science and a revised chapter on linear equations.
"synopsis" may belong to another edition of this title.
From the Back Cover:
Master linear algebra with Schaum's—
* Use detailed examples to solve problems * Brush up before tests * Find answers fast * Study quickly and more effectively * Get the big picture without poring over lengthy textbooks
Schaum's Outlines give you the information your teachers expect you to know in a handy and succinct format——fast! And Schaum's are so complete, they're perfect for preparing for graduate or professional exams.
Inside, you will find: * A bridge between computational calculus and formal mathematics * Clear explanations of eigenvalues, eigenvectors, linear transformations, linear equations, vectors, and matrices * Solved problems that relate to the field you are studying * Easy-to-understand information, perfect for pre-test review
If you want top grades and a thorough understanding of linear algebra, this powerful study tool is the best tutor you can have!
Seymour Lipschutz, Ph.D. (Philadelphia, PA), is presently on the Mathematics faculty at Temple Univeristy. He has written more than 15 Schaum's Outlines. Marc Lipson, Ph.D. (Philadelphia, PA), is on the mathematical faculty of the University of Georgia. He is co-author of Schaum's Outline of Discrete Mathematics.
Book Description McGraw-Hill Inc.,US, 1974. Book Condition: Good. S.I.ed. Ships from the UK. Former Library book. Shows some signs of wear, and may have some markings on the inside. Bookseller Inventory # GRP704567380366715
Book Description McGraw-Hill Inc.,US 01/0470843813
Book Description McGraw-Hill Inc.,US 01/0470843813
Book Description McGraw-Hill Inc.,US, 1974. Paperback. Book Condition: Good. Schaum's Outline of Theory and Problems of Linear Algebra70843813 | 677.169 | 1 |
In this short expository article our aim is to make students familiar with the concepts of categories, functors and natural transformations. Without using any set theoretic approach and only using axioms we are going to describe categories.
Click here to download the full article.... Higher Secondary Education Council(AHSEC) is responsible for conducting the 12th standard examinations for the students studying in the state board schools in Assam. Some of the AHSEC 2015 question papers are available for download below.
English, Alternative English, Physics, Chemistry, Mathematics, Biology
(Featured Image Source...
The Assam Higher Secondary Education Council is responsible for conducting the 12th standard examinations for the students studying in the state board schools in Assam. The 2015 question paper in mathematics is available for download below.
Click here for the question paper....6730" align="alignleft" width="166"] Source : Shutterstock.[/caption]
Sets, Relations and Functions are of utmost importance in the study
of mathematics. Most of the students find these topics difficult and
hence they tend to avoid studying these in Higher Secondary level. As
a result they face a lot of conceptual...
Prof. Vijaykumar Ambat from Cochin has shared with us his latest brochure on mathematical Olympiads which contains an wealth of information not only about Olympiads but about various other things that might be useful to school and college students. The brochure can be downloaded here.
[ad#ad-2]...
The Assam Academy of Mathematics has recently conducted the state level mathematics Olympiad for various categories. The question paper for the Class 9 and 10 category can be found here.
We thank Upam Sarmah for the question paper.
[ad#ad-2]... | 677.169 | 1 |
Math for Merchandising: A Step-by-Step Approach (3rd Edition)
Evelyn C. Moore
ISBN: 0131107348;
Издательство: Prentice Hall
Страниц: 368
This book takes users step by step through the concepts of merchandising math. It is organized so that the chapters parallel a career path in the merchandising industry. The book begins with coverage of fundamental math concepts used in merchandising and progresses through the forms and math skills needed to buy, price, and re-price merchandise. Next readers learn the basics of creating and analyzing six-month plans. The final section of the book introduces math and merchandising concepts that are typically used at the corporate level. For individuals pursuing a career in merchandising. | 677.169 | 1 |
Maths Practice 1000+ (New Syllabus)
Maths Practice 1000+ is a series of 6 books. Each book is thoughtfully organised to systematically introduce the pupil to a range of well-designed questions of the 2 main components in examinations - Short-answer questions and structured questions. The practices aim to effectively establish and reinforce pupils' understanding of basic mathematical concepts. By attempting this series, the pupil will be better equipped to handle tests and examinations | 677.169 | 1 |
Description:
About this title:
Synopsis: The Big Ideas Math student edition Algebra 1, Geometry and Algebra 2 textbooks mirror the pedagogical philosophy that made the middle school Big Ideas Math books so successful. Each lesson begins with an Essential Question, followed by Explorations. Once the inquiry section is completed, students begin the direct instruction Lesson, helping them to reason and make sense of their answers based on the knowledge they gained during discovery.
Book Description Big Ideas Learning. Hardcover. Book Condition: Very Good. 1608408396 MULTIPLE COPIES AVAILABLE. This book is in very nice condition and may show minor shelf wear, contain a school stamp, sticker or class set number on the inside our outside cover. This book may also contain some minor highlighting and other markings. Bookseller Inventory # SKU00006477
Book Description Big Ideas Learning. Hardcover. Book Condition: Fine. 1608408396 MULTIPLE COPIES AVAILABLE. This book is in excellent condition and may contain a school sticker, stamp or a class set number on the inside or outside cover. 100% guaranteed fast shipping!. Bookseller Inventory # SKU00006510
Book Description HOUGHTON MIFFLIN HARCOURT. Hardcover. Book Condition: Fair. 1608408396. Bookseller Inventory # Z1608408396Z4
Book Description Big Ideas Learning. Hardcover. Book Condition: New. 1608408396 MULTIPLE COPIES AVAILABLE. New book may have school stamps or class set numbers on the side but was not issued to a student. 100% guaranteed fast shipping!!. Bookseller Inventory # SKU00006476
Book Description HOUGHTON MIFFLIN HARCOURT. Hardcover. Book Condition: Very Good. 160840839608408396Z208408396GO | 677.169 | 1 |
1. Mathematical Tools
• We use math in almost every problem we solve. As a result the more relevant topics of mathematics are summarized here.
• This is not intended for learning, but for reference.
1.1 Introduction
• This section has been greatly enhanced, and tailored to meet our engineering requirements.
• The section outlined here is not intended to teach the elements of mathematics, but it is designed to be a quick reference guide to support the engineer required to use techniques that may not have been used recently.
• For those planning to write the first ABET Fundamentals of Engineering exam, the following topics are commonly on the exam.
quadratic equation
straight line equations: slop and perpendicular
conics, circles, ellipses, etc.
matrices, determinants, adjoint, inverse, cofactors, multiplication
limits, L'Hospital's rule, small angle approximation
integration of areas
complex numbers, polar form, conjugate, addition of polar forms
maxima, minima and inflection points
first order differential equations: guessing and separation
second order differential equation: linear, homogeneous, non-homogeneous, second order
1.1.1 Constants and Other Stuff
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• A good place to start a short list of mathematical relationships is with greek letters
• The constants listed are amount some of the main ones, other values can be derived through calculation using modern calculators or computers. The values are typically given with more than 15 places of accuracy so that they can be used for double precision calculations.
1.1.2 Basic Operations
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• These operations are generally universal, and are described in sufficient detail for our use.
• Basic properties include,
1.1.2.1 - Factorial
• A compact representation of a series of increasing multiples.
1.1.3 Exponents and Logarithms
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• The basic properties of exponents are so important they demand some sort of mention
• Logarithms also have a few basic properties of use,
• All logarithms observe a basic set of rules for their application,
1.1.4 Polynomial Expansions
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• Binomial expansion for polynomials,
1.2 Functions
1.2.1 Discrete and Continuous Probability Distributions
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• The Binomial distribution is,
• The Poisson distribution is,
• The Hypergeometric distribution is,
• The Normal distribution is,
1.2.2 Basic Polynomials
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• The quadratic equation appears in almost every engineering discipline, therefore is of great importance.
• Cubic equations also appear on a regular basic, and as a result should also be considered.
• On a few occasions a quartic equation will also have to be solved. This can be done by first reducing the equation to a quadratic,
1.2.3 Partial Fractions
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• The next is a flowchart for partial fraction expansions.
• The partial fraction expansion for,
• Consider the example below where the order of the numerator is larger than the denominator.
• When the order of the denominator terms is greater than 1 it requires an expanded partial fraction form, as shown below.
• We can solve the previous problem using the algebra technique.
1.2.4 Summation and Series
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• The notation
is equivalent to
assuming
and
are integers and
. The index variable
is a placeholder whose name does not matter.
• Operations on summations:
• Some common summations:
for both real and complex
.
for both real and complex
. For
, the summation does not converge.
1.3 Spatial Relationships
1.3.1 Trigonometry
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• The basic trigonometry functions are,
• Graphs of these functions are given below,
• NOTE: Keep in mind when finding these trig values, that any value that does not lie in the right hand quadrants of Cartesian space, may need additions of ±90° or ±180°.
• Now a group of trigonometric relationships will be given. These are often best used when attempting to manipulate equations.
• These can also be related to complex exponents,
1.3.2 Hyperbolic Functions
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• The basic definitions are given below,
• some of the basic relationships are,
• Some of the more advanced relationships are,
• Some of the relationships between the hyperbolic, and normal trigonometry functions are,
1.3.2.1 - Problems
Problem 1.1 Find all of the missing side lengths and corner angles on the two triangles below,
1.3.3 Geometry
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• A set of the basic 2D and 3D geometric primitives are given, and the notation used is described below,
• A general class of geometries are conics. This for is shown below, and can be used to represent many of the simple shapes represented by a polynomial.
1.3.4 Planes, Lines, etc.
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• The most fundamental mathematical geometry is a line. The basic relationships are given below,
• If we assume a line is between two points in space, and that at one end we have a local reference frame, there are some basic relationships that can be derived.
• The relationships for a plane are,
1.4 Coordinate Systems
1.4.1 Complex Numbers
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• In this section, as in all others, 'j' will be the preferred notation for the complex number, this is to help minimize confusion with the 'i' used for current in electrical engineering.
• The basic algebraic properties of these numbers are,
• We can also show complex numbers graphically. These representations lead to alternative representations. If it in not obvious above, please consider the notation above uses a Cartesian notation, but a polar notation can also be very useful when doing large calculations.
• We can also do calculations using polar notation (this is well suited to multiplication and division, whereas Cartesian notation is easier for addition and subtraction),
• Note that DeMoivre's theorem can be used to find exponents (including roots) of complex numbers
• Euler's formula:
• From the above, the following useful identities arise:
1.4.2 Cylindrical Coordinates
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• Basically, these coordinates appear as if the Cartesian box has been replaced with a cylinder,
1.4.3 Spherical Coordinates
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• This system replaces the Cartesian box with a sphere,
1.5 Matrices and Vectors
1.5.1 Vectors
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• Vectors are often drawn with arrows, as shown below,
• Cartesian notation is also a common form of usage.
• Vectors can be added and subtracted, numerically and graphically,
1.5.2 Dot (Scalar) Product
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• We can use a dot product to find the angle between two vectors
• We can use a dot product to project one vector onto another vector.
• We can consider the basic properties of the dot product and units vectors.
1.5.3 Cross Product
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• First, consider an example,
• The basic properties of the cross product are,
• When using a left/right handed coordinate system,
• The properties of the cross products are,
1.5.4 Triple Product
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1.5.5 Matrices
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• Matrices allow simple equations that drive a large number of repetitive calculations: as a result they are found in many computer applications.
• A matrix has the form seen below,
• Matrix operations are available for many of the basic algebraic expressions, examples are given below. There are also many restrictions: many of these are indicated.
• The eigenvalue of a matrix is found using,
1.5.6 Solving Linear Equations with Matrices
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• We can solve systems of equations using the inverse matrix,
• We can solve systems of equations using Cramer's rule (with determinants),
1.5.7 Problems
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Problem 1.2 Perform the matrix operations below.
Problem 1.3 Perform the vector operations below,
Problem 1.4 Solve the following equations using any technique,
1.6 Calculus
• NOTE: Calculus is very useful when looking at real systems. Many students are turned off by the topic because they "don't get it". But, the secret to calculus is to remember that there is no single "truth": it is more a loose collection of tricks and techniques. Each one has to be learned separately, and when needed you must remember it, or know where to look.
1.6.1 Single Variable Functions
• L'Hospital's rule can be used when evaluating limits that go to infinity.
• Some techniques used for finding derivatives are,
1.6.1.2 - Integration
• Some basic properties of integrals include,
• Some of the trigonometric integrals are,
• Some other integrals of use that are basically functions of x are,
• Integrals using the natural logarithm base 'e',
1.6.2 Vector Calculus
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• When dealing with large and/or time varying objects or phenomenon we must be able to describe the state at locations, and as a whole. To do this vectors are a very useful tool.
• Consider a basic function and how it may be represented with partial derivatives.
• Gauss's or Green's or divergence theorem is given below. Both sides give the flux across a surface, or out of a volume. This is very useful for dealing with magnetic fields.
• Stoke's theorem is given below. Both sides give the flux across a surface, or out of a volume. This is very useful for dealing with magnetic fields.
1.6.3 Differential Equations
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• Solving differential equations is not very challenging, but there are a number of forms that need to be remembered.
• Another complication that often occurs is that the solution of the equations may vary depending upon boundary or initial conditions. An example of this is a mass spring combination. If they are initially at rest then they will stay at rest, but if there is some disturbance, then they will oscillate indefinitely.
• We can judge the order of these equations by the highest order derivative in the equation.
• Note: These equations are typically shown with derivatives only, when integrals occur they are typically eliminated by taking derivatives of the entire equation.
• Some of the terms used when describing differential equations are,
ordinary differential equations: if all the derivatives are of a single variable. In the example below 'x' is the variable with derivatives.
first order differential equations: have only first order derivatives,
second order differential equations: have at least on second derivative,
higher order differential equations: have at least one derivative that is higher than second order.
• Note: when solving these equations it is common to hit blocks. In these cases backtrack and try another approach.
• linearity of a differential equation is determined by looking at the dependant variables in the equation. The equation is linear if they appear with an exponent other than 1.
1.6.3.1 - First Order Differential Equations
• These systems tend to have a relaxed or passive nature in real applications.
• Examples of these equations are given below,
• Typical methods for solving these equations include,
guessing then testing
separation
homogeneous
1.6.3.1.1 - Guessing
• In this technique we guess at a function that will satisfy the equation, and test it to see if it works.
• The previous example showed a general solution (i.e., the value of 'C' was not found). We can also find a particular solution.
1.6.3.1.2 - Separable Equations
• In a separable equation the differential can be split so that it is on both sides of the equation. We then integrate to get the solution. This typically means there is only a single derivative term.
1.6.3.1.3 - Homogeneous Equations and Substitution
• These techniques depend upon finding some combination of the variables in the equation that can be replaced with another variable to simplify the equation. This technique requires a bit of guessing about what to substitute for, and when it is to be applied.
1.6.3.2 - Second Order Differential Equations
• These equations have at least one second order derivative.
• In engineering we will encounter a number of forms,
homogeneous
non-homogeneous
1.6.3.2.1 - Linear Homogeneous
• These equations will have a standard form,
• An example of a solution is,
1.6.3.2.2 - Non-homogeneous Linear Equations
• These equations have the general form,
• to solve these equations we need to find the homogeneous and particular solutions and then add the two solutions.
1.6.4 Other Calculus Stuff
• The Taylor series expansion can be used to find polynomial approximations of functions.
1.7 Numerical Methods
• These techniques approximate system responses without doing integrations, etc.
1.7.1 Approximation of Integrals and Derivatives from Sampled Data
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• This form of integration is done numerically: this means by doing repeated calculations to solve the equation. Numerical techniques are not as elegant as solving differential equations, and will result in small errors. But these techniques make it possible to solve complex problems much faster.
1.7.2 Euler First Order Integration
• We can also estimate the change resulting from a derivative using Euler's equation for a first order difference equation.
1.7.3 Taylor Series Integration
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• Recall the basic Taylor series,
• When h=0 this is called a MacLaurin series.
• We can integrate a function by,
1.7.4 Runge-Kutta Integration
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• The equations below are for calculating a fourth order Runge-Kutta integration.
1.7.5 Newton-Raphson to Find Roots
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• When given an equation where an algebraic solution is not feasible, a numerical solution may be required. One simple technique uses an instantaneous slope of the function, and takes iterative steps towards a solution.
• The function f(x) is supplied by the user.
• This method can become divergent if the function has an inflection point near the root.
• The technique is also sensitive to the initial guess.
• This calculation should be repeated until the final solution is found.
1.8 Laplace Transforms
• The Laplace transform allows us to reverse time. And, as you recall from before the inverse of time is frequency. Because we are normally concerned with response, the Laplace transform is much more useful in system analysis.
• The basic Laplace transform equations is shown below,
1.8.1 Laplace Transform Tables
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• Basic Laplace Transforms for operational transformations are given below,
• A set of useful functional Laplace transforms are given below,
• Laplace transforms can be used to solve differential equations.
1.9 Fourier Series
• These series describe functions by their frequency spectrum content. For example a square wave can be approximated with a sum of a series of sine waves with varying magnitudes. | 677.169 | 1 |
Digital Design Exercises for Architecture Students teaches you the basics of digital design and fabrication tools with creative design exercises, featuring over 200 illustrations, which emphasize process and evaluation as key to designing in digital mediums.
Optimization in Practice with MATLAB® provides a unique approach to optimization education. It is accessible to both junior and senior undergraduate and graduate students, as well as industry practitioners. | 677.169 | 1 |
Why Us
Most Used Algebra Formulas Cheat Sheet
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Even if you are the very best in your class it does not mean that you are able to remember every single formula that you will need to use. It can often take a huge amount of searching through your books and your notes to find the specific algebra formula help that you need for your assignments, course work or even your exam. An easier way is to use a simple pre algebra formulas cheat sheet to provide simple and quick access to all of the formulas that you are likely to need.
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No matter what help you need with algebra just contact our experts today for tuition or even just download our algebra formulas cheat sheet for free! | 677.169 | 1 |
Basics
MathML Basics
This document illustrates some basic MathML constructions. It is tailored to display correctly with just the Symbol font. You can view its source. However, MathML documents tend to be verbose and you might get lost trying to locate a MathML fragment with the usual view source. This demo has been made to illustrate the following aspects. You can right-click on any math fragment of interest throughout this document. The context menu won't show up. Rather, the math fragment will zoom, and if you right-click a second time, you will see the MathML WYSIWYG markup of the fragment, and if you right-click again a third time, the fragment will revert to its initial state. This tri-state mode is aimed at limiting conflicts with other agents that compete for the mouse.
With MathML, one can build sets such as (go on, right-click any of these equations to experiment the zoom) {0,1,2,3,4} or {⌊ab⌋|a2+b2≤3}, write calculus dydx=1y2, form rather complicated expressions limn→N(1+1n)n−eN, k=∂2z∂x2∂2z∂y2-(∂2z∂x∂y)2(1+(∂z∂x)2+(∂z∂y)2)2, write vector equations Y=aX+b, etc.
Notice how the mathematics appear in the main flow of text and respond as you resize the window. You can also make displayed equations, such as the following ones:
x→maps toy=fn(x)=(1+1xn)n∫abf(x)dx=b-a6[f(a)+4f(a+b2)+f(b)]-(b-a)54!5!f(4)(η),a≤η≤b|x|={-xifx<0x otherwise You can also typeset 2D mathematical constructs such as matrices. The following example shows the i-th step of the multiplication of a matrix A by a vector x (notice how ai1,...,ain,x1 are on the same baseline, other alignments are possible): i-th row[a11a12a13...a1n:::...:ai1ai2ai3...ain::::an1an2an3...ann][x1x2x3:xn]
In Mozilla, MathML runs inside the main browser. So it responds to other browser operations such as the zoom (try View -> Text Zoom), and you can do links a2+b2=c2, apply stylistic effects a2+b2=c2, or use color a2+b2=c2 in very strange ways p(x)q(x)=a0+a1x+a2x2+...+an-1xn-1b0+b1x+b2x2+...+bn-1xn-1.
You can also do other weird and risky things which are not portable, bongo warns, such as mixing MathML with other markups lizard+bongo=∫abdx+
For more information about MathML in Mozilla, see the MathML Project Page. There are links to more samples, screenshots and instructions on how to download fonts for various platforms. These fonts are required to view other examples beyond the basic constructions illustrated | 677.169 | 1 |
Pages
Tuesday, July 24, 2012
July 24th Email to Parents
Below is the content of the information email I sent out to parents on July 24th for your reference.
Welcome to Algebra! I want to share a few things with you about my Algebra class now, and then you can expect another email the first week of August that will give you some more information as well as some information I'll ask you to share with your student. As was mentioned when you requested "Flipped Algebra," there are many reasons why broadband access to the Internet is going to be very helpful for your student, but two of the most important are access to some video instruction and access to pre-assessments that your student will take before formal assessments. If you're interested, there is much more information and examples listed on the information page. The AHS Math Department has created aMath Skills Assessment that all incoming freshmen need to complete in order to prepare them to be successful in mathematics at AHS. Even if your student is not a freshmen, they should complete this skills assessment and work on any skills they are having difficulty with. Our first formal assessment will be within the first few days of school over these skills. Students who are still struggling with these skills will then need to devote extra time - and their unscheduled hours - to master these skills in the first few weeks of school. Since those first few weeks are both exciting and extremely busy, it would be much better to use some of their time during the summer when things are more relaxed to make sure they have mastered these pre-algebra skills. While you obviously don't need to go shopping just yet, I wanted to share the materials your student will need for my class now. That way if you come across any great sales you can purchase early, and perhaps you might just want to purchase early since back-to-school time can be pretty busy and stressful for a lot of folks. Here's what your student will need for my class:
3-Ring Binder - Probably a 1-inch binder will do. For organizational purposes, it's going to be very helpful to have 3 dividers in that binder as well.
Notebook Paper - Your student will be taking notes, working problems, and sketching diagrams on a daily basis, probably several hundred sheets would be good to start with.
Graph Paper - This is Algebra, so graph paper is a must. Probably about 100 sheets would be a good amount to start with.
Writing Utensils - The old standby #2 pencil is going to come in handy. Some students will also find either colored pencils or pens (like those 4-color pens) handy in order to better take notes and illustrate their thinking.
Calculator - The only requirement is that your student have a scientific calculator of some sort. While graphing calculators have many advantages, they are not required - especially considering your student will have a laptop. There are many pieces of software and browser plug-ins available for their laptop, see this link for more information.
Even though they'll have their laptop, they'll most likely still want to have a separate calculator, as they won't always be able to use their laptops on assessments. A great, relatively inexpensive scientific calculator available in many stores as well as online is the TI-36 Pro (Scientific, ~$18)
Optional USB Flash Drive - For all of their classes, a USB flash drive will come in handy at times for students to transport files among various computers.
Optional Headphones - For all their classes, headphones/earbuds that can be plugged into the audio port on a computer are helpful.
I'm hopeful that you find this information helpful. If you have any questions, please let me know. Expect another email from me around the first week of August. Thanks for your time and I'm looking forward to having your student in my class this fall. Sincerely, Karl Fisch Arapahoe High School | 677.169 | 1 |
Students work in random groups to build teams and comfort with VNPS. Consolidate with discussion on taking a problem....using a model (formula, graph, table, etc,) determining an answer. Assign homework set 1
Grade 9 Orientation day. Some students were away. Did not start first lesson.
5
Sept
Fri
Characteristics
Characteristics of functions
Properties of Functions
Polygraph & Sketch
Polygraph to start and generate characteristic discussion. Give a graph and have them pick out all of the characteristics ... put on recording sheet --- Desmos activity. Have then sketch some basic functions with these properties -- Desmos. Give them one of the graphs from alex's set. (different from rest of class) have them draw on chart paper. List all characteristics. Use as a gallery walk. Show graph, cover the key characteristics...flip up when ready to check
Show a degree...arbitrary measure. Show Geogebra sketch. Ask for how many will go around. take guesses. Then show that half a radian is half the arc length divded by the radius. therefore the angle is calculated by the arc length by the radius. Show the conversions
We'll take a note on the key properties of the graphs of sine and cosine using the two desmos sketches. Students will work through the Match My Trig Function Activity. End class by sketching some by hand
Took up homework...Ss worked at their own pace through the modified Burning Daylight activity. They struggled through solving the equations and no one made it to the average and instantaneous rates of change problems. Still assigned some for homework
Started with Desmos Activity and finished with practicing problems from the book.
page 425 #6,8,9. Page 436 #17
89
Fri
20
All
Combining Functions: Dividing Functions
Graph tanx, secx, cscx plus a few more.
page 435 #4,8,9
90
Mon
23
Review
Exam Review
Review rates of change
In stations studens will complete 3 or 4 rates of change problems. Each station is a different function type. Give the sum of the 4 answers. Let them determine which of their problems need fixing if any. | 677.169 | 1 |
Meta
Goal 1:
To develop skills needed in order to read, write, recognize and appreciate a (good) mathematical proof. The corresponding assessment will be based on student performance on the two in-class exams, the final exam, and select homework assignments.
Goal 2:
To develop fluency in using basic logic and set theory, including the language of functions and relations. The extent to which this goal is achieved will be determined by student performance on the second in-class exam, the final exam, and select homework assignments.
Goal 3:
To gain familiarity with certain areas of mathematics which could be labeled as being "discrete": combinatorics, number theory, graph theory. The extent to which this goal is achieved will be assessed by means of the first in-class exam, the final exam, the graph theory quiz, and select homework assignments.
Goal 4:
To situate the practice of mathematics within its larger intellectual and social context. The corresponding assessment will rely on student essays at the end of the semester. | 677.169 | 1 |
Pages
Saturday, March 3, 2012
Basic maths
In this i am going to describe you about the functions.A functon is nothing but the one which connects it.
ONE ONE FUNCTION:
A one one function is the one in which each element is mapped with only one element.
ONTO FUNCTION:
A onto function is the one in which the all the elements of codomain must be mapped.There is no condition that all domain elements should be mapped
BIJECTION:
A bijection is the one which is one one and onto that is both domain and co domain elements are mapped | 677.169 | 1 |
Written for students taking a second or third year undergraduate course in mathematics or computer science, this book is the ideal companion to a course in enumeration. Enumeration is a branch of combinatorics where the fundamental subject matter is numerous methods of pattern formation and counting. An Introduction to Enumeration provides a comprehensive and practical introduction to this subject giving a clear account of fundamental results and a thorough grounding in the use of powerful techniques and tools.
Two major themes run in parallel through the book, generating functions and group theory. The former theme takes enumerative sequences and then uses analytic tools to discover how they are made up. Group theory provides a concise introduction to groups and illustrates how the theory can be used to count the number of symmetries a particular object has. These enrich and extend basic group ideas and techniques.
The authors present their material through examples that are carefully chosen to establish key results in a natural setting. The aim is to progressively build fundamental theorems and techniques. This development is interspersed with exercises that consolidate ideas and build confidence. Some exercises are linked to particular sections while others range across a complete chapter. Throughout, there is an attempt to present key enumerative ideas in a graphic way, using diagrams to make them immediately accessible. The development assumes some basic group theory, a familiarity with analytic functions and their power series expansion along with some basic linear algebra.
undergraduate combinatorics course. Overall, it is wonderful...I like how your book actually teaches something, beyond a loose tour of combinatorics that many other books offer. The ability to fluidly go back and forth between recurrences and generating functions, that sort of thing...It trades the broadest focus for specific techniques, in that way reminding me what's fun in say, an ODE course. My students actually learn something they couldn't do before, it makes them feel smart so they like the course " (Prof Dave Bayer, Barnard College, NY, USA)
"This work is a very basic, short introduction to combinatorial enumeration techniques. Camina (Univ. of East Anglia, UK) and Lewis (The Mathematical Association, UK) properly explain the fundamental concepts … . There are 20-25 exercises per chapter … . The numerical answers to the majority of these exercises are included at the end. … It may be useful for students who need very basic enumeration skills … . Summing Up … . Lower- and upper-division undergraduates." (M. Bona, Choice, Vol. 49 (4), December, 2011)
"This book is written as an introduction to enumeration for second or third year undergraduate students in mathematics or computer science. Its theme is counting, using finite or infinite series … . The development is interspersed with exercises, linked to particular sections, or covering a complete chapter. Solutions are provided at the end of the book." (Andreas N. Philippou, Zentralblatt MATH, Vol. 1230, 2012)
"This is an introductory text on counting and combinatorics that has good coverage … . It is aimed at a sophomore or higher level and has few prerequisites beyond power series. … There are numerous exercises, and all have brief solutions in the back. … It is unusual to see a book that combines such extensive coverage with so few prerequisites." (Allen Stenger, The Mathematical Association of America, August, 2011) | 677.169 | 1 |
graphic tools, image processing, and music.
Although Mathematica 7 is a highly advanced computational platform, the recipes in this book make it accessible to everyone -- whether you're working on high school algebra, simple graphs, PhD-level computation, financial analysis, or advanced engineering models.
Learn how to use Mathematica at a higher level with functional programming and pattern matching
Delve into the rich library of functions for string and structured text manipulation
Learn how to apply the tools to physics and engineering problems
Draw on Mathematica's access to physics, chemistry, and biology data
Get techniques for solving equations in computational finance
Learn how to use Mathematica for sophisticated image processing
Process music and audio as musical notes, analog waveforms, or digital sound samples
Sal Mangano has been developing software since the days Borland Turbo C and has worked with an eclectic mix of programming languages and technologies. Sal worked on many mission-critical applications, especially in the area of financial-trading applications. In his day job, he works mostly with mainstream languages like C++ and Java so he chooses to play with more interesting technology whenever he gets a chance.
Sal's two books (XSLT Cookbook and Math Mathematica Cookbook) may seem to be an odd pair of technologies for a single author but there is a common theme that reflects his view at what makes a language powerful. Both Mathematica and XSLT rest on the idea of pattern matching and transformation. They may use these patterns in different ways and transformations to achieve different ends but they are both good at what they do and interesting to program in for a common reason. Sal's passion for these languages and ideas comes through in both these cookbooks. He also likes to push technologies as far as they can go and into every nook and cranny of application. This is reflected in the wide mix of recipes he assembled for these books.
Sal has a Master's degree in Computer Science from Polytechnic University.
Review:
"For those willing to spend the time, effort and money, Mathematica Cookbook is a worthy purchase for the discerning Mathematica user." --Mike Riley - Dr Dobbs Code Talk
The Mathematica Cookbook does a good job of showing the wide range of capabilities of the Mathematica program... --Jerry Pournelle, Chaos Manor, The User's Column, August 2010, Column 360
[Mathematica CookBook] supplies a number of very nice examples with which to extend user expertise. --John A. Wass, Ph.D., Scientific Computing'Reilly Media, Inc, USA. Paperback. Condizione libro: new. BRAND NEW, Mathematica Cookbook, Salvatore Mangano,'ll find the recipes in "Mathematica Cookbook" extremely useful and informative. With key support from Mathematica's "Mathematica Cookbook" is ideal for professionals and hobbyists from all walks of life. Codice libro della libreria B9780596520991 | 677.169 | 1 |
How to Learn Math?
In order to learn math you have to, create the time for learning, get to know the vocabulary, get atleast two reference books on the set theory, tackle the subjects and their what is required, slowly progress through the mathematics, and never be too afraid to ask for help when you don't know something. | 677.169 | 1 |
The Ejs Harmonics and Fourier Series model displays the sum of harmonics via a Fourier series to yield a new wave. The amplitude of each harmonic as well as the phase of that harmonic can be changed via sliders. In addition, several pre-set functions can be chosen to display. You can modify thisA quick reference application for Fourier Series Properties DSP theory. Includes Discrete Time Fouries Series (DTFS) and Continuous Time Fourier Series (CTFS). App History V2.0 Second version. Bugs fixed. Application is courtesy of Quarinos Corporation. Requirements: Windows Phone 8.1,...
The inversion of Laplace transforms is performed using two methods: (1) the Zakian method and (2) the Fourier series approximation. Results are in agreement with analytical solutions. The Zakian method presents problems for transcendental functions. The Fourier series gives better results when...
Game bilingual ( English and Spanish ) for infants and children six months and older who has 5 goals : Learn to identify and associate the basic shapes : circle , triangle, square, rectangle and rhombus. Identify, associate and learn the natural numbers Identify, associate and learn theDisplays graphs of algebraic functions in a variety of forms. These include polar and cartesian co-ordinates, parametric and intrinsic functions. A wide range of functions are built-in, from simple trig and hyperbolic functions to things such as the ceil and gamma functions. On-screen HTML help... | 677.169 | 1 |
math.js has built-in support for symbolic computation (CAS). It can parse expressions into an expression tree and do algebraic operations like simplification and derivation on the tree.
It's worth mentioning an excellent extension on math.js here: mathsteps, a step-by-step math solver library that is focused on pedagogy (how best to teach). The math problems it focuses on are pre-algebra and algebra problems involving simplifying expressions.
The function accepts either a string or an expression tree (Node) as input, and outputs a simplified expression tree (Node). This node tree can be transformed and evaluated as described in detail on the page Expression trees. | 677.169 | 1 |
Shape Theory
Shape Theory by J. M. Cordier This in-depth treatment uses shape theory as a 'case study' to illustrate situations common to many areas of mathematics, including the use of archetypal models as a basis for systems of approximations. It offers students a unified and consolidated presentation of extensive research from category theory, shape theory, and the study of topological algebras.A short introduction to geometric shape explains specifics of the construction of the shape... | 677.169 | 1 |
Tag Archives: Math Education
There has been an ongoing call in mathematics education for students to be engaging in problem solving and collaborative groupwork. Although, many instructors find that when they put students in groups, some students seem disengaged and we may start to … Continue reading →
About a week after their first test, during the first round of midterms of the semester for undergraduate students, mid-semester complacency came swooping into my classroom. My once invigorated students suddenly started going through the motions of class rather than … Continue reading →
The Inquiry-Oriented (IO) Approach and Advanced Mathematical Thinking (AMT) processes play an important role in improving undergraduate math education. IO approach and AMT processes act as a new movement of modern math education based on the methods used in math … Continue reading →
Generalization and abstraction both play an important role in the minds of mathematics students as they study higher-level concepts. In the second chapter of the Springer book Advanced Mathematical Thinking, Tommy Dreyfus defines generalization as the derivation or induction from something particular to something … Continue reading →
A guest post by Jenna Jensen: Erik Stern and Karl Schaefer discuss the cross-curricular possibilities with math and the art of movement and dance in their video, Math Dance. I think it is appropriate to ask in what ways can … Continue reading →
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What resources are available for Algebra 2 by McDougal Littell online?
A:
Quick Answer
The McDougal Littell website allows students, teachers and parents to have access to lesson help and extra exercises. In order for the website to be used properly, users need to make sure they have Adobe Acrobat reader installed on their computer.
The book is broken down into chapters and students can chose the one they are interested in or need help with. The chapter page has additional links as well for students and teachers both. Students can access test practices or a problem of the week for practice. Teachers can find lesson resources and answer keys to the additional material found on the website as well as lesson planners. | 677.169 | 1 |
9780077534004
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Summary
Table of Contents
Chapter 1: Whole Number: How to Dissect and Solve Word Problems
Chapter 2: Fractions
Chapter 3: Decimals
Chapter 4: Banking
Chapter 5: Solving for the Unknown: A How to Approach to Solving Equations | 677.169 | 1 |
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