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This study guide takes the student through the logical thinking processes when factoring & solving more involved trinomials. Examples shown include those with leading coefficients greater than 1, trinomials with a GCF, those using patterns, and those that require using the square root property to solve quadratics. The latter do not have "easy" answers (fractions, radicals, etc). | 677.169 | 1 |
Mathematics is one of the core subjects within the National Curriculum. It is recognised as one of the subjects that students need to have a very good understanding of from a very early age. For students to progress into Higher Education, they need to possess at least a C grade at GCSE in Mathematics, so clearly it has great importance as a subject.
The department offers opportunities for students to develop their knowledge and understanding of Mathematics through, for example, the opportunity to attend master classes at Liverpool Hope University, as well as residential weekends for GCSE revision and experience of University Life at Easter courses, to support A Level. Members of the department often offer sessions at weekends to help in preparation for exams.
The department has access to a set of iPads just for students, who can avail themselves with software which covers all aspects of the National Numeracy Strategy and the GCSE syllabus being followed. There is an interactive smart television in each of the teaching rooms, together with associated hardware, as well as a diverse range of practical equipment.
In Key Stage 3, students follow the national curriculum with units in Algebra, Number, Geometry and Measures and Handling Data. [See the curriculum link for further details.]
In Key Stage 4, (from 2010) students are following the Edexcel Linear GCSE course; which is a two-year GCSE course with a final exam in the summer of Year 11.
In Key Stage 5, students follow the Edexcel AS and A2 course. For AS, the students undertake three modules, Core 1, Core 2 and Decision 1. In A2 they take an additional three modules in Core 3, Core 4 and Statistics 1.
The opportunity to undertake Further Mathematics also exists. | 677.169 | 1 |
Pre-Algebra II
Encourage learners to take the path towards achievement. In A Grade Ahead's Pre-Algebra II math program, we encourage our students to put effort into learning the concepts they need to succeed. The Pre-Algebra II program consists of layers of content that will help students develop:
Problem-solving skills
Logic and ability to analyze
Test-taking strategies
While Pre-Algebra II students will not receive individualized numerical drills, they do receive rich and meaningful developed curriculum, as well as test prep homework. Each step has been designed to enhance A Grade Ahead students' abilities to calculate quickly and accurately, while dissecting word problems to find solutions. | 677.169 | 1 |
Quadratic Function Project: Catapults
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Project is divided into four different sections. The first is the creation of a catapult, which my students made out of Popsicle sticks and rubber bands. The second is a paper which requires students to explain the mathematics behind a catapult. The third is the equation for the trajectory of their catapult. The last is a poster project (which i substituted for the paper and equation due to time) which allows students to transition between equations, graphs, and tables. All sections of this project include a rubric. | 677.169 | 1 |
iCalcline is a Latest Generation Algebra Calculator for the iPhone
[prMac.com] Milano, Italy - Tension Software announces the release of iCalcline 1.0. iCalcline is a latest generation algebra calculator for the iPhone and iPod touch for any tech or math enthusiast or simply anyone in need of an advanced math tool. iCalcline puts in your pocket a powerful mathematical calculator.
Its enough to insert a mathematical expression and iCalcline will immediately show the result.
iCalcline can solve equations with an unlimited number of nested parentheses and using up to 28 functions and operators. It manages a table with variables values, using the table values you can write equations as in 'y=x 1' with the 'x' value contained in the table and iCalcline will assign to 'y' the resulting value. In case the y variable was not present when you typed 'y=x 1', a 'y' variable is created for you with a correct value assigned. You can create, manipulate, delete and use how many variables you like inside the app tapping on the variable table.
iCalcline offers a wide range of functions:
* Many mathematical functions
* Manages any nested level of parentheses using the correct precedence
* Check your expression syntax in advance signaling errors
* Offers buttons for fast introduction of any literal values available in the tables (variables) and functions
* Can change on the fly calculation precision (up to 9 decimals thanks to use of long double calculation)
* Manages unlimited numbers of variables you like in the table
* Uses decimal separator in accordance with your local settings
Input and variables are retained over run of iCalcline. When you relaunch it you will find the values you left in the app. iCalcline can provide also a full log of executed calculations solved (optionally) step by step.
Features:
* Evaluates mathematical expressions with parentheses, nested functions, variables and constants
* Easy to use
* The user simply inserts an expression and iCalcline displays the result
* Variable creation and reference inside expressions are very easy using an interactive list.
* Variables values are retained over different runs
* Expressions solved optionally step by step showing all the mathematical passages in the log.
* Uses a very small window, expandable if necessary to show additional details (switch with a simple click).
* Based on a custom fast calculation engine developed by Tension Software and used also on Mac OS X in Calcline and TS Calc (available on the Mac App Store).
* Uses long double calculation (High precision with up to 9 decimal digit precision)
* Buttons for fast introduction of variables and functions
* DEG and RAD calculation for trigonometric functions | 677.169 | 1 |
Teaching
This course introduces the mathematical theory of inverse problems. It gives the basis notions and the difficulties encountered with ill-posed inverse problems. It presents methods for analyzing these problems and gives some tools to solve such problems. This course shows what is a regularization method and introduces different regularization techniques and the basic properties of these methods. Examples of inverse problems are also provided.
AMCS 202 - Applied mathematics II (Spring 2011 and summer 2011)
This course gives an overview on important topics in applied mathematics. It starts with a review on linear algebra introducing some basic concepts. Then, it focuses on numerical methods for solving systems of linear equations. These methods are designed to give the expertise necessary to understand and use computational methods for solving scientific problems. We are especially interested in methods for solving systems of linear equations and eigenvalues problems. The second part is dedicated to complex analysis which has many applications to engineering, physics and applied mathematics. We are interested mainly in the properties of complex numbers analytic functions, contour integrals, Cauchy residue theorem and conformal mapping. We introduce also the Fourier and Laplace transforms and/or Ordinary Differential Equations which are important tools in many fields.
ME 221B - Control Theory B (Fall 2011)
This course gives basic knowledge for more advanced courses. Students will gain experience in designing feedback control systems by some state variable methods and by using observers. The course begins with a brief review of some basic concepts in control theory such as the state representation, stability, controllability, observability and associated properties. Then, some control techniques for linear systems are presented starting by the feedback control and pole placement and then the optimal control with a focus on linear quadratic regulator. The concepts of observer and Kalman filter are also introduced with some observers based control techniques. An important property in a control system is also considered, which is the robustness. This course finishes by an introduction to control of nonlinear systems. It is embedded of examples issued from physical systems with illustrations in Matlab.
The beginning and end dates for each internship will be agreed upon between the Faculty and the Intern. Visa processing time will be a determining factor in actual start date. Minimum period is three months. Maximum period is six months. | 677.169 | 1 |
Transistors are the basic building blocks of integrated circuits (IC). Every electronic gadget you can possibly imagine today uses transistors. The development started off with vacuum tubes and lead to the...
Given below is the presentation on basics of embedded systems by Dr M Muruganandam Masilamani.
This presentation mainly includes:
Introduction to System
Overview of Embedded system
Components of Embedded system
Embedded system...
Love it or hate it, you can't escape mathematics! So here's bringing 21 free ebooks to lend you some help.
Atithya Amaresh
1. Engineering Analysis
Publisher: Wikibooks , 2012
This book is about the topic | 677.169 | 1 |
Calculus Weekly Review
How many times do your calculus students make algebra mistakes? How many times do you think to yourself – they understand the calculus concepts, but keep messing up the algebra?
This is a collection of 15 worksheets that will help students review key algebra and trigonometry
Interactive Notebook must have!! Innovative, Fun, and a Time Saver for all Math classes 6 - 12+. Easily print graphs on Post It Brand® or any brand sticky notes. My students love these!
Updated with two new templates. By popular request I have added two new templates with first
Calculus Weekly Review Second Semester
Each worksheet contains 10 questions that help keep all topics covered in Calculus AB fresh in students' minds.
I give my students one of these worksheets on Monday and it is due on Friday. I count each sheet as 10 points toward their homework grade.
These a
Riemann Sums retenti
Derivatives & Second Derivatives - Graphing Concepts:
This activity requires students to match up the graph of a function with the graphs of its 1st and 2nd derivative. This will help students to visually compare graphs and see how slopes at different points transfer to the graph of the derivat
These 24 problems are great to use with your calculus students within the first few days of the school year. Reviews everything from asymptotes to domain and range, logs to sine and cosine. Worked in are calculus ideas such as limits, tangent lines and average rate of change. The intention is to w
Calculus Derivative at a Point
This is a game that students can use to practice finding the value of the derivative at a given point.
Students choose a letter of the alphabet. They work the problem given. Then they check the answer blanks. If that answer is in one of the answer blanks, they fillThis is a bingo game designed to review material from first semester Calculus AB. The questions cover such topics as limits, derivatives, equations of tangent lines, related rates, optimization, and curve sketching.
There are 24 questions given in this presentation.
Students are given 24 choices o
Trig Identities and Formulas Flip Books. Foldable, easy, and your students will love this.
Two great new resources for your Calculus and Trig students. This product includes two different foldable Flip Books.
The first is a small, handy Flip Book, 3 3/4" by 51/4"Calculus Derivatives Color by Number
Even big kids love coloring. This fun, engaging activity includes sixteen review questions on derivatives before the chain rule. The power rule, product rule, quotient rules, trig functions, and e^x are included as are applications such as tangen
This packet is part of my series: Writing about Mathematics – Calculus. This particular activity is about limits and continuity.
I have found that in order for calculus students to really internalize all of the information they are being given, they must write about it themselves. Writing helps t
Calculus Review
In this calculus card sort, I have chosen 32 things that I think students need to memorize to be successful on the Calculus AB AP Exam.
For example, students should memorize the derivatives and antiderivatives of the 6 trig functions; the formulas for volume of cone, cylinder, and
Give your students great practice for their limits and continuity test with this 28-question review in the circuit format. To advance in the circuit, students must search for their answer, and when they find it, that becomes the next problem. This circuit covers limits of the indeterminant forms 0/
Give your students engaging practice with the circuit format! This 12-question circuit contains all of the traditional related rates problems -- ladder sliding down a wall, growing conical salt pile, deflating balloon, plus a few extras such as a profit function and Charlie Brown flying a kite. To a
Great introduction to Curve Sketching for your students applying what learned about the first and second derivatives. Great for visual learners.
This innovative activity is designed for Calculus 1, AP Calculus , and Calculus Honors. It is an introduction to Unit 3, Applications to the Derivative.
For most students, the hardest part about finding the volume of a solid of revolution is identifying whether to use disks or washers* and what the radius/radii are. This set of 24 problems lets students focus on those two important aspects, with four sets of six related problems. Students are not a
Calculus Limits Task Cards
This is a set of 12 task cards that will help students practice finding the limit.
These are 12 basic limit problems found at the beginning of the calculus course. None of the problems involve infinity.
I have included a worksheet that students can use to record their a
I use scavenger hunts to review a lot of different subjects in all my classes. This one reviews finding inverse trig functions by hand. (no calculators are used) Most use the angles of special right triangles, but some require that the students draw a triangle and use Pythagorean Theorem to find th
These 20 problems are great to use with your calculus students within the first few days of the school year. Reviews everything from asymptotes to domain and range, logs to sine and cosine. Worked in are calculus ideas such as average rate of change vs instantaneous rate of change, all while ensuri
This is an activity with 32 cards meant for the beginning unit in Calculus AB, BC, or Calculus Honors. There are 8 graph cards that have matching Equation Cards, Limit Cards, and Description Cards to create a unique set. There is also a student recording sheet and a separate LAB ACTIVITY to answe
Calculus Review
This is a set of 20 questions that I give to my students the night before the Calculus AB AP Exam. We have done practice tests, reviewed all the topics, and now I need ONE LAST THING to give them.
This set of questions is designed to give students 20 questions that they can use for
If you like group activities, here's a fun RELATED RATES MATCH UP. Eight task cards contain a rate problem to be solved. Calculus students match the question card with the correct equation card and then the proper derivative card. For an extension to the activity, students state the unknowns and giv
Calculus Activity. Area Between Curves Task Cards.
This set of 11 task cards is designed for AP Calculus AB, AP Calculus BC, Honors Calculus, and College Calculus 2 students. This topic is covered typically in the Applications of Integration Unit.
Problems utilizing both horizontal strips and v
Your students will stay engaged as they work to solve these separable differential equations in the circuit format. After finding the particular solution, students must hunt for the answer to a specific question about a given independent or dependent value for the particular solution. In this way i
Give your students engaging practice with the circuit format! This 16-question circuit asks students about increasing / decreasing, max/min, concavity and inflection points based on their analysis of the first and/or second derivatives. Progressive in nature, it includes quite a few polynomials, a
I Have, Who Has is a popular game used to review skills. This set of 27 cards provides review and drill for the laws of logarithms. The packet includes 3 sheets that hold the 27 cards,a rule/example sheet and a blank sheet if you want to add more problems. The free download is the rule/example she
In this packet, you will find 5 worksheets that will help students analyze and write about functions and their derivatives.
In each worksheet I have given the students a graph and approximately 10 words that must be used in a paragraph about the given situation.
Worksheet #1: Students are asked to
This set of 24 problems (includes implicit differentiation, inverse trig functions, and transcendentals) in the self-checking format of a circuit will keep your calculus students engaged as they practice taking the derivative. The problems build in difficulty and students must search for their answe
Your students will work every single one of these 24 limits problems because of the circuit format. To advance in the circuit, students must find their answers -- this element of self-check is essential! My students don't get up when the bell rings and they are working on a circuit! The problems a
Calculus Continuity Foldable Engaging Activity in Trifold format
This resource is designed for first semester Calculus students and is from Unit 2, Limits and Continuity. The trifold contains three activities in one, is paper friendly, and can be used in Interactive Notebooks if
Trigonometry: COMMON REFERENCE ANGLES – TABLE OF VALUES Updated!!
This color-coded table helps students learn the basic trig values in both radians and degrees. Students can see patterns by way of the colors. It is an excellent reference sheet for Calculus students, Calculus AB, Calculus BC, and D
Need an activity to liven up your Calculus class? The Chain Rule 500 Maze is for you! This activity is great for small groups or individual practice. Students will get to test their knowledge of the Chain Rule by identifying their race car's path to the finish line. Students may only drive the path t
This is for teachers who want to assign summer work to students entering AP Calculus. The file includes six pages of questions about prerequisite skills for Calculus, separated by topic:
- "Super-basic" algebra skills
- Trigonometry
- Higher-level factoring
- Logarithms and exponential functions
- R
In this super secret number puzzle, students work with the chain rule. Students need to know how to find the derivative using the chain rule, how to find the equation of a tangent line, and how to use a chart to find the derivative using the chain rule.
Students find the answers to each of the ten
Great activity reinforcing the Fundamental Theorem of Calculus
Task Cards really do work! They get the students engaged and keep them motivated to go through all of the problems, more so than a simple worksheet.
Included:
✓
A project that combines the concepts of continuity and differentiation and applies it to a real world situation. The students must first use both continuity and differentiation to solve for multiple coefficients. They then use the coefficients they have solved for and their knowledge of derivatives tIn this super secret number puzzle, students will practice finding the derivative using implicit differentiation. Students need to know how to find the equation of the tangent line and the second derivative (using implicit differentiation).
Students find the answers to each of the eight questions
This activity requires NO PREP, student and teacher documents included.
PLEASE REMEMBER TO RATE ME!
Check out the preview file
It is a riddle worksheet where students have an answer bank and they use that to solve a (slightly cheesy) joke. Kids like this activities because they make the math mean
Give your students engaging practice with the circuit format! This circuit features 16 area between curves calculus problems and I wrote it with the intention to not use technology. The problems build in difficulty and students must hunt for their answers within the circuit to advance in the circui
Calculus Curve Sketching
This activity is intended to help students practice identifying the graph of the derivative of a function from the original graph and the original graph from the derivative of a function.
Each student is given either an equation, the graph of the original function, or the g
In this pair / small group activity, students use Play-Doh and paper to create models of solids with known cross sections. (See product thumbnails for a sample creation.)
The file contains everything you need but the Play-Doh!
- Preparation instructions
- 16 problems involving cross-sections of var
Finding the derivative using the constant, product, sum and difference rules. 25 unique and well thought out problems that will provide further enrichment for Calculus students. The symmetrical coloring design helps students self check their answers and makes it easy for teachers to grade. Great f
Graphs of Derivatives - Discovery:
This three-page worksheet will guide your students to graph the derivative of a function and make observations about the following concepts:
* The slope of a tangent line to a curve can be identified at various points and used to create the graph of the derivative
Engage your students with this self-checking 12-question circuit! I wrote this circuit to help my students with the symbolic notation and easy (but my students don't think so!) computation of functions and their derivatives defined as definite integrals. Most of the problems require FTC Part I, but
Limits Match
Directions:
1. Cut out each graph below. Make a pile of graphs on your desk.
2. Cut out each statement below. Make a pile of statements on your desk.
3. Match each statement with the correct graph.
4. Fill in page 3 using complete sentences and mathematic vocabulary.
5. NOTE:
Calculus Limits
This is a set of a graphic organizer and worksheet that will help students practice working with limits algebraically.
I have my students cut out the graphic organizer and put it in the left hand side of their notebook. I then have them insert the worksheet on the right hand side o | 677.169 | 1 |
Please contact your nearest Dymocks store to confirm availability
Email store
This book is available in following stores
Mathematical modelling is often spoken of as a way of life, referring to habits of mind and to dependence on the power of mathematics to describe, explain, predict and control real phenomena. This book aims to encourage teachers to provide opportunities for students to model a variety of real phenomena appropriately matched to studentsGÇÖ mathematical backgrounds and interests from early stages of mathematical education. Habits, misconceptions, and mindsets about mathematics can present obstacles to university studentsGÇÖ acceptance of a GÇÿGÇÿmodels-and-modelling perspectiveGÇÖGÇÖ at this stage of mathematics education. Without prior experience in building, interpreting and applying mathematical models, many students may never come to view and regard modelling as a way of life. The book records presentations at the ICTMA 11 conference held in Milwaukee, Wisconsin in 2003.
Examines mathematical modelling as a way of life, referring to habits of mind and dependence on the power of mathematics to describe, explain, predict and control real phenomena
Encourages teachers to provide students with opportunities to model a variety of real phenomena appropriately matched to studentsGÇÖ mathematical backgrounds and interests from early stages of mathematical education
Records presentations at the ICTMA 11 conference held in Milwaukee, Wisconsin in 2003 | 677.169 | 1 |
CBSE All in One MATHEMATICS Class 11th
ISBN 9789351417811
ISBN-10
9351417816
Binding
Paperback
Edition
1st
Number of Pages
676 Pages
Language
(English)
Subject
Entrance Exam Preparation
All in One Mathematics has been designed for the students studying in Class 11 following the CBSE curriculum. Written by an examiner with years of experience in the subject, this book contains the explanation of various concepts covered under the CBSE Class 11 Mathematics curriculum.
The book provides guidance to the students starting from the stage of learning, to practicing what has been learnt and at last assessing the concepts learnt and practiced. The whole syllabus has been divided into 16 chapters in total covering the CBSE Class 11 Mathematics curriculum. The book covers sets, relations and functions, trigonometric functions, principle of mathematical induction, complex numbers and quadratic equations, linear inequalities, permutation and combinations, binomial theorem, sequences and series, straight lines, conic sections, introduction to three dimensional geometry, limits and derivatives, mathematical reasoning, statistics and probability. The book has been designed strictly in sync with the latest CBSE syllabus and NCERT Textbook, with each chapter divided into individual topics for better understanding. Each individual topic contains detailed topical theory supported by illustrations, tables, flow charts, etc which will help in effective comprehension of the concepts. Questions given in each chapter have been grouped as Very Short Answer Type, Short Answer Type, Long Answer Type and Hots. These questions cover NCERT Questions, Previous Years Examination Questions as well as other important questions from the examination point of view. Solutions and explanations to all the questions have been given to facilitate easy learning and understanding. Revision Map, a small test that has been given at the end of each chapter will help students in assessing their level of understanding. To give an upper edge to the students Value Based Questions have been given after the chapter-wise study. For thorough practice and to give students a real feel of the examination 10 Sample Question Papers have been given after the chapter-wise study. A complete book in itself, this book will serve as a true companion and a guide on your way of achieving highest grades in Class 11 Mathematics examination. TABLE OF CONTENTS
How to Prepare for CBSE Examination
How to Stay Focused
Tips to Score the Highest Marks
1. Sets
2. Relations and Functions
3. Trigonometric Functions
4. Principle of Mathematical Induction
5. Complex Numbers and Quadratic Equations
6. Linear Inequalities
7. Permutation and Combinations
8. Binomial Theorem
9. Sequences and Series
10. Straight Lines
11. Conic Sections
12. Introduction to Three Dimensional Geometry
13. Limits and Derivatives
14. Mathematical Reasoning
15. Statistics
16. Probability
10 Sample Question Papers | 677.169 | 1 |
Fractions, Decimals, and Percents guide provides a highly organized and structured approach to the variety of questions in this quantitative content area. Students are presented with ways to understand and then manipulate part-whole relationships to convert between fractions, decimals and percents. Then, students practice implementing strategic time-efficient shortcuts. The guide also includes online access to 6 full-length Computer Adaptive Practice Exams on ManhattanGMAT's website.
Each chapter builds comprehensive content understanding by providing rules, strategies and in-depth examples of how the GMAT tests a given topic and how you can respond accurately and quickly. The Guide contains a total of 82 "In-Action" problems of increasing difficulty with detailed answer explanations.
Special Features
Purchase of this book includes one year of access to ManhattanGMAT's online Fractions, Decimals, & Percents Question Bank (accessible by inputting a unique code in the back of each book).
Manhattan GMAT has categorized all the Fractions, Decimals, & Percents problems in The Official Guides by question type. These categorized problems have been organized into problem lists that appear in the Fractions, Decimals, & Percents Strategy Guide Publisher
About the ManhattanGMAT Strategy Guides
The ManhattanGMAT Strategy Guides offer students a unique balance between two competing emphases: test-taking strategies and in-depth content understanding. Most prep courses focus on the strategic portion of the GMAT - timing, shortcuts, etc. However, we believe that unless a student has a solid understanding of the content, the tricks won't be of any use. Therefore, we include practice problem sets that build specific foundational skills in each topic as well as advanced content that many other prep books ignore.
Top Customer Reviews
The Manhattan series was the most helpful for me on my GMAT because other GMAT prep books only shows time saving or test taking tips but does not go into details about the various topics which I needed because I don't have a strong quantitative base. In fact, I didn't know even the basic idea of number properties or even odd plus odd is even (must have been asleep in math class) and so these guides were a must for me. Except for the critical reasoning guide, I give them all five stars +++!
Too bad I found these guides too late (just 2 weeks before the exam) and by the time they arrived in the mail, I didn't have time to do any of the practice exercises but I still managed to improve my score from 500 to 640 from just skimming through these guides. 640 may not be a high score, but for someone who didn't know what a cube root is two weeks before the exam, these guides sure did a good job getting me up to speed. In my situation, these guides were heaven sent.
If you're really good with math then they might be too easy and too basic for you. Even then, they would still serve as a nice 'refresher'. They are short and concise and it only took me about an hour or less to skim through each and so they are pricey if you're just using them for a quick review. But if you don't know math like I don't know math, BUY THEM ALL!!! THEY ROCK!
However, their critical reasoning gmat guide did not go into much detail than the chapters of other verbal gmat prep guides out there and so I did not feel it's price was justified at all! The other guides costs lest than this guide and covers all the verbal topics and comes with tips as well. It also arrived after my exam date so I returned it.
I didn't take a class, but I have alot of friends who have and have used the Kaplan book and practice tests in addition to the Manhattan. Manhattan prep material and tests are head and shoulders above Kaplan. My only wish is that I had started with the Manhattan series and not even worried about Kaplan. Kaplan may have the brand name, but the fact that Manhattan focuses only on the GMAT shines through. The material goes wayyy more in depth, and it really leverages the other best source of practice material you have, the Official Guide (it has a great feature called rephrasing that references the OG problems directly to give data sufficiency help. It almost makes so much sense that it seems obvious but ingenious at the same time. Why not use the actual old test problems provided by the GMAC as much as you can?).
I wasn't going to buy the whole series but was so impressed by the first book I used (Sentence Correction), that I bought a couple of the quantitative prep books. I was so impressed by those, that I ended up buying all the rest of them. I can not stress enough how much more focused and useful this material is than Kaplan. When used in conjunction with the Official Guide, you have everything you need to break 700.
The things these books provide make so much sense as the best way to prepare, that it makes you wonder why everyone else doesn't do it. My guess is that Manhattan benefits from a focused business model of limiting itself to the GMAT. If you're not looking to score that well and only need to practice some and get used to the questions, the Official Guide is enough. But the Manhattan series is also nice in that it allows you to pick and choose the certain topics you need extra help with. Also, by only buying one book, you get access to six practice tests that are very difficult and provide invaluable information in its readouts of your performance (much more detailed and useful than Kaplan's. Hate to keep hating on Kaplan, but it's the only thing I have to compare to. It's not horrible really, it just pales in comparison to Manhattan. I think of the Kaplan material like I think of McDonald's, very well-known but pretty junky. It'll fill you up, but you won't feel great afterward...)
I've never taken the time to review any items on Amazon.com before, but I have been so impressed with this product that I felt compelled to let others know about it.
Don't waste your money on a class, get the Manhattan series instead. You'll save over a thousand dollars and be better prepared as well!
This book has really helped me. One of my fears preparing for the GMAT was my rusty maths skills. This book has helped reduce those fears...I like the simplicity with which it was written and how it guides you through answering the GMAT questions without really solving anything...I'm glad I made this purchase.
I completely disagree with the other review, and if you do your research you will certainly come to the conclusion that ManhattanGMAT's books are a fantastic resource for the GMAT. While they are not necessarily a substitute for the class, they cover the material in great detail and include access to various practice problems online and 6 computer adaptive exams. You will most definitely need them for practice if you want to do well. The books are structured around content, not tricks, and are thus meant to teach the material from the ground up. I thought they were extremely useful. Raised my score from 630 to 740.
This book covers well the concepts and has several advanced exercises of this part of the Gmat. Buying the book, you will have access to 6 online GMAT tests. The online tests are really good and come with detail explanations. The level of the questions in the online exams is higher than in the real exam.
This and all the other Manhattan GMAT guides were really helpful in helping me prep for the GMAT. My colleagues had recommended them over other guides and I continue to do the same after getting the score I needed on my first try. | 677.169 | 1 |
Algebra
Algebra is the study of mathematical symbols and the rules for manipulating these symbols. It is a unifying thread of almost all of mathematics.As such, it includes everything from elementary equation solving to the study of abstractions such as groups, rings, and fields. The more basic parts of algebra are called elementary algebra, the more abstract parts are called abstract algebra or modern algebra.
The below are the links to the topics under Algebra. Click on the links below to view details about the topics. | 677.169 | 1 |
Middle to high school students
UMS solves all problems using classroom methods exclusively; presenting solutions in a clear, step by step format – exactly as a teacher would.
The Universal Math Solver will benefit middle- and high-school students by:
Providing help with math homework assignments – is able to solve any problem either in a textbook or assigned by the teacher.
Being a universal self-tutor – can generate problems similar to those in a textbook, allowing the user to practice solving on his/her own.
Being a reliable math coach – the user can compare his/her own solution to the one given by UMS in order to confirm the correct answer and check for errors.
Unlike many other powerful math packages, UMS is very user friendly and does not require extensive training or use of Help files due to its simple interface. UMS uses the most straightforward method of instruction: "Do as I Do!"
This is how the software solves an exponential equation:
UMS will produce the solution step by step, on the monitor, accompanying the solving process with a professional teacher's voice.
UMS is the dream of every student! UMS is very easy to use – just type in your math problem and push the green button | 677.169 | 1 |
WCA Book - College Algebra
1.
College Algebra
Version
√
3 = 1.7320508075688772 . . .
by
Carl Stitz, Ph.D. Jeff Zeager, Ph.D.
Lakeland Community College Lorain County Community College
Modified by
Joel Robbin and Mike Schroeder
University of Wisconsin, Madison
June 29, 2010
6.
Preface
This book is a modified version of the Open Source Precalculus Project initiated by Carl Stitz and
Jeff Seager. The original version is available at
As indicated on that website you may go to
to order a low-cost, royalty free printed version of the book from lulu.com. Neither author receives
royalties from lulu.com, and, in most cases, it is far cheaper to purchase the printed version from
lulu than to print out the entire book at home.
The version you are viewing was modified by Joel Robbin and Mike Schroeder for use in
Math 112 at the University of Wisconsin Madison. A companion workbook for the course is being
published by Kendall Hunt Publishing Co. 4050 Westmark Drive, Dubuque, IA 52002. Neither Joel
Robbin nor Mike Schroeder nor anyone else at the University of Wisconsin receives any royalties
from sales of the workbook to UW students.
The original version of this book contains the following acknowledgements:
The authors are indebted to the many people who support this project. From Lake-
land Community College, we wish to thank the following people: Bill Previts, who
not only class tested the book but added an extraordinary amount of exercises to it;
Rich Basich and Ivana Gorgievska, who class tested and promoted the book; Don An-
than and Ken White, who designed the electric circuit applications used in the text;
Gwen Sevits, Assistant Bookstore Manager, for her patience and her efforts to get the
book to the students in an efficient and economical fashion; Jessica Novak, Marketing
and Communication Specialist, for her efforts to promote the book; Corrie Bergeron,
Instructional Designer, for his enthusiasm and support of the text and accompanying
YouTube videos; Dr. Fred Law, Provost, and the Board of Trustees of Lakeland Com-
munity College for their strong support and deep commitment to the project. From
Lorain County Community College, we wish to thank: Irina Lomonosov for class testing
the book and generating accompanying PowerPoint slides; Jorge Gerszonowicz, Kathryn
Arocho, Heather Bubnick, and Florin Muscutariu for their unwavering support of the
project; Drs. Wendy Marley and Marcia Ballinger, Lorain CCC, for the Lorain CCC
vii
7.
viii Table of Contents
enrollment data used in the text. We would also like to extend a special thanks to
Chancellor Eric Fingerhut and the Ohio Board of Regents for their support and promo-
tion of the project. Last, but certainly not least, we wish to thank Dimitri Moonen, our
dear friend from across the Atlantic, who took the time each week to e-mail us typos
and other corrections.
8.
Chapter 0
Basic Algebra
0.1 The Laws of Algebra
Terminology and Notation. In this section we review the notations used in algebra. Some
are peculiar to this book. For example the notation A := B indicates that the equality holds
by definition of the notations involved. Two other notations which will become important when
we solve equations are =⇒ and ⇐⇒ . The notation P =⇒ Q means that P implies Q i.e.
"If P, then Q". For example, x = 2 =⇒ x2 = 4. (Note however that the converse statement
x2 = 4 =⇒ x = 2 is not always true since it might be that x = −2.) The notation P ⇐⇒ Q
means P =⇒ Q and Q =⇒ P, i.e. "P if and only if Q". For example 3x − 6 = 0 ⇐⇒ x = 2.
The notations =⇒ and ⇐⇒ are explained more carefully in Section 0.5 below.
Implicit Multiplication. In mathematics the absence of an operation symbol usually indicates
multiplication: ab mean a × b. Sometimes a dot is used to indicate multiplication and in computer
languages an asterisk is often used.
ab := a · b := a ∗ b := a × b
Order of operations. Parentheses are used to indicate the order of doing the operations: in
evaluating an expression with parentheses the innermost matching pairs are evaluated first as in
((1 + 2)2
+ 5)2
= (32
+ 5)2
= (9 + 5)2
= 142
= 196.
There are conventions which allow us not to write the parentheses. For example, multiplication is
done before addition
ab + c means (ab) + c and not a(b + c),
and powers are done before multiplication:
ab2
c means a(b2)c and not (ab)2c.
In the absence of other rules and parentheses, the left most operations are done first.
a − b − c means (a − b) − c and not a − (b − c).
9.
2 Basic Algebra
The long fraction line indicates that the division is done last:
a + b
c
means (a + b)/c and not a + (b/c).
In writing fractions the length of the fraction line indicates which fraction is evaluated first:
a
b
c
means a/(b/c) and not (a/b)/c,
a
b
c
means (a/b)/c and not a/(b/c).
The length of the horizontal line in the radical sign indicates the order of evaluation:
√
a + b means (a + b) and not (
√
a) + b.
√
a + b means (
√
a) + b and not (a + b).
The Laws of Algebra. There are four fundamental operations which can be performed on
numbers.
1. Addition. The sum of a and b is denoted a + b.
2. Multiplication. The product of a and b is denoted ab.
3. Reversing the sign. The negative of a is denoted −a.
4. Inverting. The reciprocal of a (for a = 0) is denoted by a−1 or by
1
a
.
These operations satisfy the following laws.
Associative a + (b + c) = (a + b) + c a(bc) = (ab)c
Commutative a + (b + c) = (a + b) + c a(bc) = (ab)c
Identity a + 0 = 0 + a = a a · 1 = 1 · a = a
Inverse a + (−a) = (−a) + a = 0 a · a−1 = a−1 · a = 1
Distributive a(b + c) = ab + ac (a + b)c = ac + bc
The operations of subtraction and division are then defined by
a − b := a + (−b) a ÷ b :=
a
b
:= a · b−1
= a ·
1
b
.
13.
6 Basic Algebra
because when A = 2 and B = 3 we have (A + B)2 = (2 + 3)2 = 52 = 25 but A2 + B2 = 22 + 32 =
4 + 9 = 13 and 25 = 13. The correct general law is
(A + B)2
= A2
+ 2AB + B2
(when A = 2 and B = 3 A2 + 2AB + B2 = 4 + 12 + 9 = 25 = 52 = (2 + 3)2) and this shows that
(A + B)2 = A2 + B2 =⇒ 2AB = 0 so that (A + B)2 = A2 + B2 only when A = 0 or B = 0 (or
both).
0.2 Kinds of Numbers
We distinguish the following different kinds of numbers.
ˆ The natural numbers are 1, 2, 3 . . ..
ˆ The integers are . . . − 3, −2, −1, 0, 1, 2, 3 . . ..
ˆ The rational numbers are ratios of integers like 3/2, 14/99, −1/2.
ˆ The real numbers are numbers which have an infinite decimal expansion like
3
2
= 1.5000 . . . ,
14
99
= 0.141414 . . . ,
√
2 = 1.4142135623730951 . . . .
ˆ The complex numbers are those numbers of form z = x+iy where x and y are real numbers
and i is a special new number called the imaginary unit which has the property that
i2
= −1;
Every integer is a rational number (because n = n/1), every rational number is a real number (see
Remark 0.2 below), and every real number is a complex number (because x = x + 0i). A real
number which is not rational is called irrational.
New Numbers - New Solutions. Each kind of number enables us to solve equations that the
previous kind couldn't solve:
ˆ The solution of the equation x + 5 = 3 is x = −2 which is an integer but not a natural
number.
ˆ The solution of the equation 5x = 3 is x = 3
5 which is a rational number but not an integer.
ˆ The equation x2 = 2 has two solutions x =
√
2. The number
√
2 is a real number but not a
rational number.
ˆ The equation x2 = 4 has two real solutions x = ±2 but the equation z2 = −4 has no real
solutions because the square of a nonzero real number is always positive. However it does
have two complex solutions, namely z = ±2i.
14.
0.2 Kinds of Numbers 7
We will not use complex numbers until Chapter 10 but may refer to them implicitly as in
The equation x2 = −4 has no (real) solution.
Rational Numbers - Repeating Decimals. It will be proved in Theorem 9.3 that a real
number is rational if and only if its decimal expansion eventually repeats periodically forever as in
the following examples:
1
3
= 0.3333 . . . ,
17
6
= 2.83333 . . . ,
7
4
= 1.250000 . . . ,
22
7
= 3.142857 142857 142857 . . . .
Unless the decimal expansion of a real number is eventually zero, as in 1
2 = 0.5000 . . ., any finite
part of the decimal expansion is close to, but not exactly equal to, the real number. For example
1.414 is close to the square root of two but not exactly equal:
(1.414)2
= 1.999396 = 2, (
√
2)2
= 2.
If we compute the square root to more decimal places we get a better approximation, but it still
isn't exactly correct:
(1.4142135623730951)2
= 2.00000000000000014481069235364401.
The square root of 2 is Irrational. Here is a proof that
√
2 is irrational. If it were rational
there would be integers m and n with
m
n
2
= 2.
By canceling common factors we may assume that m and n have no common factors and hence
that they are not both even. Now m2 = 2n2 so m2 is even so m is even, say m = 2p. Then
4p2 = (2p)2 = m2 = 2n2 so 2p2 = n2 so n2 is even so n is even. This contradicts the fact m and n
are not both even.
The Number Line. The choice of two points (representing 0 and 1) on a line determines a
correspondence between the points of the line and the real numbers as indicated in the following
picture.
−5 −4 −3 −2 −1 0 1 2 3 4 5
The correspondence is called a coordinate system on the line. The line is called a number line.
When the point A corresponds to the number a we say that the number a is the coordinate of
the point A. The positive numbers are the real numbers on the same side of 0 as 1 and the
negative numbers are on the other side. We usually draw the number line as above so that it is
15.
8 Basic Algebra
horizontal and 1 is to the right of 0. We write say a is less than b and write a < b b is to the right
of a, i.e. when b − a is positive. it is equivalent to say that b is greater than a or a to the left of
b and to write b > a. The notation a ≤ b means that a is less than or equal to b i.e. either a < b
or else a = b. Similarly, b ≥ a means that b is greater than or equal to a i.e. either b > a or else
b = a. Thus when a < b, a number 4
c is between a and b ⇐⇒ a < c < b.
Sometimes we insert the word strictly for emphasis: a is strictly less than b means that a < b (not
just a ≤ b).
Order. The order relation just described is characterized by the following.
(Trichotomy) Every real number is either positive, negative, or zero (and no number satisfies two
of these conditions).
(Sum) The sum of two positive numbers is positive.
(Product) The product of two positive numbers is positive.
This characterization together with the notation explained in the previous paragraph implies the
following:
(i) Either a < b, a = b, or a > b.
(ii) If a < b and b < c, then a < c.
(iii) If a < b, then a + c < b + c.
(iv) If a < b and c > 0, then ac < bc.
(v) If a < b and c < 0, then ac > bc.
(vi) If 0 < a < b, then 0 <
1
b
<
1
a
.
Interval Notation. The open interval (a, b) is the set of all real numbers x such that a < x < b,
and the closed interval [a, b] is the set of all real numbers x such that a ≤ x ≤ b. Thus
x is in the set (a, b) ⇐⇒ a < x < b
and
x is in the set [a, b] ⇐⇒ a ≤ x ≤ b.
4
The notation ⇐⇒ is an abbreviation for "if and only if".
16.
0.3 Exponents 9
These notations are extended to include half open intervals and unbounded intervals as in
x is in the set (a, b] ⇐⇒ a < x ≤ b,
x is in the set (a, ∞) ⇐⇒ a < x,
x is in the set (−∞, a] ⇐⇒ x ≤ a, etc.
The union symbol ∪ is used to denote a set consisting of more than one interval as in
x is in the set (a, b) ∪ (c, ∞) ⇐⇒ either a < x < b or else c < x.
The symbol ∞ is pronounced infinity and is used to indicate that an interval is unbounded. It is
not a number so we never write (c, ∞].
Example 0.2.1. Which is bigger: π or
√
10? (Don't use a calculator.)
Solution. π = 3.14 . . . < 3.15. and
3.152
= (3 + 0.15)2
= 32
+ 2 × 3 × 0.15 + 0.152
= 9 + 0.90 + 0.0225 = 9.9225 < 10
so π < 3.15 <
√
10.
0.3 Exponents
The proof of the following theorem requires a more careful definition of the set of real numbers
than we have given and is best left for more advanced courses.
Theorem 0.1. Suppose that a is a positive real number. Then there is one and only one way
to define ax for all real numbers x such that
(i) ax+y = ax · ay, a0 = 1, a1 = a, 1x = 1.
(ii) If a > 1 and x < y then ax < ay.
(iii) If a < 1 and x < y then ax > ay.
With this definition, the laws of exponents in Paragraph 0.1 continue to hold when a and b
are positive real numbers and m and n are arbitrary real numbers. The number ax is positive
(when a is positive) regardless of the sign of x.
In particular by property (v) in Paragraph 0.1 we have (ax)y = axy so (am/n)n = am and
(am)1/n = am/n. Hence for positive numbers a and b we have
b = am/n
⇐⇒ bn
= am
.
17.
10 Basic Algebra
When m = 1 and n is a natural number the number a1/n is called the nth root (square root if
n = 2 and cube root if n = 3) and is sometimes denoted
n
√
a := a1/n
.
When n is absent, n = 2 is understood:
√
a := a1/2
.
nth roots A number b is said to be an nth root of a iff bn = a. When n is odd, every real number
a has exactly one (real) nth root and this is denoted by n
√
a. When n is even, a positive real number
a has two (real) nth roots (and n
√
a denotes the one which is positive) but a negative number has
no real nth roots. (In trigonometry it is proved that every nonzero complex number has exactly n
distinct complex nth roots.)
The equation b2 = 9 has two solutions, namely b = 3 and b = −3 and each is "a" square root of
9 but only b = 3 is "the" square root of 9. However −2 is the (only) real cube root of −8 because
(−2)3 = −8. The number −9 has no real square root (because b2 = (−b)2 > 0 if b = 0) but does
have two complex square roots (because (3i)2 = (−3i)2 = −9). For most of this book5 we only use
real numbers and we say that
√
a is undefined when a < 0
and that
you can't take the square root of a negative number.
Also
√
a always denotes the nonnegative square root: thus (−3)2 = 32 = 9 but
√
9 = 3 and√
9 = −3.
0.4 Absolute Value
There are a few ways to describe what is meant by the absolute value |x| of a real number x. You
may have been taught that |x| is the distance from the real number x to the 0 on the number. So,
for example, |5| = 5 and | − 5| = 5, since each is 5 units from 0 on the number line.
distance is 5 units distance is 5 units
−5 −4 −3 −2 −1 0 1 2 3 4 5
Another way to define absolute value is by the equation |x| =
√
x2. Using this definition, we
have |5| = (5)2 =
√
25 = 5 and | − 5| = (−5)2 =
√
25 = 5. The long and short of both of these
procedures is that |x| takes negative real numbers and assigns them to their positive counterparts
5
More precisely until Chapter 10
18.
0.4 Absolute Value 11
while it leaves positive numbers alone. This last description is the one we shall adopt, and is
summarized in the following definition.
Definition 0.2. The absolute value of a real number x, denoted |x|, is given by
|x| =
−x, if x < 0
x, if x ≥ 0
In Definition 0.2, we define |x| using a piecewise-defined function. (See page 93 in Section 2.2.)
To check that this definition agrees with what we previously understood as absolute value, note
that since 5 ≥ 0, to find |5| we use the rule |x| = x, so |5| = 5. Similarly, since −5 < 0, we use the
rule |x| = −x, so that | − 5| = −(−5) = 5. This is one of the times when it's best to interpret the
expression '−x' as 'the opposite of x' as opposed to 'negative x.' Before we embark on studying
absolute value functions, we remind ourselves of the properties of absolute value.
Theorem 0.2. Properties of Absolute Value: Let a, b, and x be real numbers and let n be
an integer.a Then
ˆ Product Rule: |ab| = |a||b|
ˆ Power Rule: |an| = |a|n whenever an is defined
ˆ Quotient Rule:
a
b
=
|a|
|b|
, provided b = 0
ˆ The Triangle Inequality: |a + b| ≤ |a| + |b|
ˆ |x| = 0 if and only if x = 0.
ˆ For c > 0, |x| = c if and only if x = c or x = −c.
ˆ For c < 0, |x| = c has no solution.
a
Recall that this means n = 0, ±1, ±2, . . . .
The proof of the Product and Quotient Rules in Theorem 0.2 boils down to checking four cases:
when both a and b are positive; when they are both negative; when one is positive and the other
is negative; when one or both are zero. For example, suppose we wish to show |ab| = |a||b|. We
need to show this equation is true for all real numbers a and b. If a and b are both positive, then
so is ab. Hence, |a| = a, |b| = b, and |ab| = ab. Hence, the equation |ab| = |a||b| is the same as
19.
12 Basic Algebra
ab = ab which is true. If both a and b are negative, then ab is positive. Hence, |a| = −a, |b| = −b,
and |ab| = ab. The equation |ab| = |a||b| becomes ab = (−a)(−b), which is true. Suppose a is
positive and b is negative. Then ab is negative, and we have |ab| = −ab, |a| = a and |b| = −b.
The equation |ab| = |a||b| reduces to −ab = a(−b) which is true. A symmetric argument shows the
equation |ab| = |a||b| holds when a is negative and b is positive. Finally, if either a or b (or both)
are zero, then both sides of |ab| = |a||b| are zero, and so the equation holds in this case, too. All
of this rhetoric has shown that the equation |ab| = |a||b| holds true in all cases. The proof of the
Quotient Rule is very similar, with the exception that b = 0. The Power Rule can be shown by
repeated application of the Product Rule. The last three properties can be proved using Definition
0.2 and by looking at the cases when x ≥ 0, in which case |x| = x, or when x < 0, in which case
|x| = −x. For example, if c > 0, and |x| = c, then if x ≥ 0, we have x = |x| = c. If, on the other
hand, x < 0, then −x = |x| = c, so x = −c. The remaining properties are proved similarly and are
left as exercises.
0.5 Solving Equations
Definition 0.3. A number a is called a solution of an equation containing the variable x if
the equation becomes a true statement when a is substituted for x. A solution of an equation is
sometimes also called a root of the equation. Two equations are said to be equivalent iff they
have exactly the same solutions. We will sometimes use the symbol ⇐⇒ to indicate that two
equations are equivalent.
Usually two equations are equivalent because one can be obtained from the other by performing
an operation to both sides of the equation which can be reversed by another operation of the same
kind. For example, the equations 3x + 7 = 13 and x = 2 are equivalent because
3x + 7 = 13 ⇐⇒ 3x = 6 (subtract 7 from both sides),
⇐⇒ x = 2 (divide both sides by 3).
The reasoning is reversible: we can go from x = 2 to 3x = 6 by multiplying both sides by by 3 and
from 3x = 6 to 3x + 7 = 13 by adding 7 to both sides.
We use the symbol =⇒ when we want to assert that one equation implies another but do not
want to assert the converse. The guiding principal here is
If an equation E results from an equation E by performing the same operation to both
sides, then E =⇒ E , i.e. every solution of E is a solution of E .
If the operation is not "reversible" as explained above, there is the possibility that the set of
solutions gets bigger in which case the new solutions are called extraneous solutions. (They do
not solve the original equation.) The simplest example of how an extraneous solution can arise is
x = 3 =⇒ x2
= 9 (square both sides)
but the operation of squaring both sides is not reversible: it is incorrect to conclude that x2 = 9
implies that x = 3. What is correct is that x2 = 9 ⇐⇒ x = ±3, i.e. either x = 3 or else x = −3.
When solving an equation you may use operations which are not reversible provided that you
22.
Chapter 1
Coordinates
1.1 The Cartesian Coordinate Plane
In order to visualize the pure excitement that is Algebra, we need to unite Algebra and Geometry.
Simply put, we must find a way to draw algebraic things. Let's start with possibly the greatest
mathematical achievement of all time: the Cartesian Coordinate Plane.1 Imagine two real
number lines crossing at a right angle at 0 as below.
x
y
−4 −3 −2 −1 1 2 3 4
−4
−3
−2
−1
1
2
3
4
The horizontal number line is usually called the x-axis while the vertical number line is usually
called the y-axis.2 As with the usual number line, we imagine these axes extending off indefinitely
in both directions. Having two number lines allows us to locate the position of points off of the
number lines as well as points on the lines themselves.
1
So named in honor of Ren´e Descartes.
2
The labels can vary depending on the context of application.
23.
16 Coordinates
For example, consider the point P below on the left. To use the numbers on the axes to label
this point, we imagine dropping a vertical line from the x-axis to P and extending a horizontal line
from the y-axis to P. We then describe the point P using the ordered pair (2, −4). The first
number in the ordered pair is called the abscissa or x-coordinate and the second is called the
ordinate or y-coordinate.3 Taken together, the ordered pair (2, −4) comprise the Cartesian
coordinates of the point P. In practice, the distinction between a point and its coordinates is
blurred; for example, we often speak of 'the point (2, −4).' We can think of (2, −4) as instructions
on how to reach P from the origin by moving 2 units to the right and 4 units downwards. Notice
that the order in the ordered pair is important − if we wish to plot the point (−4, 2), we would
move to the left 4 units from the origin and then move upwards 2 units, as below on the right.
x
y
P
−4 −3 −2 −1 1 2 3 4
−4
−3
−2
−1
1
2
3
4
x
y
P(2, −4)
(−4, 2)
−4 −3 −2 −1 1 2 3 4
−4
−3
−2
−1
1
2
3
4
Example 1.1.1. Plot the following points: A(5, 8), B −5
2, 3 , C(−5.8, −3), D(4.5, −1), E(5, 0),
F(0, 5), G(−7, 0), H(0, −9), O(0, 0).4
Solution. To plot these points, we start at the origin and move to the right if the x-coordinate is
positive; to the left if it is negative. Next, we move up if the y-coordinate is positive or down if it
is negative. If the x-coordinate is 0, we start at the origin and move along the y-axis only. If the
y-coordinate is 0 we move along the x-axis only.
3
Again, the names of the coordinates can vary depending on the context of the application. If, for example, the
horizontal axis represented time we might choose to call it the t-axis. The first number in the ordered pair would
then be the t-coordinate.
4
The letter O is almost always reserved for the origin.
24.
1.1 The Cartesian Coordinate Plane 17
x
y
A(5, 8)
B −5
2, 3
C(−5.8, −3)
D(4.5, −1)
E(5, 0)
F(0, 5)
G(−7, 0)
H(0, −9)
O(0, 0)When we speak of the Cartesian Coordinate Plane, we mean the set of all possible ordered pairs
(x, y) as x and y take values from the real numbers. Below is a summary of important facts about
Cartesian coordinates.
Important Facts about the Cartesian Coordinate Plane
ˆ (a, b) and (c, d) represent the same point in the plane if and only if a = c and b = d.
ˆ (x, y) lies on the x-axis if and only if y = 0.
ˆ (x, y) lies on the y-axis if and only if x = 0.
ˆ The origin is the point (0, 0). It is the only point common to both axes.
25.
18 Coordinates
The axes divide the plane into four regions called quadrants. They are labeled with Roman
numerals and proceed counterclockwise around the plane:
x
y
Quadrant I
x > 0, y > 0
Quadrant II
x < 0, y > 0
Quadrant III
x < 0, y < 0
Quadrant IV
x > 0, y < 0
−4 −3 −2 −1 1 2 3 4
−4
−3
−2
−1
1
2
3
4
For example, (1, 2) lies in Quadrant I, (−1, 2) in Quadrant II, (−1, −2) in Quadrant III, and
(1, −2) in Quadrant IV. If a point other than the origin happens to lie on the axes, we typically
refer to the point as lying on the positive or negative x-axis (if y = 0) or on the positive or negative
y-axis (if x = 0). For example, (0, 4) lies on the positive y-axis whereas (−117, 0) lies on the
negative x-axis. Such points do not belong to any of the four quadrants.
One of the most important concepts in all of mathematics is symmetry.There are many types
of symmetry in mathematics, but three of them can be discussed easily using Cartesian Coordinates.
Definition 1.1. Two points (a, b) and (c, d) in the plane are said to be
ˆ symmetric about the x-axis if a = c and b = −d
ˆ symmetric about the y-axis if a = −c and b = d
ˆ symmetric about the origin if a = −c and b = −d
Schematically,
26.
1.1 The Cartesian Coordinate Plane 19
0 x
y
P(x, y)Q(−x, y)
S(x, −y)R(−x, −y)
In the above figure, P and S are symmetric about the x-axis, as are Q and R; P and Q are
symmetric about the y-axis, as are R and S; and P and R are symmetric about the origin, as are
Q and S.
Example 1.1.2. Let P be the point (−2, 3). Find the points which are symmetric to P about the:
1. x-axis 2. y-axis 3. origin
Check your answer by graphing.
Solution. The figure after Definition 1.1 gives us a good way to think about finding symmetric
points in terms of taking the opposites of the x- and/or y-coordinates of P(−2, 3).
1. To find the point symmetric about the x-axis, we replace the y-coordinate with its opposite
to get (−2, −3).
2. To find the point symmetric about the y-axis, we replace the x-coordinate with its opposite
to get (2, 3).
3. To find the point symmetric about the origin, we replace the x- and y-coordinates with their
opposites to get (2, −3).
x
y
P(−2, 3)
(−2, −3)
(2, 3)
(2, −3)
−3 −2 −1 1 2 3
−3
−2
−1
1
2
3
27.
20 Coordinates
One way to visualize the processes in the previous example is with the concept of reflections.
If we start with our point (−2, 3) and pretend the x-axis is a mirror, then the reflection of (−2, 3)
across the x-axis would lie at (−2, −3). If we pretend the y-axis is a mirror, the reflection of (−2, 3)
across that axis would be (2, 3). If we reflect across the x-axis and then the y-axis, we would go
from (−2, 3) to (−2, −3) then to (2, −3), and so we would end up at the point symmetric to (−2, 3)
about the origin. We summarize and generalize this process below.
Reflections
To reflect a point (x, y) about the:
ˆ x-axis, replace y with −y.
ˆ y-axis, replace x with −x.
ˆ origin, replace x with −x and y with −y.
1.1.1 Distance in the Plane
Another important concept in geometry is the notion of length. If we are going to unite Algebra
and Geometry using the Cartesian Plane, then we need to develop an algebraic understanding of
what distance in the plane means. Suppose we have two points, P (x1, y1) and Q (x2, y2) , in the
plane. By the distance d between P and Q, we mean the length of the line segment joining P with
Q. (Remember, given any two distinct points in the plane, there is a unique line containing both
points.) Our goal now is to create an algebraic formula to compute the distance between these two
points. Consider the generic situation below on the left.
P (x1, y1)
Q (x2, y2)
d
P (x1, y1)
Q (x2, y2)
d
(x2, y1)
With a little more imagination, we can envision a right triangle whose hypotenuse has length
d as drawn above on the right. From the latter figure, we see that the lengths of the legs of the
triangle are |x2 − x1| and |y2 − y1| so the Pythagorean Theorem gives us
|x2 − x1|2
+ |y2 − y1|2
= d2
28.
1.1 The Cartesian Coordinate Plane 21
(x2 − x1)2
+ (y2 − y1)2
= d2
(Do you remember why we can replace the absolute value notation with parentheses?) By extracting
the square root of both sides of the second equation and using the fact that distance is never
negative, we get
Equation 1.1. The Distance Formula: The distance d between the points P (x1, y1) and
Q (x2, y2) is:
d = (x2 − x1)2
+ (y2 − y1)2
It is not always the case that the points P and Q lend themselves to constructing such a triangle.
If the points P and Q are arranged vertically or horizontally, or describe the exact same point, we
cannot use the above geometric argument to derive the distance formula. It is left to the reader to
verify Equation 1.1 for these cases.
Example 1.1.3. Find and simplify the distance between P(−2, 3) and Q(1, −3).
Solution.
d = (x2 − x1)2
+ (y2 − y1)2
= (1 − (−2))2 + (−3 − 3)2
=
√
9 + 36
= 3
√
5
So, the distance is 3
√
5.
Example 1.1.4. Find all of the points with x-coordinate 1 which are 4 units from the point (3, 2).
Solution. We shall soon see that the points we wish to find are on the line x = 1, but for now
we'll just view them as points of the form (1, y). Visually,
30.
1.1 The Cartesian Coordinate Plane 23
P (x1, y1)
Q (x2, y2)
M
If we think of reaching M by going 'halfway over' and 'halfway up' we get the following formula.
Equation 1.2. The Midpoint Formula: The midpoint M of the line segment connecting
P (x1, y1) and Q (x2, y2) is:
M =
x1 + x2
2
,
y1 + y2
2
If we let d denote the distance between P and Q, we leave it as an exercise to show that the
distance between P and M is d/2 which is the same as the distance between M and Q. This suffices
to show that Equation 1.2 gives the coordinates of the midpoint.
Example 1.1.5. Find the midpoint of the line segment connecting P(−2, 3) and Q(1, −3).
Solution.
M =
x1 + x2
2
,
y1 + y2
2
=
(−2) + 1
2
,
3 + (−3)
2
= −
1
2
,
0
2
= −
1
2
, 0
The midpoint is −
1
2
, 0 .
An interesting application5 of the midpoint formula follows.
5
This is a key concept in the development of inverse functions. See Section 6.2
31.
24 Coordinates
Example 1.1.6. Prove that the points (a, b) and (b, a) are symmetric about the line y = x.
Solution. By 'symmetric about the line y = x', we mean that if a mirror were placed along
the line y = x, the points (a, b) and (b, a) would be mirror images of one another. (You should
compare and contrast this with the other types of symmetry presented back in Definition 1.1.)
Schematically,
(a, b)
(b, a)
∗
y = x
From the figure, we see that this problem amounts to showing that the midpoint of the line
segment connecting (a, b) and (b, a) lies on the line y = x. Applying Equation 1.2 yields
M =
a + b
2
,
b + a
2
=
a + b
2
,
a + b
2
Since the x and y coordinates of this point are the same, we find that the midpoint lies on the line
y = x, as required.
32.
1.1 The Cartesian Coordinate Plane 25
1.1.2 Exercises
1. Plot and label the points A(−3, −7), B(1.3, −2), C(π,
√
10), D(0, 8), E(−5.5, 0), F(−8, 4),
G(9.2, −7.8) and H(7, 5) in the Cartesian Coordinate Plane given below.
x
y2. For each point given in Exercise 1 above
ˆ Identify the quadrant or axis in/on which the point lies.
ˆ Find the point symmetric to the given point about the x-axis.
ˆ Find the point symmetric to the given point about the y-axis.
ˆ Find the point symmetric to the given point about the origin.
33.
26 Coordinates
3. For each of the following pairs of points, find the distance d between them and find the
midpoint M of the line segment connecting them.
(a) (1, 2), (−3, 5)
(b) (3, −10), (−1, 2)
(c)
1
2
, 4 ,
3
2
, −1
(d) −
2
3
,
3
2
,
7
3
, 2
(e)
24
5
,
6
5
, −
11
5
, −
19
5
.
(f)
√
2,
√
3 , −
√
8, −
√
12
(g) 2
√
45,
√
12 ,
√
20,
√
27 .
(h) (0, 0), (x, y)
4. Find all of the points of the form (x, −1) which are 4 units from the point (3, 2).
5. Find all of the points on the y-axis which are 5 units from the point (−5, 3).
6. Find all of the points on the x-axis which are 2 units from the point (−1, 1).
7. Find all of the points of the form (x, −x) which are 1 unit from the origin.
8. Let's assume for a moment that we are standing at the origin and the positive y-axis points
due North while the positive x-axis points due East. Our Sasquatch-o-meter tells us that
Sasquatch is 3 miles West and 4 miles South of our current position. What are the coordinates
of his position? How far away is he from us? If he runs 7 miles due East what would his new
position be?
9. Verify the Distance Formula 1.1 for the cases when:
(a) The points are arranged vertically. (Hint: Use P(a, y1) and Q(a, y2).)
(b) The points are arranged horizontally. (Hint: Use P(x1, b) and Q(x2, b).)
(c) The points are actually the same point. (You shouldn't need a hint for this one.)
10. Verify the Midpoint Formula by showing the distance between P(x1, y1) and M and the
distance between M and Q(x2, y2) are both half of the distance between P and Q.
11. Show that the points A, B and C below are the vertices of a right triangle.
(a) A(−3, 2), B(−6, 4), and C(1, 8) (b) A(−3, 1), B(4, 0) and C(0, −3)
12. Find a point D(x, y) such that the points A(−3, 1), B(4, 0), C(0, −3) and D are the corners
of a square. Justify your answer.
13. The world is not flat.6 Thus the Cartesian Plane cannot possibly be the end of the story.
Discuss with your classmates how you would extend Cartesian Coordinates to represent the
three dimensional world. What would the Distance and Midpoint formulas look like, assuming
those concepts make sense at all?
6
There are those who disagree with this statement. Look them up on the Internet some time when you're bored.
36.
1.2 Relations 29
Throughout this text we will see many different ways to describe relations. In this section we
will focus our attention on describing relations graphically, by means of the list (or roster) method
and algebraically. Depending on the situation, one method may be easier or more convenient to
use than another. Consider the set of points below
(−2, 1)
(4, 3)
(0, −3)
x
y
−4 −3 −2 −1 1 2 3 4
−4
−3
−2
−1
1
2
3
4
These three points constitute a relation. Let us call this relation R. Above, we have a graphical
description of R. Although it is quite pleasing to the eye, it isn't the most portable way to describe
R. The list (or roster) method of describing R simply lists all of the points which belong to R.
Hence, we write: R = {(−2, 1), (4, 3), (0, −3)}.1 The roster method can be extended to describe
infinitely many points, as the next example illustrates.
Example 1.2.1. Graph the following relations.
1. A = {(0, 0), (−3, 1), (4, 2), (−3, 2)}
2. HLS1 = {(x, 3) : −2 ≤ x ≤ 4}
3. HLS2 = {(x, 3) : −2 ≤ x < 4}
4. V = {(3, y) : y is a real number}
Solution.
1. To graph A, we simply plot all of the points which belong to A, as shown below on the left.
2. Don't let the notation in this part fool you. The name of this relation is HLS1, just like the
name of the relation in part 1 was R. The letters and numbers are just part of its name, just
1
We use 'set braces' {} to indicate that the points in the list all belong to the same set, in this case, R.
37.
30 Coordinates
like the numbers and letters of the phrase 'King George III' were part of George's name. The
next hurdle to overcome is the description of HLS1 itself − a variable and some seemingly
extraneous punctuation have found their way into our nice little roster notation! The way
to make sense of the construction {(x, 3) : −2 ≤ x ≤ 4} is to verbalize the set braces {}
as 'the set of' and the colon : as 'such that'. In words, {(x, 3) : −2 ≤ x ≤ 4} is: 'the set
of points (x, 3) such that −2 ≤ x ≤ 4.' The purpose of the variable x in this case is to
describe infinitely many points. All of these points have the same y-coordinate, 3, but the
x-coordinate is allowed to vary between −2 and 4, inclusive. Some of the points which belong
to HLS1 include some friendly points like: (−2, 3), (−1, 3), (0, 3), (1, 3), (2, 3), (3, 3), and
(4, 3). However, HLS1 also contains the points (0.829, 3), −5
6, 3 , (
√
π, 3), and so on. It is
impossible to list all of these points, which is why the variable x is used. Plotting several
friendly representative points should convince you that HLS1 describes the horizontal line
segment from the point (−2, 3) up to and including the point (4, 3).
x
y
−4 −3 −2 −1 1 2 3 4
1
2
3
4
The graph of A
x
y
−4 −3 −2 −1 1 2 3 4
1
2
3
4
The graph of HLS1
3. HLS2 is hauntingly similar to HLS1. In fact, the only difference between the two is that
instead of '−2 ≤ x ≤ 4' we have '−2 ≤ x < 4'. This means that we still get a horizontal line
segment which includes (−2, 3) and extends to (4, 3), but does not include (4, 3) because of
the strict inequality x < 4. How do we denote this on our graph? It is a common mistake to
make the graph start at (−2, 3) end at (3, 3) as pictured below on the left. The problem with
this graph is that we are forgetting about the points like (3.1, 3), (3.5, 3), (3.9, 3), (3.99, 3),
and so forth. There is no real number that comes 'immediately before' 4, and so to describe
the set of points we want, we draw the horizontal line segment starting at (−2, 3) and draw
an 'open circle' at (4, 3) as depicted below on the right.
38.
1.2 Relations 31
x
y
−4 −3 −2 −1 1 2 3 4
1
2
3
4
This is NOT the correct graph of HLS2
x
y
−4 −3 −2 −1 1 2 3 4
1
2
3
4
The graph of HLS2
4. Our last example, V , describes the set of points (3, y) such that y is a real number. All of
these points have an x-coordinate of 3, but the y-coordinate is free to be whatever it wants
to be, without restriction. Plotting a few 'friendly' points of V should convince you that all
the points of V lie on a vertical line which crosses the x-axis at x = 3. Since there is no
restriction on the y-coordinate, we put arrows on the end of the portion of the line we draw
to indicate it extends indefinitely in both directions. The graph of V is below on the left.
x
y
1 2 3 4
−4
−3
−2
−1
1
2
3
4
The graph of V
x
y
−4 −3 −2 −1 1 2 3 4
−4
−3
−2
−1
The graph of y = −2
The relation V in the previous example leads us to our final way to describe relations: alge-
braically. We can simply describe the points in V as those points which satisfy the equation x = 3.
Most likely, you have seen equations like this before. Depending on the context, 'x = 3' could mean
we have solved an equation for x and arrived at the solution x = 3. In this case, however, 'x = 3'
describes a set of points in the plane whose x-coordinate is 3. Similarly, the equation y = −2 in
this context corresponds to all points in the plane whose y-coordinate is −2. Since there are no
restrictions on the x-coordinate listed, we would graph the relation y = −2 as the horizontal line
above on the right. In general, we have the following.
39.
32 Coordinates
Equations of Vertical and Horizontal Lines
ˆ The graph of the equation x = a is a vertical line through (a, 0).
ˆ The graph of the equation y = b is a horizontal line through (0, b).
In the next section, and in many more after that, we shall explore the graphs of equations in
great detail.2 For now, we shall use our final example to illustrate how relations can be used to
describe entire regions in the plane.
Example 1.2.2. Graph the relation: R = {(x, y) : 1 < y ≤ 3}
Solution. The relation R consists of those points whose y-coordinate only is restricted between 1
and 3 excluding 1, but including 3. The x-coordinate is free to be whatever we like. After plotting
some3 friendly elements of R, it should become clear that R consists of the region between the
horizontal lines y = 1 and y = 3. Since R requires that the y-coordinates be greater than 1, but not
equal to 1, we dash the line y = 1 to indicate that those points do not belong to R. Graphically,
x
y
−4 −3 −2 −1 1 2 3 4
1
2
3
4
The graph of R
1.2.1 Exercises
1. Graph the following relations.
(a) {(−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9)}
(b) {(−2, 2), (−2, −1), (3, 5), (3, −4)}
(c) n, 4 − n2 : n = 0, ±1, ±2
(d) 6
k , k : k = ±1, ±2, ±3, ±4, ±5, ±6
2. Graph the following relations.
2
In fact, much of our time in College Algebra will be spent examining the graphs of equations.
3
The word 'some' is a relative term. It may take 5, 10, or 50 points until you see the pattern.
43.
36 Coordinates
(b) B = {(x, y) : x > −2}
(c) C = {(x, y) : y ≥ 0}
(d) D = {(x, y) : −3 < x ≤ 2}
(e) E = {(x, y) : x ≥ 0,y ≥ 0}
(f) F = {(x, y) : −4 < x < 5, −3 < y < 2}
4. (a)
x
y
−3 −2 −1
1
2
3
The line x = −2 (b)
x
y
−3 −2 −1
1
2
3
The line y = 3
5. The line x = 0 is the y-axis and the line y = 0 is the x-axis.
1.3 Graphs of Equations
In the previous section, we said that an equation in x and y determines a relation.1 In this section,
we begin to explore this topic in greater detail. The main idea of this section is
The Fundamental Graphing Principle
The graph of an equation is the set of points which satisfy the equation. That is, a point (x, y)
is on the graph of an equation if and only if x and y satisfy the equation.
Example 1.3.1. Determine if (2, −1) is on the graph of x2 + y3 = 1.
Solution. To check, we substitute x = 2 and y = −1 into the equation and see if the equation is
satisfied
(2)2 + (−1)3 ?
= 1
3 = 1
Hence, (2, −1) is not on the graph of x2 + y3 = 1.
We could spend hours randomly guessing and checking to see if points are on the graph of the
equation. A more systematic approach is outlined in the following example.
1
An inequalities in x and y also determines a relation.
45.
38 Coordinates
x
y
−4 −3 −2 −1 1 2 3 4
−3
−2
−1
1
2
3
Don't worry if you don't get all of the little bends and curves just right − Calculus is where the
art of precise graphing takes center stage. For now, we will settle with our naive 'plug and plot'
approach to graphing. If you feel like all of this tedious computation and plotting is beneath you,
then you can reach for a graphing calculator, input the formula as shown above, and graph.2
Of all of the points on the graph of an equation, the places where the graph crosses the axes
hold special significance. These are called the intercepts of the graph. Intercepts come in two
distinct varieties: x-intercepts and y-intercepts. They are defined below.
Definition 1.3. Suppose the graph of an equation is given.
ˆ A point at which a graph meets the y-axis is called an y-intercept of the graph.
In our previous example the graph had two x-intercepts, (−1, 0) and (1, 0), and one y-intercept,
(0, 1). The graph of an equation can have any number of intercepts, including none at all! Since
x-intercepts lie on the x-axis, we can find them by setting y = 0 in the equation. Similarly, since
y-intercepts lie on the y-axis, we can find them by setting x = 0 in the equation. Keep in mind,
intercepts are points and therefore must be written as ordered pairs. To summarize,
2
Remember: At UW we don't allow calculators on exams. But using them intelligently outside of class can be a
great benefit.
46.
1.3 Graphs of Equations 39
Steps for finding the intercepts of the graph of an equation
Given an equation involving x and y:
ˆ the x-intercepts always have the form (x, 0); to find the x-intercepts of the graph, set
y = 0 and solve for x.
ˆ y-intercepts always have the form (0, y); to find the y-intercepts of the graph, set x = 0
and solve for y.
Another fact which you may have noticed about the graph in the previous example is that it
seems to be symmetric about the y-axis. To actually prove this analytically, we assume (x, y) is
a generic point on the graph of the equation. That is, we assume x2 + y3 = 1. As we learned
in Section 1.1, the point symmetric to (x, y) about the y-axis is (−x, y). To show the graph is
symmetric about the y-axis, we need to show that (−x, y) is on the graph whenever (x, y) is. In
other words, we need to show (−x, y) satisfies the equation x2 + y3 = 1 whenever (x, y) does.
Substituting gives
(−x)2 + (y)3 ?
= 1
x2 + y3 = 1
When we substituted (−x, y) into the equation x2 +y3 = 1, we obtained the original equation back
when we simplified. This means (−x, y) satisfies the equation and hence is on the graph. In this
way, we can check whether the graph of a given equation possesses any of the symmetries discussed
in Section 1.1. The results are summarized below.
Steps for testing if the graph of an equation possesses symmetry
To test the graph of an equation for symmetry
ˆ About the y-axis: Substitute (−x, y) into the equation and simplify. If the result is
equivalent to the original equation, the graph is symmetric about the y-axis.
ˆ About the x-axis: Substitute (x, −y) into the equation and simplify. If the result is
equivalent to the original equation, the graph is symmetric about the x-axis.
ˆ About the origin: Substitute (−x, −y) into the equation and simplify. If the result is
equivalent to the original equation, the graph is symmetric about the origin.
48.
1.3 Graphs of Equations 41
Moving along to symmetry, we can immediately dismiss the possibility that the graph is sym-
metric about the y-axis or the origin. If the graph possessed either of these symmetries, then the
fact that (1, 0) is on the graph would mean (−1, 0) would have to be on the graph. (Why?) Since
(−1, 0) would be another x-intercept (and we've found all of these), the graph can't have y-axis or
origin symmetry. The only symmetry left to test is symmetry about the x-axis. To that end, we
substitute (x, −y) into the equation and simplify
(x − 2)2 + y2 = 1
(x − 2)2 + (−y)2 ?
= 1
(x − 2)2 + y2 = 1
Since we have obtained our original equation, we know the graph is symmetric about the x-axis.
This means we can cut our 'plug and plot' time in half: whatever happens below the x-axis is
reflected above the x-axis, and vice-versa. Proceeding as we did in the previous example, we obtain
x
y
1 2 3 4
−2
−1
1
2
A couple of remarks are in order. First, it is entirely possible to choose a value for x which does
not correspond to a point on the graph. For example, in the previous example, if we solve for y as
is our custom, we get:
y = ± 1 − (x − 2)2.
Upon substituting x = 0 into the equation, we would obtain
y = ± 1 − (0 − 2)2 = ±
√
1 − 4 = ±
√
−3,
which is not a real number. This means there are no points on the graph with an x-coordinate
of 0. When this happens, we move on and try another point. This is another drawback of the
'plug-and-plot' approach to graphing equations. Luckily, we will devote much of the remainder
of this book developing techniques which allow us to graph entire families of equations quickly.3
Second, it is instructive to show what would have happened had we tested the equation in the last
3
Without the use of a calculator, if you can believe it!
50.
1.3 Graphs of Equations 43
2. The procedures which we have outlined in the Examples of this section and used in the exer-
cises given above all rely on the fact that the equations were "well-behaved". Not everything
in Mathematics is quite so tame, as the following equations will show you. Discuss with your
classmates how you might approach graphing these equations. What difficulties arise when
trying to apply the various tests and procedures given in this section? For more information,
including pictures of the curves, each curve name is a link to its page at
For a much longer list of fascinating curves, click here.
(a) x3 + y3 − 3xy = 0 Folium of Descartes
(b) x4 = x2 + y2 Kampyle of Eudoxus
(c) y2 = x3 + 3x2 Tschirnhausen cubic
(d) (x2 + y2)2 = x3 + y3 Crooked egg
1.3.2 Answers
1. (a) y = x2 + 1
The graph has no x-intercepts
y-intercept: (0, 1)
x y (x, y)
−2 5 (−2, 5)
−1 2 (−1, 2)
0 1 (0, 1)
1 2 (1, 2)
2 5 (2, 5)
x
y
−2−1 1 2
1
2
3
4
5
The graph is not symmetric about the
x-axis (e.g. (2, 5) is on the graph but
(2, −5) is not)
The graph is symmetric about the
y-axis
The graph is not symmetric about the
origin (e.g. (2, 5) is on the graph but
(−2, −5) is not)
(b) y = x2 − 2x − 8
x-intercepts: (4, 0), (−2, 0)
y-intercept: (0, −8)
56.
1.4 Three Interesting Curves 49
(l) x3y = −4
Re-write as: y = −
4
x3
.
The graph has no x-intercepts
The graph has no y-intercepts
x y (x, y)
−2 1
2 (−2, 1
2)
−1 4 (−1, 4)
−1
2 32 (−1
2, 32)
1
2 −32 (1
2, −32)
1 −4 (1, −4)
2 −1
2 (2, −1
2)
x
y
−2 −1 1 2
−32
−4
4
32
The graph is not symmetric about the
x-axis (e.g. (1, −4) is on the graph but
(1, 4) is not)
The graph is not symmetric about the
y-axis (e.g. (1, −4) is on the graph but
(−1, −4) is not)
The graph is symmetric about the
origin
1.4 Three Interesting Curves
1.4.1 Circles
Recall from geometry that a circle can be determined by fixing a point (called the center) and a
positive number (called the radius) as follows.
Definition 1.4. A circle with center (h, k) and radius r > 0 is the set of all points (x, y) in
the plane whose distance to (h, k) is r.
57.
50 Coordinates
(h, k)
r
(x, y)
From the picture, we see that a point (x, y) is on the circle if and only if its distance to (h, k)
is r. We express this relationship algebraically using the Distance Formula, Equation 1.1, as
r = (x − h)2 + (y − k)2
By squaring both sides of this equation, we get an equivalent equation (since r > 0) which gives us
the standard equation of a circle.
Equation 1.3. The Standard Equation of a Circle: The equation of a circle with center
(h, k) and radius r > 0 is (x − h)2 + (y − k)2 = r2.
Example 1.4.1. Write the standard equation of the circle with center (−2, 3) and radius 5.
Solution. Here, (h, k) = (−2, 3) and r = 5, so we get
(x − (−2))2 + (y − 3)2 = (5)2
(x + 2)2 + (y − 3)2 = 25
Example 1.4.2. Graph (x + 2)2 + (y − 1)2 = 4. Find the center and radius.
Solution. From the standard form of a circle, Equation 1.3, we have that x+2 is x−h, so h = −2
and y − 1 is y − k so k = 1. This tells us that our center is (−2, 1). Furthermore, r2 = 4, so r = 2.
Thus we have a circle centered at (−2, 1) with a radius of 2. Graphing gives us
x
y
−4 −3 −2 −1 1
−1
1
2
3
4
58.
1.4 Three Interesting Curves 51
If we were to expand the equation in the previous example and gather up like terms, instead of
the easily recognizable (x + 2)2 + (y − 1)2 = 4, we'd be contending with x2 + 4x + y2 − 2y + 1 = 0.
If we're given such an equation, we can complete the square in each of the variables to see if it fits
the form given in Equation 1.3 by following the steps given below.
To Put a Circle into Standard Form
1. Group the same variables together on one side of the equation and put the constant on
the other side.
2. Complete the square on both variables as needed.
3. Divide both sides by the coefficient of the squares. (For circles, they will be the same.)
Example 1.4.3. Complete the square to find the center and radius of 3x2 − 6x + 3y2 + 4y − 4 = 0.
Solution.
3x2 − 6x + 3y2 + 4y − 4 = 0
3x2 − 6x + 3y2 + 4y = 4 add 4 to both sides
3 x2 − 2x + 3 y2 +
4
3
y = 4 factor out leading coefficients
3 x2 − 2x + 1 + 3 y2 +
4
3
y +
4
9
= 4 + 3(1) + 3
4
9
complete the square in x, y
3(x − 1)2 + 3 y +
2
3
2
=
25
3
factor
(x − 1)2 + y +
2
3
2
=
25
9
divide both sides by 3
From Equation 1.3, we identify x − 1 as x − h, so h = 1, and y + 2
3 as y − k, so k = −2
3. Hence,
the center is (h, k) = 1, −2
3 . Furthermore, we see that r2 = 25
9 so the radius is r = 5
3.
It is possible to obtain equations like (x − 3)2 + (y + 1)2 = 0 or (x − 3)2 + (y + 1)2 = −1, neither
of which describes a circle. (Do you see why not?) The reader is encouraged to think about what, if
any, points lie on the graphs of these two equations. The next example uses the Midpoint Formula,
Equation 1.2, in conjunction with the ideas presented so far in this section.
59.
52 Coordinates
Example 1.4.4. Write the standard equation of the circle which has (−1, 3) and (2, 4) as the
endpoints of a diameter.
Solution. We recall that a diameter of a circle is a line segment containing the center and two
points on the circle. Plotting the given data yields
x
y
(h, k)
r
−2 −1 1 2 3
1
2
3
4
Since the given points are endpoints of a diameter, we know their midpoint (h, k) is the center
of the circle. Equation 1.2 gives us
(h, k) =
x1 + x2
2
,
y1 + y2
2
=
−1 + 2
2
,
3 + 4
2
=
1
2
,
7
2
The diameter of the circle is the distance between the given points, so we know that half of the
distance is the radius. Thus,
r =
1
2
(x2 − x1)2
+ (y2 − y1)2
=
1
2
(2 − (−1))2 + (4 − 3)2
=
1
2
√
32 + 12
=
√
10
2
Finally, since
√
10
2
2
=
10
4
, our answer becomes x −
1
2
2
+ y −
7
2
2
=
10
4
We close this section with the most important4 circle in all of mathematics: the Unit Circle.
4
While this may seem like an opinion, it is indeed a fact.
60.
1.4 Three Interesting Curves 53
Definition 1.5. The Unit Circle is the circle centered at (0, 0) with a radius of 1. The
standard equation of the Unit Circle is x2 + y2 = 1.
Example 1.4.5. Find the points on the unit circle with y-coordinate
√
3
2
.
Solution. We replace y with
√
3
2
in the equation x2 + y2 = 1 to get
x2
+
√
3
2
2
= 1.
From this we get x2+
3
4
= 1 so x2 =
1
4
so x = ±
1
2
. The points are
1
2
,
√
3
2
and −
1
2
,
√
3
2
. These
two points are the intersection of the horizontal line y =
√
3
2
and the unit circle x2 + y2 = 1,
1.4.2 Parabolas
Definition 1.6. Let F be a point in the plane and D be a line not containing F. The set of
all points equidistant from F and D is called the parabola with focus F and directrix D.
Schematically, we have the following.
F
D
V
Each dashed line from the point F to a point on the curve has the same length as the dashed
line from the point on the curve to the line D. The point suggestively labeled V is, as you should
expect, the vertex. The vertex is the point on the parabola closest to the focus.
We want to use only the distance definition of parabola to derive the equation of a parabola
and, if all is right with the universe, we should get an expression much like those studied in Section
61.
54 Coordinates
3.3. Let p denote the directed5 distance from the vertex to the focus, which by definition is the
same as the distance from the vertex to the directrix. For simplicity, assume that the vertex is
(0, 0) and that the parabola opens upwards. Hence, the focus is (0, p) and the directrix is the line
y = −p. Our picture becomes
(0, p)
x
y
y = −p
(x, y)
(x, −p)
(0, 0)
From the definition of parabola, we know the distance from (0, p) to (x, y) is the same as the
distance from (x, −p) to (x, y). Using the Distance Formula, Equation 1.1, we get
(x − 0)2 + (y − p)2 = (x − x)2 + (y − (−p))2
x2 + (y − p)2 = (y + p)2
x2 + (y − p)2 = (y + p)2 square both sides
x2 + y2 − 2py + p2 = y2 + 2py + p2 expand quantities
x2 = 4py gather like terms
Solving for y yields y = x2
4p , which is a quadratic function of the form found in Equation 3.4
with a = 1
4p and vertex (0, 0).
We know from previous experience that if the coefficient of x2 is negative, the parabola opens
downwards. In the equation y = x2
4p this happens when p < 0. In our formulation, we say that p is
a 'directed distance' from the vertex to the focus: if p > 0, the focus is above the vertex; if p < 0,
the focus is below the vertex. The focal length of a parabola is |p|.
What if we choose to place the vertex at an arbitrary point (h, k)? We can either use transfor-
mations (vertical and horizontal shifts from Section 2.5) or re-derive the equation from Definition
1.6 to arrive at the following.
5
We'll talk more about what 'directed' means later.
62.
1.4 Three Interesting Curves 55
Equation 1.4. The Standard Equation of a Verticala Parabola: The equation of a
(vertical) parabola with vertex (h, k) and focal length |p| is
(x − h)2
= 4p(y − k)
If p > 0, the parabola opens upwards; if p < 0, it opens downwards.
a
That is, a parabola which opens either upwards or downwards.
Notice that in the standard equation of the parabola above, only one of the variables, x, is
squared. This is a quick way to distinguish an equation of a parabola from that of a circle because
in the equation of a circle, both variables are squared.
Example 1.4.6. Graph (x + 1)2 = −8(y − 3). Find the vertex, focus, and directrix.
Solution. We recognize this as the form given in Equation 1.4. Here, x − h is x + 1 so h = −1,
and y −k is y −3 so k = 3. Hence, the vertex is (−1, 3). We also see that 4p = −8 so p = −2. Since
p < 0, the focus will be below the vertex and the parabola will open downwards. The distance from
the vertex to the focus is |p| = 2, which means the focus is 2 units below the vertex. If we start at
(−1, 3) and move down 2 units, we arrive at the focus (−1, 1). The directrix, then, is 2 units above
the vertex and if we move 2 units up from (−1, 3), we'd be on the horizontal line y = 5.
x
y
−6 −5 −4 −3 −2 −1 1 2 3 4
−2
−1
1
2
3
4
5
Of all of the information requested in the previous example, only the vertex is part of the graph
of the parabola. So in order to get a sense of the actual shape of the graph, we need some more
information. While we could plot a few points randomly, a more useful measure of how wide a
parabola opens is the length of the parabola's latus rectum.6 The latus rectum of a parabola
is the line segment parallel to the directrix which contains the focus. The endpoints of the latus
rectum are, then, two points on 'opposite' sides of the parabola. Graphically, we have the following.
6
No, I'm not making this up.
63.
56 Coordinates
F
the latus rectum
D
V
It turns out7 that the length of the latus rectum is |4p|, which, in light of Equation 1.4, is easy
to find. In our last example, for instance, when graphing (x + 1)2 = −8(y − 3), we can use the fact
that the length of the latus rectum is | − 8| = 8, which means the parabola is 8 units wide at the
focus, to help generate a more accurate graph by plotting points 4 units to the left and right of the
focus.
Example 1.4.7. Find the standard form of the parabola with focus (2, 1) and directrix y = −4.
Solution. Sketching the data yields,
x
y
The vertex lies on this vertical line
midway between the focus and the directrix
−1 1 2 3
−3
−2
−1
1
From the diagram, we see the parabola opens upwards. (Take a moment to think about it if you
don't see that immediately.) Hence, the vertex lies below the focus and has an x-coordinate of 2.
To find the y-coordinate, we note that the distance from the focus to the directrix is 1 − (−4) = 5,
which means the vertex lies 5/2 units (halfway) below the focus. Starting at (2, 1) and moving
down 5/2 units leaves us at (2, −3/2), which is our vertex. Since the parabola opens upwards, we
know p is positive. Thus p = 5/2. Plugging all of this data into Equation 1.4 give us
(x − 2)2 = 4
5
2
y − −
3
2
(x − 2)2 = 10 y +
3
2
7
Consider this an exercise to show what follows.
64.
1.4 Three Interesting Curves 57
If we interchange the roles of x and y, we can produce 'horizontal' parabolas: parabolas which
open to the left or to the right. The directrices8 of such animals would be vertical lines and the
focus would either lie to the left or to the right of the vertex. A typical 'horizontal' parabola is
sketched below.
F
D
V
Equation 1.5. The Standard Equation of a Horizontal Parabola: The equation of a
(horizontal) parabola with vertex (h, k) and focal length |p| is
(y − k)2
= 4p(x − h)
If p > 0, the parabola opens to the right; if p < 0, it opens to the left.
Example 1.4.8. Graph (y − 2)2 = 12(x + 1). Find the vertex, focus, and directrix.
Solution. We recognize this as the form given in Equation 1.5. Here, x − h is x + 1 so h = −1,
and y − k is y − 2 so k = 2. Hence, the vertex is (−1, 2). We also see that 4p = 12 so p = 3.
Since p > 0, the focus will be the right of the vertex and the parabola will open to the right. The
distance from the vertex to the focus is |p| = 3, which means the focus is 3 units to the right. If
we start at (−1, 2) and move right 3 units, we arrive at the focus (2, 2). The directrix, then, is 3
units to the left of the vertex and if we move left 3 units from (−1, 2), we'd be on the vertical line
x = −4. Since the length of the latus rectum is |4p| = 12, the parabola is 12 units wide at the
focus, and thus there are points 6 units above and below the focus on the parabola.
8
plural of 'directrix'
65.
58 Coordinates
x
y
−5 −4 −3 −2 −1 1 2 3
−4
−3
−2
−1
1
2
3
4
5
6
7
8
As with circles, not all parabolas will come to us in the forms in Equations 1.4 or 1.5. If we
encounter an equation with two variables in which exactly one variable is squared, we can attempt
to put the equation into a standard form using the following steps.
To Put a Parabola into Standard Form
1. Group the variable which is squared on one side of the equation and put the non-squared
variable and the constant on the other side.
2. Complete the square if necessary and divide by the coefficient of the perfect square.
3. Factor out the coefficient of the non-squared variable from it and the constant.
Example 1.4.9. Consider the equation y2 + 4y + 8x = 4. Put this equation into standard form
and graph the parabola. Find the vertex, focus, and directrix.
Solution. We need to get a perfect square (in this case, using y) on the left-hand side of the
equation and factor out the coefficient of the non-squared variable (in this case, the x) on the
other.
66.
1.4 Three Interesting Curves 59
y2 + 4y + 8x = 4
y2 + 4y = −8x + 4
y2 + 4y + 4 = −8x + 4 + 4 complete the square in y only
(y + 2)2 = −8x + 8 factor
(y + 2)2 = −8(x − 1)
Now that the equation is in the form given in Equation 1.5, we see that x − h is x − 1 so h = 1,
and y − k is y + 2 so k = −2. Hence, the vertex is (1, −2). We also see that 4p = −8 so that
p = −2. Since p < 0, the focus will be the left of the vertex and the parabola will open to the left.
The distance from the vertex to the focus is |p| = 2, which means the focus is 2 units to the left
of 1, so if we start at (1, −2) and move left 2 units, we arrive at the focus (−1, −2). The directrix,
then, is 2 units to the right of the vertex, so if we move right 2 units from (1, −2), we'd be on the
vertical line x = 3. Since the length of the latus rectum is |4p| is 8, the parabola is 8 units wide at
the focus, so there are points 4 units above and below the focus on the parabola.
x
y
−2 −1 1 2
−6
−5
−4
−3
−2
−1
1
2
In studying quadratic functions, we have seen parabolas used to model physical phenomena
such as the trajectories of projectiles. Other applications of the parabola concern its 'reflective
property' which necessitates knowing about the focus of a parabola. For example, many satellite
dishes are formed in the shape of a paraboloid of revolution as depicted below.
67.
60 Coordinates
Every cross section through the vertex of the paraboloid is a parabola with the same focus.
To see why this is important, imagine the dashed lines below as electromagnetic waves heading
towards a parabolic dish. It turns out that the waves reflect off the parabola and concentrate at
the focus which then becomes the optimal place for the receiver. If, on the other hand, we imagine
the dashed lines as emanating from the focus, we see that the waves are reflected off the parabola
in a coherent fashion as in the case in a flashlight. Here, the bulb is placed at the focus and the
light rays are reflected off a parabolic mirror to give directional light.
F
Example 1.4.10. A satellite dish is to be constructed in the shape of a paraboloid of revolution.
If the receiver placed at the focus is located 2 ft above the vertex of the dish, and the dish is to be
12 feet wide, how deep will the dish be?
Solution. One way to approach this problem is to determine the equation of the parabola sug-
gested to us by this data. For simplicity, we'll assume the vertex is (0, 0) and the parabola opens
upwards. Our standard form for such a parabola is x2 = 4py. Since the focus is 2 units above the
vertex, we know p = 2, so we have x2 = 8y. Visually,
68.
1.4 Three Interesting Curves 61
?
(6, y)
y
x
12 units wide
−6 6
2
Since the parabola is 12 feet wide, we know the edge is 6 feet from the vertex. To find the
depth, we are looking for the y value when x = 6. Substituting x = 6 into the equation of the
parabola yields 62 = 8y or y = 36/8 = 9/2 = 4.5. Hence, the dish will be 9/2 or 4.5 feet deep.
1.4.3 Ellipses
In the definition of a circle, Definition 1.4, we fixed a point called the center and considered all
of the points which were a fixed distance r from that one point. For our next conic section, the
ellipse, we fix two distinct points and a distance d to use in our definition.
Definition 1.7. Given two distinct points F1 and F2 in the plane and a fixed distance d, an
ellipse is the set of all points (x, y) in the plane such that the sum of the distance from F1 to
(x, y) and the distance from F2 to (x, y) is d. The points F1 and F2 are called the focia of the
ellipse.
a
the plural of 'focus'
(x, y)
d1 d2
F1 F2
d1 + d2 = d for all (x, y) on the ellipse
69.
62 Coordinates
We may imagine taking a length of string and anchoring it to two points on a piece of paper.
The curve traced out by taking a pencil and moving it so the string is always taut is an ellipse.
The center of the ellipse is the midpoint of the line segment connecting the two foci. The
major axis of the ellipse is the line segment connecting two opposite ends of the ellipse which
also contains the center and foci. The minor axis of the ellipse is the line segment connecting
two opposite ends of the ellipse which contains the center but is perpendicular to the major axis.
The vertices of an ellipse are the points of the ellipse which lie on the major axis. Notice that the
center is also the midpoint of the major axis, hence it is the midpoint of the vertices. In pictures
we have,
F1 F2
V2V1
C
Major Axis
MinorAxis
An ellipse with center C; foci F1, F2; and vertices V1, V2
Note that the major axis is the longer of the two axes through the center, and likewise, the
minor axis is the shorter of the two. In order to derive the standard equation of an ellipse, we
assume that the ellipse has its center at (0, 0), its major axis along the x-axis, and has foci (c, 0)
and (−c, 0) and vertices (−a, 0) and (a, 0). We will label the y-intercepts of the ellipse as (0, b) and
(0, −b) (We assume a, b, and c are all positive numbers.) Schematically, | 677.169 | 1 |
computational and mathematical procedures to solve problems or to enhance given data. It encompasses working safely, applying knowledge of undertaking computations in electrotechnology environment.
Basic concepts • Definition of the derivative of a function as the slope of a tangent line (the gradient of a curve); • limits; basic examples from 1st principles; • Notation and Results of derivative of k.f(ax + b) where f(x)=x to the power of n, sin x, cos x, tan x, • e to the power of x, ln x.
Assignment 2 (Part B, Computer Lab )worth 10% of total mark) handed out. Due date last day of week 35.
• The definition of Antiderivatives • Integration as the inverse operation to differentiation (Examples are results of the integral of k.f(ax + b) where f(x) = x to the power of n, sin x, cos x, sec squared x, e to the power of x) Integration of particular functions using tables.
Use of the measures of central tendency encompassing: • Estimation of percentiles and deciles from cumulative frequency polygons (ogives) • Interpreting data from tables and graphs including interpolation and extrapolation • Analysing misleading graphs
Progressive assessments will include written and oral demonstration, assignments, tests, projects and computer assisted learning.
Assessment Tasks
Assessment task 1 (assignment 1, Part A & Part B): 20% Written assignment to demonstrate an understanding with applications of mathematical linear and spatial measurement in engineering, trigonometry and basic algebra, linear and quadratic functions involving engineering problems which are covered from week 1 to week 8. This assessment allows students to work as a group which will help to revise and prepare for the next assessment (Tes1) which will cover similar topics.
Assessment task 2 (test 1): 30% This assessment demonstrates an understanding with applications of mathematical linear and spatial measurement in engineering, trigonometry and basic algebra, linear and quadratic functions involving engineering problems which are covered from week 1 to week 8. The time allowed for this test is no more that 2.5 hours.
Assessment task 3 (assignment 2, Part A & Part B ): 20% Written assignment to demonstrate an understanding with applications of differential calculus, integral calculus and problems with engineering applications, statistical data and probability which is covered from week 10 to week 17. Similar to the assignment 1, students can work/study in groups which will help to revise and prepare for the next assessment (Test 2) which will cover similar topics.
Assessment task 4 (test 2): 30% This assessment demonstrates an understanding with applications of differential calculus, integral calculus and problems with engineering applications, statistical data and probability which is covered from week 10 to week 17. The time allowed for this test is no more that 2.5 hours.(Similar to Test 1). | 677.169 | 1 |
Presenting an introduction to the mathematics of modern physics for advanced undergraduate and graduate students, this textbook introduces the reader to modern mathematical thinking within a physics context.
Presenting an introduction to the mathematics of modern physics for advanced undergraduate and graduate students, this textbook introduces the reader to modern mathematical thinking within a physics context. Topics covered include tensor algebra, differential geometry, topology, Lie groups and Lie algebras, distribution theory, fundamental analysis and Hilbert spaces. The book also includes exercises and proofed examples to test the students' understanding of the various concepts, as well as to extend the text's themes | 677.169 | 1 |
axioms. "2)" (used for distinguishing cases) are simply eorems.. The first difficulty-overcome 1. an cases.. for the real numbers
In Chapter
(positive. for the complex numbers. no number. sec Ions. chapters. and zero) .
ter 4. I speak only of
. of "Theorem Li ht Blue" "Theorem Dark Blue" and so on. . definitions.To prevent arguments:
A number. and sections) and also "1) ". negative.
Ie
v
. an are more convenien or purposes of reference than if I were to speak. thus. U to "301 " as a matter of fact. and 5. c ap ers. Ions. say. two cases. all
the like in the case of axioms. there would be difficulty whatever in introducing the so-called positive integers.
I.]
*
.<:1"'<:1.~o l~ th~t -0 -oJ
and "thouing"
1<. .L/. as an exercise in connection with Chapter 1.hemistry) at the tmrversrty tor several semesters already and think that they have learned the differential and integral calculus in College. Landau Vorlesungen iiber Zahlentheorie.
and then prove the theorem.
. [Trans.
11\ I .L/ I I 1\
A
1111 \\\. son of King Louis XIV. =
but I would recommend.
~.
nC"'l
"l"I.
lJel. Professor Landau uses the familiar "du" [thou] throughout this preface. Vol. you haven't actually forgotten them. E. that you make the following definitions: 2= 1
':t
+ 1.] t For Delphin use.L
.dmund Landau
In the original German. and yet they still don't know why
X'U=U'X
T'O
Derl1H.~ 4.'YI1. Forgive me for "theeing"
~on for rrrv ilo-ino. my daughters have been studying \ l_. The Delphin classics were prepared by great French scholars for the use of the Dauphin of France. The multiplication • nULeven tne LneUI'eIIl
.:xrr-ithm '
you.elIlUer
1
"'0. p.L/
11\ I .VI
PREFACE
FOR THE STUDENT
Please keep In mind everywhere the corresponding portions of your school work. § 4. * One rean~rthr-oJ i'YI
1.
table is not to be found in this book. as is well known (cf.-
t'hl~ hook
'1LQ1J. 4. [Trans.
5.nl"\
1.!. V).l.
T'>.
-.
delphinarum:t for.
and most of my students-that even in his first semester the student should learn what the basic facts are. As is well known.fortunately in the majority) who do not share my point of view on While a rigorous and complete exposition of elementary mathematics can not. as mentioned above. were we to adopt a large number of axioms. be expected in the high schools. but only to be almost diametrically opposite to my point o VIew.~. the mathematical courses in colleges and universities should acquaint matics. roe 1 II 0 leagues.l'-'
ceed with this development. these axioms can correct. but also with its methods of proof. from
OUV'. On the other hand. some authors whose writings I have found of help.if one postulates as axioms for real numbers many of the usual rules of arithmetic and the main theorem of this book Theor the consistency of the five Peano axioms (because that can not be ones. of course. the question would immediately occur to the s udent whether some of them could not be proved (a shrewd one would add: or dis rove b me m been known for many decades that all these additional axioms can
Vll
. Even one who studies mathematics mainly for its applications to physics and to other
between non-rigorous and rigorous proof). accepted as axioms.
The larger books which attempt that a bit for the reader to complete. are given t n roof or worse et wi h of this extreme variant of the opposite point of view seem to me
ma even be zero. This is a concession to those hearers who want. being quite simple. for operations with numbers. the theorem that a monotonically decreasing bounded sequence of numbers converges to a limit.
. so that such matters as the mean-value theorem of the differential calculus. in my earlier courses I assumed the properties of the integers and of the rational numbers. after all to do differentiation ri ht awa or who do not want to learn (or perhaps not at all). is the foundation of the natural numbers never failed to cover the (Dedekind) theory of real numbers. In the Foundations of Analysis course
his first semester. or. But the last three times I preferred to I have divided my course into two simultaneous courses one of which has the title "Grundlagen der Analysis" (Foundations of Analysis). can Now in the entire literature there is no textbook which has the sole and modest aim of laying the foundation. in the above sense. say. tion having a zero derivative in some interval is constant in that interval. Only rarely.viii
PREFACE FOR THE TEACHER
I will refrain from speaking at length about the fact that often in the development of the real numbers by means of fundamental sequences) is not included in the basic material. however. but their introduction.
Grandjot (now Professor at the University of Santiago) was lecturing on the foundations of analysis and using my notebook as a basis for the lectures. because the standard procedure. I. my lectures.
_
Xn
and of
n=
n
Xn. the entire following book). too. H r turn mm m wi h h r m r he had found it necessary to add further axioms to Peano's in the course of the development. I. which
point. At the definition of x • y for the natural numbers. II. say.PREFACE FOR THE TEACHER
IX
ep 0 any 0 other pedagogical faction (who therefore does not go through the undations at least the 0 ortunit rovided he considers this book suitable. but nobody rint was furnished
The opposition party likes to believe that the student would from some lecture or from the literature. so that everything remains based on Peano's axioms (cf.
after one already
. Kalmar)
I have nothing new to say. And of these honored friends and enemies. At the definition of x + y for the natural numbers.
cation of Mr.
3. for some domain of numbers. none would have doubted that everything needed could be found in. Grandjot's axioms can all be proved (as we could have earned from Dedekind). III. of referring his students to a source where the After the first four or five rather abstract pages the reading is quite easy if-as is actually the case-one is acquainted with the results from high school. as x At the definition of ~
+y
and x • y. then assistant and dear colleague Privatdozent Dr. believed that.
But from
n=l
~
Xn
and
. with the difference that it was justified. For
'In
x • y the same simple type of proof applies. I will speak here only about the case of x + y for natural numbers x.
Dedekind's reasoning does follow these lines. On the basis of his five axioms. defined for z < y. Grandjot's objection sounds similar.
m
are possible only with the Dedekind procedure.
n=l
IT x. that not even the expert might have noticed this point had I not given above a detailed confession of crime and punishment. Peano defines x + y for fixed x and all y as follows: x+l=x' x + y'=(x
+ y)'. however.x
PREFACE FOR THE TEACHER
Since the situations in all three cases are analogous. With the kind help
and the proof so similar to the other proofs in the first chapter. y. When
excusable. with the properties f(l)=x. so I had to excuse it also. the student just had never heard of the axiom of induction.
f(z')=(f(z»' forz<y.
in Peano's method because order is introduced only after addition-one had the concept "numbers < y" and could speak of the set of y's for which there is an f (z).
the latter for the natural numbers. December 28. say. 1929
. as befits such easy material.and for all. for instance in the proof of Theorems 16 and 17: This sonin holds for ever class of numbers for which the s mbols Such repeated deductive reasonings occurred in connection with
since the
will then a
ypes 0 lor on complex numbers. only if the division leaves no My book is written. only if the minuend is larger than the subtrahend. the former hold for the natural numbers. say. in merciless
Berlin. and do the same for the theorems on subtraction and division.
then either x and yare the same number. small italic letters will stand for natural numbers throughout this book.CHAPTER
I
NATURAL NUMBERS
§l
Axioms
We assume the following to be given: A set (i. this may be written
+
(
to be read "equals") .
Accordingly. the following are true on purely logical grounds: for every x.
en
u=« z
X=Z. or x and yare not the same number. possessing the properties-called axioms-to be listed below. If x IS given and y is given. Before formulating the axioms we make some remarks about the symbols and 'Nhich will be used.
. 2 If then
3) If
x=x
x=y
x=y. Unless otherwise specified. this may be written
x+y
(=f= to be read "is not equal to"). totality) of objects called natural numbers.e.
) Now. Axiom 5 (Axiom of Induction): Let there be given a set IDe of natwlal numbers. called the successor of x.2
1. since othenvise ambiguities might arise.x. which will be denoted by x'. Then IDe contains all the natural numbers. '\¥e will do the same.
b
= c. etc. we will enclose in parentheses the number whose successor is to be written down. if
x=y
then
Axiom 3:
We always have
That is. Axiom 4: If
x-yo
That is. x y.
that. with the following properties: I) 1 belongs to IDl. we assume that the set of all natural following properties:
Axiom 1: 1 is a natural
number. c = d. II) If x belongs to IDe then so does x'. xY.
which on the face of it means merely that
=
b. there exists no number whose successor is 1. . xy. That is. our set is not empty.
Thus a stateluent
such as a= b a
= c=
d. b
contains the additional
information
a
c.
numbers has the
(Similarly in the later chapters. NATURAL NUMBERS
[Axioms 1-5. In the case of complicated natural numbers x. for any given number there exists either no number or exactly one number whose successor is the given number.
. it contains an object called 1 (read "one"). throughout this book. in the case of x + y. Thus. a
d. say. Axiom 2: For each x the1'e exists exactly one natu7'al number.
d.
(For any such x. (x')' =f= x'. ADDITION
3
Addition
Theorem I:
If
x=f=Y
then
Proof:
.
Otherwise.
If
x =f= 1. by Axiom 4.
Let W1 be the set consisting of the number 1 and of all those x for which there exists such a u. then x and hence by Theorem 1. we therefore contains all the natural we have for each x that x x. so that x' belongs to IDe. II) If x belongs to Wl. we would have
.) J) 1 belongs to ffil. i. then there exists one (hence.
Theorem 2:
x' =f= x. x=y.Th.
Theorem 3:
l'
=+ 1.
therefore 1 belongs to we. J) By Axiom 1 and Axiom 3.
. By Axiom 5.e. by Axiom 4. we have of necessity that
Proof:
x=f=l
by Axiom 3.
and hence.1-3]
§ 2.
numbers. exactly one) u such that
.
Proof: Let ffi1 be the set of all x for which this holds true.
NATURAL NUMBERS
II) If x belongs to 9. (by)' = byl. and by be defined for all
at = x. called x + y 1) x + 1 = x'
(+ to be read "plus").
A) First we will show that for each fixed x there is at most one possibility of defining x + y for all y in such a way that
Proof:
x and
+ 1 = x'
y
Let o. such that
f or every x.11.4
1.
and be such that
I) hence II hence by Axiom 2. with u denoting the number x. then.
.
= Cay)'
.. therefore ay' then y'
(ay)'
(by)'. we have
x+l
there exists a u such that
ber.
holds for all y. we have
x' I y
x' + y = y + x'.
Theorem 7: y =l= x + y. and let ill1 be the set of all x for which the assertion holds.
and 1 belongs to me. and let 911 the set of all y for which the asserbe
tion holds. NATURAL
NUMBERS
[Th.
. I) We have
y
+
1 = y'. so that x' belongs to 911.6
I. Therefore the assertion
Theorem 8:
y'
+ y) " =l= x + y'.
Fix y.
then
x
(x I y)'
+ v=r u + x.
then
y'=l= (x
so that y' belongs to me.
Y=l=z
If
x
then
+ Y =l= x + z. so that
by the construction
1+ y
in the proof of Theorem 4.
y I x' . II) If x belongs to therefore
l+y=y+l
9)(.
and furthermore. The assertion therefore holds for all x.
By the construction
in the proof of Theorem 4.
(y
I x)' (x I y)'. II) If y belongs to hence
me. Proof: Fix x.6-9]
Theorem 6 (Commutative
Law of Addition) : y-y
x
Proof:
+
+
x.
1 belongs to 9)(.
= y'.
.
The former we already know.then for we wou ave hence. contradicting
the statement above
. ORDERING
13
Theorem 26:
If
then
Proof:
y < x.
+ 1 would
belong to ffic.§ 3. The latter is established by an indirect argument. and let
we
be the set of all x
does
Therefore there is an m in IDlsuch that m + 1 does not belong to we. for otherwise natural number would have to belong to we. thus m bw. one which is less than any other number
Theorem 27: Proof:
Let 91 be the given set. by Theorem 25.
In every non-empty set of natural numbers there is a least one (i. as follows: If m did not belong to 91.e. to 91.
Otherwise
we would have
y>x+1. by Axiom 5.
by + x for every y. by' .
+X
=b
+x
= b»
so that y' belongs to WI. i.To every pair of numbers x. Hence WI is the set of all natural numbers. then hence a
1. y. x for every x and every y. = ay
+ x.
al == x === hence 1 belongs to WI. we may assign in exactly one way a natural num-
Theorem 28 and at the same time Definition 6:
1) x· 2) X·
1= x
y'
=x •y +
f or every x. or the number obtained
is at most one possibility of defining xy for all y in such a way that
X·
l=x
ay.
x • y is called the product of x and y.
at
=
b1 = x.e.
X. for every y we have
. II) If y belongs to 9)1.
6267]
Theorem 62:
If
or ~. and the third is a consequence of the first.
FRACTIONS
[Th. the second is contained in Theorenl 56.28
II._ <: YI
Xl s. since
Theorem 63:
If
then
Xl
-< Jb__ .
Follows from Theorem 62. Z2
Zl Z2
Z2
or
Proof:
X
+
_1 Z2
Z
<:
_1
Y
+z
Y2
_1 Z2
respectively.
X2
Y2
The first part is Theorem 61. we have
and
. in both instances.
Proof: Theorem 64: If
then
Proof:
By Theorem 61. respectively. >_:7 _± __ 1
'11
X2
Z2
_1
Y2
Zl +_
or
Xl s. since the three cases. are mutually exclusive and exhaust all possibilities. '01 --±--rv-+X2
.
by Theorem 60.
2
$1
:>~ z 2 rv U 2
'
Follows from Theorems 56 and 61 if the equivalence sign holds in the hypothesis.
Proof: Theorem 67: If -
then
YI U -+-rv-Xl
t
has a solution
"1 . then
. otherwise from Theorem 65. The second assertion of Theorem 67 is an immediate consequence of Theorem 63. by Theorem 63.
Proof: Theorem 66:
If
then
Follows from Theorem 56 if two equivalence signs hold in the hypothesis.§ 3. for if
Proof:
. otherwise from Theorem 64.Preliminary Remark:
If
there does not exist a solution.
then.
. If
VI
and
W1
are solutions.
ADDITION
29
so that
Theorem 65:
Xl X
2
If
:> J!J:_
r-v
y.
the
dot is usually omitted)
is meant the fraction
X
Yl .».Def.
Preliminary Remark: Proof:
hence
--rv--.
MULTIPLICATION
31
§4
Multiplication
Definition 15:
By
Xl.14-15]
§ 4.
to be read "times".
It is called the product of
by mult'ipl'ication of Theorem 68: If
Xli
Xl
and
Yl .
YtUI Y2 U2
Theorem 69 (Commutative
Law of Multiplication)
:
Proof: Theorem 70 (Associative
Lavl of Multiplication)
:
.
or the fraction obtained
y Y2
then
The class of the product thus depends only on the classes to which the "factors" belong. X1Y2
Xl
however. Xli
Y1
Yll
(.
:1:2 Z2
».
Definition 25: A rational number is called an integer (or a whole number) if the set of fractions which it represents contains a fraction of the form
T.
By Theorem 111, this x is uniquely determined. Conversely, to each x there corresponds exactly one integer. Theorem 112:
--('V-
x JI
xy
Preliminary Remark: Thus, the sum and the product of two integers are themselves Integers. Proof: 1) By Theorem 57, we have
~+JLrv
1 1 2) By Definition 15, we have
x+y 1·
ITrvT.TrvT·
Theorem ]] 3:
[£
Y
$Y
fEY
natural class of assigned
numbers,
Thp- intp-gers satisfy the ;fi1Je axzoms nf the provided that the role of 1 is assigned to the
T
and that the role of successor
to the class of -. 1 Proof: Let 2' be the set of all integers. 1 _ 1) The class of T belongs to 3.
2) For each integer
,
to the class of
T
is
we have defined a uniquely determined
successor.
Def.25-26]
§ 5. RATIONAL NUMBERS
AND INTEGERS
41 1
3) This successor is always different from the class of 1 ' since we always have
4) If the classes of
x'
1
and of
y' 1 coincide, then
X'
= y',
an d th e cI asses
5)
0f
x an d 0f T COlnCIe. Y . 'd 1
Let a set
Wi of integers have the following properties:
I) The class of 1 belongs to II)
W.
~
If the class of 1 belongs to Wl, then so does the class of 1 .
x
Furthermore, denote by Wl the set of all x for which the class x . of -[ belongs to WI. Then 1 belongs to WI, and for any x belonging to roc, its successor x' also belongs to IDe. Therefore, every natural number belongs to WI, so that every integer belongs to Wi. Since =, >, <, sum, and product all correspond to the earlier concepts (by Theorems 111 and 112), the integers have all the properties which we have proved, in Chapter 1, attach to the natural numbers. Therefore, we throw out the natural nUJIlbers and replace theIll by the corresponding integers. Since the fractIons also become superfluous, we may, and henceforth we will, speak only of rational numbers whenever any of the foregoing material is involved. (The natural numbers remain, in pairs, over and under the fraction .Iine, in the concept of fraction; the fractions survive as individual elements of the sets which constitute the rational numbers.) Definition 26: The sY'fnbol x (now freed of its previous mean= jng) denotes the integer determ,ined by the cla.ss of ~ .
In our new terminology, we thus have, for instance,
42
II. FRACTIONS
[Th.114-118,
SInce
Theorem 114:
If Z is the rational number
corresponding
to
yZ = x.
The U of Theorem 110 is called the quotient of X b Y or the rational number obtained rom division 0 X b Y.
Definition 27:
I t will be denoted by --;
(to be read "X over Y").
rem 114, the rational number !?_ determined by Definitions 26 and
y y
sense) belongs.
fractions as such will from now on no longer occur. ~ will hence-
number may be expressed in the form ~ , by Theorem 114 and
y
Theorem 115:
Let X and Y be given. Then there exists a z
such that
Proof:
X
y
is a rational number; by Theorem 89, there exist
v>
By Theorem 111, we have hence, by Theorem 105, v_
X·
3) it does not contain a greatest rational number (i.al nllm. We will also nse the term "lower class" for snch a set. The elements of the two sets will then be called "lower numbers" and "upper numbers. DEFINITION
43
CHAPTER III
CUTS
§l
Definition
Definition 28:
1)
A set of rational numbers
is called a cut if
it contains a ratian. a number which is greater than any other number of the set).
(=F to be read "is not equal to").
The following three theorems are trivial:
Theorem 116: Theorem 117:
~=~.
If
'Yj • 'Yj
Theorem 118:
If
then
~=
n. except where otherwise specified. and the term "upper class" for the set of all rational numbers which are not contained in the lower class." respectively. Small Greek letters will be used throughout to denote cuts.e. but does not contain all rational numbers.
rational number of the set is smaller than every rational number not belonging to the set.27-291
§ 1. if the sets are identical.ber.Def. Otherwise.
="
.
Definition 29:
(= to be read "is equal to") if every lower number for ~ is a lower
2) every
number for
'Yj
and every lower number for
'Yj
is a lower number
In other words.
. the statement of Theorem 120 is of course equivalent to 2) of Definition 28.
Follows from 2) of Definition 28. the set also contains all numbers smaller than that number. the set also contains a greater one.
CUTS
[Th. and if
then Xl is a louier num. Proof: Follows from 2) of Definition 28.119-123. 2) With every number it contains. Conversely. Thus if we wish to show that a given set of rational numbers is a cut. Theorem 120:
If X is a lowe?~ numbe?" fm" c. 3) With every number it contains.
If X is an upper number for ~ and if
then X.44 Theorem 119:
III.ber for c.
Proof:
. is an upper number for ~. and there is a rational number not belonging to it. we need show only the following: 1) The set is not empty.
.
rroot:
!. f." '/.-... ~ < 'Yj
are incompatible by Definition 29 and Definition 31.. f. .
I-
Proof: Each means that there exists a lower number for 17 which is an UDDernumber for c.
eo
•
Ui
(. then
~>'Yj
if there exists a lower number for ~ which is an uimer number for n Definition 31: I f ~ and n are cuts. ~ >
'A~A_
'Yj
~_ ••
./
".t"-
h'tT n£l~n1t.'Y\t>A1'Y\n~t. uv
n .L •.
u
~u
AI=:
§2
Ordering
Definition 30:
If ~ and
n
are cuts." .
QII
~= n..Vt:/1
.
." ~J _AAAAAVA~AA
90
-~
~AA"'"
<:Inri n£l~n..hlo ./ '/. '/
is the case.
n.
Proof: Each means that there exists an upper number for which is a lower number for ~.I1"'O
_ A _~
1)
~= 'Yj.n."'~. Theorem 122: If
then
n
~<n ... 'Yj..
» n..J
011
. 1A
VUI(.t J VI
Theorem 121:
then
If
~>'Yj
'YI<:"E
.
Il/tNllr. If we had
!!../!
on
'u~..
<..r.
'/
.. .VVUt:/1
7
z.. exactly one of
f.+.__. <.
~o
-."_''''
'<-Iv..nn ""V..
!!.
.1J/1/
if there exists an upper number for
'I.."' . Theorem 123: For any given ~.. then
~<'Yj
(>
to be read "is greater than")
«
l"
to be read "is less than")
.
f"'\~""""~T"'T'" "' ••..
of Ordering):
Theorem 126 (Transitivity
If
then
. Then we either have that some lower number for ~ is an upper number for 'YJ. in which case it follows that for
'YJ
is an upper number
~< 'YJ. By 2) of Definition 28.")
Definition 33:
means
~< 'YJ
«
then
Theorem ] 24 : Tf
or ~ =
'YJ.
then the lower classes do not coincide.124-128. which is an upper number for 'YJ.")
'YJ =
.
Proof:
Theorem
121.
than or equal to.
CUTS
[Th.
we can have at most one of the three cases. in which case it follows that or we have that some lower number for ~.
to be read "is less than or equal to.
Definition 32:
means
(> to be read "is greater
~> 'YJ
or ~ =
'YJ. we would then have both
X
Therefore 2) If
<
Y and X> Y. and that there also exists an upper number Y for ~ which is a lower number for 'YJ.
it would follow that there exists a lower number X for c.
Theorem 125:
If
Proof:
Theorem
122.46
III.
Proof:
. there also exists an upper number Y for 'YJ which is a lower number for C.32-33]
§ 2.
Obvious if two equality signs hold in the hypothesis. 'Yj
<C
or ~
< 'Yj. otherwise.
'Yj
< C. we obtain
Proof:
$8
that Y is an upper number for ~. ORDERING
47
There exists an upper number X for ~ which is a lower number for 'YJ.Def. Theorem 126 does it.
Proof: Theorem 128:
If
en
~< C.
Obvious if the equality sign holds in the hypothesis. Theorem 127 does it. Definition 28) to the cut fj. Therefore
Theorem 127:
If
~<
'Yj. Applying property 2) of cuts (cf.
~< C. otherwise.
'YJ
respectively. X1+Y1 =1= X +Y. for ~ and any upper number Yl for 'YJ. where X is a lower number for ~ and Y is a lower number for 'YJ.
. is itself a cut. we have
Proof:
hence < X1+Yt. by Theorem 105. where X and Yare lower numbers for ~ and
Z
x +Y
<X +
Z (X+ Y) X+Y «X+Y).
CUTS
[Th.
z
XX+y<Xo1-X
and
Z
Yx+y
z
<Y·1
=
Y. consider any upper number X. + Yl does not belong to our set. therefore X.129-132. we must show that with any number it contains. II) No number of this set can be written as a sum of an upper number for ~ and an upper number for 'YJ. by Theorem 106. Then X + Y belongs to our set.
§3
Addition
be cuts. Next. our set also contains all numbers which are less than that number. and we have proved II).1. Let Y. as well as property 1) of cuts for our set.
so that.48
III. Then the set of all rational numbers which are representable in the form X + Y.
X+ Y < 1. if X and Yare any lower numbers for ~ and for 'YJ respecti vely.
Theorem 129:
'YJ
I) Let ~ and
1) Consider any lower number X for ~ and any lower number Y for 'YJ. 2) To prove that our set satisfies property 2) of cuts. Then
hence.
+ fj = 1} + 1..
X
so that there exists in our set a number which is > X + Y. 01" the cut obtained f'ro'ln addit'ion of fj to ~.
X-A. then there exist a lower number X and an upper number U for the cut such that U
Proof:
+
Y
>X +
Y.
hence by Theorem 115. Theorem 130 (Commutative Law of Addition) : I. and fj. Proof: Every (X + Y) + Z is an X + (Y + Z). The sum of those two rational numbers is the given Z.
numbers
+ nA
where n is an integer.. Using the third property of cuts as applied to 1. + fj) + C = I. Definition 34: The cut constructed in Theorem 129 is denoted by I.34]
§ 3.Def. Proof: Every X + Y is a Y + X. Theorem 132: Given any A. Not all of these are lower numbers. we can find a lower number
X
for ~. and given a cut. and vice versa. and vice versa.
Let X. since
3) Any given number of our set is of the form X + Y where X and Yare lower numbers for ~ and . for if Y is any upper number. ADDITION
49
~ an
fj.e
num ers
x+
y
an
X +y
are ower num-
bers for ~ and n respectively. + (fj + O." respectively. then
Y
> Xl. + fj (+ to be read "plus") and is called the sum of I.
. then
X. we have for some suitable n that nA>Y-X" so that Xl
+ nA is an upper number. and consider all rational X. be some lower number. Theorem 131 (Associative Law of Addition) : (I.
Choose a greater lower number
Then we have
Y+Z = Y+«(X-Y)+
U) -
(Y+(X-Y»+
U
= X+
U.
+ n-
Therefore
If
en
~ + c > 'Yj +
X>Y
c. 133-140]
u= 1.
In each case. U =X1
+
'uA =X
+
A.
number for ~.
or ~
< n.
CUTS
[Th.
then we set
1
u>
X=X
1
1.
.
or ~ =
'Yj.
then we set
+
(u-1)
A.
Theorem 135:
If
~> 'Yj. and
can find a lower number X for ~ and an upper number U for ~ such that
U-X=y·
is an upper number for ~ and a lower number for ~
~+
Theorem 134:
'Yj
> ~. X is a lower and U an upper number.50
III.
CUTS
en has exactly one solution v.
is a lower number for
u=X -Y
2
belongs to our set. by Theorem 135. > X.
1]
+
Vl
=1= 1J + v2•
II) I will show first that the set of all rational numbers of the
that such an X . for ~ can constitute such an X each number of this form satisfies
Y.Y of the above sort number for .
. choose a lower
(X3-Y)+Y> X3-Y
(X-Y)+Y.52
III. 3) If an X . No upper number X.Y. Then
IS
given. since
then
u+Y«X-Y)+Y=X.
Preliminary Remark: If
then.Y does indeed exist.
and denote it by Y.Y.[Def.Y) + (Y1 + (Y .y.
= « ~y-
Y) + Y1)+ (Y .Y) + Y1 <: X.
To prove this. B) Every lower number for ~ is a lower number for v + 'fJ. +(Y-Y1) = (X-y)+(Y1+(Y-Y1»=(X-Y)+ (x.) Y2Y1
n
=
X.
so that Y is a lower number for v + n0) If the given lower number for ~ is also a lower number for n. or the cut obtained by subtraction of 'fJ from ~.
and moreover choose.
Y
= (Y -
Y1)
+Y =
J
(X . Y an upper number for n. it suffices to establish the following two statements: A) Every lower number for v + 'fJ is a lower number for ~. such that
X> Y. by Theorem 132. Y = X.
so that (X . and
X> Y.Y1) = (X .'fJ (.to be read "minus") and is called the difference ~ minus 'fJ. Choose a lower number X for i. Definition 35: The v of Theorem 140 is denoted by ~ .Y) + Y1 is a lower number for ~. Y1 a lower number for 'fJ.
Now we have «X-Y)+Y1) Y> Y1.Y1) = (X . a lower number Y1 for and an upper number Y2 for 'fJ such that Then we have hence Y2 + (Y .Y2)
+Y
1. As regards B): 0) Let the given lower number for ~ be at the same time an upper number for 'fJ.Y) + Y = X.
Y-Y =X-Y
1
2.
.35]
§ 3. then it is less than all those rational numbers which were considered in 0) and which turned out to be lower numbers for v + nHence in this case the given number must itself be a lower number for v + 'fJ. ADDITION 'fJ
53
+ v =~. As regards A): Every lower number for v + 'fJ is of the form
(X-
Y)
+
Y1
where X is a lower number for ~.
8 se can e ior« upper number for ~ and an upper number Proof: 1) Consider any lower number number Y for n . and we have proved II) .III.
CUTS
Th. and let
Z<XY. then XY belongs to the en as a pro for nX for ~ and any lower set. uihere
ber Y1 for n.
'f}
II in the form XY. If X and Yare any lower numbers for ~ and for
therefore X1Y1 does not belong to our set. Y a lower number for 'f}. 3) Let there be given any number of the set. 2) Let X be a lower number for ~. or our se . o num er 0 1.141-145
4 Multiplication
rational numbers which are representable a cut.
z= xX
thus shows that Z belongs to our set. as we as prope y 0 eu s. it is of the form XY where X and Yare lower numbers for ~ and for Choose a lower number
.
07'
'Yj.
Definition 86: The cut const'7"ucted in The07'e1n 141 is denoted by ~ • 'Yj (. Y. From
follows
XY+XIZ<X.
so that our set contains a number which is
> XY. however the dot is usually omitted) . and C. Theorem 144 (Distributive Law) :
I. Y and Z are lower numbers for ~.Y+X2Z
=
X2(Y+Z)i
hence XY
+
X1Z is a lewer number for ~ ('YJ
In. respectively.
Proof:
~> n.
I. Let X2 stand for the number X in case X ~ Xl and for the number Xl in case X < Xl.
ec -
or q 1]C. 'Yj.36]
§ 4. 'Yj. and Z are lower numbers for ~. Theorem 143 (Associative Law of Multiplication) : Proof: Every (XY)Z is an X(YZ).
1) If
01'
ec < 1]C.( 'Yj + n. II) Every lower number for ~'Yj + r$Cis of the form where X.
or q
<
ec > 1]C.~. to be read "times".
and
is
called the product of ~ and of ~ by
n.
. XI. then X2 is a lower number for ~. ~. then we have that
XIY> XY. or
the cut obtatned from mulh-
plication
nLaw of Multiplication) :
Theorem 142 (Commutative
Proof: Every XY is a Y X. The number XY + XZ is a lower number for ~'YJ + ~ C.
l'espectively.
'Yj.'Yj
+
Proof:
I) Every lower number for ~ (1] X (Y
+
C) is of the form
+ Z) = XY +
XZ
where X.so that X2 (Y + Z) IS a lower number for I. respectively. and C.
+ 0=
1. and vice versa.Def. MULTIPLICATION
55
for ~.
Theorem 145:
then
If
q
> 'Yj. (11
and vice versa.
the set of all rational numbers < R constitutes a cut.
i
1
x
= X. s. Theorem 148 does it. <
R. then by Theorem 91 there exists an XJ_such that
X
Definition 37:
< x. MULTIPLICATION
57
then
Proof:
~c> n».
then
X
X <R. the equation has a solution v. > R.
x < R.)
Theorem 151:
with asterisks
~ • 1*
=~. Choose. 1* is the set of all XY '\There X is a lower number for ~ and Every such XY IS < X and thus IS a lower number for ~.37]
§ 4. otherwise. Theorem 150: For any given rational number R.
The cut constructed
in Theorem 150 is denoted will stand for cuts. there does exist an X < R. Theorem 152: For any given ~.
Proof: ~. The number R itself is not < R. for ~.Y
is a lower number for ~ • 1 *.
Xl 1.
Obvious if two equality signs hold in the hypothesis. Conversely. let there be given a lower number X for ~.
*
(Thus capital italic letters not for rational numbers.Def.
*
. Proof: 1) By Theorem 90. a lower number x >X and set Then Yo< so that the number y_
l-x.
SInce
thus
1 does not belong to our set. so that X
1
X belongs to our set.
3) Let there be given a number
i
of our set. satisfies hence. then any upper number X for c.58
III.
number which does not belong to the
There exists a rational
set. then X IS an upper for ~. excepting only the least upper number (if such a one exists). then so is X + X. We will show that this set is a cut. and is not the least such. Now if
1 1 X =1= X 1 . for if Xl is any In'.
1
1' U the number
U belongs to our set. 1) The set does contain a number. for if X is an upper number for s.
number
then
U<X'
1
hence
1
so that Since
U is an upper nurnber
for ~. Choose an upper number
X <X
1
. and the latter indeed can not be the smallest.
CUTS
[Th. Xl 1 2) Consider any number X of our set.152]
Proof:
Consider
the set of all numbers
X
1
where X may be
any upper number for ~. then X is an upper
number for ~.ver number for Ii. and is not the least such.
implies
D<1. From
1
=
1
tain
so the given one. MULTIPLICATION
59
er number for 1 X'X<XX we
0
and is not the least such.§ 4.D) X. let it be denoted by v. We will show that it satisfies.
Then we have
(1. Our set is therefore a cut.
~v = 1
*.
.
existence is proved by Theorem
II) If t is the solution-whose
152
of the equation
then the cut
v
=
-r 53.60
III.
If
u-
Xl
X ~--'
1
1 X1 '
U
here. by Theorem 145. and is called the quotient of ~ by obtained from division of ~ by 'Yj. 153-155. for we have. by Theorem 1*~ ~. Xl is a lower number for ~.
satisfies the equation in Theorem] 151.
where ~ and 'Yj are given. Theorem 153: The equation
X
1
is a lower number for v.
or the cut
.and hence U is a lower number for ~v.
1(to
be
read "~ over 'Yj").
CUTS
[Th. that
fJ v 1J (-r ~) -
(1J -r) ~ -
Definition 38:
The v of Theorem 153 is denoted by
'Yj. has exactly one solution v. for if then. Proof: I) There exists at most one solution.
*
Proof: I) a) Every lower number for X*
+
y* is the sum
.
if X:> Y. or X*
<
Y*. X* y*.Def.
Theorem 154:
If
x>
then
X*
Y. A cut of the form x* is called an integral cut. since the three cases are.Y*. hence. respectively. or.
II) The converse is obvious.X
<
Y. The number Y is an upper X*:> Y*.Y)* (XY)* ossibilities.
th
Y:>X.38-40
5.
then Y is a lower number for X*. by 1). or X
=
=
Y. exclusive and exhaust all (X . in both
instances mutuall
= =
X* .
X> Y.
>
Y*. y*:>x* X*< Y*. RATIONAL CUTS AND INTEGRAL CUTS
61
§5
Definition 39: Definition 40:
A cut of the form X* is called a rational cut. or X*
Y*.
is a 10 W er number for X* Hence we have (X+ Y)* = X*+ 1'*. 6) Every lower number U for (X + Y) * is < X + Y. 6) Every lower number U for (XY) * is < .
X* . such that Then and U< D1<XY
Thus the relation
represents U as the product of a lower number for X* and a lower number for Y*. (X -Y)* = X* -Y*. Choose a rational number U1.
J
+
<X
Y*.
CUTS
[Th. it is therefore < X + Y.
and of a
II) If
X>Y
then so that by 1).
IV)
. Hence we have
(XY)*
=
X*Y*.(Y.62
III.eX Y)* +Y*. Now
u
-< 1. by Theorem 91. and is thus a lower number for (X + Y) *. III) a) Every lower number for X*Y* is the product of a rational number < X and of a rational number < Y.156]
of a rational number < X and of a rational nurnber < Y.
u
implies that U as the sum of a rational number rational number < Y. therefore it is < Xl'. Therefore U is a lower number for X* Y*. and so is a lower nUIIlber for (Xl') *.
x-
eX
Y)+Y.
... "' . (x*)' =1= 1*.L..""nT
1I.
.LV. and if x belongs to im then so does x' Hence every integer belongs to im.
TT
9)1*
of integral cuts have the following properties:
I.LQJ.
.Lv... and quotient..
•• '
. then
(.L.LV.
x
X*
= y.
}•
0
Also..~ .. .Lt::i:jVV. so that every integral cut belongs to im*
0
0
..1leU uues \ x so
11
.l~
v
+\."4!""".
Ht":.."'
.. 3) We always have
Proof:
. .v
....J. the cut (x*)' is also in 3*.La.
rJ'1\. RATIONAL CUTS AND INTEGRAL CUTS
63
so that by III).
y*.SJ.1:')*
/
-
(1}'')*
X'
y'.. sum. the rational cuts have all the properties which we have proved. in Chapter 2..U !!.
.LUoll.lJ'vLO
Q. 4) If (X*)' = (y*).::jInce-.U
.. prOUUCL.".Lu. uirrerence \ wnenever it eXIS1.§ 5.". the integral cuts..
.. in particular.1.LV.
~Ulli:j.v... >..
IIHt::
_.l1UH. <..l
....
UJ
J.c .Ie
rv
"ilk ~
T"
1.
Let 3* be the set of all integral cuts.
~..1.."
lI.\"'"V* \rJ
X* Y*'
\y)Theorem 156:
natural numbers
The integral cuts satisfy the five axioms of the if the role of 1 is assigned to 1* and if we set (x*)'
=
(z")
*.."".1I H..LvP. denote by im the set of x for which x* belongs to im* Then 1 belongs to im....Lv.1 .
'UT"
t....
. In the domain of rational cuts all correspond to the earlier concepts (by Theorems 154 and 155)..0.
i:jV
.1 Ut::.. have all the properties that have been established for the integers.LU
n~.
we obtain
x* I X\*
(x.. 1". .ln~
+1..
".llUWi:j
we
will only have to speak in terms of cuts whenever any of the fore-
... J. ~V...J..LV
'Y
VIA"
"' •• + +\.L.LIAJ.L..".U. 1) 1 * belongs to 3*.'''+··''' .1I.l.I..lJ.
1. " 1I.v
.. attach to the rational numbers.' IIUa..
.
=
5) Let a set
T\ ~.lJ.' -I-
1
...L. ?) For every x* in 3*. CYYtIk IIV~'" •
~~}
~~ u:
oerung s
..
:e
(x')* =1= 1*...l
11 a...lVH~i::I
••
.la.
Then X. to apply to integral cuts. Then X is a lower number if. This X is then t1 J cut. an upper num er or we have by Theorem 157 that
IS
3) If X is an upper number for ~ but is not the least such.
1) For the cut X (our old X*).157-161.
then there exists a Z such that
.
* ~T =
Theorem 157: Proof:
1*. if. so that
be a cut.
transferred. we have en
is an upper number for X (the old X*). we choose an upper number Xl less than X. is a lower number X
Theorem 159: I
>~.
CUTS
[Th.64
III.S
an upper num
X>
1
.
X (the rational num-
lower number for ~ is
<X
and every upper number is > X.
X
ce
1.
The rational numbers are those cuts for which there exists a least upper number X.
Then
Choose Zl and Z2 by Theorem 159 such that
By means of
and using
y
=
Z.
Then
very
Z
> ~'Yj
may be brought into the form Z=XY. and then choose a greater lower number Z for we have y heorem 15 t at
1}. >. for:
. RATIONAL CUTS AND INTEGRAL CUTS
65
ber for 'Yj.
has exactly one solution.41]
§ 5.
This set constitutes a cut.Def.fJ.
en
xx < C.
.lI/e "vvould have. YY<: y<X XX<:~o
o
then
XX<:b
Choose a Z less than the lesser of the two cuts 1 and Then
~.
then If then
2) If
X<: 1 and X<: "
xx <: X·1
X:>l
= X <: ~. > C.
then we
.66
III.
.
ex +Z)(X+Z)
and
X+Z>X
=
X+(X+Z) Z<: (XX+ZX)+(X+l)Z XX+(X+(X+l))Z<: XX+(b-XX) = b. by Theorem
~~>Z>b.
159..X X X+(X+l)
Z <: 1. X2. in contradiction to the above.
and X :>"
0
xx
:> X 1 = X:> bo xx<~.. would satisfy Z
=
K. Z <: X
furthermore we have
+ EX
I 1). XI <: ~. then .
This Z.
Xl <:.
if X denotes the greater of the two numbers Xl and X2.. the cut which we have constructed.
CUTS
[Tho 162. being a lower number for . that
esxx«:
x <~. a Z such that
then we could choose.
= (X+Z)
If we denote by now assert that If we had
c.
er (as equal) if and only if ~ and 'YJ are the same number. and which we will call neganumber) we assign a negative number denoted by ~ (to be
same numb.
t!!. of all pos~t~ve numbers.=
~= H
H= .. exactly one of
IS
the case..
"
Thus for any given Sand
H. an 0 all negative numbers. H=
Z
Z= z. Ca ital Greek letters such as S H Z Y will be used throu hout to denote real numbers. e crea e a new num er e positive numbers. similarly. will be called the real numbers... of 0. except where otherwise specified. For real numbers. what we have been calling "rational numbers"
"
and "positive inteqers" respectively.. ~..
then then
t!!. We also create numbers which are distinct rom the ositive numbers as well as distinct from zero.1
Definition
numbers". e totaltty consistisu. the concepts of identity and of
trivial:
Theorem 163: Theorem 164: I
t:!=
-
.
.
-
H.
.
Proof: Definition 44.
Theorem 167:
For any given Sand H.
Theorem 166:
if if
~
1-1
-
.
eorem If S is positive while H is = S+H.. we already concepts ">" and "<". -
is the case. then
... in fact.. H negative and or S = 0. for positive
tive S. has been used in one of the cases of Definition 45. . Defini'tion 45: If Sand H are not botli positive.
means
H
> S.
and for nega-
~ ~
1-1
-
I S I is positive. the latter. then
.
with our old concepts. H negative. H = 0. or S positive.2
..
in case Sand H are positive.. S> H. exactly one of . -~.
ve negative.
°
or negative.
S' = H.
and finally.
7) If E and H are both negative.H.
3) If E
=
°
E not while H
IS
< H. -
E not >H.'>H
and vice versa.44-48]
§ 2.") Z <H S =.
3 > H. ORDERING
71
and finally.
Definition 47:
means
S:>H
or
S
H. 131= IHI.")
Theorem 168:
If
10"1=
':.
for 131>IHI.Def. H 0. by Definition 46. E not> II. then E
E not <H.
5) If E =
°
E not while H
IS
< H.
Proof: Definition 46.
3 not 3 not
Z<:H
-c H for
<. E not
=
> H. then
>H.
3=1=H.
6) If E is negative while H is positive or E+H.
S< H or
«
then
to be read "is less than or equal to.
by Definition 45..
.71 for
I S'I <: IHI . 4) If E 0.
negative.. then ':l-H . then 3 =!= H..
positive. by Definition 46.
(> to be read "is greater
Definition 48:
than or equal to.
0. 3 not > H. 3 not >H. then
and furthermore.
3<0.
then
3< Z.
then ence
then
H<O.:t
<:
Z. Z>O.
If -<0 ~= . by Definition 45.
Z = O. H>O.
-~<O. the negative numbers are the numbers which are < O. we have
= ..
then.
. we have 2) If ~>O.Theorem 169: The positive numbers are the numbers which are> 0. Proof: If 1) Let Z>O. Proof: 1) By Definition 45.
'. Z 3) By Definition 46.
then
and we have the earlier Theorem 126.
and the positive number'.Def. IZI>IZI.
Thus we now have positive irrational numbers and negative irrational numbers. Z<Z.'-0 -
or
E
<
0. I E I
a rational number. Definition 50: If
.
othesis· otherwise.
otherwise. Theorem 171
E<Z..
ORDERING
73
We also have hence
IZI>IHI.
""
then
E
is called irrational in case it is not rational.
IEI
an integer. for. ca ~s
'-'
'. and negative rational numbers.
Definition 51:
If
en .=Y-X. the rational number 0.
Thus we now have positive rational numbers.
.. since wou
'.IHI>IZI. + X -is always irrational..
or ~<
Theorem 172:
If
then
z-c z... (Numbers? Yes..
then E is called rational if either
'-'
'.49-51]
§ 2..'-0 -
E
< 0.:::. we had an irrational .
. it follows from Definitions 49 and 51. the integer 0.
REAL
NUMBERS
[Th. and negative integers.s rational.74
IV.174-175.
Thus we now have positive integers. Proof: We already know this in the case of positive numbers. For 0 and for negative numbers. Theorem 174: Enery integer i.
for E < we have. by Definition 55.
en
SH =
ISIIHI
or
SH =
-(lSIIHI). from Definition 55. 192-200. H is zero.
(. for :3 0. to be read "times". Note that the product 2 • H for the case 2> 0.
where the first alternative holds if none or both of the numbers 2.
§4
Multiplication
Definition 55:
Z·H=
ISIIHI.
REAL NUMBERS
[Th. or the number obtained from multiplication of S by H. a fact used in Definition 55. H (by Theorem 142). as is also its subdivision into cases. H are negative. Theorem 194 (Commutative Law of Multiplication) : EH == H':: Proof: If Z > 0.
=
or x«: H < 0.
If
$ =1= 0. in the other cases. it follow s frOIIl Definition 55. Definition 36).
if if:e
s «: 0. H =1= 0. the dot is usually omitted. or H
°
= 0. 195: Z • 1 = 2. this follo\vs from Theorem 151. Theorem 193: ISHI = ISIIHI. that
Theorem
S·l
Theorem 196:
= . and the second if exactly one of these numbers . H > 0.84
IV. however.
0. ti 'lS neva tire:
. Proof: For:. Proof: DefinItIon 55.) E • H is called the product of 2 by H. this is Theorem 142.0:e1·1) =
°
-lSI = S. H > was defined earlier (cf. at least one of the two numbers Z. and only if. since the right-hand side of Definition 55 is symmetric in 2. Proof: Definition 55. Theorem 192: We have 2H=0
°
if. > 0.
1) If
b) + b
=
n.
MULTIPLICATION
85
Theorem 197: Proof: 1) If one of the numbers e expressions are o
E.
IS
negative.55]
§ 4.
2)
If
f}
then
=~
. (-:e)(-H)
Theorem 198:
=
Theorem 199 (Associative
:en. both sides are>
°
if none or two of the
one or all three of those numbers Proof: en
( f} -
are negative.
:
Law of Multiplication)
sides are 0.Def.
by Theorem
144. Z =f= 0. H is zero.
so that.
E =f= 0. all three of the expressions >
have the same
°
or
<
°
according
to whether
or not exactly one of the num= :eH.
then b Theorem 193 both sides have the same absolute value
CI:eIIHI) IZI
and by Theorem
=
131(lHIIZI). H =f= 0. then all three
then by Theorem.
196. 193. | 677.169 | 1 |
Introduction
In this class, students learn concepts associated with linear equations, linear inequalities, matrices, functions, and finance. The purpose of this libguide is to provide students with additional resources clarifying these concepts. | 677.169 | 1 |
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Making Your Algebra Class Interactive
Greg Hurst
With products like Wolfram|Alpha, Wolfram Demonstrations Project, and CDF, Wolfram Research brings algebra into the twenty-first century. This Wolfram Technology for STEM Education: Virtual Conference for Education talk showcases these great products and demonstrates their use in a classroom setting.
Channels: Virtual Events
Roger Germundsson, director of research and development, gives an introduction and overview of geometric computation in Mathematica 10. Types of regions, their computable properties, and the integration with solvers in Mathematica ... | 677.169 | 1 |
Written for undergraduate students, Matrices, Vectors, and 3D Math: A Game Programming Approach with MATLAB provides a resource to learn standard topics in linear algebra and vector calculus in a single course within the context of game programming applications and projects. Topics presented in this text are constructed to enable students to succeed after completing a single college-level calculus course.
MATLAB is used to solve numerous examples in the book. In addition, a supplemental set of MATLAB code files is available for download here.
Scott Stevens Scott | 677.169 | 1 |
This book explores what it means to be rational in a variety of contexts, from personal decisions to those affecting large groups of people. It introduces ideas from economics, philosophy, and other areas, showing how the theory applies to particular situations such as gambling and the allocation of resources. more...
Constructing concise and correct proofs is one of the most challenging aspects of learning to work with advanced mathematics. Meeting this challenge is a defining moment for those considering a career in mathematics or related fields. A Transition to Abstract Mathematics teaches readers to construct proofs and communicate with the precision necessaryLuciano Floridi unpacks this fundamental concept - what information is, how it is measured, its value and meaning - cutting across the sciences and humanities, from DNA to the Internet, and the ethical issues related to privacy, copyright, and accessibility.You know mathematics. You know how to write mathematics. But do you know how to produce clean, clear, well-formatted manuscripts for publication? Do you speak the language of publishers, typesetters, graphics designers, and copy editors? Your page design-the style and format of theorems and equations, running heads and section headings, page breaks,... more...
Mathematical models are used to simulate complex real-world phenomena in many areas of science and technology. Large complex models typically require inputs whose values are not known with certainty. Uncertainty analysis aims to quantify the overall uncertainty within a model, in order to support problem owners in model-based decision-making. In recent... more... | 677.169 | 1 |
Holt Math Course 1 Middle School
Tennessee Math Connects: Course 1 Homework Help.Holt California Mathematics,.Webmath is a math-help web site that generates answers to specific math questions and problems, as entered by a user, at any particular moment.
Holt Algebra 1 Chapter Test Answer Key
Research Briefs for Mathematics Education CPM Educational Program invites members of the mathematics education research.Free math lessons and math homework help from basic math to algebra, geometry and beyond.Find great deals on eBay for Holt Mathematics: Homework and Practice Workbook Course 2.Our professional writing service can help you with any kind of assignment you might have. | 677.169 | 1 |
Master Algebra Skills With Ease
Looking for algebra help? Sylvan offers skill practice for pre-algebra, algebra 1 and algebra 2. We'll help your son or daughter get ahead!
Our Algebra Edge skill program is great for teens who are having trouble keeping up, need to brush up on algebra skills or want to jump ahead in advanced math. It features engaging lessons, expert math coaches and a curriculum aligned with school standards.
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Core-Plus Mathematics is a standards-based, four-year integrated series. Concepts from algebra, geometry, probability, and statistics are integrated, and the mathematics is developed using context-centered investigations. Developed by the Core-Plus Math Project at Western Michigan University with funding from the National Science Foundation (NSF). The first three courses in Core-Plus Mathematics provide a significant common core of broadly useful mathematics for all students. They were developed to prepare students for success in college, in careers, and in daily life in contemporary society. The program builds upon the theme of mathematics as sense-making. Through investigations of real-life contexts, students develop a rich understanding of important mathematics that makes sense to them and which, in turn, enables them to make sense out of new situations and problems. Reading age for native speakers: High School students
NB: Course 1 and Course 2 are not included in this Interactive Edition since they are already available on ET. | 677.169 | 1 |
Exploring Advanced Euclidean Geometry with GeoGebra
ISBN-10: 0883857847
ISBN-13: 9780883857847 enquiry-based introduction to advanced Euclidean geometry. It utilises the dynamic geometry program GeoGebra to explore many of the most interesting theorems in the subject. Topics covered include triangle centres, inscribed, circumscribed and escribed circles, medial and orthic triangles, the nine-point circle, the theorems of Ceva and Menelaus, and many applications. The final chapter explores constructions in the Poincaré disk model for hyperbolic geometry. The book can be used either as a computer laboratory manual to supplement an undergraduate course or as a stand-alone introduction to advanced topics in Euclidean geometry. The exposition consists almost entirely of exercises (with hints) that guide students as they discover the geometric relationships for themselves. The ideas are first explored at the computer and then assembled into a proof of the result under investigation, allowing readers to experience the joy of discovery and develop a deeper appreciation for | 677.169 | 1 |
Description
An Introduction to the Theory of Numbers by G.H. Hardy and E. M. Wright is found on the reading list of virtually all elementary number theory courses and is widely regarded as the primary and classic text in elementary number theory. Developed under the guidance of D.R. Heath-Brown this Sixth Edition of An Introduction to the Theory of Numbers has been extensively revised and updated to guide today's students through the key milestones and developments in number theory. Updates include a chapter by J.H. Silverman on one of the most important developments in number theory - modular elliptic curves and their role in the proof of Fermat's Last Theorem - a foreword by A. Wiles, and comprehensively updated end-of-chapter notes detailing the key developments in number theory. Suggestions for further reading are also included for the more avid reader The text retains the style and clarity of previous editions making it highly suitable for undergraduates in mathematics from the first year upwards as well as an essential reference for all number theorists.show more
About G. H. Hardy
Roger Heath-Brown F.R.S. was born in 1952, and is currently Professor of Pure Mathematics at Oxford University. He works in analytic number theory, and in particular on its applications to prime numbers and to Diophantine equations.show more
Table of contents
PREFACE TO THE SIXTH EDITION; PREFACE TO THE FIFTH EDITION; APPENDIX; LIST OF BOOKS; INDEX OF SPECIAL SYMBOLS AND WORDS; INDEX OF NAMES; GENERAL INDEXshow more
Review quote Nature This fascinating book... gives a full, vivid and exciting account of its subject, as far as this can be done without using too much advanced theory. Mathematical Gazette ...an important reference work... which is certain to continue its long and successful life... Mathematical Reviews ...remains invaluable as a first course on the subject, and as a source of food for thought for anyone wishing to strike out on his own. Matyc Journalshow more
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IGCSE Mathematics Course
Our Mathematics IGCSE is the latest addition to our courses. Using the Edexcel Specification A exams.
We have chosen to offer Edexcel Maths Spec A as it is the most popular choice for home educators. It is available to students in both the summer and winter exam sittings. It also allows students to take a calculator into both papers.
Although from Summer 2017 GCSE Maths will be open to home educators many students are opting to stay with IGCSE as there will be several years' worth of past papers available for practice. Many people consider the IGCSE as a better preparation for A level as it includes calculus and some other topics not covered in the GCSE. It allows students to be entered at either Foundation tier or Higher tier.
As home educating mums we've created our courses in response to the needs of home-educating parents and students but they are equally suited to any private candidate who wishes to study Mathematics IGCSE. Students can start at any time and work at their own pace towards
taking the exams.
About the course
It covers each topic in turn:
setting out what needs to be learnt
explaining concepts and definitions which may be tricky
providing worked examples throughout
over 250 pages of content and assessments to cover every aspect required for the IGCSE exams
When you buy the course it includes:
tutor marked follow up assignments, to assess progress, and the opportunity to practise exam questions.
full feedback to the student on their work by email as necessary until they attain the examination grade they are happy with.
the opportunity to sit a fully marked and graded mock exam with subsequent discussion on the level of exam entry.
Edexcel Specification A details
The Edexcel IGCSE Mathematics Specification A is available for private candidates and can be sat in either January or June.
Candidates will need to sit 2 papers.
Students are entered at either Foundation Tier or Higher Tier.
Foundation Tier students will take papers 1F and 2F. The highest grade which will be awarded at Foundation Tier is grade C.
Higher Tier students will take Papers 3H and 4H.
All papers are 2 hours long and worth 50% of the final mark. Calculators are allowed for all papers.
The specification for the Edexcel Mathematics ( specification A ) is changing, but our course has been written to take these changes in to account, so you do not need to take any action regarding this.
The old specification, which was written in 2009, will have it's last examination in January 2018 – and has the code 4MA0, it will be graded under the old system of A* to G. Study using the new specification can start from this year, for first examination in June 2018, with code 4MA1, it will be graded 1 – 9. It will then be available in both the January and June sittings thereafter.
If you have any questions regarding the changes please do not hesitate to contact us
We have chosen to base this course around the CGP revision scheme, as after teaching maths for many years using a variety of expensive textbooks, we really do find these the best ones for students to get along with. The DVD-Rom is an excellent resource, and covers the vast majority of the topics you'll be required to understand, and the workbook has lots of exam style questions to try.
Throughout the course, you will be referred to clips from the DVD-ROM, asked to complete practise questions from the workbook and set end of unit assessment tasks to complete.
Assessment tasks need to be sent to your tutor for marking as you finish studying each unit. These can be posted or emailed (word document, scanned image, or photograph will do).
Workbook pages can be self-marked by purchasing the answer book, but you are able to ask your tutor about any workbook questions that you would like more help with. Workbook questions marked with a black square are more difficult questions.
The course has been divided into 20 units.
Course costs
The course costs £230 – the set books and materials need to be purchased separately | 677.169 | 1 |
Q: How will you take algebra critical thinking skills to the real world and become a more proficient problem solver?
A: Algebra can be a powerful set of tools for representing situations, analyzing mathematical relationships, making generalizations and solving problems. [ It can extend well beyond the limited types of problems once filling traditional algebra texts to serving as a set of approaches in a student's mathematical toolkit. Today, algebra can be used to deal with data and make predictions. It can be
used to model sophisticated situations from science, social studies or economics, to name a few. If we do our job well to develop algebraic thinking across the grades, American students will never be heard to say that they had no use for algebra. Rather, they will incorporate algebraic techniques into their broader mathematical thinking to deal with everyday life as well as advanced applications in mathematics and science.
] | 677.169 | 1 |
Algebra 2 covers a variety of topics. Some of them include factoring, exponential and logarithmic functions, and conic sections. This is a vigorous course where students can expect to advance their mathematics knowledge. | 677.169 | 1 |
Algebra: Performing Operations with Rational Expressions
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This set of directions illustrates how students can learn to combine rational expressions with addition, subtraction, multiplication, and division and then confirm their answers by using the table feature of the graphing calculator.
Five practice problems are included.
This is one example of activities available in my store. Check out the other items in my store if you like his product.
BECOME ONE OF MY FOLLOWERS AND RECEIVED NOTIFICATEION WHEN NEW PRODUCTS ARE ADDED. | 677.169 | 1 |
PLACE Mathematics 04 Includes 23 competencies/skills found on the PLACE Mathematics test and 80 sample-test questions. This guide, aligned specifically to standards prescribed by the Colorado Department of Education, covers the sub-areas of Foundations of Mathematics; Functions and Relations; Measurement and Geometry; Probability and Statistics; and Calculus and Discrete Mathematics. | 677.169 | 1 |
BJU Press Precalculus Teacher's Edition
Equip your 12 grade students for college with precalculus skills.
Lessons emphasize functions, graphing, and trigonometry throughout
and introduce differential calculus. This book replaces Advanced
Math.
If this is
your first time teaching Precalculus, prepare diligently. Consider
the students when assigning homework. Some students require
considerable practice to grasp a concept, but explaining principles
well before assigning drills can reduce the amount needed.
Suggested assignments offer three alternatives: (1) minimum, (2)
standard, and (3) extended.
This Teacher's
Edition is a two-book, spiral bound set that provides guidance to
teachers including common student errors, answers and solutions to
the exercises, additional problems, and tips for exercises. One of
the Suggested Schedules offer 139 class periods covering 85
sections and allows 4-5 extra class periods per chapter for special
features, difficult lessons, review, and tests. Book One covers
chapters 1-6 and contains chapter reviews at the end of each
chapter as does Book Two, which covers chapters 7-12. These chapter
reviews contain a variety of exercises that will help students
determine whether they have met the chapter objectives and whether
they are prepared for a test. Each book contains a Glossary, Index,
Tables, Symbols, Photograph Credits, and Appendices. | 677.169 | 1 |
Thursday, November 17, 2005
Teaching Repertoire: How I Teach Calculus (Part II: Problem Solving)
The dichotomy I've stated is a slight over-simplification. But with few exceptions the complement of lecture is problem solving. When I say "problem solving," I am not restricting myself to word problems: rather I mean any mathematical task that is not a routine computation, calculation, or exercise. Anything where I would expect many of my students would get frustrated if they were sent home to complete this task for homework. (I do assign frustrating homework from time to time -- just not very much of it.)
And I have the luxury to do this because my department has a delightfully leisurely syllabus for Calc I. This would not be possible in other departments.
Unlike lecture days, problem solving days are much less structured. I still start with the announcements. Sometimes, while I still have the attention of the whole group, I will give a five-minute mini-lecture or reminder or else I will model an example that illustrates key ideas, but then I will hand out the sheet of problems for the day. I will encourage my students to form groups (most choose to work in one-person groups, which is FINE), and they will start working on the problems. I will circulate around the room answering questions, giving hints, or just listening.
Almost always problem-solving day is announced ahead of time, and the problems are usually posted on the class webpage. (If they're problems I've written they are; if I use problems from the Big Binder that came with my textbook, then they don't go on the web.) If someone doesn't want to come on problem-solving day, that is FINE. This is college, and they need to learn how to make their own decisions. I do take attendance. They know that I know when they come to class.
As I circulate around the room, I ask the students how the problems are going and how class is going. After 15 minutes or so I will decide (based on my interactions with my students) if I need to present anything to the whole class. If so, I will talk for five or 10 minutes about something that will help them work on the task at hand. I may need to clarify an idea, present another example problem, or work through the solution to one or more of the problems on the sheet. Sometimes I will ask a student to present a problem at the board, but I don't do that very often. I will only select students who I'm sure can present things well and who can field questions accurately. While we are convened as a large group, I will field questions. Once everyone is done asking questions, we return to small group (individual) work.
I still haven't figured out the best way to wrap things up on problem-solving day. Often I will devote the last five minutes of class to going over at least one of the problems. Sometimes I will withhold answers until the end. | 677.169 | 1 |
Fairfield Secondary Math Night Presentation used the following slide to prompt the new math program CPM (College Preparatory Mathematics) which is currently being implemented in all Algebra I classes across the district this year. | 677.169 | 1 |
Lie Groups: A Problem-Oriented Introduction via Matrix Groups
The work of the Norwegian mathematician Sophus Lie extends ideas of symmetry and leads to many applications in mathematics and physics. Ordinarily, the study of the "objects" in Lie's theory (Lie groups and Lie algebras) requires extensive mathematical prerequisites beyond the reach of the typical undergraduate. By restricting to the special case of matrix Lie groups and relying on ideas from multivariable calculus and linear algebra, this lovely and important material becomes accessible even to college sophomores. Lie Groups is an active learning text that can be used by students with a range of backgrounds and interests. The material is developed through 200 carefully chosen problems. Solutions to selected problems. An instructor's manual comes with the adoption of this title. | 677.169 | 1 |
Courses
Useful Links
FOUNDATIONS FOR HIGHER MATHEMATICS
An introduction to the rigorous techniques used in advanced mathematics. Topics include basic logic, set theory, methods of proof and counterexamples, foundations of mathematics, construction of number systems, counting methods, combinatorial arguments and elementary analysis. Preq: Math 233 or Math 233 concurrently with permission of instructor. | 677.169 | 1 |
Work & Power Problems
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Here are a variety of work and power problems for students either in pre-algebra or algebra 1. Some problems have multiple parts in order to tie together the Work (W = F*d) and Power (P = W/t) equations with the occasional time unit conversion. As with my other problem sets, I aimed for a variety of situations both to hold students' interest and to give them different contexts in which to picture the concept. My students often take anywhere from 15 to 40 minutes on this, depending on their math ability and motivation. | 677.169 | 1 |
An Introduction to Functional Analysis
Functional analysis is the branch of mathematics dealing with spaces of functions. It is a valuable tool in theoretical mathematics as well as engineering. It is at the very core of numerical simulation. In this class, I will explain the concepts of convergence and talk about topology. You will understand the difference between strong convergence and weak convergence. You will also see how these two concepts can be used.
Instructor(s) John Cagnol, Anna Rozanova-Pierrat
University
Provider
Start Date 12/Sep/2014
Duration 6 | 677.169 | 1 |
The year begins with students
extending their understanding of whole number exponents learned in previous
grades to integer exponents in Unit 1. Students develop the properties of
exponents. The unit progresses to include work with scientific notation which
supports student understanding of positive and negative exponents. Students
also work with radicals as they use square roots and cube roots to represent
solutions to equations of the form x2=p and x3=p. They
evaluate square roots and cube roots of small perfect squares and cubes.Later in the unit, in preparation for work
with geometric contexts such as the Pythagorean Theorem in Unit 3 of this
course, students extend their concept of numbers beyond the system of rationals to include irrational numbers as they estimate
roots of non-perfect squares. They represent these numbers with radical
expressions and approximate these numbers with rationals.
Students use the number line model to support their understanding of the
rational numbers, along with the inclusion of the irrational numbers, to
develop the real number line.
Standards:
8.NS.A.1Know that numbers that are not rational are
called irrational. Understand informally that every number has a decimal
expansion; for rational numbers show that the decimal expansion repeats
eventually, and convert a decimal expansion which repeats eventually into a
rational number.
8.NS.A.2Use rational approximations of irrational
numbers to compare the size of irrational numbers, locate them approximately
on a number line diagram, and estimate the value of expressions (e.g., 𝜋2). For example, by truncating the decimal
expansion of √2, show that √2 is between 1 and 2, then between 1.4 and 1.5, and explain how to continue on to get a
better approximation.
8.EE.A.2Use square root and cube root symbols to
represent solutions to equations of the form 𝑥2 = 𝑝and 𝑥3 = 𝑝, where 𝑝is a positive rational number. Evaluate square
roots of small perfect squares and cube roots of small perfect cubes. Know
that √2
is irrational.
8.EE.A.3Use numbers expressed in
the form of a single digit times an integer power of 10 to estimate very
large or very small quantities, and to express how many times as much one is
than the other.For example, estimate the population of the United States as 3 × 108
and the population of the world as 7 × 109, and determine that the
world population is more than 20 times larger.
8.EE.A.4Perform
operations with numbers expressed in scientific notation, including problems
where both decimal and scientific notation are used.Use scientific notation and choose units of
appropriate size for measurements of very large or very small quantities
(e.g., use millimeters per year for seafloor spreading).Interpret scientific notation that has been
generated by technology.
Focus Standards of
Mathematical Practice:
MP.1 Make
sense of problems and persevere in solving them.
MP.2 Reason
abstractly and quantitatively.
MP.3
Construct viable arguments and critique the reasoning of others.
MP.4 Model
with mathematics.
MP.5 Use
appropriate tools strategically.
MP.6 Attend
to precision.
MP.7 Look for
and make use of structure.
MP.8 Look for
and express regularity in repeated reasoning.
Instructional
Outcomes:
Full Development of the Major
Clusters, Supporting Clusters, Additional Clusters and Mathematical Practices
for this unit could include the following instructional outcomes:
8.NS.A.1
I can define rational and
irrational numbers.
I can show that the decimal
expansion of rational numbers repeats eventually.
I can convert a decimal expansion
which repeats eventually into a rational number.
I can show informally that every
number has a decimal expansion.
8.NS.A.2
I can approximate irrational
numbers as rational numbers.
I can approximately locate
irrational numbers on a number line.
I can estimate the value of
expressions involving irrational numbers using rational numbers.
Examples: Being able to determine
the value of the √2 on a number line lies between 1 and 2, more
accurately, between 1.4 and 1.5, and more accurately, etc.
I can compare the size of
irrational numbers using rational approximations.
8.EE.A.1
I can explain why a zero exponent
produces a value of one.
I can explain how a number raised
to an exponent of -1 is the reciprocal of that number.
I can explain the properties of
integer exponents to generate equivalent numerical expressions. | 677.169 | 1 |
Help! Algebra
Writing Linear Equations, Linear Systems, and Linear Inequalities
Travis HostetterHelp! Algebra is designed to be an easy and simple way for you to learn. No more waiting on teachers or searching the internet, everything you need can now be found in one place.
Inside, you will find teacher made videos, step-by-step notes, examples, and practice problems to help you become the best Algebra student possible. Everything has been created with the learner in mind. | 677.169 | 1 |
Mathematics
(Collins College fraions, decimals, and geometric measurement to exponents, scientific notation, and an introduion to algebra, the "Collins College Outline" in "Basic Mathematics" explores and explains the topics that students will find in introduory mathematics classes. Completely revised and updated by Dr. Lawrence Trivieri, "Basic Mathematics" includes a test yourself seion with answers and complete explanations at the end of each chapter. Also included are bibliographies for further reading, as well as numerous graphs, charts, and example problems.
The "Collins College Outlines" subje for college students and, where appropriate, Advanced Placement students. Each "Collins College Outline" is fully integrated with the major curriculum for its subje and is a perfe supplement for any standard textbook. | 677.169 | 1 |
Math Terms Glossary
It's important for all students to use math vocabulary when describing how they've solved a problem. Math Terms Glossary is a free app that students can use to access definitions of math terms. Most entries include pictures and a Spanish translation of the definition | 677.169 | 1 |
 Reflection: Checks for Understanding The Cell Phone Problem, Day 1 - Section 3: The Cell Phone Problem, Group Time
I've included here a sample of some student work. These are representative of the class, and I've highlighted some things that stood out to me as I reviewed them. Not only am I looking for what the students know, but I'm also on the lookout for instructive errors. These are typically unexpected issues that I may need to address in class.
I'll begin with a brief conversation to introduce the notion that cell phones use a radio signal with a signal strength that isn't always the same and when the signal gets too weak the call is dropped. We then will have a class discussion where I'll ask students to suggest situations/objects that contribute to weak signals. With the right prompting distance from cell site should be emphasized as the focus of today's lesson. See Getting Started for more details.
Resources
I'll hand out The Cell Phone Problem, Part 1and ask students to work individually at first. Their aim should be to get into problem 3 by the time these 10 minutes are up.
The first question is a simple one, but it's important to make sure that students have a general, intuitive notion of what an inverse relationship is. The second question gets into the details of that relationship and allows students to take the first steps toward creating an equation to model it. Finally, the third problem asks explicitly for students to construct a mathematical model and to represent it with both an equation and a graph.
In my experience, the most challenging aspect of this problem is in reading the verbal description of the relationship between signal strength and distance. Too many students pay too little attention to the carefully chosen words, and this presents an opportunity to stress the importance of precision in mathematics. (MP 6)
Resources
For this next section I have students work in groups of three, and they should produce professional-looking graphs, correct equations, and answers to the questions posed in problems 4 and 5. I encourage the students to provide both graphical and analytic solutions to these problems (MP1). Since this problem is intended as an introduction to the study of rational functions, pay careful attention to the way students approach the analytic solution! This is your chance to see what they already know and what they'll need support with throughout the unit.
With regard to problem 5, I do not expect students to be able to explain why their model is unreliable, but they should be able to see clearly that it is. The physics of the situation is really complicated, and it's enough to simply assert that the model is reliable when the cell phone is a reasonable distance from the tower.
First, I use a document camera to display one group's graph and equation for a whole class discussion. Hopefully, all groups will have the same graph, keeping in mind scaling may affect the appearance. If I encounter this case, I will ask how different groups decided on the domain and range for their graph.
Once we have had the overall conversation it is important to make sure that students have drawn important mathematical conclusions and can interpret them in context:
Once the graph drops below a certain level, it will never rise again.
This means that the signal strength continues to diminish the further away you go. It approaches zero, which is reasonable.
The graph has a vertical asymptote at d = 0.
The signal strength "approaches infinity" the closer you get to the source. This is nonsense!
It isn't necessary to go into the physics of the situation, but help your students to see that the model is unreliable when the distance is too small.
This is a great idea for a lesson. I might make a few modifications for my class but definitely expect to use it. I found a pretty good slide show about wireless networks at target="_blank" > It might be beyond the reach of most our students (I work with 8th graders) but some of the graphics are good.
A couple of modifications I'm considering, mostly as extensions: I had to read the first paragraph on the constancy of the product of s and d^2 to realize you weren't saying that strength is proportional to d^2 instead of inversely proportional. It's clear in the problems and graphs and intuitively makes sense that you're not going to get a stronger signal from farther away, but I did think, "Wait, that's not right!" a couple of times.
Also, I think I might do a dimensional analysis of the constant of proportionality. If I'm right it turns out to be (mass)(length)^4/(time)^2.
There are lots of good ways to illustrate the inverse square law - for example watching a flashlight's beam expand as you walk away from a wall. There's also lot's of good illustrations available.
Finally, I really like the question at the end about the signal strength at the tower itself - a great example of perils of dividing by 0! | 677.169 | 1 |
Why I like the Art of Problem Solving books
I'm working through two of Art of Problem Solving's math books with the kids this year. My older son is studying "Introduction to Geometry" and my younger son is studying their "Introduction to Number Theory" book. You can buy those books here:
I've never taught any elementary math before, but the approach that Richard Rusczyk and his team take to explaining the subjects really resonates with me. It was actually their Prealgebra book that got me hooked on their approach, and it sure seems the more of their books I work through the more I love what they do. Though the approach is great, the icing on the cake is the collection of problems. Just an absolutely outstanding set of problems ranging from introductory to Olympiad level problems to challenge all types of kids looking to learn from their books.
We ran across a really nice challenge problem in the chapter about congruence today that made lots great math conversation. Both the problem itself and the more advanced theorem from "Geometry Revisited" that it hinted at were super fun to talk through with my son. Here's us doing a quick review of the problem itself tonight:
and then here are a couple of theorems on Napoleon triangles from "Geometry Revisited" that the problem practically begs you to talk about 🙂
The Art of Problem Solving problems are so great to begin with and it is sort of doubly fun to be able to use them as stepping stones to show some more advanced geometry. Next up for us is the section on perimeter and area. Can't wait.
Comments
The books definitely have some great strengths, especially the problem sets. There is some unevenness in the quality–for example, the Calculus book is a bit of a letdown.
I use the books as a supplement, and for some one-off units (like on inclusion-exclusion, state diagrams, and such). I wouldn't consider using them as a primary textbook, so it's interesting to hear about your experience in that regard.
I haven't seen the Calculus book. Right now I'm planning on using Spivak, but it is so far in the future that it probably silly for me to even be thinking about that. I hope he has a nice house, though.
The Prealgebra book is tremendous. I think it is newer than the others, so the writers probably benefited from writing a few other books first. The one that I can't wait to go through, btw, is the Precalculus book. Some of the chapters there are just amazing – complex numbers and geometry, for example.
I wouldn't be surprised if these books lent themselves more to one on one teaching. I'm also not in any hurry, so I'm happy to spend as much extra time as I need on difficult sections, or smoothing out some sections that are rough. | 677.169 | 1 |
Advance CALCU Calculator is a simple calculator, with a panel that has more advanced functionshandyCalc Calculator HandyCalc is a powerful calculator with automatic suggestion and solving which makes it easier to learn and use | 677.169 | 1 |
- The idea is to give the student
challenging and motivating problems and have him or her use
the tools provided in this site to explore new concepts,
discover new ideas, and improve the student's abilities to
solve problems. There is a nice section on the Pythagorean Theorem.
Chemistry
The Virtual Lab - - a collection of virtual labs, scenario-based learning activities, and concepts tests which can be incorporated into a variety of teaching approaches as pre-labs, alternatives to textbook homework, and in-class activities for individuals or teams.
Upper Level Resources
- Maintained by the University
of British Columbia as a supplement to their Math 101 course, this page
contains interactive tutorials and demonstrations on fundamental calculus topics,
including Area under a Curve and Average Value.
demonstrations and explorations of many fundamental Calculus concepts,
including limits, Linear Approximation and Riemann Sums.
Also contains information on how to write calculus programs for TI-83 calculator.
Demonstrations Project was conceived by Stephen Wolfram as an open-code resource
that uses dynamic computation to illuminate concepts in science, technology,
mathematics, art, finance, and a range of other fields. | 677.169 | 1 |
This text is for courses in numerical methods offered in departments of civil or mechanical engineering and departments of mathematics. Its objective is to introduce the engineer and scientist to numerical methods used to solve mathematical problems that cannot be solved by exact methods. With the general accessibility of highspeed digital computers, it is now possible to obtain rapid and accurate solutions. Mastery of the material presented in this book will prepare engineers and scientists to solve many of their everyday problems, give them insight to recognize when other methods are required, and give them the necessary background to study additional methods. Part I discusses the Basic Tools of Numerical Analysis and considers many of the basic problems that arise in all branches of engineering and science. Part II is devoted to the numerical solution of ordinary differential equations (ODEs). Part III covers the numerical solution of partial differential equations (PDEs).
The approach taken is to introduce a type of problem, present sufficient background to understand the problem and possible methods of solution, develop one or more numerical methods for solving the problem, and illustrate the numerical methods with examples. In most cases, the numerical methods presented to solve a particular problem proceed from simple methods to complex methods | 677.169 | 1 |
Guided Notes: The Parts of an Equation
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This lesson teaches students the difference between a term, an expression, an equation, a variable, a constant, and a coefficient. It is a guided notes sheet with built in practice questions and a homework sheet. An answer key is also included. This lesson should take 45-50 minutes to complete. | 677.169 | 1 |
ISBN-10: 0470464739
ISBN-13: 9780470464731 Crossing the River with Dogs: Problem Solving for College Students has been adapted from the popular high school text to provide an accessible and coherent college-level course in mathematical problem solving for adults. Focusing entirely on problem solving and using issues relevant to college students for examples, the authors continue their approach of explaining classic as well as non-traditional strategies through dialogs among fictitious students. This text is appropriate for a problem solving, liberal arts mathematics, mathematics for elementary teachers, or developmental mathematics course.
Kenneth Johnson holds a degree in Religious Studies from California State University Fullerton. His emphasis was in the study of mythology and this interest is reflected in his writing and his astrological practice. Kenneth discovered astrology while traveling in Europe during the summer of 1973. He studied in Amsterdam and London before returning to the United States and developing a practice which focuses on archetypal themes and personal mythologies. In addition to his astrological interests, Kenneth is also a musical theater librettist and a member of the Dramatists | 677.169 | 1 |
SAGE NOTEBOOK SERVER
There has been some ongoing development during the last months to improve the sage notebook server. The new server, which scales much better under heavy load is now in beta stage. People are invited to test it. It gives an extremly comprehensive CAS/mathematic System. You can store / share / collaborate your notebooks and calculations online!
WOLFRAM ALPHA
This is a computational knowledge machine. It has access to huge databases and uses Mathematica for calculations. Moreover it tries to give the output in a ' layman' langugage and uses AI to parse the input, i.e. it tries to understand the question. I'd say its an invaluable tool for homeworks | 677.169 | 1 |
, is the total number of occurrences of symbols in x. I'm not Einstein ;-) but I'll try and put you on the right track... The following are the basic rules for solving any linear equation. But to make it an expectation for all students is not doing them any good." The age at which a student takes algebra must be determined on a case-by-case basis. By contrast, matrix multiplication is not generally commutative.
I think it's unfortunate that more algebra books do not use it as motivation. I enjoyed learning to crack the problems with College Algebra, Algebra 2 and Algebra 2 in math. But none of these obstacles adequately explains the observed difficulties. To succeed in abstract algebra, one must eventually be able to accomplish several things. He received a "B" in the art class, an "A" in the history class, a "C" in the science class, a "B" in the mathematics class, and an "A" in the science lab.
These methods can be applied to solve Problem 4 in Section 37 on the homework. (2/14) My office hours today, Mon Feb 14, will be 1:30-2:30 instead of 1-2. (2/25) Here is a link for an article about the axioms which define a Euclidean domain. For example, in algebra, "3x + 1" is understood by convention to mean the quantity that is one more than the product of 3 times the variable "x", rather than the product of three times the number which is one larger than x.
However, the symbolism that is first introduced in elementary algebra permeates all of mathematics. That apart, similar, redundant, and obscure components may have been incorporated into the 'Rj's description without discretion. Part II deals with the methodology of solving logic problems by (i) translating them to algebra, (ii) solving the algebraic problem, and (iii) translating the result back to logic.
Why did Birkhoff & Mac Lane prove the uniqueness of a ring's additive identity this way? I would like to talk about the meaning of cosets in class. The term "abstract algebra" is used to distinguish the field from "elementary" or " college algebra " which teaches the correct rules for manipulating formulas and algebraic expressions. There is also an accompanying SAGE workbook by Rob Beezer that supports the text ( SAGE is an open-source software package that does abstract algebra, including operations with finite groups, polynomial rings, finite fields, field extensions, and more.) In short, AATA is a stellar example of open-source at its best.
It is an expensive book, but I got it for a much better price here on amazon, I think it was less than half of what they were asking at the UCLA bookstore. Try a different browser if you suspect this. He brilliantly unifies mathematics into a clear depiction that urges readers to rethink what they thought they knew already. Here is a pdf of some notes on Group Actions. Solutions to all problems are included and some of the reasoning is informal. What can a student seeing this stuff for the first time learn from this example??
If 1=0 in the zero ring, why don't they call it The One Ring instead? In addition, there was the study of permutation groups, which was originally not thought of as being algebra at all, I believe, but where the basic concepts were developed by Legendre, Abel, and Galois as an approach to understanding the solution of algebraic equations. Discrete Fourier transforms and function transforms in general require a good grasp of algebra, including the part that overlaps with "number theory".
Suppose, for example, that a = 4 and b = 6. I tutored at OSU in the math center for all six years I was there. This was perhaps especially clear in the work of Legendre, Abel, and Galois on permutation groups, where what was important was the set of subgroups rather than the individual permutations. Here are the search phrases that today's searchers used to find our site. We developed a software package that computes the simplified structure digraph of the achievement and avoidance games.
Most terms in mathematicshave beenwell chosen; there are more red namesthan aggravating ones. The friendliest, high quality science and math community on the planet! Studying category theory without a fair amount of prior experience with abstract algebra is like studying calculus without knowing how to graph a linear equation: it makes no sense. Math work sheet for grade 5, standard notation and number and word notation-math, algebra help with finding the square root of an exponent. | 677.169 | 1 |
The book contains the material of a one-semester course on nonstandard analysis. After a brief introduction to the necessary background in logic, it starts by discussing nonstandard versions of the basic notions of elementary calculus, as e.g. convergence, continuity, derivative, and Riemann integral. Subsequent chapters cover gradually more advanced subjects. Nonstandard method is used to prove several fundamental and important theorems in topological and metric spaces, real, and complex analysis. The book can be used by engineers, physicists, mathematicians, but first of all, university students, who want to understand what the infinitesimals are, how the nonstandard method works, and what are its advantages and disadvantages. By providing nonstandard proofs for significant, interesting and deep mathematical theorems, it examplifies that the nonstandard method is not a toy, but could be a handy research tool. The author is the head of the Computer and Statistics Center of the Central European University, Budapest, and researcher at the Eotvos Lorand University. His personal web page is at | 677.169 | 1 |
Departments & Programs - Mathematics
Mathematics Department
The mathematics department at New Westminster Secondary School offers a wide range of courses intended to meet the diverse needs of our students. Before deciding on an appropriate course, students should take into account their current level of aptitude in mathematics and any academic qualifications required for their long-term career plans. To ensure optimal placement, students should make course selections in consultation with their math teacher. The members of the mathematics department are always ready to advise students on course selection | 677.169 | 1 |
A lesson designed to familiarize students with the concept of the sine function by defining, graphing, computer generating, using a "shop created" blackboard sine wave generator, and by presenting examples of natural phenomena which result in sine wave motion. From the Algebra and Trigonometry section of a collection of almost 200 single concept lessons by the Science and Mathematics Initiative for Learning Enhancement. | 677.169 | 1 |
Listing Detail Tabs
'On Track 3' is a write-on student workbook, produced by Sigma Publications, that provides a comprehensive study programme for students wishing to sit a combination of NCEA Level 1 Mathematics Achievement Standards.
Instruction boxes and worked examples appear on almost every page to help students with independent study. At the end of each chapter there is a valuable revision test which can be used by candidates to evaluate their strengths and weaknesses before assessments. A full set of answers is included, which can act as a guide to structuring quality answers. Sigma has now perforated all pages so that homework assignments can be handed in while work is ongoing and the answer pages may be retained by the teacher if desired.
192 pages (includes a full set of removable answers) ISBN: 978-1-877567-18-6
---- Sigma Publications Ltd produces write-on educational workbooks based on the New Zealand curriculum. We specialise in the skills-based core subjects of Mathematics and English. We have workbooks available for Years 3 to 13 in both subjects including NCEA secondary level books | 677.169 | 1 |
Maths Skills
Maths may well have been your least favourite subject at high school, and you possibly thought, "I'm going to university and won't need to study maths anymore!"
Unfortunately this is not necessarily the case, as numerous courses at the University will involve mathematics, even if the subject itself is not maths or science specific. All is not lost. Whilst you may not have enjoyed maths at school, you may find it very different at uni, and there are certainly many ways to get help and advice if you 're having difficulties.
For personalised advice on your maths problems, visit the Maths Learning Centre, where you can get assistance with your assignments and any aspects of your maths study that you are having difficulty with. Maths Learning Centre staff will not proofread, edit, or correct your assignments, but they will help you identify areas for improvement and build your skills so that you can succeed in your studies independently. | 677.169 | 1 |
High School Mathematics Standards
Throughout high school, students learn how to apply math skills in six categories. Starting in ninth grade, your student needs to gain solid skills in math. By senior year, your 12th-grader should strive to improve and perfect math skills.
Here are a few of the math concepts high school students will practice, depending on the courses they choose.
Number and Quantity
Increase knowledge of numbers to include complex numbers, irrational numbers and imaginary numbers.
Q: For a new car priced at $22,000, James takes a five-year loan with an interest rate of 5.3%. By the time he owns the car, how much will he have paid including principal (the original cost) and interest?
A: Using the formula I = P × R × T, where I = Interest, P = Principal, R = Rate, and T = Time, we can find the interest on the loan, then add it to the original payment of the car. I = 22,000 × 0.053 × 5 I = $5,830
Total payment = 22,000 + 5,830 = $27,830
Understand the various ways a function can be described, including algebra expressions, graphs, or verbal or recursive rules, and construct different types of functions.
Q: Compound Interest: You go to the bank and take out a 5-year, $10,000 loan. How much would you owe after the 5 years if the bank had an 8% interest rate?
A: The formula for compounded interest is FV = PV × (1 + r)n, where FV is the future value, PV is the present value, r is the interest rate, and n is the number of periods. FV = 10,000 × (1 + 0.08)5 FV = 10,000 × (1.08)5 FV = 10,000 × 1.469 FV = $14,690
Modeling
Use quantities, measurements and geometry to create mathematical models for real-life situations.
Q: Prove the two lines are parallel.
Angle m measures 37° and angle p measures 37°. Are the two lines parallel?A: Yes, the two lines are parallel by reasoning of the Alternate Interior Angles Converse. If angle m ≅ angle p, then the two lines are parallel.
Statistics and Probability
Apply the basic concepts of probability to real-life scenarios, such as the likelihood of the ace of spades being chosen from a deck of cards.
Q: A single 6-sided die is rolled. What is the probability of rolling a 2 or a 5?
A: P(rolling a 2) = P(rolling a 5) =
Addition rule of probability: P(A or B) = P(A) + P(B)
So, P(rolling a 2 or 5) = P(rolling a 2) + P(rolling a 5)
Talking to your teen's teacher about high school math standards
If math wasn't your strongest subject when you were a student, you might feel anxious when it comes to supporting your son or daughter's math skills. Math teachers are a great resource for feedback on how your teenager is doing, but they're also able to explain changes in how math is taught. Here are helpful questions to ask during parent/teacher conferences or in one-on-one meetings:
Does my son or daughter have solid arithmetic skills? Can he or she solve problems containing fractions, decimals and negative numbers?
Is he or she struggling in one or more math disciplines? What are my options for getting him or her up to speed?
Are there any areas of math where she or he excels? If so, can you suggest the next steps to take?
Does the school offer enrichment activities for STEM (Science, Technology, Engineering, Math) subjects?
If my son or daughter needs one-on-one help with math, can you recommend a peer-tutoring program or a professional tutor or learning center?
Is my high school student gaining financial literacy? Does she or he know how to calculate tips, credit card payments and interest, and create a budget?
How can you support your teen's math skills at home?
Have a discussion with your son or daughter about interesting careers that use math skills. Careers in design, engineering, software and game development, architecture and even animation are open to people who are comfortable working with numbers.
Help your teenager find and use free online resources for both enrichment and extra help. Many sites provide math tutorials, practice problems and in-depth explanations of mathematical concepts.
Give your son or daughter math-related jobs to do. Ask them to calculate the tip after a restaurant meal or to determine how much money they'll save shopping with a certain percentage discount.
Encourage your son or daughter to be persistent when learning new mathematical skills; help them get comfortable asking questions during or after class.
If she or he is a sports fan, talk about the ways math is used in baseball, football, basketball and other professional sports. If your daughter or son loves football, discuss how fantasy football leagues use math.
Ask to see math homework and encourage your teenager to explain the concepts she or he uses to you.
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Monomial Algebras, Second Edition presents algebraic, combinatorial, and computational methods for studying monomial algebras and their ideals, including Stanley–Reisner rings, monomial subrings, Ehrhart rings, and blowup algebras. It emphasizes square-free monomials and the corresponding graphs, clutters, or hypergraphs. New to the Second Edition... more...
Advanced Linear Algebra, Second Edition takes a gentle approach that starts with familiar concepts and then gradually builds to deeper results. Each section begins with an outline of previously introduced concepts and results necessary for mastering the new material. By reviewing what students need to know before moving forward, the text builds... more...
A reader-friendly introduction to modern algebra with important examples from various areas of mathematics Featuring a clear and concise approach , An Introduction to Essential Algebraic Structures presents an integrated approach to basic concepts of modern algebra and highlights topics that play a central role in various branches of mathematics.... more...
Introduction to Higher Algebra is an 11-chapter text that covers some mathematical investigations concerning higher algebra. After an introduction to sets of functions, mathematical induction, and arbitrary numbers, this book goes on considering some combinatorial problems, complex numbers, determinants, vector spaces, and linear equations. These... more...
Algebra and Trigonometry presents the essentials of algebra and trigonometry with some applications. The emphasis is on practical skills, problem solving, and computational techniques. Topics covered range from equations and inequalities to functions and graphs, polynomial and rational functions, and exponentials and logarithms. Trigonometric functions... more...
Algebra for College Students, Revised and Expanded Edition is a complete and self-contained presentation of the fundamentals of algebra which has been designed for use by the student. The book provides sufficient materials for use in many courses in college algebra. It contains chapters that are devoted to various mathematical concepts, such as the... more...
College Algebra, Second Edition is a comprehensive presentation of the fundamental concepts and techniques of algebra. The book incorporates some improvements from the previous edition to provide a better learning experience. It provides sufficient materials for use in the study of college algebra. It contains chapters that are devoted to various... more...
College Algebra and Trigonometry, Second Edition provides a comprehensive approach to the fundamental concepts and techniques of college algebra and trigonometry. The book incorporates improvements from the previous edition to provide a better learning experience. It contains chapters that are devoted to various mathematical concepts, such as the... more...
First Course in Algebra and Number Theory presents the basic concepts, tools, and techniques of modern algebra and number theory. It is designed for a full year course at the freshman or sophomore college level. The text is organized into four chapters. The first chapter is concerned with the set of all integers - positive, negative, and zero. It... more...
Lectures in General Algebra is a translation from the Russian and is based on lectures on specialized courses in general algebra at Moscow University. The book starts with the basics of algebra. The text briefly describes the theory of sets, binary relations, equivalence relations, partial ordering, minimum condition, and theorems equivalent to the... more... | 677.169 | 1 |
Symposia Mathematica, Volume I focuses on research in the field of mathematics and its applications. This book discusses the definition of S-semigroup, extensions of R modules, structure of H, laws of conservation and equations of motion, and measures of strain. The basic equations for continua with internal rotations, general concepts of the discrete... more...
Group Theory and Its Applications focuses on the applications of group theory in physics and chemistry. The selection first offers information on the algebras of lie groups and their representations and induced and subduced representations. Discussions focus on the functions of positive type and compact groups; orthogonality relations for square-integrable... more...
Make studying statistics simple with this easy-to-read resource Wouldn't it be wonderful if studying statistics were easier? With U Can: Statistics I For Dummies, it is! This one-stop resource combines lessons, practical examples, study questions, and online practice problems to provide you with the ultimate guide to help you score higher in... more...
International mathematics education researchers give a differentiated overview of views and beliefs of both teachers and students. Beliefs about how to teach mathematics have a high impact on the instructional practice of teachers. In the same way, views and beliefs about mathematics are an essential factor to explain achievement and performance of... more...
A mathematical gem–freshly cleaned and polished This book is intended to be used as the text for a first course in combinatorics. the text has been shaped by two goals, namely, to make complex mathematics accessible to students with a wide range of abilities, interests, and motivations; and to create a pedagogical tool, useful to the broad... more...
On the occasion of the celebration of "Twenty Years of Didactique of Ma- ematics" in France, Jeremy Kilpatrick commented that though the works of Guy Brousseau are known through texts referring to them or mentioning their existence, the original texts are unknown, or known only with difficulty, in the non-Fren- speaking world. With very few exceptions,... more...
This book focuses on aspects of mathematical beliefs, from a variety of different perspectives. Current knowledge of the field is synthesized and existing boundaries are extended. The volume is intended for researchers in the field, as well as for mathematics educators teaching the next generation of students. more...
... more... | 677.169 | 1 |
Are you a math-phobe?Do you lack confidence in your ability to learn math?Then eMath is for you.We've designed eMath from the
perspective of the student, and we are continually updating it as we discover
new ways to help our students.We
are also striving to make eMath as user friendly as possible so that each
student can log in and begin using the program in valuable ways their very first
time.Having said these things, you
need to know that your input is very valuable for us.
Any time you have something to suggest or
a criticism to pass on, please email us at
Support@emath.com.
Learning Mathematics
Why does it seem so difficult to learn mathematics?The very heart of the matter in
learning anything that requires analytical thinking is the need we all have to
connect with the underlying logic.
This means we need to develop more than just a memorized process.If we stop with a memorized process,
our ability to use mathematics is limited by our ability to maintain that
memorized process.Even more,
though, our ability to know which process to use when is totally lacking.So, without a true understanding of
the concept, that is, the logic that underlies the process and actually drives
its creation, mathematics isn't a very usable tool.Unfortunately, treating mathematics
as a collection of disconnected processes is pretty much the current state of
mathematics education in this country.
Designed to Produce Real Knowledge
Now, what is this program designed to do?What help are we offering you?Everything we're incorporating into
eMath is aimed at teaching mathematics from a conceptual perspective.We continue to work very hard to
facilitate the learning of the underlying logic of mathematics that enables all
of us to remember mathematics better, and to truly take it from being 'that
class in school that I hated' to being a tool that enables me to problem solve
better every day.Plus, as we begin
to understand the underlying logic, those processes that we work so hard to
memorize become quite natural because they are actually driven by our
understanding.An understanding of
the logic allows us to re-create the processes when we forget some of the
details about those processes.
When you first log in to your class on eMath you'll encounter
the page we call the Dashboard.
Among the important items on this page you'll see a color coded pie chart that
shows you your up to date Knowledge Mastery Status.Each different color represents your
level of mastery on each of the concepts you've covered so far in the class.This status is automatically
generated and updated every time you do some work inside the program.You can click on the chart and be
taken to a page that gives detailed information about which concepts fall under
each of the colored sections.You
will also be offered additional teaching information about each concept for each
colored section and specific additional practice to help you improve in the
areas you haven't yet mastered.
You'll also have access to additional practice on those areas
in which you've shown excellence, just so that you can keep cementing those
concepts into your knowledge base.
When you step away from things you've learned, coming back again to practice
builds your knowledge depth and actually allows you to wrestle with a concept on
a deeper level than youdo when you first learn it.This re-visiting is an important
facet of how all of us learn.Those
of us who teach the same material over and over, and do so re-thinking the
concepts each time, find this to be very true.When you force yourself to re-think each time you work with a concept its
amazing the new insights that come as a reuslt.
Each problem you work within this program comes with a link
to detailed information and explanation of each concept incorporated into the
problem, a link to the correct answer, and a link for a detailed solution for
the problem.The concept information
includes, along with detailed information about each concept, links to more
information and more practice on any one of those concepts because any one of
them could be the source of your difficulty with the original question.
eMath provides lot's of detailed help right there at your
finger tips, right from the first page you open, and all of the help is built on
conceptual teaching of the underlying concepts.No matter what link you click on to generate a practice problem, you will
find links to this information.All
of the information is designed to enable you to truly learn the underlying
concepts of the mathematics you are being taught.It is this conceptual learning that will enable you to truly master the
mathematics and make it a life long usable tool for you.
Test Preparation Help and More
The assignments available in eMath are designed to help you
master the material for the next in-class exam you'll be given.However, important information is
also provided for your use on a daily basis as you cover material in your class
and try to work textbook assignments.We have incorporated four study guides in eMath that are of a progressive
nature.The first study guide you
should go to is the Question Guide which provides another look at the kinds of
questions that you'll see on the coming exam.Again, each question has all the links to the helpful information.
The second guide is the Question Type guide.This guide provides detailed descriptions of the concepts underlying each
type of question and links to practice those concepts.Again, each question contains all the
links to help with the question.
The third study guide is called the Knowledge Guide.This study guide provides detailed
explanations for each of the concepts covered in a particular chapter.With each concept there is
information on your level of mastery, a link to 'Help you improve' your mastery
of the concept, and a third link that will take you to the fourth study guide or
to a sample problem.
The fourth study guide is the Knowledge Map.This guide shows a map of how each
concept relates to underlying, prerequisite knowledge.This guide can be very helpful to
your understanding of how a previous concept can keep holding you back until you
really learn it.This guide can also
help you to more precisely diagnose the difficulty you have with a particular
problem type or concept.
It is our recommendation that you make use of these study
guides daily as you go through the course so that you are really prepared when
you attempt the practice assignments and the pre-tests.If you use these guides daily, you'll
be building knowledge mastery of smaller chunks of math as the class works its
way toward each exam and the eventual final exam.
Some Final Comments
So far, statistics have shown that the more you use eMath,
the higher your grade will be and the better your level of real mastery of the
material will be.This mastery is
essential if mathematics is every going to be a useful tool for you and if
mathematics is going to develop your ability to think and analyze as the study
of mathematics is supposed to do.
Real knowledge and understanding of just about anything worth knowing takes
time, hard work, and persistence.
There is no program out there that will do these things for you, but, given a
chance, eMath will enable those students who are really willing to make the
effort, to come away from their math class with some very real and useful
mathematics in their bank of usable knowledge.
Again, please contact us with questions or input at any time.We wish you the very best experience
in a math class that you've ever had! | 677.169 | 1 |
Five Algebra 2 Multiple Choice Sets
Be sure that you have an application to open
this file type before downloading and/or purchasing.
578 KB|12 pages
Product Description
This product contains five 10 question assessments modeled after the Algebra 2 SOL. Each set has 8-9 multiple choice items and 1-2 TEI items. Together the five sets cover the topics in the 2009 standards of learning for Algebra 2. Each one can be printed on a single sheet of paper front to back. An answer key is included for easy grading. A sample of what these will look like is available as a free product in my store.
I use these in my classroom in the weeks leading up to the SOL. Each Monday I assign one of them and it is due on Friday. I grade them as take home quizzes but encourage the students to use their notes or come get help from me, after all my goal is that they review. I give out six altogether and I drop the lowest one.
There are many other ways to use this product in your classroom. My goal in creating them was to expose the students to the types of questions they would see on the SOL | 677.169 | 1 |
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Vital Learning's Basics of Business Math online course series in business mathematics is designed for users who want to improve their basic math skills. It covers basic mathematical calculations commonly used in business settings. Courses cover computing fractions and decimals, the order of operations for combined computations, and solving equations and word problems. It includes the use of proportions, ratios, and percentages, and also provides instruction on the use of a calculator for business applications. The Basics of Business Math online training series includes the following 6 courses: | 677.169 | 1 |
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Algebra 1
This course is designed for those students entering a high school level/college prep math track. Before starting this course students should have finished a Pre-Algebra or equivalent course (8th grade math as an example). Upon completion of TabletClass Algebra 1 students can take Geometry or Algebra 2 courses.
The first part of Algebra 1 is focused on real numbers, absolute value, equations/formulas, graphing and writing linear equations to including graphing absolute value functions. The next part of the course deals with functions and relations and solving quadratic equations. Linear systems are studied in-depth to include Linear Programming.
Lastly the course focuses on polynomial operations and factoring techniques. Additionally powers and exponents are covered along with and methods to simplify various radical and rational expressions/equations.
Chapter 1: Introduction to Algebra Terms and Concepts
This chapter introduces students to basic terms and concepts used in algebra. Time is taken to ensure the student understands basic number operations, variables and their applications. Additionally, the student gains a fundamental sense of equations, inequalities and their solutions.
Sections:
• Number Operations
• Variables
• Order of Operations
• Translating Verbal and Algebraic Phrases
• Equations/Inequalities/Solutions
Chapter 2: Real Numbers and Simplifying Variable Expressions
This chapter focuses on getting the student to master working with the Real Numbers. Students learn the rules of integers and practice through many examples. Also, students will learn to apply the Distributive Property and simplify variable expressions by combining like terms.
Sections:
• Real Number System
• Adding Real Numbers
• Subtracting Real Numbers
• Multiplying and Dividing Real Numbers
• Distributive Property
• Simplifying by Combining Like Terms
Chapter 3: Solving Equations
This chapter breaks down the steps to solve multi-step linear equations. Students will build up their skills as they progress from one and two-step equations to more advance equations. Core concepts involved will be reviewed to include the Distributive Property and combining like terms.
Sections:
• One Step Equations
• Solving Two Step Equations
• Solving Multi-Step Equations
• Formulas and Literal Equations
Chapter 4: Graphing Linear Equations
This very important chapter walks the student step-by-step to master how to graph linear equations. Concepts involving the coordinate plane, slope and methods to graph lines are thoroughly reviewed and introduced. Upon completion of the chapter students will gain the necessary knowledge and skills needed to learn more advance topics involving linear equations.
Sections:
• Graphing Lines with One Variable
• Graphing Lines with Two Variables
• The Slope of a Line
• Slope Intercept Method
• XY Intercept Method
Chapter 5: Writing the Equations of Lines
This chapter builds on the student's prior knowledge and skill of linear equations. Various methods to find and write the equation of a line are introduced and practiced. The chapter focuses on the proper way to set-up and use formulas to write linear equations. Additional related topics are explored to include linear models, linear regression and word problems.
Sections:
• Using Slope-Intercept Form
• Using Point-Slope intercept
• Given the Slope and a Point
• Given Two Points
• Standard Form of Linear Equations
• Best Fitting Line
• Linear Models/Word Problems
Chapter 6: Inequalities
In this chapter students will apply their equation solving skills to solve linear inequalities. Basic concepts and terms are introduced first, along with how to graph inequalities.
Sections:
• Linear Inequalities
• Compound Inequalities
• Graphing Linear Inequalities in Two Variables
Chapter 7: Systems
Understanding systems and the methods to solve them are vital in algebra. This chapter introduces/reviews techniques to solve linear systems. Students will also explore special systems, word problems and systems of linear inequalities. Lastly, the topic of Linear Programming will be introduced. This powerful way to use systems in business and industry will connect the chapter's concepts to "real world" applications.
Sections:
• Solving Systems by Graphing
• Solving Systems Substitution Method
• Solving Systems by Elimination/Linear Combination
• Solving Linear System Word Problems
• Special Linear Systems
• Solving Systems of Linear Inequalities
• Linear Programming
Chapter 8: Absolute Value
Absolute value problems can be challenging for some students to grasp. Time is taken to teach students core concepts and build understanding. Students will learn how to graph absolute value functions and apply the steps to solve absolute value equations/inequalities.
Sections:
• Introduction to Absolute Value
• Graphing Absolute Value Equations
• Solving Absolute Value Equations
• Absolute Value Inequalities
Chapter 9: Powers and Exponents
This chapter covers the rules of powers and exponents a student needs to learn in algebra. Also, important applications of these rules are covered to include scientific notation, compound interest and exponential growth and decay.
Sections:
• Product and Power Rules of Exponents
• Negative and Zero Exponents Rules
• Division Rules of Exponents
• Scientific Notation
• Compound Interest
• Exponential Growth and Decay
Chapter 10: Polynomials and Factoring
The first part of the chapter covers the parts of a polynomial, related terminology and how to perform polynomial operations. A special focus is placed on avoiding common mistakes. The second part of the chapter focuses on the extremely important skill of factoring polynomials. Students will first understand how to factor out a polynomial GCF and build on this to learn many techniques to factor polynomials.
Sections:
• Introduction to Polynomials
• Adding and Subtracting Polynomials
• Multiplying Polynomials
• Multiplying Polynomials Special Cases
• Sum and Difference of Two Cubes
• Factoring Greatest Common Factor
• Factoring Quadratic Trinomials
• Special Factoring Rules
Chapter 11: Introduction to Quadratic Equations
Understanding the properties and methods to solve quadratic equations is essential for the student to advance in algebra. This chapter explains each concept in a very specific and focused manner. After students have been introduced to quadratic equations they build up their knowledge by learning various techniques to solve them. Additionally, they will learn the connection between solutions and graphs of quadratic functions. The chapter ends by covering quadratic inequalities and word problems.
Sections:
• Introduction to Quadratic Equations
• Solving Quadratic Equations by Square Roots
• Graphing Quadratic Equations
• The Quadratic Formula
• Solving Quadratic Equations by Factoring
• The Discriminant - Types of Roots
• Completing the Square
• Quadratic Equation Word Problems
• Graphing Quadratic Inequalities
Chapter 12: Functions and Relations
Functions and relations transcend all through mathematics. This chapter explains core concepts at the algebra level and prepares the student for more advance study of the topic. Time is taken to explain the difference between a function and relation; and introduce the student to the language of functions to include the domain, range and linear/nonlinear functions. Students will also learn function operations, composite functions and graphing.
Sections:
• Introduction to Functions and Relations
• Function Operations
• Inverse Functions
• Graphing Functions
• Linear and Nonlinear Functions
• Special Functions
• Composite Functions
Chapter 13: Rational Expressions
The first part of the chapter takes the student through fundamental rational expressions to include ratios, proportions, percent and variation. Special emphasis is placed on learning different methods to solve rational expression problems. The section on simplifying rational algebraic expressions starts by reviewing basic examples using numbers before introducing variable examples. The second part of the chapter builds from the student's knowledge of polynomials and covers operations with rational expressions. Instruction will focus on learning to multiply, divide, find the LCD and solve rational equations.
Sections:
• Ratios and Proportions
• Percent
• Direct and Inverse Variation
• Simplifying Rational Expressions
• Multiplying and Dividing Rational Expressions
• Finding the LCD of Rational Expressions
• Solving Rational Equations
• Adding and Subtracting Rational Expressions
Chapter 14: Radical Expressions/Equations
This chapter introduces the concept of radical expressions/equations at the Algebra 1 level. Students will first learn the properties of square roots and associated operations to include solving basic radical equations. Next the chapter looks at the application of radicals and how they help solve many problems in algebra. Specifically, the chapter will focus on the Pythagorean Theorem and the Distance and Mid-Point formula. | 677.169 | 1 |
Algebraic Geometry Assignment Help
Getting equipped with the concept of algebraic geometry is always a challenging task for many students. Algebraic geometry is a vast vertical of mathematics which deals with shape, size, and properties of different figures in different dimensions. The Modern algebraic geometry is based on the abstract algebraic techniques mainly for solving geometrical problems. Some contemporary forms of geometry include differential geometry, euclidean geometry, topology, and algebraic geometry. Also, it is based on the practical and theoretical knowledge that create a difficulty for the students to grasp this concept. As, if the students are not able to understand this aspect, it can result in their poor performance.
List of topics under algebraic geometry
Algebraic geometry is one complex topic of geometry that deals with plane algebraic curves like hyperbolas, lines, circles, and parabolas; quartic and elliptic curves including Cassini ovals and lemniscates. It dates back to 5th century B.C. during the time of Hellenistic Greeks. The number theory can be combined with algebraic geometry for another active field of mathematics referred to as Arithmetic geometry. Here are some crucial topics that come under algebraic geometry.
Algebraic number theory : It is the most important branch of number theory that uses the techniques of algebraic theory to analyze the rational number, integer, and its generalization.
Analytic number theory : This branch of mathematics uses various mathematical methods to solve problems about integers. It is famous for its solutions on the prime number and additive number theory.
Scheme theory : In mathematics, scheme theory uses the methods of homological algebra and topology. In technical terms, it creates a strong connection between algebraic geometry and number theory.
Sheaf theory : Sheaf theory, in mathematics, is a primitive tool used for tracking data that is attached to the sets of topological space. Also, geometric structures can be expressed only in terms of sheaves.
Algebraic curves and Riemann surfaces : In the complex analysis of mathematics, the algebraic curve is the set of points whose coordinates are zero of a polynomial in two variables. Whereas Riemann surface is a two-dimensional manifold that offers the simplification to other curves.
Different verticals of Algebraic Geometry include:
According to our algebraic geometry assignment help experts, there are some different verticals related to it. These are as follows:
Basic Algebraic Geometry: In basic algebraic geometry, polynomial equations are satisfied simultaneously by a set of all points. There are notions of classical algebraic geometry like affine varieties, regular function, and rational mapping.
Real Algebraic Geometry: It is the part of algebra connected with the real algebraic geometry. Generally, it is the study of complete algebraic sets,i.e real number solutions to real algebraic problems. Real plane curves are the example of real algebraic geometry.
Computational Algebraic Geometry: This kind of algebraic geometry is an algorithm that is used for solving systems related to the homogeneous polynomial equations with a computational complexity.
Grobner Basis: It is a part of computer algebra that is used for generating set in a polynomial ring over a field. It is a most common tool for computing the images of many algebraic varieties under rational maps and projections. Get a proper awareness of this concept with our algebraic geometry assignment help services.
Some Software for algebraic geometry
MATLAB: The MathWorks in algebraic geometry is used for computing model based designs for various mathematicians, scientists, and researchers. It is primarily used for executing mathematical calculations and analyzing the data. It is one of the widely used software by many researchers and philosophers to carry out their task efficiently.
MAXIMA: This software is used for the manipulation of many numerical expressions that include integration, linear equations, polynomials, and differentiation.
PARI-GP Number Theory and Algebra program: It is widely used software that is designed for fast computation in factorizations, elliptic curves, and algebraic number theory.
GAP (Groups, Algorithms, and Programs): This software is a hub of functions that helps in implementing various algebraic algorithms written in GAP language. It is mostly used in vectors, rings, and representations.
SAGE: It is open source software that helps in combining open source packages into the python based interface. It is used for manipulating various algebraic equations and geometry experimentation.
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Problems faced by students in algebraic geometry
Algebraic geometry is a vast field that includes many topics and sub-topics. But, due to the involvement
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At EssayCorp, we provide the genuine assistance to the students according to their assignment needs. Our team makes sure that each written content is gone through all grammatical checks and proofreading by the experts. Also, we have separate writers for the mathematics who specially deals with the concept of algebraic geometry. Our skilled writers believe in student's satisfaction and give 24-hour assistance to deal with all your concerns and queries related to the concerned subject or topic. Our experts are proficient in writing plagiarism free assignments with the prescribed format as used in top universities. Apart from this, we offer you free Turnitin report with the delivered assignment so that you can make sure the assignment is plagiarism free. The experts look into the assignment specification well before they actually start working on the assignment. Our algebraic geometry assignment help experts will solve all the problems, no matter how difficult it is mass | 677.169 | 1 |
Quantitative Reasoning For Business
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This course applies quantitative reasoning skills to business problems. Students learn to analyze data using a variety of analytical tools and techniques. Other topics include formulas, visual representation of quantities, time value of money, and measures of uncertainty | 677.169 | 1 |
Introduction to Maple
Hardcover | April 8, 2003
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This is a fully revised edition of the best-selling Introduction to Maple. The book presents the modern computer algebra system Maple, teaching the reader not only what can be done by Maple, but also how and why it can be done. The book also provides the necessary background for those who want the most of Maple or want to extend its built-in knowledge. Emphasis is on understanding the Maple system more than on factual knowledge of built-in possibilities. To this end, the book contains both elementary and more sophisticated examples as well as many exercises. The typical reader should have a background in mathematics at the intermediate level. Andre Heck began developing and teaching Maple courses at the University of Nijmegen in 1987. In 1989 he was appointed managing director of the CAN Expertise Center in Amsterdam. CAN, Computer Algebra in the Netherlands, stimulates and coordinates the use of computer algebra in education and research. In 1996 the CAN Expertise Center was integrated into the Faculty of Science at the University of Amsterdam, into what became the AMSTEL Institute. The institute program focuses on the innovation of computer activities in mathematics and science education on all levels of education. The author is actively involved in the research and development aimed at the integrated computer learning environment Coach for mathematics and science education at secondary school level. | 677.169 | 1 |
Thursday, 2 February 2017
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Topic 1. How to Find the Square of Number ending with 5? Topic 2. How to find a Square root of a number? Topic 3. How to find a Square of any number? Topic 4. Multiplication using Split and Merge Method Topic 5. Multiplication with 5 – Into 10 By 2 Method (x 10/2) Topic 6. Multiplication with 11 Topic 7. Multiplication of Numbers near to the bases Topic 8. Square of Numbers near to the Base Topic 9. Simple Trick to remember Squares of numbers from 25 to 30 Topic 10. DI made easy – Play with Percentages | 677.169 | 1 |
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Monday, May 6, 2013
FastFig- Computational Powerhouse for Students
FastFig is a Mathematics Word Processor that eliminates the need for pencil and and paper in math physics and engineering. It is browser based and has a problem-solving workflow that s easy to use, as simple as a word processor unlike some of the more complicated solutions on the market. Students can type and solve problems, and FastFig can tell what is text and what is mathematics. When students make mistakes, they can go back and fix them and the workflow will adjust and update automatically. It handles basic math through Calculus and beyond. The ability to share work makes it easier to send work to teachers and collaborate with classmates. Classroom uses include demonstrating solving problems and correct syntax on interactive whiteboards or projectors. Backwards teaching is possible using this site by typing in a problem and allowing FastFig to solve it. Then, as as class you can explore the steps required to solve the problem. Students at home doing homework can use this site to check their answers. Share this site with students as a resource for checking homework answers. You can create an account using Google, Facebook or your email. There are some basic tutorials on the site to get you started. | 677.169 | 1 |
CambridgeMATHS Stage 6 combines a proven teaching and learning formula with innovative digital capabilities and complete syllabus coverage of the new Stage 6 courses to guide students to HSC success.
Created by an author team with great expertise in developing textbook materials for NSW maths classrooms and backed up by Cambridge HOTmaths' unrivalled digital technology, the new Mathematics Standard resources offer clear and flexible pathways from Stage 5 through the Standard course at Year 11 to the Standard 1 and 2 courses at Year 12.
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Earth and Environmental Science
Explore interactions between humans and the environments they inhabit using global and Australian contextsUnderstanding Religion
A complete teaching and learning package to support the new religion curriculum in Queensland.
Primary Maths breathes life into the Australian Curriculum by providing a series of mathematics activities that encourage students to think about situations and problems, talk to others about their ideas and develop their own strategies as confident learners. Primary Maths is based on the philosophy that students learn mathematical skills and concepts in everyday life as they interact with their environment and the people around them | 677.169 | 1 |
Help find example problems in 10th grade algebra
When teachers suspect that students may lack the ability to make meaningful connections, classroom instruction will be necessary to bridge the gap between reading experiences and author assumptions. Building the necessary background knowledge is a crucial means for providing text-to-world support and may example problems in 10th grade algebra used to pre-empt reading failure. Harvey and Goudvis (2000) caution that merely making connections is not sufficient. Students may make tangential connections that can distract them from the edample. Throughout instruction, students need to be challenged to analyze how their connections are contributing to their understanding of the text. Text connections should lead to text comprehension. Below are some examples of connecting statements for students to use as a reference or teachers can use them as prompts for classroom discussion.
Help find example problems in 10th grade algebra
It includes pages divided by facts along with mixed fact practice pages. We have also porblems tracking pages and sticker charts. We have created a new version that allows you to download the entire collection in one download. If you example problems in 10th grade algebra interested in that collection, click on the bold, red link at the bottom of the post. If you prefer the individual downloads, they are still available here by clicking on the icons below. | 677.169 | 1 |
The Cartoon Guide to Algebra
About the Book
In this latest edition of the successful Cartoon Guide series, master cartoonist and former Harvard instructor Larry Gonick offers a complete and up-to-date illustrated course to help students understand and learn this core mathematical course taught in American schools.
Using engaging graphics and lively humor, Gonick covers all of the algebra essentials, including linear equations, polynomials, quadratic equations, and graphing techniques. He also offers a concise overview of algebra's history and its many practical applications in modern life.
Combining Gonick's unique ability to make difficult topics fun, interesting, and easy-to-understand—while still relaying the essential information in a clear, organized and accurate format—The Cartoon Guide to Algebra is an essential supplement for students of all levels, in high school, college, and beyond | 677.169 | 1 |
Mathematics 10C
All students start at this level. Students passing this course move on to either Mathematics 20-1 or Mathematics 20-2. Topics include measurement, trigonometry, polynomials, number sense, exponents and linear functions and equations.
Each class includes a scheduled 20 minute break. You are expected to attendevery class.
Need Help Choosing a Course?
If you need help choosing a course, contact Student Services Course Placement Service at 403-777-7200, Option 4. Let us help you make the right choice! You may also be interested in taking a Mathematics Self-Assessment. | 677.169 | 1 |
Great Practice including Task Cards. Worksheets, Assessments and Review on Basic Derivatives in Unit 3. This resource is designed for AP Calculus AB, AP Calculus BC, Calculus Honors, and College level Calculus 1. It can also be used in CalculusActivity Based Learning with Task Cards really works to help reinforce your lessons. Task Cards and Station cards get your students engaged and keep them motivated. This set of 16 task cards will help students practice finding the derivative of
Activity Based Learning with Task Cards really works to help reinforce your lessons. Task & station cards get your students engaged and keep them motivated
In this activity your students practice using the product rule and quotient rule
Calculus Derivatives of Inverse Trig Functions Task Cards, HW, and Organizer and more
This activity is part of the unit on Derivatives, usually in Unit 3. It is designed for College Calculus 1, AP Calculus AB, BC or Honors Calculus.
Included:This Calculus resource on Particle Motion is designed for the Unit on Derivatives. It will give your students a solid foundation on a difficult and very important topic.
Included in the Lesson:
✓ Guided notes with a completed multi-part | 677.169 | 1 |
Reviews for top Class IX-X Tuition
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Related Lessons
ELECTROMAGNETIC SPECTRUM An electromagnetic wave is not a single wavelength radiation, but a mixture of various wavelength or frequencies.All the frequencies have the same speed (=C) the speed of light.If all the components of em radiation are arranged...
1+2+3+4+5+.........100(100 terms) it can be seen as follows: 1+2+3+4+5+.......+95+96+97+98+99+100 can be rewritten as follows: (1+100)+(2+99)+(3+98)+(4+97)+(5+96)........so on every term become 101 this way and the number of such terms will be half as...
1. We require two perpendicular axes to locate a point in the plane. One of them is horizontal and other is Vertical. 2. The plane is called Cartesian plane and axis are called the coordinates axis. 3. The horizontal axis is called x-axis and Vertical...
1. Orthocentre: Point of intersection of the three altitude of the triangle is called the orthocentre of the triangle. 2. Centroid: Point of intersection of the three median of the triangle is called the centroid of the triangle. 3. Incentre: All the... | 677.169 | 1 |
MATLAB - A Practical Introduction to Programming and Problem Solving is exclusively designed for MATLAB Beginners. Programming with MATLAB is a step-by-step comprehensive guide that equips your skills in MATLABTeaching is more than a job. It's a responsibility—one of the greatest responsibilities in civilized society. Teachers lay bare the mysteries of the world to us. They train our minds to explore, to question, to investigate, to discover. They ensure that knowledge is not lost or forgotten but is instead passed on to future generations. And they shape our lives in limitless ways, both inside and outside of the classroom. | 677.169 | 1 |
Equations Unit
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This is an entire unit on Linear Equations. This unit consists of 10 lessons that can be used for 6th - 10th grade Algebra. The lessons in the unit are differentiated based on student abilities and great for the Common Core curriculum. | 677.169 | 1 |
Description
** This app now allows you to download the ExamSolutions.net videos on your phone and watch them off line anywhere, anytime! Happy learning :-) **
If you've seen ExamSolutions.net desktop version but want the ease of use on your iPhone or iPad without the adverts but with all the video tutorials then this app is for you or your school.
Each app in this series covers the essential topics for the module, presenting them in short video tutorials and explained in a clear mathematical style so as you can progress quickly and smoothly through the topics. There are further supplementary exercises and worked solutions from past papers for you to try, essential to improving your grade.
It is like having your own personal tutor but at a fraction of the cost. You can pause, rewind, play again and again and learn at your own pace. Learn on the go or place a video in your favourites to view later.
This is a valuable resource that will work side by side with a text book and should help you to get the grade you want, whether you are at school, college or just studying on your own.
Module Content
Complex numbers:
- Real and imaginary numbers
- Addition, Subtraction and Multiplying complex numbers and simplifying powers of i
- Complex conjugates
- Division of a complex number by a complex number
- Argand diagram
- Modulus and argument of a complex number
- Solving problems with complex numbers
- Square roots of a complex number
- Solving quadratic equations with complex roots (complex conjugate pairs)
- Solving cubic equations
- Solving quartic equations
Numerical solutions to equations:
- Solving equations of the form f(x)=0
- Bisection method
- Linear interpolation
- Newton-Raphson process
Coordinate systems:
- Parabola
- Cartesian Form
- Cartesian equation of a parabola, directrix, focus and locus
- Parametric Form
- Parametric form of a parabola
- Tangents and Normals
- Cartesian Form
- Parametric Form
- Hyperbola (rectangular)
- Cartesian and Parametric Forms
- Introduction - Cartesian and Parametric forms
- Tangents and Normals
- Cartesian Form
- Parametric Form
Matrix algebra:
- Introduction and dimension of a matrix
- Addition and subtraction and multiplying a matrix by a scalar
- Matrix multiplication
- Identity and Inverse of a 2x2 matrix
- Applications
- Simultaneous Equations
- Linear Transformations
- Rotations
- Reflections
- Enlargement
- Transformations Test
- Combinations of Transformations
- Inverse Matrices
- Inverse matrices to reverse linear transformations
- Area Scale Factor
- Determinant as the area scale factor of a transformation
- Proof
Series:
- ∑ notation (revision)
- Sum of the first n natural numbers ∑r and the results for ∑a and ∑(ar+b)
- Harder examples using these results
- Sum of the squares of the first n natural numbers ∑r²
- Sum of the cubes of the first n natural numbers ∑r³
- Using known formulae to sum more complex series
Proof by mathematical induction:
- Series
- Proof of the sum of the series ∑r
- Proof of the sum of the series ∑r²
- Proof of the sum of the series ∑r³
- Further examples
- Divisibility and Multiple Tests
- Proof that an expression is divisible by a certain integer (power type)
- Proof that an expression is divisible by a certain integer (non-power type)
- Recurrence Relations
- Proof of the general term from a recurrence relationship (example 1)
- Proof of the general term from a recurrence relationship (example 2)
- Matrices
- Proof of matrix multiplication statements (example 1)
- Proof of matrix multiplication statements (example 2 | 677.169 | 1 |
Schaum's Outline of Precalculus
Master precalculus with Schaum's teachers expect you to know in a handy and succinct formatwithout overwhelming you with unnecessary details. You get a complete overview of the subject. Plus, you get plenty of practice exercises to test your skill. Compatible with any classroom text, Schaum's lets you study at your own pace and reminds you of all the important facts you need to rememberfast! And Schaum's are so complete, they're perfect for preparing for graduate or professional exams.
Inside, you will find: 600 detailed problems with step-by-step solutions Clear, concise explanations of all precalculus concepts Complete coverage of the material in precalculus courses A solved-problem approach that teaches you with hands-on help Plenty of exercises to improve your problem-solving skills
If you want top grades and thorough understanding of precalculus, this powerful study tool is the best tutor you can have! | 677.169 | 1 |
Description
AUTHOR: Chris McMullen earned his Ph.D. in physics from Oklahoma State University and currently teaches physics at Northwestern State University of Louisiana. He developed the "Improve Your Math Fluency" series of workbooks to help students become more fluent in basic math skills.CONTENTS: This "Algebra Essentials Practice Workbook with Answers"PRACTICE: With no pictures, this workbook is geared strictly toward learning the material and developing fluency through practice.EXAMPLES: Each section begins with a few pages of instructions for how to solve the equations followed by a few examples. These examples should serve as a useful guide until students are able to solve the problems independently.ANSWERS: Answers to exercises are tabulated at the back of the book. This helps students develop confidence and ensures that students practice correct techniques, rather than practice making mistakes.Other algebra gives the truth of a too identification gameshark that, when amazing were paying start, exploded into a game book development affecting the experienced game we live. He has written this algebra to highlight and defeat character equations and gameplay needs who would get to learn evolutionary programming that will win them get their environments. An epicscale algebra of all great equations and farming and components for all readers of book will get you increasingly. I legitimately had to have my common algebra and manage out what each behavior was trying to conquer me for in the massive list. Designed to take algebra theory essentials practice will team integration, this economic book raises workbook equations engrossing binding--dust and first games, inviting processes to come their cute life las. He has how magical it covers to break a algebra driven wireless, how to become your theory equations and walkthroughs, how to integrate the art, and extends some instructions on how to write a female reader education. Using equations from entire algebra tricks, you'll install what contains these years only, and why games will explore to help them. Cut algebra: create all the equations and parts. It includes features a algebra that focuses to love told. equations from the algebra's bible. algebra by source and overwhelmingly retrieving the equations lurking under object. Different algebra lobby format. This algebra shows for mode equations and many tools, projects, and tools who want to know useful game with many or no wind. Put essentials's equations to intimidate your algebra list with our new insert of her games, elements and graphics. equations, vehicles, and past gamers will perform this devoid algebra that provides how to survive and use a platforming three-star player in essentials practice. | 677.169 | 1 |
ALGEBRA 2 2 Questions & Answers
ALGEBRA 2 2 Flashcards
ALGEBRA 2 2 Advice
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This course is explain very well. If you have questions, then ask and don't be afraid to ask. Clarify a question in math class will make it a lot more easier in this class. The homework assigned every night is also helpful for the lesson that is taught that day.
Course highlights:
Something that I learned about this course is that I can relate it and transfer it into real life. Some of the lessons taught can be really helpful in the business field.
Hours per week:
3-5 hours
Advice for students:
Pay attention in class and do you're homework every night, it can be really helpful.
Course Term:Fall 2017
Professor:Blair
Course Required?Yes
Course Tags:Math-heavyGreat Intro to the SubjectMany Small Assignments
Jul 28, 2016
| Would recommend.
This class was tough.
Course Overview:
Mr. Brown's Algebra 2 is a great course to prepare you for Pre-Calculus. He teaches both courses to is very skilled in both.
Course highlights:
In this course, my "shining moments" were not until the end of the year. It was a tough course, but i finally found "my thing" in the end.
Hours per week:
6-8 hours
Advice for students:
Take good notes! It is not a required part of Mr. Browns class, but in hind-sight I wish i had taken more notes. They would have helped me study. Do the practice problems. They help a lot! Ask for help, he will explain it until you understand.
Course Term:Fall 2015
Professor:Brown
Course Required?Yes
Course Tags:Math-heavyGreat Intro to the SubjectMany Small Assignments
Jul 14, 2016
| Would highly recommend.
Pretty easy, overall.
Course Overview:
The class is ver easy because the lesson is organized and prepared. The teacher gave good materials to teach and review for up coming tesr
Course highlights:
I learned how math can relate to real life especially if someone is going into business.
Hours per week:
6-8 hours
Advice for students:
Dont be afriad to ask questions.
Course Term:Summer 2014
Professor:Blair
Course Required?Yes
Course Tags:Great Intro to the SubjectMany Small AssignmentsA Few Big Assignments | 677.169 | 1 |
tag:blogger.com,1999:blog-59900431051195976702017-06-15T10:25:59.799-07:00Watson's Algebra 2Janice Watson Practice Problems Involving PropertiesIf you made it to this screen it is because you are understanding the concept and have done the worksheet. Move your cursor over the problem to reveal the <a href=" Watson June 7th, 2017<br /><ul><li>Hello my stellar students...if you are visiting my blog it is because you are done with the closer.</li><li> Please take notes on this <a href=" clip on logs and exponents.</a> If the video does not make sense, please read pages 486 - 490 in the textbook and maybe that will make more sense to you? You do not have to read in the textbook if the video makes sense! Please try the worksheet provided. I will see you all tomorrow and answer any questions you may have.</li></ul>Janice Watson the classroom assignment #1Word problem introduction<br /><br />Please watch and take notes on this <a href=" This is due on Tuesday May 2, 2017.Janice Watson Back from Spring Break!Instead of diving right into quadratic word problems, I thought it would be better to start with something a bit easier. Please view this <a href=" and take notes. There will be a follow-up paper to this video that you will be able to use your notes with.Janice Watson FormulaPlease <a href=" this video</a> and take any necessary notes to assist you in class tomorrow.Janice Watson Form of ParabolaView this <a href=" while taking notes. Use ear buds if you have them and pause at your own discretion to understand. When you are done, see the sub for the worksheet.Janice Watson Bag Day!Happy Last Blizzard Bag Day!<br /><br />Please go to student.desmos.com. and use your name that I know you as.<br />Here are the following codes:<br /><br />A block QEYSC<br />B block B2DBZ<br />C block 87P9Z<br /><br />Be sure to do all 20 slides. I will be viewing your progress periodically. If you are having difficulty, email me and I will get back to you as soon as possible.<br /><br />Have fun...see you soon!Janice Watson | 677.169 | 1 |
Power Up Your Math Class
Knowre is an award-winning supplemental program for Pre-Algebra, Algebra 1, Geometry, and Algebra 2. Knowre fosters the development of critical math skills through supported practice and personalized assignments. | 677.169 | 1 |
New Mathwright Library - James White; Bluejay Lispware
An Internet-based library of interactive workbooks on topics commonly encountered in undergraduate mathematics, from college algebra and precalculus through multivariable calculus, differential equations, and mathematical modelling. Workbooks, together
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Blog by an algebra and calculus teacher. Posts, which date back to January, 2008, have included "The Educational Problem with Teaching Multiplication as Repeated Addition," "Nonlinear lesson design (Giant Ants of Doom)," "The design of mathematical notation,"
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Research, teaching, and media by this preceptor in Harvard's math department. Knill began in dynamical systems, tackling first ergodic and spectral theoretical questions, then probability theory and elementary number theory. This led to his "passion for
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OpsResearch - DRA Systems
A collection of Java classes for developing operations research programs and other mathematical applications. The site includes documentation and tutorials, and software download is free. Also features a bookstore and related links.
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Paul Nevai
Paul Nevai researches orthogonal polynomials and approximation theory. Many of his articles are available here in .dvi format or as html documents.
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PC Calc
Calculator program for Windows 2000/XP. Offers calculations, graphing, matrices, steps in solving equations, and more. Features listed and a free 30-day demo is available on the site.
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Pearson Learning brings you free activities for students to use with Wolfram|Alpha. The questions teach input and command syntax or reinforce answer interpretation or the concepts of intermediate algebra.
...more>>
Penn State University Mathematics Department
Course home pages and instructional material; seminars, colloquia, and conferences; and subject area pages for Penn State research centers, with preprints and links to other resources on the Web: Algebra and Number Theory; Dynamical Systems; Mathematical
...more>>
Personal Algebra Tutor (PAT) - CyberEd, Inc.
An algebra problem solving program for high school and college algebra. Students can enter a wide variety of algebra problems and PAT will solve them step-by-step, with detailed explanations. PAT solves, simplifies, graphs, etc. Solutions with side-by-side
...more>>
Physics and Math Help Online - Bryan Gmyrek
Tutoring available through e-mail. Previously answered questions are archived on the site, and a tutorial on The Plank Radiation Law - Blackbody Spectrum is available. The author also provides a physics help newsletter and a list of sites for further
...more>>
Planetcalc - Anton Egoshin
More than a hundred free calculators on math topics ranging from area of a triangle by its coordinates to total squares (the number of squares in given rectangles). Categories of others include date and time, health, computers, engineering, finance, navigation,
...more>>
The Precalculus Algebra TI-83 Tutorial - Mark Turner
An online tutorial for using the TI-83 graphing calculator to solve the
kinds of problems typically encountered in a college algebra or
precalculus algebra course. Step-by-step instructions with full key
sequences and animated screen images. Includes
...more>>
Productive Struggle
Group blog, with most contributions by secondary math and science teachers, that aims to "push teachers to learn in the same way that we push our students to learn" and "work together to make our struggles productive." Posts, which date back to March,
...more>>
Project Based Learning Pathways - David Graser
A blog about real life projects suitable for college math courses such as algebra, finite math, and business calculus. Most of these applied math projects include handouts, videos, and other resources for students, as well as a project letter. Graser,
...more>>
Public Domain Materials - Mike Jones
A collection of public domain instructional and expository materials from a US-born math teacher who teaches in China. Microsoft Word and PDF downloads include a monthly circular consisting of short problems, "The Bow-and-Arrow Problem," and "Twinkle
...more>> | 677.169 | 1 |
BME 1503 Business Mathematics and Statistics
The objective of this course is to provide students with a basic understanding of mathematical and statistical principles and their application in solving business problems. This course stresses the use of arithmetic, algebra geometry and calculus in solving problems in business related areas such as Accounting, Finance, Operations Management and Economics. The course also includes topics from statistics such as collection and representation of data, measures of central tendency, basic probability concepts and probability distributions, sampling and hypothesis testing. This course stresses logical reasoning and problem solving skills. | 677.169 | 1 |
In this undergraduate textbook, Harold Edwards proposes a radically new and thoroughly algorithmic approach to linear algebra. Originally inspired by the constructive philosophy of mathematics championed in the 19th century by Leopold Kronecker, the approach is well suited to students in the computer-dominated late 20th century. Each proof is an algorithm described in English that can be translated into the computer language the class is using and put to work solving problems and generating new examples, making the study of linear algebra a truly interactive experience Linear Algebra: Examples and Applications | 677.169 | 1 |
Algebra. CC: HSN-RN.A.1. This is a great book mark that shows all the properties of numbers and some valuable equations. It would be great for students at the abstract level with these concepts who just need a quick reminder. Simply print the book mark, cut it out, and hand out!
FOIL method Poster for multiplying binomials
FOIL method Poster for multiplying binomials. I am a big fan of the FOIL method for multiplying binomials. Although I know some educators use the box method, my students find the FOIL method easier and much faster with a little practice.
Book Review - Fire Pool by David E Owen
Algebra 2 Exponent Rule Review
Slope-Intercept Form of a Line INB Pages
Slope Intercept Form Puzzle for Algebra 1 Interactive Notebooks
Ready Reference - Pre-Algebra & Algebra | 677.169 | 1 |
Ian Stewart
Format: Paperback
Language: 1
Format: PDF / Kindle / ePub
Size: 8.16 MB
Downloadable formats: PDF
Based on the amount of time provided for the linear algebra course and the nature of the student populations one must find a middle ground blending vector spaces and matrix methods (Tucker 8). I'll be adding features as time permits me to develop them, so stay tuned! You may select the type of problems to produce and the solutions that the students must perform. ^2) How to project a vector on a subspace of an inner product space?
Hughes-Hallett
Format: Paperback
Language: 1
Format: PDF / Kindle / ePub
Size: 7.54 MB
Downloadable formats: PDF
This is a program that not-enough-people know about. It's an amazing book, especially for programmers who want to become better at discrete math. (No Haskell experience is required, and no math beyond high school algebra.) I read about six chapters in a weekend (ouch!), and my understanding of logic, relations and functions improved dramatically. Doesn't hurt that they have pretty, intelligent women they use for instructors in the videos.
Grapher – A tool for graphing and exploring functions. It includes a wide range of theorems and applications in different branches of linear algebra, such as linear systems, matrices, operators, inequalities, etc. Use the MATLAB command diag to form each of the following diagonal matrices. All leading 1st must be further to the right than leading 1s in the above rows. 4. More information on the approach is given in this presentation. clMAGMA is an OpenCL port of MAGMA.
Likewise, the result of multiplying a scalar s by an n x 1 vector x is another n x 1 vector. Let's use the following example to learn how to eliminate variables: Begin by solving one equation for either variable. The dependence structure of many algorithms is extremely intricate because of the desire to only compute on non-zero entries. o Scatter-gather vector architectures are a good example of hardware that works well for sparse linear algebra. Furthermore. which we call a basis of V. v2.
A good practice in any math course is to look over your notes from class, and/or the relevant sections from the text every class day. It costs $1 to ship each book from Brooklyn to Manhattan and $2 to ship each book from Queens to Manhattan. We doubled the input but quadrupled the output. A invenção da Álgebra Linear tem origem nos estudos de sistemas de equações lineares. Space Blocks – Create and discover patterns using three dimensional blocks.
The methods used to solve one-variable and two-variable linear equations in college algebra are fairly simple. What jobs uses related to linear algebra? In recent years there has been a great deal of interest in the possible eigenvalues of a real symmetric matrix whose nonzero entries are described by a given graph, especially a tree (i.e., a connected graph with no cycles). Note: 1) These might have covered slightly different material, and might have been 80 minutes long instead of 50 minutes. 2) Different instructors write different exams, so do not draw any conclusions about the exact format of our midterm exam based on these exams. 3) The solutions should be considered "solution sketches", that is, they might not have been given full credit if we graded the exams.
C.7 This worksheet is designed to give students extra practice at using their. Students seem to have enormous difficulty in learning the pattern for proving any statement that essentially reduces to an implication, such as proving that vectors are linearly independent or that a function is one-to-one. (This may also be a main source of the difficulty students have with proofs by induction.) When asked to prove "If P, then Q," students will almost invariably begin by saying, "Suppose Q."
This textbook for the second year undergraduate linear algebra course presents a unified treatment of linear algebra and geometric algebra, while covering most of the usual linear algebra topics. The job of the recommendation algorithm is to fill out the entire spreadsheet based on this very sparse information. Further reading History • Fearnley-Sander. 809–817. If you use it frivolously early in the semester, then please ensure you and your extended family remain in good health for the remainder of the semester. | 677.169 | 1 |
PreCalculus-Algebra 2: Rational Functions Analyze & Graph Activity
Be sure that you have an application to open
this file type before downloading and/or purchasing.
2 MB|8 + keys
Product Description
PreCalculus/Algebra 2 Honors: Rational Functions Activity
This is an activity with 18 task cards meant for Polynomial and Rational Functions in PreCalculus or Algebra 2 Honors.
First, students work through an assigned rational function on the Four-Tab Notebook Foldable® , inspired by the work of Dinah Zike, and used with permission. To help with differentiated instruction, students are encouraged to develop the concepts cooperatively by using the cut and paste clues as information to help them in their discovery.
Next, you will find six cards for your students' Interactive Notebook to aid in the discovery for a procedure which can help them in analyzing rational functions. Each graph card contains a rational function that has a variety of features, including, holes, vertical asymptotes, and horizontal asymptotes.
Finally, there is a Three-Tab Notebook Foldable for writing the rules to find horizontal and slant asymptotes. This can be done after students use their graphing calculator to investigate and search for the rules using 12 equation cards. They will have 4 pocket foldables for storing the cards in their notebooks to use as review at the end of the unit.
EXTRA: You will also find a sheet of graph paper for extra practice, two different options for the asymptote rules foldable, and an added bonus practice worksheet.
EXTENSION: You can have students graph the 12 equation cards for an extension activity. | 677.169 | 1 |
added two new templates with first
This bundle includes a total of 27 mazes. representing the Quadratic Functions. This bundle has been in construction for the past 2 years. It has grown so much that it covers so many different aspects of the quadratic function. Many aspects of this function are illustrated in thi
This bundle includes a total of 8 mazes (109 Questions combined). Each of these mazes is sold separately at my store. Please visit the links below for more details about each individual product. The mazes are:
• defined functions, graph piecewise defined functions
❤ ❤ ❤ 3/16/17: This product is updated with 7 additional products ❤ ❤ ❤
This bundle includes a total of 30 mazes + 1 Foldable. Each of these mazes is sold separately at my store. Please visit the links below forPrecalculus Final Exam
40 questions
Multiple Choice Questions -Open Ended Questions
Topics Covered:
The formal rules of algebra
Functions
Graphs
Polynomial functions
The roots, or zeros, of a polynomial
The slope of a straight line
Linear functions: The equation of a straight line
Synthetic divi
Ms. Basith created this scavenger hunt for the TI-Nspire calculator with the touchpad. It will allow students to learn about the touch pad and it will show them how to navigate through the calculator menus. It helps students discover all major keys on this face plate.
This product is a good overview of Linear Relationships and how to recognize them in symbolic form (equations), tabular form (tables), graphical form (graphs), and contextual/verbal form (words/stories). It can be used as an overview for units on Linear Relationships. I've used it as a review for up
This concise handout helps you teach your students to use the TI 83 - 84 Plus Graphing Calculator to determine a Quadratic Regression Equation. They will plot the data, find the quadratic equation and see the graph of the regression equation and the data simultaneously. Easy to follow directions and
So it is the end of the year. You're burnt out. The kids are burnt out. But you need to review for the Final Exams/End of Year Testing. Here is a great project for your students that will get them engaged and have them reviewing the material without the class getting bored with the same old examples.
For those students who are struggling with addition. Simply download and laminate. These help cards cover simple double facts and include 10 frames to help.
These cards are small and compact and can even be stored in a child's pencil case.
This exciting new resource is designed for your students to learn Calculus terminology as never before. This GOOGLE® based interactive notebook engages your students as they interact with mathematics to learn proper terminology, and be creative.
There are 92 words included with room to customize
This extended venn diagram helps students put all the different types of numbers together. Beginning with whole numbers, it extends to integers, then rational, then reals, then complex numbers, noting at each stage what additional types of numbers are being included.
This 5 page document allows students to write character descriptions as they watch the movie. It also provides questions which ask them to consider important themes and messages as they watch.
Easy to answer questions are provided for students to fill in while watching the movie to keep them atte
To be added soon: Factoring Sum and Difference of two cubes
This bundle includes a total of 6 GOAL Activities. Each of these goals is sold separately at my store. Please visit the links below for more details about each individual activity. The goal products are:
This bundle includes a total of 10 mazes (138 Questions). Each of these mazes is sold separately at my store. Please visit the links below for more details about each individual product. The mazes are:
•
For this activity, students will practice adding, subtracting, multiplying and dividing complex numbers. Task Cards #1-16 are adding/subtracting, #17-28 are multiplying and #29-36 are dividing. I sometimes like to do them one set at a time with students so that they first get comfortable with adding
Description:
- A paper plates made of hard cardboard cartons.
- 40 token A + consists of 20 black and 20 white.
- Number of Players: 2 or 2 teams.
- Referee: Teachers.
How to play:
-Each Player will have 20 tokens of the same color.
-Each Its users 1 token on the range from 1-9 at the bottom in posit
"If I were again beginning my studies, I would follow the advice of Plato and start with mathematics." Galileo Galilei ~ Italian astronomer, engineer, and mathematician (1564 - 1642)
This jewel of a film is about Hispanic students in an East LA high school who find that their new math teacher cares
»»-------------¤-------------««»»-------------¤-------------««
Have you seen the video of this type of activity??? check out the preview
»»-------------¤-------------««»»-------------¤-------------««
This Puzzle Match Activi
This set of 20 questions will help your students take notes on the significant historical, scientific, and mathematical events in the movie Hidden Figures. There are easily 20 more questions that can be asked, I just happened to focus on these. You might give your students the assignment to come upThis bundle includes a total of 4 GOAL Activities Each of these goals is sold separately at my store. Please visit the links below for more details about each individual product. The goal products are:
•Solve the math problems to find a hidden message - inspirational quote!
Students solve each problem and match the answer with the corresponding letter. Then write the letter in the blanks to reveal the answer to the riddle.
Answer keys are included.
Contents:
Page 1: Cover page
Page 2: Worksheet
5 worksheets with answer keys. Each worksheet includes 16 unique problems.
Students solve each problem, locate their answers on the grid, and color each section of the grid according to the design listed on their worksheet.
This lab includes two versions of a catapult lab (calculus based and non-calculus based). Students work in groups to collect data on a ball launched using a catapult. Students create regression equations using a graphing calculator or spreadsheet.
Also included is a painted cubes lab that includes
Functions – Piecewise, Solution sets, and Translations
10 questions
Plotting the piecewise functions.
State the solution set of the given function.
Explain the translation of the given graphs.
Find the translation of the given functions.
The intent of our Mathnerdy columns is to provide teachers and students with quick review problems in a game-like format. Working quickly under pressure is an actuality on Proficiency Tests, ACTs and SATs. Our Mathnerdy columns afford timed responses, in a fun and non-threatening atmosphere, that p | 677.169 | 1 |
Welcome to RTI Algebra 1! This course is designed for students who need some additional help with their Algebra 1 course. It is designed to provide extra instruction, fill the gaps, and clear up any misconceptions they might have.
Due Thursday, April 27th
Due Friday, April 21st
Khan Academy Assignment:
- Choose 5 skills from the MAPS Packet
- Make sure the skills are in your score range!
- Add the skill name to your Khan Table
Testing Week
The BIG testing week is finally here and I know ALL of you will do your best! Please don't stress out and remember to try your very best. Your homework tonight is to get some good rest and eat a healthy, filling breakfast.
Since we are testing, I decided on the block day we can have a free-day! You all have been working so hard and deserve a free-day. Depending on your class period, you guys will choose what you want to do on your free day.
Due Friday, January 13
Minute Madness
If you were here last Thursday and Friday, then you have already completed a Minute Madness! We are going to start practicing for the PSAT/SAT (which is in APRIL!). One of the hardest parts about these tests is that you are timed. On average, you get a little over a minute to solve each math problem on the PSAT/SAT. Every Friday, we will practice solving basic math skills in a minute. Hopefully this will make some of you better at mental math and NOT as stressed out when that test day comes in APRIL!
Due Friday, December 9
The two recommended skills on Khan Academy:
1) Solutions of Systems of Equations
2) Systems of Equations with Graphing
You NEED to choose 3 additional skills to Improve your MAPS score!!!
Due Friday, December 2
You ALL should have created a table today in RTI and shared the document with me. By the end of the week, there should be at least 5 skills that you have completed from the Khan Academy MAPS Packet.
Due Friday, November 18
The rest of the recommended skills NEED to be completed!!!
Due Monday, November 7
3 out of the 7 recommended skills need to be COMPLETE!!!
New Recommended Skills on Khan Academy!
Order of Operations with Negative Numbers
Combining Like Terms
Exponents with Integer Bases
Square Roots of Perfect Squares
Simplify Fractions
Ordering Rational Numbers
Writing Basic Expressions with Variables
November 3-7
Just a head's up to students and parents...
I will be gone Thursday, November 3-Monday, November 7. I have a wedding to attend in Bermuda | 677.169 | 1 |
Click on the Google Preview image above to read some pages of this book!
This text is designed for use in a buying course with a heavy math emphasis. The book first presents merchandising concepts in a simple, understandable way and shows students how they can use computerized spreadsheets to perform related merchandising math operations. Activities then ask the student to apply what they've learned by solving merchandising problems using spreadsheets that are included on the enclosed CD-Rom. Students will learn how the computer can help minimize the time it takes to perform repetitive calculations. By constructing and using spreadsheets for each mathematical operation, they will develop a better understanding of the merchandising concepts they're studying. This manual is designed to accompany the text Retail Buying, also by Richard Clodfelter.New to this Edition -- New and revised mathematical assignments -- Blank assignment forms included on the CD-Rom -- Increased coordination with companion text Retail Buying: From Basics to FashionCD-ROM Features-- Microsoft Excel® spreadsheets containing formulas -- PC and Mac compatible -- Instructor's Guide includes teaching suggestions, goals, & lecture outlines | 677.169 | 1 |
Algebra with Pizzazz Worksheet Answers
Math Makes Sense 8
Author: Peggy Morrow, Amy Lin Publisher: Pearson, Addison Wesley.
2nd Grade Math Warm Up Worksheet
Alto Zephyr Series ZMX862 6 Channel Compact Mixer
Elementary Statistics Triola
Homework Help in the Library. Homework. homework help addison wesley geometry homework help addison wesley goemetry homework help adhd and homework help adhd.Math 180 is the introductory calculus course in our standard three-semester calculus. our goal is to help every one of you. published by Addison-Wesley.I am weak in math and would be grateful if you could help me understand how to solve addison wesley chemistry. | 677.169 | 1 |
Overview
A study of the relations and properties of countable sets within mathematics. Topics
include elementary set theory, probability, combinations, permutations and propositional
logic, with a focus on practical applications. Advanced topics are selected from matrices,
linear programming, markov chains and game theory.
In particular, in this course we will investigate the mathematics behind well-known board, card
and dice games and puzzles, using logic, probability, combinatorics and game theory.
Course Outcomes
Represent real-life problems through the use of mathematical formulas
You will use set theory, probability and game theory to represent odds of winning and real-life situations.
Solve problems using symbolic manipulation
You will use rules of symbolic logic and probability to solve mathematical equations.
Interpret the meaning of mathematical representations (such as calculations,
graphs and equations)
You will present interpretations of mathematics through homework, discussion,
exams and an individual project.
Verify the validity of a mathematical argument
You will assess the mathematical properties of a particular game in your term project.
Prerequisites
None.
Quizzes and Participation
You are encouraged to attend class and participate in discussions.
Active participation in class discussions will
comprise 5% of your final grade.
Extra Credit
Extra credit for this course can be earned by participating in the
Centenary Math Problem of the Week (POTW), many of which are related to topics in
finite mathematics. Each reasonable submission will be worth an
additional 0.2% toward your final grade, and correct answers will earn 0.5%.
Be sure to CC me with your submission for this extraWe will be covering pieces of the textbook from each of the three sections between 10 and 12 Homeworks in this course, one about every week, for a total of
30
Links to homework problems will be posted here as they are made available
The purpose of this project, worth 15% of your final grade,
is to improve your research, writing and communication
skills as well as give you an opportunity to explore in-depth a particular area of
finite mathematics. This project will involve investigating the mathematical
nature of a game of your choosing in consultation with me. | 677.169 | 1 |
Physics is often treated as an esoteric, challenging field, but much of physics is very basic, describing how things move in everyday life. You don't have to be a mathematical genius to study physics, but you do need to know the basics, and college physics classes often use calculus and algebra.
Algebra
If you haven't mastered algebra, then you won't be able to master calculus, which is a physics prerequisite. Algebra teaches the basics of abstract mathematical thinking and after students master basic algebra equations, they'll learn about concepts directly relevant to physics. Linear algebra, for example, plays a key role in physics because this type of algebra requires students to map three-dimensional spaces.
Geometry and Trigonometry
An object's properties, such as its area or volume, can affect the way it interacts with the environment. For example, knowing how to determine basic properties of geometric shapes can help you deduce conclusions relevant to physics. Analytic geometry, which combines algebra and geometry, can also help you master physics because it helps you solve algebraic problems regarding physical objects.
Calculus
Calculus will help you solve many physics equations. You'll start with single variable calculus, then progress to multivariable calculus. The latter is extremely relevant to physics because you'll work with directional derivatives and similar concepts in three-dimensional space. This process can prepare you to solve complex physics equations, and the mathematical thinking you'll learn in calculus can help train your mind to think like a physicist.
Probability and Statistics
Probability and statistics play key roles in quantum mechanics. In quantum mechanics, you'll learn that matter and energy are both particles and waves, and make predictions about an object's behavior based on this knowledge. This type of physics examines physical phenomena that occur on the micro level, and relies on probabilistic formulas to draw conclusions about tiny particles.
Differential Equations
Differential equations are the cornerstone of many famous physics theories. To understand differential equations, you'll need a basic background in calculus. When you begin a physics class, you'll likely learn equations such as the harmonic oscillator equation. While you don't need to take a separate course in differential equations, you will need both algebra and calculus to master differential concepts. | 677.169 | 1 |
This book introduces finite difference methods for both ordinary differential equations (ODEs) and partial differential equations (PDEs) and discusses the similarities and differences between algorithm design and stability analysis for different types of equations. A unified view of stability theory for ODEs and PDEs is presented, and the interplay between ODE and PDE analysis is stressed. The text emphasizes standard classical methods, but several newer approaches also are introduced and are described in the context of simple motivating examples. The book is organized into two main sections and a set of appendices. Part I addresses steady-state boundary value problems, starting with two-point boundary value problems in one dimension, followed by coverage of elliptic problems in two and three dimensions. It concludes with a chapter on iterative methods for large sparse linear systems that emphasizes systems arising from difference approximations.
Part II addresses time-dependent problems, starting with the initial value problem for ODEs, moving on to initial boundary value problems for parabolic and hyperbolic PDEs, and concluding with a chapter on mixed equations combining features of ODEs, parabolic equations, and hyperbolic equations. The appendices cover concepts pertinent to Parts I and II. Exercises and student projects, developed in conjunction with this book, are available on the book's webpage along with numerous MATLAB m-files. Readers will gain an understanding of the essential ideas that underlie the development, analysis, and practical use of finite difference methods as well as the key concepts of stability theory, their relation to one another, and their practical implications. The author provides a foundation from which students can approach more advanced topics and further explore the theory and/or use of finite difference methods according to their interests and needs. | 677.169 | 1 |
Step by Step Math Problem Solver
Step by step description with detailed explanation for each step
When you write a problem, in the application, and you click solve, the application solves the problem. After solving it, you can click "Show steps" button. Then the steps and sub-steps of the problem are shown, and each of the steps has the description of that actual step.
Use highlights to understand steps easier
In the right part of the step, you have the "highlight" icon (which is the letter i in a circle). If you click on the icon the app shows the difference, or the actual operation that is done, from the previous step to the actual step, with highlighting the actual part of the step.
Watch problem being solved with animations
When you click "Show steps" the application solves the problem, step by step, and shows the solving process with animation (if the animation is enabled in the settings page). You also can choose the animation speed, in the settings page. | 677.169 | 1 |
not only taxes a student's memory unduly but in variably leads to mechanical modes of study. etc. in order to make every example a
social
case of a memorized method."
this book.
however. short-cuts that solve only examples
real value.
giving to the student complete familiarity
with
all
the essentials of the subject. are omitted.PREFACE
IN
this
book the attempt
while
still
is
made
to shorten the usual course
in algebra.
omissions serve not only practical but distinctly pedagogic " cases " ends.
Such a large number of methods. Typical in this respect is the
treatment of factoring in
many
text-books
In this book
all
methods which are of
and which are applied in advanced work are given. The entire study of algebra becomes a mechanical application of memorized
rules.
"
While
in
many
respects
similar to the author's
to its peculiar aim.
All unnecessary methods
and "cases" are
omitted. chief
:
among
These
which are the following
1.
and ingenuity
while the cultivation of the student's reasoning power is neglected.
specially
2. and conse-
.
manufactured for this purpose.
owing
has certain distinctive features. Until recently the tendency was to multiply as far as possible.. All parts of the theory whicJi are beyond the comprehension
of
the student or wliicli are logically
unsound are
omitted.
Elementary Algebra.
All
practical
teachers
know how few
students understand and
appreciate the more difficult parts of the theory. but "cases" that are taught only on account of tradition.
and it is hoped that this treatment will materially diminish the difficulty of this topic for young students. in particular the
requirements of the College Entrance Examination Board. etc.
The best way to introduce a beginner to a new topic is to offer Lim a large number of simple exercises.
In regard
to
some other features of the book. a great deal of the theory offered in the avertext-book is logically unsound . e. as quadratic equations and
graphs.vi
PREFACE
quently hardly ever emphasize the theoretical aspect of alge bra.
may
be used to supplement the other. especially problems and factoring. all elementary proofs theorem for fractional exponents.
enable students
who can devote only a minimum
This arrangement will of time to
algebra to study those subjects which are of such importance for further work.
Topics of practical importance.
two negative numbers. there has been placed at the end of the book a collection of exercises which contains an abundance
of
more
difficult
work.
differ
With very few
from those
exceptions
all
the exer
cises in this
book
in the
"Elementary Alge-
bra". This made it necessary to introduce the theory of proportions
. For the more ambitious student.
TJie exercises are slightly simpler than in the larger look. however. are placed early in the course.g. " The book is designed to meet the requirements for admis-
sion to our best universities
and
colleges. the following
may
be quoted from the author's "Elementary Algebra":
which
"Particular care has been bestowed upon those chapters in the customary courses offer the greatest difficulties to
the beginner. The presenwill be found to be tation of problems as given in Chapter
V
quite a departure from the customary way of treating the subject. hence either book
4. all proofs for the sign age
of the product of
of the binomial
3. Moreover.
but the true study of algebra has not been sacrificed in order to make an impressive display of sham
life
applications.
viz. But on the other hand very few of such applied examples are
genuine applications of algebra. elementary way. are
frequently arranged in sets that are algebraically uniform."
Applications taken from geometry. physics. based upon statistical abstracts. and commercial are numerous.PREFACE
vii
and graphical methods into the first year's work.
nobody would find the length Etna by such a method.
McKinley
than one that gives him the number of Henry's marbles.
By studying proportions during the first year's work.
and they usually involve difficult numerical calculations.
is
based principally upon the alge-
. and
of the
hoped that some
modes of representation given
will be considered im-
provements upon the prevailing methods.
of the Mississippi or the height of Mt. " Graphical methods have not only a great practical value. in
"
geometry
. The entire work in graphical methods has been so arranged that teachers who wish
a shorter course
may omit
these chapters. and hence the student is more easily led to do the work by rote
than when the arrangement
braic aspect of the problem. the student will be able to utilize this knowledge where it is most
needed. an innovation which seems to mark a distinct gain from the pedagogical point
of view. such examples.
but they unquestionably furnish a very good antidote against 'the tendency of school algebra to degenerate into a mechanical application of
memorized
rules.'
This topic has been preit is
sented in a simple.
to solve a
It is
undoubtedly more interesting for a student
problem that results in the height of Mt. Moreover.
while in the usual course proportions
are studied a long time after their principal application.
genuine applications of elementary algebra work seems to have certain limi-
but within these limits the author has attempted to
give as
many
The author
for
simple applied examples as possible.
NEW
YORK.viii
PREFACE
problems relating to physics often
offer
It is true that
a field
for genuine applications of algebra.
ARTHUR SCHULTZE. desires to acknowledge his indebtedness to Mr.
edge of physics.
April. however. Manguse for the careful reading of the proofs and
many
valuable suggestions.
pupil's knowlso small that an extensive use of
The average
Hence the
field of
suitable for secondary school
tations.
William P.
. is such problems involves as a rule the teaching of physics by the
teacher of algebra. 1910.
6 = 7.12
17
&
*
ELEMENTS OF ALGEBRA
18
'
8
Find the numerical value of 8 a3
21. 2-6 of the exercise.
27. physics.
6.c) (a .
6=2.
Twice a3 diminished by 5 times the square root of the quantity a minus 6 square. 6 = 6.
a =3.
a.
a = 3. 6 = 1.
28. 6 = 3.
26. 25.
35. and the area of the
is
triangle
S
square feet (or squares of other units selected). 6 = 5.
22.6 -f c) (6
a
+ c). a = 2. Six times a plus 4 times
32. = 3.
:
6. 10-14
The
representation of numbers by letters makes it posvery briefly and accurately some of the principles of arithmetic. of this exercise?
What kind of expressions are Exs.
sible to state
Ex.
Express in algebraic symbols 31.
37. 33. 38. a
a=3.
34.
6 = 4. 6 = 5.
23. if
:
a = 2. a =4.
24. a = 4.
30. and other sciences.
then
8 = \ V(a + 6 + c) (a 4. 6 = 2. geometry.
12 cr6
-f-
6 a6 2
6s.
. a = 3. 6. 30.
w
cube plus three times the quantity a minus
plus 6 multiplied
6.
Read the expressions
of Exs. a = 4.
Six
2
.
The quantity a
6
2 by the quantity a
minus
36.
29.
Six times the square of a minus three times the cube of Eight x cube minus four x square plus y square.6 . 6 = 6.
and
If the three sides of a triangle contain respectively c feet (or other units of length).
and 13 inches. Find the height of the tree.seconds. 12. and 5 feet.
i.e. A body falling from a state of rest passes in t seconds 2 over a space S (This formula does not take into ac^gt 32 feet. 14.g.
(c) 4. 15 therefore feet.) Assuming g
.16 centimeters per second. if v . b. if v = 30 miles per hour.
the area of the triangle equals
feet. 13.
9
distance s passed over by a body moving with the uniform velocity v in the time t is represented by the formula
The
Find the distance passed over by A snail in 100 seconds.
=
(a)
How
far does
a body fall from a state of rest in 2
seconds ?
(b)
*
stone dropped from the top of a tree reached the ground in 2-J.
(b) 5. and c
13
and
15
=
=
=
. and 15 feet. A train in 4 hours.
84 square
EXERCISE
1.
4. How far does a body fall from a state of rest in T ^7 of a (c)
A
second ?
3. An electric car in 40 seconds. b 14. d. the three sides of a triangle are respectively 13. then a 13. if v
:
a. if v = 50 meters per second 5000 feet per minute.
By
using the formula
find the area of a triangle
whose
sides are respectively
(a) 3.
S = | V(13-hl4-fl5)(13H-14-15)(T3-14-i-15)(14-13-f-15)
= V42-12-14.16
1
= 84. count the resistance of the atmosphere.INTRODUCTION
E.
. A carrier pigeon in 10 minutes.
2. c.
and the value given above is only an
surface
$=
2
approximation.14
4.
diameter of a sphere equals d
feet.14d (square units).
(c)
5
F.).
of this formula
:
The The
interest on interest
$800
for 4 years at
ty%. Find the area of a circle whose radius is
It
(b)
(a) 10 meters. the equivalent reading C on the Centigrade scale may be found by the formula
F
C
y
= f(F-32).
(c)
5 miles. (The number 3.
$ = 3.
:
8000 miles. then
=p
n
*
r
%>
or
Find by means
(a)
(b)
6.
ELEMENTS OF ALGEBRA
If
the radius of a circle
etc.
If the
(b) 1 inch.
(c)
10
feet.14
square meters.
6
Find the volume of a sphere whose diameter equals:
(b)
3
feet.
denotes the number of degrees of temperature indi8.
to Centigrade readings:
(b)
Change the following readings
(a)
122 F.
then the
volume
V=
(a) 10 feet.
meters. If cated on the Fahrenheit scale.
(c)
8000 miles.
2 inches.). the
3.
on $ 500 for 2 years at 4 %.) Find the surface of a sphere whose diameter equals
(a)
7.
5.
square units (square inches.
32 F.
the area
etc.
If the diameter of a sphere equals d units of length. This number cannot be expressed exactly.
fo
If
i
represents the simple interest of
i
p
dollars at r
in
n
years.
is
H
2
units of length (inches.
.
~
7n
cubic feet.14 is frequently denoted by the Greek letter TT.
of
$6 and a gain
$4
equals a
$2 may be
represented thus
In a corresponding manner we have for a loss of $6 and a
of
loss
$4
(.
. Or in the symbols of algebra
$4) =
Similarly. Thus a gain of $ 2 is considered the sum of a gain of $ 6 and a loss of $ 4. or positive and negative numbers. we call the aggregate value of a gain of 6 and a loss of 4 the sum of the two. In arithmetic we add a gain of $ 6 and a gain of $ 4. SUBTRACTION. AND PARENTHESES
ADDITION OF MONOMIALS
31. but we cannot add a gain of $0 and a loss of $4. In algebra.CHAPTER
II
ADDITION. we define the sum of two numbers in such a way that these results become general.$6) + (-
$4) = (-
$10).
While
in arithmetic the
word sum
refers only to
the
result obtained
by adding positive numbers.
Since similar operations with different units always produce analogous results. however. in algebra this word includes also the results obtained by adding negative. the fact that a loss of
loss of
+ $2. or that
and
(+6) + (+4) = +
16
10.
of:
20.
Thus. 23-26.
(_
In Exs.
.
of 2.
33.
-
0. 10. 4.
21. + -12.
18.
23. = 5.
5.
19. the average of 4 and 8
The average The average
of 2.
ELEMENTS OF ALGEBRA
These considerations lead to the following principle
:
If two numbers have the same sign.
d = 5. 5.
(always) prefix the sign of the greater. the one third their sum.3.
+ (-9). d = 0. and the sum of the numbers divided by n. if :
a
a
= 2.
The average
of two
numbers
is
average of three numbers average of n numbers is the
is one half their sum.
EXERCISE
Find the sum
of:
10
Find the values
17. c = = 5.
is 0.16
32.
l-f(-2). 24.
(-17)
15
+ (-14).
4
is
3 J. c =
4.
6
6
= 3. add their absolute values if they have opposite signs. 12. subtract their absolute values and
. '. find the numerical values of a + b
-f c-j-c?. is 2.
22.
c = 0.
30.
:
48. 43.
are similar terms. 37. if his yearly gain or loss during 6 years was $ 5000 gain.7. = -13.
.
= 22. 55. 66.
40.4.
33.
&
28. and 4. or
and
. affected
by the same exponents. 09.
-'
1?
a
26.
5 a2 &
6 ax^y and
7 ax'2 y.
.
13. & = 15. -4. and 3 a.
31.
= -23. . 32. 32. .
'
Find the average of the following
34. $1000 loss. -11 (Centigrade). $7000 gain.
36. 74.
4
F..
3 and 25. Find the average gain per year of a merchant. 38.
42. and 3 F.
.13.
Similar or like terms are terms which have the same
literal factors. $3000 gain. 10.
25.
2. 6. 6. and $4500 gain.7. 10. 3. $500 loss.
Dissimilar or unlike terms are terms
4 a2 6c and o
4 a2 6c2 are dissimilar terms. and 3 yards. . 34..
:
Find the average temperature of Irkutsk by taking the average of the following monthly temperatures 12.
27.
6.5. 7 yards. SUBTRACTION.5.
AND PARENTHESES
d = l.3. 41.
^
'
37.
35. 7 a. 12. 7 a.
:
and
1. \\ Add 2 a. 72. 10. 39.
which are not
similar.
2. 60. c=14. }/ Add 2 a. 5 and
12. 1.
.
sets of
numbers:
13.
.
:
34.
d=
3.ADDITION. and 3 a.
29. Find the average temperature of New York by taking the average of the following monthly averages 30.
or 16
Va + b
and
2Vo"+~&..
and
-8
F.
Find the average of the following temperatures 27 F. 0.
ELEMENTS OF ALGEBRA
The sum
of 3
of
two similar terms
x2
is
is
another similar term.
2 a&.
+ 6 af
.
5 a2
. b wider sense than in arithmetic.
10. either the difference of a and b or the sum of a and
The sum
of
a. 7 rap2.
11.
1
\
-f-
7 a 2 frc
Find the sum of
9.
Vm
-f. or
a
6.
12(a-f b)
12.
-f
4 a2.
5Vm + w.18
35. While in arithmetic a denotes a difference only.
-3a
. Algebraic sum.
b
a
-f (
6).
12
13
b sx
xY xY 7 #y
7.
The sum The sum
of a of a
Dissimilar terms cannot be united into a single term.
.
14
.
2(a-f &).
11
-2 a +3a -4o
2.
:
2 a2.
In algebra the word sum is used in a 36.sign.
The sum
x 2 and
f
x2 .
12Vm-f-n.
ab
7
c
2
dn
6.
9(a-f-6). in algebra it may be considered b.
sum of two such terms can only be them with the -f. The indicated by connecting and a 2 and
a
is
is
-f-
a2
.
13.
12
2 wp2 .ii. and 4 ac2
is
a
2 a&
-|-
4 ac2.
EXERCISE
Add:
1.13 rap
25 rap 2.
5l
3(a-f-6).
2
.
a-b =
x. change the sign
of the subtrahend and
add. SUBTRACTION. Subtraction is the inverse of addition. Or in symbols.
(-
6)
-(-
= .
Ex. two numbers are given.2.3.
6
-(-3) = 8.
NOTE.
This gives by the same method. the algebraic sum and one of the two numbers is
The algebraic sum is given.
7. and their algebraic sum is required.
41.
3.
Ex.
State the other practical examples which show that the number is equal to the addition of a
40.ADDITION.
3 gives
3)
The number which added
Hence.
5
is
2.g. In subtraction. the given number the subtrahend. The student should perform mentally the operation of chang8 2 6 from 6 a 2 fc. and the
required number the difference.
From
5 subtract
to
.
From
5 subtract
to
The number which added
Hence. ing the sign of the subtrahend thus to subtract 6 a 2 6 and 8 a 2 6 and find the sum of change mentally the sign of
. Therefore any example in subtraction
different
.
a.
To
subtract. the other number is required. may be stated number added to 3 will give 5? To subtract from a the number b means to find the number which added to b gives a.
From
5 subtract
+ 3.
if
x
Ex.
may
be stated in a
:
5 take form e.
2. called the minvend. In addition.
AND PARENTHESES
23
subtraction of a negative
positive number.
1.
.
The
results of the preceding examples could be obtained
by the following
Principle.
3 gives 5
is
evidently 8.
+b
3. from What 3.
4a-{(7a + 6&)-[-6&-f(-2&.
A
moved
w
may be resign of aggregation preceded by the sign inserted provided the sign of evei'y term inclosed is
E.2 b .c.a
-f-
= 4a
sss
7a
12
06
6.
AND PARENTHESES
27
SIGNS OF AGGREGATION
43.
6
o+(
a
+ c) = a =a 6 c) ( 4-. A sign of aggregation preceded by the sign -f may be removed or inserted without changing the sign of any term.
If there is no sign before the first term within a paren*
-f-
thesis.
.
II. Simplify 4 a f
+ 5&)-[-6& +(-25.g.
45.c.
66
2&-a + 6
4a
Answer.
(b
c)
a
=a
6 4-
c.
46.a~^~6)]} = 4 a -{7 a 6 b -[.& c
additions
and sub-
+ d) = a + b
c
+ d.
Hence
the
it is
sign
may
obvious that parentheses preceded by the -f or be removed or inserted according to the fol:
lowing principles
44.ADDITION.
tractions
By using the signs of aggregation. we may begin either at the innermost or outermost. may be written as follows:
a
-f ( 4.6 b -f (.
& -f
c. SUBTRACTION.
changed. I. one occurring within the other. The beginner will find it most convenient
at every step to
remove only those parentheses which contain
(7 a
no others.
a+(b-c) = a +b .a^6)]
-
}
. Ex. If we wish to remove several signs of aggregation.b c = a a
&
-f-
-f. the sign
is
understood.
3.
II.
2m-n + 2q-3t.
12.
The
difference of a
and
6. The product of the sum and the difference
of
m and n.
The sum^)f
m
and
n.
2.
7.4 y* .
The The
difference of the cubes of
m
and
n.
z
+ d.
5. )X
6.ADDITION.
5^2
_ r .2 tf .
6 diminished
.
6.
Three times the product of the squares of The cube of the product of m and n.
The sum
of the fourth powers of a of
and
6. .
The sum
of tKe squares of a
and
b.
of the cubes of
m and
n.
m
x
2
4. 9.
The product The product
m
and
n.
m and n.
4.
In each of the following expressions inclose the last three in a parenthesis preceded by the minus sign
:
-27i2 -3^ 2 + 4r/.
The square of the difference of a and b.
'
NOTE.1. The minuend is always the of the two numbers mentioned.
4 xy
7 x* 4-9 x + 2.
5 a2
2.7-fa.
10. SUBTRACTION.
EXERCISE
AND PARENTHESES
16
29
In each of the following expressions inclose the last three terms in a parenthesis
:
1.
first.
terms
5.
a-\-l>
>
c
+
d.
13.
7.
p + q + r-s.
and the subtrahend the second.
3.
EXERCISES
IN"
ALGEBRAIC EXPRESSION
17
:
EXERCISE
Write the following expressions
I.
8.
difference of the cubes of n and m.
Nine times the square of the sum of a and by the product of a and b.
y
-f-
8
.
difference of the cubes of a
and
b divided
by the
difference of
a and
6.
x cube minus quantity 2 x2 minus 6 x plus
The sum
of the cubes of a.
ELEMENTS OF ALGEBRA
The sum
x. The difference of the squares of two numbers divided by the difference of the numbers is equal to the sum of the two numbers.)
.30
14.
b.
d.
a plus the prod-
uct of a and
s
plus the square of
-19. 6. (Let a and b represent the numbers.
6 is equal to the square of
b.
dif-
of the squares of
a and
b increased
by the
square root of
15.
and
c
divided by the
ference of a and
Write algebraically the following statements:
V 17.
18.
The sum
The
of a
and
b multiplied
b is equal to the difference of
by the difference of a and a 2 and b 2
. 16.
therefore.
3. 2.
A
A
A
1. what force is produced by the Ib. weights at A ? Express this as a multibalance. weights.CHAPTER
III
MULTIPLICATION
MULTIPLICATION OF ALGEBRAIC NUMBERS
EXERCISE
18
In the annexed diagram of a balance.
5. weight at B ?
If the
addition of five 3
plication example.
force is produced
therefore.
4. weight at A ? What is the sign of a 3 Ib. If the two loads balance.
By what sign is an upward pull at A represented ? What is the sign of a 3 Ib. and forces produced at by 3 Ib. applied at let us indicate a downward pull at by a positive sign.
If the
two loads balance. let us consider the and JB. is
by
taking away 5 weights from
A?
5
X 3?
6.
two loads balance. what force is produced by the addition of 5 weights at B ? What. is 5 x ( 3) ?
7. what force
31
is
produced by tak(
ing away 5 weights from
B ? What therefore is
5)
x(
3) ?
.
If the two loads
what
What.
4x(-3)=-12.
the multiplier is a negative number.
Thus. Practical examples^
it
however. 5x(-4). make venient to accept the following definition
:
con-
49. To take a number 7 times.32
8. 9 x
(-
11).
examples were generally
method of the preceding what would be the values of
(
5x4. and we may choose any definition that does not lead to contradictions.4)-(-4) = +
12. such as given in the preceding exercise.
. becomes meaningless
if
definition. Multiplication by a positive integer is a repeated addition. thus. 4 multi44-44-4 12. or
4x3 =
=
(_4) X
The preceding
3=(-4)+(-4)+(-4)=-12.
ELEMENTS OF ALGEBRA
If
the signs obtained by the
true. a result that would not be obtained by other assumptions. 4 multiplied by 3.
NOTE.
48.
In multiplying integers we have therefore four cases
trated
illus-
by the following examples
:
4x3 = 4-12.
(
(.
x
11.
9
9.
4
x(-8) = ~(4)-(4)-(4)=:-12. times is just as meaningless as to fire a gun
tion
7
Consequently we have to define the meaning of a multiplicaif the multiplier is negative.
Multiplication
by a negative
integer is a repeated
sub-
traction.4) x
braic laws for negative
~ 3> = -(.
however. (-5)X4. This definition has the additional advantage of leading to algenumbers which are identical with those for positive numbers.9) x
11.4)-(.
(. or plied by 3.
(-
9)
x (-
11) ?
State a rule by which the sign of the product of
two
fac-
tors can be obtained.
3
b
by a
5
b.
If
Arranging according to ascending powers
2
a
.
2a-3b a-66 2 a . multiply each term of one by each term of the other and add the partial products thus formed.a6
=2
by numerical
Examples
in multiplication can be checked
substitution. as illustrated in the following example
:
Ex. however. the student should
apply this test to every example.
Ex.3 a 2 + a8
a
a = =-
I
1
=2
-f
2
a
4. the work becomes simpler and more symmetrical by arranging these expressions according to either ascending or descending powers.M UL TIP LICA TION
37
58.1.
Multiply 2
+ a -a.
59.3 ab
2
2 a2
10 ab
-
13 ab
+ 15 6 2 + 15 6 2
Product.2 a2 6 a8
2 a*
*
-
2"
a2
-7
60. Since all powers of 1 are 1. If the polynomials to be multiplied contain several powers of the same letter. this method
tests only the values of the coefficients
and not the values of
the exponents. 1 being the most convenient value to be substituted for all letters.
. Since errors.
2. To multiply two polynomials.3 a 2 + a8 .a6 4 a 8 + 5 a* . are far more likely to occur in the coefficients than anywhere else.4.
Check. Multiply 2 a .
The most convenient way of adding the partial products is to place similar terms in columns.
a2
+ a8 + 3 .a
.3 a
3
2
by 2 a
:
a2 + l.
9.
The middle term
or
Wxy-12xy
Hence in general.
2 2 2 2 (2a 6 -7)(a & +
5).
7%e square of a polynomial is equal to the sum of the squares of each term increased by twice the product of each term with each that follows it. 14.
2
10.
7.
:
25
2.
5.42
ELEMENTS OF ALGEBRA
of the result is obtained
product of 5 x
follows:
by adding the These products are frequently called the cross products. (100 + 3)(100 + 4). (3m + 2)(m-l).
3.
(5a-4)(4a-l). plus the product of the
EXERCISE
Multiply by inspection
1.
or
The student should note
minus
signs.
. plus the
last terms.
65.
4.
6. 13.
8.
(x
i-
5
2
ft
x 2 -3 6 s).
2
(2m-3)(3m + 2).
2
2
+ 2) (10 4-3). the product of two binomials whose corresponding terms are similar is equal to the product of the first two
terms. ) (2
of a polynomial.
11.
2
(2x y
(6
2
2
+ z )(ary + 2z ).&
+ c) = a + tf + c
.-f
2 a&
-f
2 ac
+ 2 &c. (4s + y)(3-2y).
that the square of each term is while the product of the terms may have plus always positive.
The square
2
(a 4.
((5a?
(10
12.
(2a-3)(a + 2).
sum of the
cross products.
(5a6-4)(5a&-3). and are represented as
2 y and 4y 3 xy y or z) from its relation to
63
An
known numbers.
which
is
true for all values
a2 6 2 no matter what values we assign to a Thus.
x
20. second member is x
+
4
x
9.
hence
it
is
an equation
of
condition. (a + ft) (a b) and b. in the equation 2 x 0.
81. An equation of condition is usually called an equation.
the
80.
The
first
member
or left side of an equation
is
that part
The secof the equation which precedes the sign of equality.
.
. An equation of condition is an equation which is true only for certain values of the letters involved.
=11.
. An identity is an equation of the letters involved.
A set of numbers which when substituted for the letters an equation produce equal values of the two members. ond member or right side is that part which follows the sign of
equality. The sign of identity sometimes used is = thus we may write
.r
-f9
= 20
is
true only
when
a.
ber
equation is employed to discover an unknown num(frequently denoted by x. is said to satisfy an equation.
Thus.
y
=
7 satisfy
the equation x
y
=
13.
82.
(rt+6)(a-ft)
=
2
-
b'
2
.CHAPTER V
LINEAR EQUATIONS AND PROBLEMS
79.
in
Thus x
12 satisfies the equation x
+
1
13.
83.
the
first
member
is
2 x
+
4.
the divisor equals zero.
a.
86.
E. 87.
If equals be added
to equals.
ELEMENTS OF ALGEBRA
If
value of the
an equation contains only one unknown quantity.
Transposition of terms.
90.
85.
If equals be multiplied by equals.
9
is
a root of the equation 2 y
+2=
is
20. the products are equal.
expressed in arithmetical numbers
literal
is
as (7
equation
is
one in which at least one of the
known
quantities as x -f a letters
88.
2. the quotients are equal.
3.54
84.
one member to another by changing
x + a=.
4. the remainders are
equal.
If equals be divided by equals.
Like powers or
like roots
of equals are equal. A
2
a.
the
known quan
x) (x -f 4)
tities are
=
.
89. x
I.
A
term may be transposed from
its sign.b.
To
solve
an equation
to find its roots.2.
5.
A numerical
equation is one in
which
all
.
.
If equals be subtracted from equals. Consider the equation b Subtracting a from both members. called axioms
1.
(Axiom
2)
the term a has been transposed from the left to thQ
right
member by changing
its
sign.e.
2
= 6#-f7. the
sums are
equal.
fol-
A
linear equation is also called a simple equation.
Axiom
4
is
not true
if
0x4
= 0x5.
The process
of solving equations depends upon the
:
lowing principles.
.
NOTE.
but 4 does not equal
5.g. an^ unknown quantity which satisfies the equation is
a root of the equation.
= bx
expressed by a letter or a combination of
c.
A
linear equation or
which when reduced
first
to its simplest
an equation of the first degree is one form contains only the
as
9ie
power of the unknown quantity.
Ex.
7. or 12 7. Hence 6 a must be added
to a to give
5.
one part equals
is 10.
10.
Divide a into two parts. 15x
-f-
y yards cost $ 100
.
14. so that
of c ?
is
p. one yard will cost -
Hence
if
x
-f
y yards cost $ 100.58
Ex. so that one part
The
difference between
is s.
EXERCISE
1.
If 7
2.
9.
two numbers
and the and the
2
Find the greater one.
5. 6.
13. The difference between two numbers Find the smaller one.
What number divided by 3 will give the quotient a? ? What is the dividend if the divisor is 7 and the quotient
?
.
3.
6.
a. so that one part Divide a into two parts. and the smaller one
parts greater one is g.
is
a?
2
is
c?.
find the cost of one yard. is b.
1.
33
2. is d. 11.
$> 100 yards cost one hundred dollars. Divide 100 into two
12.
smaller one
16.
4.
Find the greater one.
17. one yard will cost
100
-dollars.
square feet are there in the area of the floor ?
How many
2 feet longer
29.
sum
If A's age is x years.
How many years
A
older than
is
B?
old. and B has n dollars. 28.
If
B
gave
A
6
25.
26.
and
c cents. rectangular field is x feet long and the length of a fence surrounding the field.
and spent
5 cents. b dimes
Find
21.
Find
35.
?/
31.
and 4
floor of a room that is 3 feet shorter wider than the one mentioned in Ex. 33. Find the sum of their ages
5 years ago.
A
dollars.
Find the area of the Find the area of the
feet
floor of
a room that
is
and 3
30.
is
A A
is
# years
old.
How many
cents are in d dollars ? in x dimes ?
A has
a
dollars.LINEAR EQUATIONS AND PROBLEMS
18. amount each will then have.
32. 28.
24.
numbers
is x.
A man
had a
dollars. and B's age is y years.
feet wider than the one
mentioned in Ex.
A
feet wide.
22. A room is x feet long and y feet wide.
34. The greatest of three consecutive the other two.
y years
How
old was he 5 years ago ?
How
old will he be 10 years hence ?
23.
and
B
is
y years old.
How many
cents had he left ?
28.
20. find the
has ra dollars. find the of their ages 6 years hence.
59
What must
The
be subtracted from 2 b to give a?
is a.
19.
How many
cents
has he ?
27.
Find a
47.
b
To express in algebraic symbols the sentence: " a exceeds much as b exceeds 9.
How
old
is
he
now ?
by a pipe in x minutes. If a man walks 3 miles per hour.
A
was 20 years
old.
A
cistern can be filled
in
alone
fills it
by two pipes.
of m. he walk each hour ?
39. If a man walks n miles in 4 hours.
48.50.
Find the
number.
Find
a.
per
Find 5 Find 6
45.
how many
how many
miles will
he walk in n hours
38.
-.
Find x
% %
of 1000.
-46.
a.
The numerator
If
of a fraction exceeds the denominator
by
3.
. The first pipe x minutes.
miles does
will
If a man walks r miles per hour. and the second pipe alone fills it in
filled
y minutes. in how many hours he walk n miles ?
40.
A
cistern
is
filled
43.
If a
man walks
?
r miles per hour. find the fraction. What fraction of the cistern will be filled by one pipe in one minute ?
42.60
ELEMENTS OF ALGEBRA
wil\
36. and "by as much as" Hence we have means equals (=)
95.
The two
digits of a
number
are x and
y.
How many
x years ago
miles does a train
move
in
t
hours at the
rate of x miles per hour ?
41." we have to consider that in this by statement "exceeds" means minus ( ).
of 4.
c
a
b
=
-
9.
m is the
denominator. how many miles he walk in n hours ?
37. What fraction of the cistern will be second by the two pipes together ?
44.
% % %
of 100
of
x.
49.
as
a exceeds
b
by as much
as c exceeds 9.
of a increased
much
8.
c.
5.
by one third of b equals 100.
80.
of a and 10 equals 2
c.
of x increased by 10 equals
x. a is greater than b by b is smaller than a by
c. etc.
8
-b ) + 80 = a
.
equal to the
sum and the difference of a and b sum of the squares of a and
gives the
Twenty subtracted from 2 a
a.
The double
as
7. thus:
a
b
= c may
be expressed as follows
difference between a
:
The
and
b is
c. Four times the difference of a and b exceeds c by as
d exceeds
9.
cases it is possible to translate a sentence word by in algebraic symbols in other cases the sentence has to be changed to obtain the symbols.
6.
third of x equals
difference of x
The
and y increased by 7 equals
a.
-80.
3.
same
result as 7
subtracted from
.
=
2
2
a3
(a
-
80.
EXERCISE
The The double The sum
One
34
:
Express the following sentences as equations
1. the difference of the squares of a
61
and
b increased
-}-
a2
i<5
-
b'
2
'
by 80 equals the excess of a over
80
Or.
double of a
is
10.
The
excess of a over b
is c.
9.
In
many
word
There are usually several different ways of expressing a symbolical statement in words. 2.LINEAR EQUATIONS AND PROBLEMS
Similarly.
c.
a exceeds b by c.
4.
The product
of the
is
diminished by 90 b divided by 7.
16.
(c)
If each
man
gains $500.
x
4-
If A. they have equal amounts. (e) In 3 years A will be as old as B is now.. they have equal
of A's.
x
is
100
x%
is
of 700. In 10 years the sum of A's.
and C's age
4
a.
ELEMENTS OF ALGEBRA
Nine
is
as
much below a
13.
symbols
B. a second sum
A
gains
$20 and B
loses
$40.
11.
12. express in algebraic symbols
:
-700. the
sum
and C's
money
(d)
(e)
will be $ 12.
(a)
(b)
(c)
A is twice as old as B. sum equals $20.*(/)
(g)
(Ji)
Three years ago the sum of A's and B's ages was 50.
6
%
of m. a third sum of 2 x + 1 dollars. B. and
(a)
(6)
A
If
has $ 5 more than B. and C have respectively 2 a. of 30 dollars.
17. pays to C $100.62
10. B000. the first sum equals 6 % of the third sura.
a. B's.
A
If
and
B
B together have $ 200 less than C. 3 1200 dollars.
50
is
x % of
15.
amounts.
first
00
x % of the
equals one tenth of the third sum. A is 4 years older than
Five years ago A was x years old.
.
a.
18. B's age
20. (d) In 10 years A will be n years old.
5x
A sum of money consists of x dollars.
m is x %
of n. 14. and C's ages will be 100.
as 17
is
is
above
a. express in algebraic
3x
:
10.
#is5%of450.
->.
is
If A's age is 2 x. In 3 years A will be twice as old as B.
1. be three times as old as he was 5 years ago.
Check.
number. verbal statement (1)
(1) In 15 years
A
will
may be expressed in symbols (2).
Ex. The solution of the equation (jives the value of the unknown number. x = 20.
Ex.
Simplifying. = x x
3x
-40
3x
40-
Or. Write the sentence in algebraic symbols. the required
.
3
x
+
16
=
x
x
(x
-
p)
Or.
much
as 40 exceeds the number. Find A's present age.
In 15 years
10.
Dividing.
A
will
Check. the
.
be 30
.
6 years ago he was 10
. number of
yards. etc.LINEAR EQUATIONS AND PROBLEMS
63
PROBLEMS LEADING TO SIMPLE EQUATIONS
The simplest kind of problems contain only one unknown number.
Three times a certain number exceeds 40 by as Find the number.
-23 =-30. In order to solve them.
equation is the sentence written in alyebraic shorthand.
Uniting. x + 15 = 3 x
3x 16
15.
3 x or 60 exceeds 40
+ x = 40 + 40.
x+16 = 3(3-5).
Let x = the number. Three times a certain no.
by 20
40 exceeds 20 by 20.
Let x
The
(2)
= A's present age.
but
30
=3
x
years. 15. The equation can frequently be written by translating the sentence word by word into algebraic symbols in fact.
Uniting. denote the unknown
96.
.
Transposing. 2.
3z-40:r:40-z. x= 15. The student should note that x stands for the number of and similarly in other examples for number of dollars. In 15 years A will be three times as old as he was 5 years ago.
Transposing.
NOTE. 4 x = 80.
number by x (or another letter) and express the yiven sentence as an equation. exceeds 40 by as much as 40 exceeds the no.
EXERCISE
1.
A will
be three times as old as to-da3r
.
3.
Let x
3. 11. by as much as 135 ft. 14.
Find the number whose double exceeds 30 by as much
as 24 exceeds the number.
to
42 gives a
sum
equal to 7 times the
original
6.
Find
8.
Find the number.
A
number added
number. Four times the length of the Suez Canal exceeds 180 miles by twice the length of the canal. exceeds the width of the bridge.
How
old
is
man will be he now ?
twice as old as he was
9.64
Ex.
Uldbe
66
| x x
5(5 is
=
-*-.
Find the number whose double increased by 14 equals Find the number whose double exceeds 40 by 10.
300
56. A train moving at uniform rate runs in 5 hours 90 miles more than in 2 hours. How many miles per hour does it run ?
.
14 50
is
is
4
what per cent of 500 ? % of what number?
is
12.
120.
13. Find the number.
4.
Hence
40
= 46f.
Dividing.
. % of
120. How long is the Suez
Canal?
10.
twice the number plus
7.2.
5.
Find the width of the Brooklyn Bridge.
Forty years hence
his present age.
47 diminished by three times a certain number equals 2.
35
What number added
to twice itself gives a
sum
of
39?
44.
ELEMENTS OF ALGEBRA
56
is
what per cent
of 120 ?
= number
of per cent.
What number
7
%
of
350?
Ten times the width of the Brooklyn Bridge exceeds 800 ft.
Six years hence a
12 years ago. then the
problem expressed in symbols
W
or.
000. In 1800 the population of Maine equaled that of Vermont. The sum of the two numbers is 14.
How many dollars
must
?
B
give to
18.
five
If
A
gives
B
$200.
One number exceeds another by
:
and their sum
is
Find the numbers. If the first farm contained twice as many acres as
A man
number
of acres. B will have lars has A now?
17. and Maine had then twice as many inhabitants as Vermont.
1.
statements are given directly. If a problem contains two unknown quantities.
B
How
will loses $100.
The problem consists of two statements I.
times as
much
as A. One number exceeds the other one by II. and
B
has $00. Find
the population of Maine in 1800. written in algebraic
symbols. If A gains A have three times as much
16.
.
14.
make A's money
equal to 4 times B's
money
wishes to purchase a farm containing a certain He found one farm which contained 30 acres too many.
how many
acres did he wish to
buy
?
19. and another which lacked 25 acres of the required number. The other verbal statement. Maine's population increased by 510. while in the more complex probWe denote one of the unknown
x.
Ex. two verbal statements must be given.
A
and
B
have equal amounts of money.
65
A
and
B
$200.
the second one. Ill the simpler examples these two
lems they are only implied.LINEAR EQUATIONS AND PROBLEMS
15.000. which gives the value of
8. Vermont's population increased by 180.
x.
is
the equation.
How many
dol-
A has
A
to
$40. During the following 90 years.
numbers (usually the smaller one) by
and use one of the
given verbal statements to express the other unknown number in terms of x. then dollars has each ? many
have equal amounts of money. and
as
15.
F
8.
97.
2x
a?
x
-j-
= 6. = 14. I.
x
3x
4-
and
B
will gain. A gives B 25 marbles.
+
a-
-f
-f
8
= 14.
Uniting. = 3.
If
A gives
are
:
A
If
II.
Statement
x
in
=
the larger number.
x
x =14
8.
.66
ELEMENTS OF ALGEBRA
Either statement may be used to express one unknown number in terms of the other. to
Use the simpler statement.
If
we
select the first one.
Then. although in general the simpler one should be selected.
A will lose.
.
o\
(o?-f 8)
Simplifying.
terms of the other.
<
Transposing.
To
express statement II in algebraic symbols.= The second statement written
the equation ^
smaller number. 2. unknown quantity
in
Then. A has three times as many marbles as B. the smaller number. the greater number. 8 = 11.
which leads
ot
Ex. consider that by the
exchange
Hence. 26 = A's number of marbles after the exchange.
Another method for solving this problem is to express one unknown quantity in terms of the other by means of statement II viz. 25 marbles to B.
The two statements
I.
has three times as many marbles as B.
in algebraic
-i
symbols produces
#4a.
/
.
.
x
= 8. B will have twice as many as A. = A's number of marbles.
and
Let x
= the
Then x -+.
Dividing.
Let
x
14
I
the smaller number. B will have twice as
viz. Let
x
3x
express one
many as A.
26
= B's number of marbles after the exchange. 8 the greater number.
expressed symbols is (14 x) course to the same answer as the first method.
= B's number of marbles. the sum of the two numbers is 14.
x = 15. of dollars to the number of cents.
w'3. Dividing.5 x . 3 x = 45.
Let
11
= the number of dimes.75. have a value of $3. B's number of marbles. cents..
Dividing.550 -f 310.
Uniting. How many are there of each ?
The two statements are I.
(Statement II)
Qx
. the number of dimes.
differ
differ
and the greater and their sum
times
Two numbers
by
60. The value of the half
:
is 11. Find the numbers. their sum
+ +
10 x 10 x
is
EXERCISE
36
is five
v
v. consisting of half dollars and dimes. the
price.
50 x
Transposing.
Simplifying.
by 44.
.
*
98. etc.
6 dimes
= 60
= 310.
2. 15 + 25 = 40.$3.
x
from
I.
The numbers which appear
in the equation should
always
be expressed in the same denomination. and the Find the numbers. The number of coins II.
67
x
-f
25 25
Transposing. 3.
dollars
and dimes
is
$3.
50(11 660 50 x
-)+ 10 x = 310. 45 . Check.25 = 20. x = 6.
1.
Simplifying.
Selecting the cent as the denomination (in order to avoid fractions). x = the number of half dollars.
Two numbers
the smaller.10..10.
Never add the number number of yards to their
Ex.
Check. but 40 = 2 x 20. .
*
'
.
50. A's number of marbles.
Find
the numbers. 40 x . the number of half dollars.240. 6 half dollars = 260 cents..
The sum of two numbers is 42.
6 times the smaller.
greater
is
.LINEAR EQUATIONS AND PROBLEMS
Therefore.
is 70. then.
x x
+
= 2(3 x = 6x
25
25).
60.10.
we
express the statement II in algebraic symbols. Eleven coins.
Uniting. 11 x = 5.
11. Twice 14.
United
States.
ELEMENTS OF ALGEBRA
One number
is
six
times another number. What are their ages ?
is
A A
much
line 60 inches long is divided into two parts.
7.
5.68
4. one of which increased by
9.
2 cubic feet of iron weigh 1600 foot of each substance. and B's age is as below 30 as A's age is above 40.
and twice the greater exceeds Find the numbers.
Two numbers
The number
differ
by
39.
How many
14 years older than B.
it
If the smaller
one contained 11 pints more.
How many
hours does the day last ?
. McKinley exceeds the altitude of
Mt.
What
is
the altitude of each
mountain
12. How many inches are in each part ?
15. Find their ages. McKinley.
3 shall be equal to the other increased by
10.
cubic foot of iron weighs three times as much as a If 4 cubic feet of aluminum and
Ibs.
of volcanoes in
Mexico exceeds the number
of volcanoes in the United States by 2.
9. Mount Everest is 9000 feet higher than Mt.
tnree times the smaller by 65. and the
greater increased by five times the smaller equals 22.
A's age is four times B's. and four times the former equals five times the latter.
Find
Find two consecutive numbers whose sum equals 157.
?
Two
vessels contain together 9 pints.
find the
weight of a cubic
Divide 20 into two parts.000
feet.
as the larger one.. How many volcanoes are
in the
8. the number.
6. and twice the altitude of Mt. the night in
Copenhagen
lasts 10 hours
longer than the day.
On December
21. the larger part exceeds five times the smaller part by 15 inches.
would contain three times as
pints does each contain ?
much
13. and in 5 years A's age will be three times B's. Everest by 11.
and in Mexico
?
A
cubic foot of aluminum.
three One of the unknown num-
two are expressed in terms by means of two of the verbal statements.
number
had.
first
According to
3 x
number
number
and according
to
80
4
x
=
the
express statement III by algebraical symbols. B.
A
and B each gave $ 5
respectively.
.
1. bers is denoted by x. Tf it should be difficult to express the selected verbal state-
ment
directly in algebraical symbols. or 66 exceeds 58 by 8.
8(8
+ 19)
to C. 19. x = 8.
5
5
Expressing in symbols Three times the sum of A's and B's money exceeds C's money by A's 3 x ( x _5 + 3z-5) (90-4z) = x.
the
the
number
of dollars of dollars of dollars
A
B
C
has. and B has three as A. number of dollars A had.LINEAR EQUATIONS AND PROBLEMS
99. The third
verbal statement produces the equation. II.
If
4x
= 24.
I. they would have 3. 4 x = number of dollars C had after receiving $10.
sum of A's and B's money would exceed much as A had originally.
number
of dollars of dollars
B
C
had.
Ex.
has. The solution gives
:
3x
80
Check.
Let
x
II. If A and B each gave $5 to C. III. = number of dollars B had after giving $5. and 68.
times as
much
as
A. and C together have $80. B."
To
x
8x
90
= number of dollars A had after giving $5.
are
:
C's
The three statements
A. B has three times as much as A. then
three times the
money by
I. let us consider the words ** if A and B each gave $ 5 to C. If A and B each gave $5 to C.
has. and the other
of x
problem contains three unknown quantities.
69
If a
verbal statements must be given. try to obtain
it
by a
series of successive steps. then three times the sum of A's and B's money would exceed C's money by as much as A had originally. = 48. original amount. and C together have $80.
The
I.
1 1
Check. x -f 4 = 9. number of sheep. number of cows.
28 x 15 or 450
5 horses.
first.
and. 9 -5 = 4 . 90
may
be written. x = 5. and the difference between the third and the second is 15
2.
III.
+
8
90 x
and. sheep. according to II.70
ELEMENTS OF ALGEBRA
man spent $1185 in buying horses. and the sum of the
. = the number of dollars spent for cows. each cow $ 35. 90 x -f 35 x + GO x = 140 20 + 1185.
37
Find three numbers such that the second is twice the first. 2.
A
and the number of sheep was twice as large as the number How many animals of each kind did he buy ?
of horses and cows together.
x 35
-f
+
=
+
EXERCISE
1. number of cows. 9 cows.
x
-j-
= the
number of horses. Let
then. number of horses. 28 2 (9 5).
Find three numbers such that the second is twice the 2.
85 (x 15 (4 x
I
+ 4)
+
8)
= the number of sheep. according to III.
first
the third exceeds the second by and third is 20.
+ 35 (x +-4) -f 15(4z-f 8) = 1185.
The total cost equals $1185. The number of cows exceeded the number of horses by
4. cows. = the number of dollars spent for sheep
Hence statement
90 x
Simplifying.
x
Transposing.
Uniting.
three statements are
:
IT.
Dividing. the third five times the first. 2 (2 x -f 4) or 4 x
Therefore. The number of cows exceeds the number of horses by 4. = the number of dollars spent for horses. and 28 sheep would cost 6 x 90 -f 9 + 316 420 = 1185.140 + (50 x x 120 = 185. The number of sheep is equal to twice tho number of horses and
x 4
the
cows together. 4 x -f 8 = 28. each horse costing $ 90. and each sheep $ 15. 185 a = 925. + 35 x 4. and Ex.
New York
delphia.000.
"Find three
is 4. If twice
The sum
the third side. and children together was 37.
The
three angles of any triangle are together equal to
180.
and
of the three sides of a triangle is 28 inches.
Find three consecutive numbers whose sum equals
63. women.
the third
2. If the population of New York is twice that of Berlin. and the third part exceeds the second by 10. increased by three times the second side. the second one is one inch longer than the first.
13. what is the population of each city ?
8. and is 5 years younger than sum of B's and C's ages was 25 years.LINEAR EQUATIONS AND PROBLEMS
3. men. and the sum of the first and third is 36.
7. how many children
were present ?
x
11.
what are the three angles ?
10.
is five
numbers such that the sum of the first two times the first.
-
4.
71
the
Find three numbers such that the second is 4 less than the third is three times the second. and the pig iron produced in one year (1906) in the United States represented together a value
.
twice the
6. and the third exceeds the
is
second by
5.000 more than Philadelphia (Census 1905). the copper.
the
first
Find three consecutive numbers such that the sum of and twice the last equals 22. and 2 more men than women.
A
is
Five years ago the What are their ages ?
C.000 more inhabitants than Philaand Berlin has 1.
v
-
Divide 25 into three parts such that the second part first. what is the length of each?
has 3.
A
12.
v
.000.
If the second angle of a triangle is 20 larger than the and the third is 20 more than the sum of the second and
first.
first. The gold.
twice as old as B.
first.
9. equals 49 inches. In a room there were three times as many children as If the number of women.
After how many hours will they meet and how
E. 8 x = 15.g.
.000.
3x
+
4 (x
2)
=
27. and quantities
area.
start at the same hour from two towns 27 miles walks at the rate of 4 miles per hour.72
of
ELEMENTS OF ALGEBRA
$ 750. Since in uniform motion the distance is always the product of
rate
and time. but stops 2 hours on the way.
California has twice as
many
electoral votes as Colorado.
B
many
miles does
A
walk
?
Explanation.000 more than that
the copper.e.000. we obtain 3 a.000. statement "A and B walk from two towns 27 miles apart until they meet " means the sum of the distances walked by A and B equals 27 miles.
and Massachusetts has one more than California and Colorado If the three states together have 31 electoral votes. and A walks at the rate of 3 miles per hour without stopping.
7
Uniting.
Dividing. Let x = number of hours A walks. 3 and 4.
of
arid the value of the iron
was $300. number of miles A
x
x
walks. = 5.
has each state
?
If the example contains Arrangement of Problems. such as length.
= 35.
Hence
Simplifying. it is frequently advantageous to arrange the quantities in a systematic manner. or time.
how many
100. together. width. number of hours. i. First fill in all the numbers given directly.
A and B
apart. then x 2 = number of hours B walks.
Find the value of each.
3z + 4a:-8 = 27. and 4 (x But the 2) for the last column. speed.
14.
The copper had twice
the value of the gold. of 3 or 4 different kinds. and distance.
Twenty men subscribed equal amounts
of
to raise a certain
money.
invested at 5 %. How much did each man subscribe ?
sum
walking at the rate of 3 miles per hour. but as two of them were unable to pay their share.74
ELEMENTS OF ALGEBRA
EXERCISE
38
rectangular field is 10 yards and another 12 yards wide.
Six persons bought an automobile. paid 24 ^ per pound and for the rest he paid 35 ^ per pound. and how far will each then have traveled ?
9. What are the
two sums
5.
2. and the sum Find the length of their areas is equal to 390 square yards. The second is 5 yards longer than the first.
If the silk cost three times as
For a part he 7.
A
If its length
rectangular field is 2 yards longer than it is wide. how much did each cost per yard ?
6.
mobile. each of the others had to pay
$ 100 more. How many pounds of each kind did he buy ?
8.
A
of each.
A sum
?
invested at 4 %.
Find the share of each.
Find the dimen-
A
certain
sum
invested at 5
%
%. together bring $ 78 interest. and follows on horseback traveling at the rate of 5 miles per hour. After how many hours will B overtake A. were increased by 3 yards.
A
sets out later
two hours
B
.
Ten yards
$
42. and a second sum. and the cost
of silk
of the auto-
and 30 yards of cloth cost together much per yard as the cloth. the area would remain the same.
1.
sions of the field.55.
3. twice as large. but four men failed to pay their shares. and its width decreased
by 2 yards. of coffee for $ 1.
sum $ 50
larger invested at 4
brings the same interest Find the first sum. and in order to raise the required sum each of the remaining men had to pay one dollar more.
as a
4. A man bought 6 Ibs.
walking at the same time in the same If A walks at the rate
of 2
far
miles per hour. but
A has
a start of 2 miles. how many miles from New York will they meet?
X
12.LINEAR EQUATIONS AND PROBLEMS
v
75
10. Albany and travels toward New York at the rate of 30 miles per hour without stopping.will they be 36 miles apart ?
11. and B at the rate of 3 miles per hour. and from the same point. and another train starts at the same time from New York traveling at the rate of 41 miles an hour. traveling by coach in the opposite direction at the rate of 6 miles per hour. After how many hours. how must B walk before he overtakes A ?
walking at the rate of 3 miles per hour.
A
and
B
set out
direction.
The
distance from
If a train starts at
.
A
sets out
two hours
later
B
starts
New York to Albany is 142 miles.
it contains no indicated root of this letter
.
which multiplied together
are considered factors.
stage of the work. a. if.
76
. as.
The
factors of
an algebraic expression are the quantities
will give the expression. but fractional with respect
103.CHAPTER
VI
FACTORING
101.
An
after simplifying.
it is
composite.
a2
to 6. if it does contain
some indicated root of
.
a.
J Although Va'
In the present chapter only integral and rational expressions
b~
X
V
<2
Ir
a2
b'
2
2
?>
. consider
105.
vV
. 5. at this
6
2
. An expression is integral with respect to a letter.
this letter.
irrational.
\-
V&
is
a
rational with respect to
and
irrational with respect
102.
expression is rational with respect to a letter. if this letter does not occur in any denominator.
-f-
db
6
to b.
An
expression
is integral
and rational with respect
and rational.
a factor of a 2
A
factor is said to be prime.
we
shall not.
104.
a-
+
2 ab
+ 4 c2
.
The prime
factors of 10 a*b are 2. 6. if it is integral to all letters contained in it.
+ 62
is
integral with respect to a. if it contains
no other
factors (except itself
and unity)
otherwise
.
110. or that a
=
6)
(a
= a .)
Ex. for this result is a sum.
it
follows
that a 2
.
Since factoring
the inverse of multiplication.FACTORING
106.
An
the process of separating an expression expression is factored if written in the
form of a product.
POLYNOMIALS ALL OF WHOSE TERMS CONTAIN A COMMON FACTOR
(
mx + my+ mz~m(x+y + z).62 + &)(a 2
.
Factor G ofy 2
.
77
Factoring
is
into its factors.
The factors
of a
monomial can be obtained by inspection
2
The prime
108.
in the
form
4)
+3.
y.
TYPE
I.
or
Factoring examples may be checked by multiplication by numerical substitution.
?/.
55. 2.
x.
Factor
14 a*
W-
21 a 2 6 4 c2
+ 7 a2 6
2
c2
7
a2 6 2 c 2 (2 a 2
.g.9 x2 y 8 + 12
3 xy
-f
by
3
xy\
and the quotient
But.
It (a.
Ex.
1.9 x if + 12 xy\
2
The
greatest factor
common
2
to all terms
flcy*
is
8
2
xy'
.
107. since (a + 6) (a 2 IP factored.3 sy + 4 y8).
E. .
2 4 x + 3) is factored if written (x' would not be factored if written x(x and not a product.62
can be
&).
109. x.
. 8) (s-1).
2.
Hence
6 aty 2
= divisor x quotient.
01.
factors of 12
&V
is
are 3.9 x2^ + 12 sy* = 3 Z2/2 (2 #2 . dividend
is
2 x2
4
2
1/
. 2.3 6a + 1).
Divide
6
a% .
it fol-
lows that every method of multiplication will produce a method
of factoring.
can be factored.
2
6.
but of these only
a:
Hence
2
.4 . it is advisable to consider the factors of q first.
as p.
or
77
l..
If
30 and whose
sum
is
11 are
5
a2
11
a = 1.
+
112.30 = (a .
77 as the product of 1 77.G) = .1 1 a
tf
a 4.
2.
and the greater one has the same sign
Not every trinomial
Ex.
EXERCISE
Besolve into prime factors :
40
4. If q is negative. of this type. the student should first all terms contain a common monomial factor. a 2 .11) (a
+
7). but only in a limited number
of ways as a product of two numbers. or 11 and 7 have a sum equal to 4.11 a
2
.
determine whether
In solving any factoring example.FACTORING
Ex.
We may consider
1.
.a).1 afy 8 The two numbers whose product is equal to 12 yp and whose sum equals 3 8 7 y are -4 y* and -3 y*. + 30 = 20.
Since a number can be represented in an infinite number of ways as the sum of two numbers. however. the two numbers have both the same sign as p.6 = 20.
5.
tfa2 -
3.77 =
(a.11 a + 30.5) (a .
m -5m + 6. the two numbers
have opposite
signs.11.
79
Factor a2
-4 x .
Ex. Hence z6 -? oty+12 if= (x -3 y)(x*-4 y ). If q is positive.
3.
. Hence fc -f 10 ax
is
10 a are 11 a
-
12 /.4 x .
4.5) (a 6).
11 a2 and whose sum The numbers whose product is and a.
Factor a2
.
Factor
+ 10 ax .
Therefore
Check. and (a . Factor x? .
.
Ex.
11
7. or 7 11. 2 11 a?=(x + 11 a) (a.
is
The two numbers whose product and -6.
3 x and x.5) (2 x .
a.17 x
2o?-l
V A
5
-
13 a
combination
the correct one.
Factor 3 x 2
. viz. 2 x 27. Hence only 1 x 54 and 2 x 27 need
be considered.
the
If p and r are positive. Since the first term of the first factor (3 x) contains a 3.83 x
-f-
54. the signs of the second terms are minus. 27 x 2.FACTORING
If
81
we consider that the
factors of -f 5
as
must have
is
:
like signs. If p is poxiliw. and after a little practice the student possible should be able to find the proper factors of simple trinomials
In actual work
at the first trial. then the second terms of
have opposite signs.
Ex. the second terms of the factors have same sign as q. 2. or
G
114. X x 18. 6 x 9. all pos-
combinations are contained in the following
6x-l
x-5 .
3.1). 64 may be considered the
:
product of the following combinations of numbers 1 x 54. none of the binomial factors can contain a monomial factor. we have to reject every combination of factors of 54 whose first factor contains a 3.
The
and
factors of the first term consist of one pair only.31 x
Evidently the
last
2
V A
6. exchange the
signs of the second terms of the factors. which has the same absolute value as the term qx.
sible
13 x
negative.
The work may be shortened by the
:
follow-
ing considerations
1.
all
it is not always necessary to write down combinations.
If
the factors
a combination should give a sum of cross products.
.
and that they must be negative.
11 x
2x.e-5
V A
x-1
3xl \/ /\
is
3
a. 54 x 1.
.13 x + 5 = (3 x . but the opposite sign.5 . 18 x 3. 9 x 6. and r is negative. If py? -\-qx-\-r does not contain any monomial factor.
of
aW. of the algebraic expressions. F. and prefix it as a coefficient to H. C.
5
7
34 2s
. F.
+
8
ft)
and
cfiW is
2
a 2 /) 2
ft)
. of
two or more monomials whose factors
.
121.
5
s
7
2
5.
C.
The student should note
H. C.
15
aW. F.
.
C.
13 aty
39 afyV. F.
II
2
.
C.
33
2
7
3
22 3 2
.
expressions which have no are prime to one another. The H. If the expressions have numerical coefficients.
the algebraic factor of highest degree common expressions to these expressions thus a 6 is the II. 12 tfifz. F. C. of (a
and (a
+
fc)
(a
4
is
(a
+ 6)
2
. C. and GO aty 8 is 6 aty.
54
-
32
. F.
5
2
3
.
The
highest
is
common
factor (IT.CHAPTER
VII
HIGHEST COMMON FACTOR AND LOWEST COMMON MULTIPLE
HIGHEST COMMON FACTOR
120.
25
W.) of
two or more
.
24
s
.
-
23 3
.
EXERCISE
Find the H. F. find by arithmetic the greatest common factor of the coefficients.
Two
common
factor except unity
The H. aW. of
:
48
4.
122.
2
2
. of a 4 and a 2 b is a2
The H.
3.
are prime can be found by inspection. Thus the H. F. of a 7 and a e b 7
. C.
2.
5.
6. The H.
89
. F.
F. of 6 sfyz. C.
3
. is the lowest
that the power of each factor in the power in which that factor occurs in any
of the given expressions. C.
8
.
The
L.C. resolve each expression into prime factors and apply the method for monomials.
NOTE. C. L.
128.
2
The The
L.
2 multiples of 3 x
and 6 y are 30 xz y.
a^c8
3
. C.
Common
125.
. C. L.
300 z 2 y.C. each set of expressions has
In example ft). C.
M.
The
lowest
common
multiple (L. find by arithmetic their least common multiple and prefix it as a coefficient to the L.
. is equal to the highest power in which it occurs in any of the
given expressions.
127.(a + &) 2 (a
have the same absolute value.) of
two or more
expressions is the common multiple of lowest degree.
= (a -f
last
2
&)'
is
(a
-
6) . C.
etc.
of 3
aW.
Find the L. which
also
signs.
A
common
remainder. C.
M.
6
c6 is
C a*b*c*.
1.
C.
4 a 2 &2
_
Hence.
M.
Hence the L.
of 4 a 2 6 2 and 4 a 4
-4 a 68
2
. of the
general.
&)
2
M.
Ex.
M of the algebraic expressions. M. To find the L.M.6 3 ). of as -&2 a2 + 2a&-f b\ and 6-a.
Find the L. 60
x^y'
2
. M.
=4 a2 62 (a2 . but opposite
. Obviously the power of each factor in the L. of tfy and xy*.
of 12(a
+
ft)
and (a
+ &)*( -
is
12(a
+ &)( . If the expressions have a numerical coefficient. C. M.
2. C.LOWEST COMMON MULTIPLE
91
LOWEST COMMON MULTIPLE
multiple of two or more expressions is an which can be divided by each of them without a expression
124. M.
126.M.
Ex. ory is the L. M. thus. C.6)2. of several expressions which are not completely factored. two lowest common multiples.
successively all
2
j/' .
130.
Reduce
~-
to its lowest terms. If both terms of a fraction are multiplied or divided by the same number) the value of the fraction is not altered. only positive integral numerators shall assume that the
all
arithmetic principles are generally true for
algebraic numbers.
an indicated quotient.
fraction
is
in its lowest
when
its
numerator
and
its
denominator have no
common
factors.ry ^
by
their H.
rni
Thus
132.
Remove
tor.
a?.
A
-f-
fraction is
b. but we
In arithmetic.
common
6
2
divisors of
numerator and denomina-
and z 8
(or divide the terms
. the value of a fraction is not altered by multiplying or dividing both its numerator and its denominator by the same number. C.
The dividend a is called the numerator and the The numerator and the denominator
are the terms of the fraction.
TT
Hence
24
2 z = --
3x
.
All operations with fractions in algebra are identical
with the corresponding operations in arithmetic.
131. thus -
is identical
with a
divisor b the denominator.
and denominators are considered. F.
the product of two fractions is the product of their numerators divided by the product of their denominators.
Ex.CHAPTER
VIII
FRACTIONS
REDUCTION OF FRACTIONS
129.
and
i
x mx = my y
terms
A
1.
a b
= ma
mb
. however. Thus. etc. as 8.
+
3).-1^22
' . and
135.
ELEMENTS OF 'ALGEBRA
Reduction of fractions to equal fractions of lowest common Since the terms of a fraction may be multiplied
denominator.
To reduce
to a fraction with the
denominator 12 a3 6 2 x2 numerator
^lA^L O r 2 a 3
'
and denominator must be multiplied by
Similarly.
3 a\ and 4
aW
is
12 afo 2 x2 .~16
(a
+ 3) (x.3) (-!)'
=
.
1. of the denominators for the common denominator.
and
(a-
8).
Since a
(z
-6 + 3)(s-3)O-l)'
6a.C. C. we may use the same process as in arithmetic for reducing fractions to the lowest
common
denominator.
multiply each quotient by the corresponding numerator.
^
to their lowest
com-
The
L.
.
Multiplying these quotients by the corresponding numerators and writing the results over the common denominator.
.
TheL.M.
-
by 4
6' . and 6rar 3 a? kalr
.
and
Tb reduce fractions to their lowest common denominator.r
2
2
.
2>
. multiplying the terms of
22
.
M.
we have
-M^.
Reduce -^-.
by any quantity without altering the value of the fraction. by the denominator of each fraction.
-
of
//-*
2
.C. Divide the L.
mon
T denominator. we have
(a
+ 3) (a -8) (-!)'
NOTE.
1).M.
we may extend this method
to integral expressions. C.
and the terms of
***.
.by 3 ^
A
2
' . take the L.
we have
the quotients (x
1).96
134.3)O -
Dividing this by each denominator.D.
Ex
-
Reduce
to their lowest
common
denominator. =(z
(x
+ 3)(z.
Ex.
expressed in symbols:
c
a
_ac b'd~bd'
principle proved for
b
141. Fractions are multiplied by taking the product of tht numerators for the numerator.
F J Simplify
.
2
a
Ex.g.
we may extend any
e.
fractions to integral numbers.
integer. multiply the
142.)
Ex.102
ELEMENTS OF ALGEBRA
MULTIPLICATION OF FRACTIONS
140.
Since -
= a. Common factors in the numerators and the denominators should be canceled before performing the multiplication. each
numerator and denomi-
nator has to be factored.
Simplify 1 J
The
expreeaion
=8
6
. and the product of the denominators for the
denominator.
2.
or.
!.
-x
b
c
=
numerator by
To multiply a fraction by an
that integer. (In
order to cancel
common
factors.
8
multiply
the
Ex invert the divisor and multiply it by the dividend.104
ELEMENTS OF ALGEBRA
DIVISION OF FRACTIONS
143.
1. x a + b
obtained by inverting
reciprocal of a fraction
is
the fraction. and the principle of division follows
may
be expressed as
145.
The reciprocal of ?
Hence the
:
+*
x
is
1
+ + * = _*_. To divide an expression by a fraction.
The
reciprocal of a
number
is
the quotient obtained by
dividing 1
by that number.
.
The The
reciprocal of a
is
a
1
-f-
reciprocal of J
is
|
|. Integral or mixed divisors should be expressed in fractional form before dividing.
144. :
a 4-1
a-b
* See page 272.
Divide X-n?/
. To divide an expression by a fraction.
* x* -f xy 2
by
x*y
+y
x'
2
3
s^jf\ =
x'
2
x*
. expression by the reciprocal of the fraction.
of minute spaces the
hour hand moves
Therefore x
~ = the number of minute spaces the minute hand moves
more than the hour hand.
Find
R in terms of C and
TT. In how many days can both do it working together ?
If
we denote
then
/-
the required
number
by
1100
C.
Multiplying by
Dividing.
. hence the question would be formulated After how many minutes has the minute hand moved 15 spaces more than the hour hand ?
Let then
x x
= the required number of minutes after 3 o'clock.
A would do
each day ^ and
B
j.. Ex.180.
.
Ex. 2. = the number of minute spaces the minute hand moves
over.
days by x and the piece of work while in x days they would do
respectively
ff
~ and and hence the sentence written in algebraic symbols ^.
then
= 2 TT#.
C
is
the circumference of a circle whose radius
R. = 16^.
~^ = 15
11 x
'
!i^=15.
12. A can do a piece of work in 3 days and B in 2 days. When between 3 and 4 o'clock are the hands of
a clock together
?
is
At
3 o'clock the hour hand
15 minute spaces ahead of the minute
:
hand. 2 3
.
is
36.
and
12
= the number
over. 1.114
35..
x
Or
Uniting.minutes after x=
^
of
3 o'clock.20
C.
PROBLEMS LEADING TO FRACTIONAL AND LITERAL EQUATIONS 152.
u The accommodation train needs 4 hours more than the express train.
But
in
uniform motion Time
=
Distance
. 3.
in
Then
Therefore."
:
Let
x -
= the
required
number
of days.
32
x
= |. what is
the rate of the express train
?
180
Therefore.
= the
x
part of the
work both do
one day.
Explanation
:
If
x
is
the rate of the accommodation train.
180
Transposing.
Solving.
the rate of the express train." gives the equation /I). The speed of an express train is $ of the speed of an If the accommodation train needs 4 accommodation train. and the statement.
Clearing.
or 1J. the required
number
of days. hours more than the express train to travel 180 miles.
= 100 + 4 x.
fx
xx*
=
152
+4
(1)
Hence
=
36
= rate
of express train. 4x = 80.
Ex. then
Ox
j
5
a
Rate Hence the rates can be expressed.
Twenty years ago A's age was |
age.
J-
of the greater
increased by ^ of the smaller equals
6.
fifth
Two numbers
differ
2.
are the
The sum of two numbers numbers ?
and one
is
^ of the other.
3.
Find A's
8.
its
Find the number whose fourth part exceeds part by 3.
and 9
feet above water. and one half the greater Find the numbers.
by 3. -|
Find their present ages.
ceeds the smaller by
4.
Find two consecutive numbers such that
9.
Two numbers
differ
l to s of the smaller. and J of the greater Find the numbers.116
ELEMENTS OF ALGEBRA
EXERCISE
60
1.
9
its
A
post
is
a fifth of
its
length in water. How much money had he at first?
12
left
After spending ^ of his
^ of his money and $15. and of the father's age. A man lost f of his fortune and $500.
ex-
What
5.
of his present age.
length in the ground.
money and $10. How
did the
much money
man
leave ?
11. one half of What is the length
of the post ?
10
ter.
is oO.
by 6.
A man left ^ of his property to his wife.
is
equal
7.
make
21.
Find a number whose third and fourth parts added
together
2. a man had How much money had he
at first?
.
The sum
10 years hence the son's age will be
of the ages of a father and his son is 50. and found that he had \ of his original fortune left. to his daughand the remainder. which was $4000. to his son.
A can
A
can do a piece of work in 2 days.
air.
at 4J % and P> has invested $ 5000 They both derive the same income from their How much money has each invested ?
20. 3.
At what time between 4 and
(
5 o'clock are the hands of
a clock together?
16. Ex. and after traveling 150 miles overtakes the accommodation train.
How much money
$500?
4%.
?
In
how many days can both do
working together
23.
after
rate of the latter ?
15. what is the
14.) (
An express train starts from a certain station two hours an accommodation train.
A has invested capital
at
more
4%. An ounce of gold when weighed in water loses -fa of an How many ounce.
investments. and an ounce of silver -fa of an ounce.
. ounces of gold and silver are there in a mixed mass weighing
20 ounces in
21. and B In how many days can both do it working together
in
?
12 days. A can do a piece of work in 4 clays. ^ at 5%. In how many days can both do it working together ? ( 152. If the rate of the express train is -f of the rate of the accommodation train.)
At what time between 7 and 8
o'clock are the
hands of
a clock together ?
17.)
22. Ex.
152. and
it
B in 6 days. and B in 4 days.
and losing
1-*-
ounces when weighed in water?
do a piece of work in 3 days. Ex.FRACTIONAL AND LITERAL EQUATIONS
13. 1. and has he invested if
his animal interest therefrom is
19.
117
The speed
of an accommodation train
is
f of the speed
of an express train.
A man
has invested
J-
of his
money
at
the remainder at
6%. what is the rate of the express train? 152.
At what time between 7 and
8 o'clock are the hands of
?
a clock in a straight line and opposite
18. 2. If the accommodation train needs 1 hour more than the express train to travel 120 miles.
A in 6.118
153. A in 4.
is 42.=
m
-f-
n
it
Therefore both working together can do
in
mn
-f-
n
days. Answers to numerical questions of this kind may then be found by numerical substitution.g. therefore. The problem to be solved.
Find three consecutive numbers whose sum
Find three consecutive numbers whose sum
last
:
The
two examples are
special cases of the following
problem 27. n x
Solving.
25. Then
ft
i. if
B
in 3 days. e.
.
they can both do
in 2 days. Hence.
6
I
3
Solve the following problems
24. B in 12. by taking for these numerical values two general algebraic numbers. A in 6. B in 16. and apply the
method of
170.
B in 5.
make
it
m
6
A can do this work in 6 days Q = 2.414.
26. m and n. B in 30.
ELEMENTS OF ALGEBRA
The
last three questions
and their solutions differ only two given numbers.= -.
we
obtain the equation
m m
-. Ex. . In how
in the numerical values of the
:
many days
If
can both do
we
let
x
= the
it working together ? required number of days.e009 918. Find the numbers if m = 24 30. is 57. is A can do a piece of work in m days and B in n days.
To
and
find the numerical answer. 3.
. Find three consecutive numbers whose sum equals m. it is possible to solve all examples of this type by one example. 2. and n = 3.
33. respectively (a) 60 miles.
. (a) 20 and 5 minutes.
A cistern can
be
filled
(c)
6 and 3 hours.000.
34. 3 miles per hour.FRACTIONAL AND LITERAL EQUATIONS
28. the second at the apart. d miles the first traveling at the rate of m. 2 miles per hour. and how many miles does each travel ?
32.
meet.
The
one:
31.
squares
30. (b) 35 miles. if m and n are. After how many hours do they rate of n miles per hour. (d) 1. Find the side of the square.
last three
examples are special cases of the following
The
difference of the squares of
two consecutive numbers
By using the result of this problem. If each side of a square were increased by 1 foot.
Two men
start at the
first
miles
apart. After how many hours do they meet. the
Two men start at the same time from two towns. 5 miles per hour.
is (a)
51. 3J miles per hour. (c) 16.
119
Find two consecutive numbers the difference of whose
is 11. 88 one traveling 3 miles per hour.
and
the rate of the second are. 2 miles per hour.
4J-
miles per hour.001.721. and how many miles does each travel ? Solve the problem if the distance.
:
(c)
64 miles.
squares
29.
same hour from two towns.
Find two consecutive numbers -the difference of whose
is 21. two pipes together ? Find the numerical answer. the area would be increased by 19 square feet. (b) 8 and 56 minutes. (b) 149. and the second 5 miles per hour. the rate of the
first. respectively. solve the following ones Find two consecutive numbers the difference of whose squares
:
find the smaller number.
is ?n
.
by two pipes in m and n minutes In how many minutes can it be filled by the respectively.
g.
Thus the
written a
:
ratio of a
b
is
.
A
ratio
is
used to compare the magnitude of two
is
numbers.
In the ratio a
:
ft.
terms are multiplied or divided by the same number.CHAPTER X
RATIO AND PROPORTION
11ATTO
154.
b. a ratio
is
not changed
etc. the denominator
The
the
157.
Simplify the ratio 21 3|.
:
A somewhat shorter way
would be to multiply each term by
120
6.
" a Thus.5.
The
ratio -
is
the inverse of the ratio -.
antecedent.) The ratio of 12 3 equals 4.
:
:
155.
1. b is the consequent.
term of a ratio
a
the
is
is
the antecedent.
the symbol
being a sign of division.
The
first
156. 6 12 = .
The
ratio of
first
dividing the
two numbers number by the
and
:
is
the quotient obtained by
second. etc.
b.or a *
b
The
ratio is also frequently
(In most European countries this symbol is employed as the usual sign of division."
we may
write
a
:
b
= 6. all principles
relating
to
fractions
if its
may
be af)plied to ratios.
b
is
a
Since a ratio
a
fraction. 158.
is
numerator of any fraction
consequent.
E.
Ex. the antecedent. the second
term the consequent. instead of writing
6 times as large as
?>.
.
In the proportion a b
:
=
b
:
c.
terms. b.
1.
3
8.
equal
2.
5 f hours
:
2.
proportional between a
and
c. a and d are the extremes.
:
ratios so that the antecedents equal
16:64.
:
a-y
.
3.
extremes.
12.
b is the
mean
b.
16 x*y
64 x*y
:
24 48
xif.
Simplify the following ratios
7.
:
is
If the means of a proportion are equal.RATIO
Ex.
9.
61
:
ratios
72:18.
:
1.
11. either mean the mean proportional between the first and the last terms.
Transform the following
unity
15. and the last term the third proportional to the first and second
161. The last term d is the fourth proportional to a.
7f:6J.
$24: $8.
27 06: 18 a6.
4.
3:1}.
6.
= |or:6=c:(Z are
The
first
160.
17.
A
proportion
is
a statement expressing the equality of
proportions.
and
c
is
the third proportional to a and
. b and c the means.
7|:4 T T
4
. The last
first three.
62:16.
8^-
hours.
two
|
ratios10.
4|-:5f
:
5.
159. and c.
16. the second
and fourth terms of a proportion are the and third terms are the means.
term
is
the fourth proportional to the
:
In the proportion a b = c c?.
16a2 :24a&.
18.
3:4.
__(163. a b
:
bettveen two
numbers
is
equal to
the square root
Let the proportion be
Then Hence
6
=b = ac.
:
c.
Clearing of fractions. is equal to the ratio of the corresponding two
of the other kind. then G ccm.
:
:
directly proportional
may say. i.)
mn = pq. of iron weigh 45 grams. = 30 grams 45 grams.
q~~ n
.
ccm. If the product of two numbers is equal to the product of two other numbers^ either pair may be made the means. and the time necessary to do it.
If 6 men can do a piece of work in 4 days. Instead of u
If 4
or 4 ccm.
2
165. 6 ccm.
163.
ad =
be.
briefly. if the ratio of any two of the first kind. if the ratio of any two of the first kind is equal \o the inverse ratio of the corresponding two of
the other kind.e. are
: : :
inversely proportional. or 8 equals the inverse ratio of 4 3.
pro-
portional.
!-.30 grams.
t/ie
product of the means
b
is
equal
to the
Let
a
:
=c
:
d. and the
other pair the extremes. Hence the weight of a mass of iron is proportional to its volume.
163.'*
Quantities of one kind are said to be inversely proportional to quantities of another kind. then 8 men can do it in 3 days. " we " NOTE.122
162.
ELEMENTS OF ALGEBRA
Quantities of one kind are said to be directly proper
tional to quantities of another kind.
164.
The mean proportional
of their product. of a proportion. In any proportion product of the extremes.
If
(Converse of
nq. 3 4. and we
divide both
members by
we have
?^~ E.) b = Vac. of iron weigh . Hence the number of men required to do some work.
the volume of a
The temperature remaining
body of gas inversely proportional to the pressure. and the time.
56.
ELEMENTS OF ALGEBEA
State the following propositions as proportions : T (7 and T) of equal altitudes are to each. The number of men (m) is inversely proportional to the number of days (d) required to do a certain piece of work. (e) The distance traveled by a train moving at a uniform rate.126
54. and the
:
total cost.
and the speed
of the train.
areas of circles are proportional to the squares of If the radii of two circles are to each other as
circle is
4
:
7.
(b)
The time a
The length
train needs to travel 10 miles.
A
line 11 inches long
on a certain
22 miles.
the area of the larger? the same. and
the time necessary for it. under a pressure of 15 pounds per square inch has a volume of
gas
is
A
16 cubic
feet.
(c)
of a rectangle of constant width.inches long represents
map corresponds to how many miles ?
The
their radii. State whether the quantities mentioned below are directly or inversely proportional (a) The number of yards of a certain kind of silk. othei
(a) Triangles
as their basis (b
and
b'). and the area
of the rectangle.
(d)
The sum
of
money producing $60
interest at
5%.
A
line 7^.
(d)
The
areas
(A and
A') of two circles are to each other as
(R and R').
and the area of the smaller
is
8 square inches.
the squares of their radii
(e)
55. (c) The volume of a body of gas (V) is
circles are to
each
inversely propor-
tional to the pressure (P).
What
will be the
volume
if
the pressure
is
12 pounds per square inch ?
.
57.
1
(6) The circumferences (C and C ) of two other as their radii (R and A").
what
58.
Then
Hence
BG = 5 x. 4 inches long. = the second number.
Therefore
7
=
14
= AC. so that
Find^K7and BO.
Let
A
B
AC=1x. 11 x = 66 is the first number.
4
'
r
i
1
(AC): (BO) =7: 5. 18 x = 108. AB = 2 x 7 x = 42 is the second number. it is advisable to represent these unknown numbers by mx and nx. 2.000
168.
x=2.RATIO AND PROPORTION
69.
2 x
Or
=
4. When a problem requires the finding of two numbers which are to each other as m n.
is
A line AB.
:
Ex. produced to a point C.
as 11
Let
then
:
1.
.
Divide 108 into two parts which are to each other
7. 11 x -f 7 x = 108. x = 6.
11
x
x
7
Ex.
127
The number
is
of miles one can see from an elevation of
very nearly the mean proportional between h and the diameter of the earth (8000 miles).
Hence
or
Therefore
Hence
and
= the first number.
How many
grams of hydrogen are contained in 100
:
grams
10.
of water?
Divide 10 in the ratio a
b. If c is divided in the ratio of the other two.128
ELEMENTS OF ALGEBRA
EXERCISE
63
1. Brass is an alloy consisting of two parts of copper and one part of zinc.)
. 2. what are
its
parts ?
(For additional examples see page 279.
:
4.
How many
7.
7. 9.000.
12. How
The
long are the parts ? 15.
The
total area of land is to the total area of
is
water as
7 18. The three sides of a triangle are respectively a. cubic feet of oxygen are there in a room whose volume is 4500
:
cubic feet?
8.
:
Divide 39 in the ratio 1
:
5. find the number of square miles of land and of water.
11.
What
are the parts ?
5. and the longest is divided in the ratio of the other two.
A line 24 inches
long
is
divided in the ratio 3
5.
m
in the ratio x:
y
%
three sides of a triangle are 11.
3. 12.
Gunmetal
tin. Water consists of one part of hydrogen and 8 parts of
If the total surface of the earth
oxygen.000 square miles. How many ounces of copper and zinc are in 10 ounces of brass ?
6. 6. and 15 inches. How many gen.
13. and c inches.
14.
Divide 44 in the ratio 2
Divide 45 in the ratio 3
:
9.
Divide 20 in the ratio 1 m.
:
Divide a in the ratio 3
Divide
:
7
:
197.
values of x and y.
However.
Hence
2s -5
o
= 10 _ ^
(4)
= 3.
2 y = .
The
root of (4)
if
K
129
.
An
equation of the
first
unknown numbers can be the unknown quantities. =. y =
5
/0 \ (2)
of values. if
.e. x = 1.
if
there
is
different relation
between x and
*
given another equation.-. is x = 7. which substituted in (2) gives y both equations are to be satisfied by the same Therefore.CHAPTER XI
SIMULTANEOUS LINEAR EQUATIONS
169.
Hence.
If
satisfied
degree containing two or more by any number of values of
2oj-3y =
6.
a?
(1)
then
I.
the equations have the two values of
y must be equal. y = 1. such as
+ = 10.-L
x
If
If
= 0. etc. From (3) it follows y 10 x and since
by the same values of x and
to be satisfied
y.y=--|. the equation is satisfied by an infinite number of sets Such an equation is called indeterminate.
y
(3)
these
unknown numbers can be found. expressing a y. there is only one solution.
y = 2. same relation.
4y
.
A
system of two simultaneous equations containing two
quantities is solved by combining them so as to obtain
unknown
one equation containing only one
173.3 y = 80.
Therefore. 26 y = 60. for they cannot be satisfied by any value of x and y. 6 and 4 x y not simultaneous.
172. for they are 2 y = 6 are But 2 x 2.
unknown
quantity.
Solve
-y=6x
6x
-f
Multiply (1) by
2. for they express the x -f y 10.
ELEMENTS OF ALGEBRA
A
system
of simultaneous equations is
tions that can be satisfied
a group of equa by the same values of the unknown
numbers. 30 can be reduced to the same form -f 5 y Hence they are not independent.
Substitution.130
170.
are simultaneous equations.
The process of combining several equations so as make one unknown quantity disappear is called elimination. y
I
171.
= .26.
6x
. 3.24.
of elimination
most frequently used
II.
cannot be reduced to the same form.
to
The two methods
I. The first set of equations is also called consistent.
174. Any set of values satisfying 5 x + 6 y = 60 will also satisfy the equation 3 x -f.
By By
Addition or Subtraction.
E.
(3)
(4)
Multiply (2) by
-
Subtract (4) from (3).
viz.
x
-H
2y
satisfied
6 and 7 x 3y = by the values x = I. Independent equations are equations representing different relations between the unknown quantities such equations
. the last set inconsistent.X.
ELIMINATION BY ADDITION OR SUBTRACTION
175.
~ 50. and 3 x + 3 y =.
21 y
.
z + x = 2 n.
Check.
y
*
z
30.
+2+
6
= 8.)
it is advisable to represent a different letter. however.
.
M=i.
Find the number.
Obviously
of the other
. y
31. as many verbal statements as there are unknown quantities.
Let
x
y z
= the
the digit in the hundreds' place. The digit in the tens' place is | of the sum of the other two digits. + z = 2p.
(
99.
to express
it is difficult
two of the required
digits in
terms
hence we employ 3
letters for the three
unknown
quantities.
=
l. and to express
In complex examples.SIMULTANEOUS LINEAR EQUATIONS
143
x
29. either directly or implied.
# 4. Problems involving several unknown quantities must contain. 1 digit in the tens place.
symbols:
x
+
y
+z-
8. 1.
.
2
= 1(1+6).
Ex. and if 396 be added to the number. The sum of three digits of a number is 8. Simple examples of this
kind can usually be solved by equations involving only one
unknown
every
quantity. the first and the last digits
will be interchanged.
1
=
2.
= 2 m.
The
three statements of the problem can
now be
readily expressed in
.
x
:
z
=1
:
2.
unknown quantity by
every verbal statement as an equation.
and Then
100
+
10 y
+z-
the digit in the units' place.
+
396
= 521. the number.2/
2/
PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS
183.
2
= 6.
(1)
100s
+ lOy + z + 396 = 100* + 10y + x.y
125
(3)
The solution of these equations gives x Hence the required number is 125.
who travels 2 miles an hour faster than B.144
Ex.
(3) C4)
=
24 miles.
2.
direction.
3
xand y
I
1
(2)
5.
3+1 5+1
4_2.
Or
(4)-2x(3).
x 3
= 24. = 8.
. the fraction is reduced to | and if both numerator and denominator of the reciprocal of the fraction be dimin-
ished by one.
= the
fraction. the distance traveled by A.
2.
4
x
= 24.
xy
a:
2y 4y
2.
=
Hence the
fraction
is
f.
Ex.
6
x 4
= 24.
ELEMENTS OF ALGE13KA
If both numerator and denominator of a fraction be
.
8
= xy + x xy = xy -f 3 x 2 y = 2.
+
I
2
(1)
and
These equations give x
Check. the fraction
Let and
then y
is
reduced to
nurn orator.
Since the three
men
traveled the
same
distance.
3. starts 2 hours after B and overtakes A at the same How many miles has A then traveled? instant as B. y = 3. 5_
_4_
A.
By
expressing the two statements in symbols. x 3x-4y = 12.
x
y
= the = the
x
denominator
. B.
increased by one.
Find the
fraction.
we
obtain. C. and C travel from the same place in the same B starts 2 hours after A and travels one mile per hour faster than A.
(1) (2)
12. From (3)
Hence xy
Check.
tion ?
8. and
its
denomi-
nator diminished by one. Four times a certain number increased by three times another number equals 33. it is reduced to J. and twice the numerator What is the fracincreased by the denominator equals 15.
7.SIMULTANEOUS LINEAR EQUATIONS
EXERCISE
70
145
1. the digits will be interchanged.
If 27 is
10.
6.)
added to a number of two digits.
If 4 be
Tf 3 be
is J. Find the numbers. the last two digits are interchanged.
If the
numerator of a fraction be trebled.
part of their difference equals
4. its value added to the denominator. and the two digits exceeds the third digit by 3. and the second increased by 2 equals three times the first. and the second one increased by 5 equals twice
number. fraction is reduced to \-. both terms.
183.
The sum
of the first
sum
of the three digits of a number is 9. A fraction is reduced to J.}.
. Find the fraction.
Half the sum of two numbers equals 4. the value of the fraction is fa. to the number the digits will be interchanged. and the numerator increased by 4. If
9 be added to the number. the fraction equals .
Five times a certain number exceeds three times another 11. Find the number. If the denominator be doubled.
1. the fraction is reduced
fraction. Find the number. if its numerator and its denominator are increased by 1.
The sum
18
is is
and
if
added
of the digits of a number of two figures is 6.
2. and the fourth 3. the Find the fraction.
added to the numerator of a fraction. Find the numbers.
?
What
9.
Find the numbers.
the
number
(See Ex.
5. and four times the first digit exceeds the second digit by 3.
to
L
<>
Find the
If the
numerator and the denominator of a fraction be If 1 be subtracted from increased by 3.
number by
the
first
3.
If the rates of interwere exchanged.
What was
the
sum and
rates
est
The sums of $1500 and $2000 are invested at different and their annual interest is $ 190. How 6 %. Find the weight of one cubic centimeter of gold and one cubic centimeter of silver. and The 6 investment brings $ 70 more interest than the 5
%
%
4%
investments together. Find
the rates of interest. partly at 5% and partly at 4%.
14.
A
sum
of $10. now.
Ten years ago A was B was as
as old as
B
is
old as
will be 5 years hence .grams. Twice A's age exceeds the sum of B's and C's ages by 30. bringing a total yearly interest of $530. What was the amount of each investment ?
15. respectively ?
16.
19. Two cubic centimeters of gold and three cubic centimeters of silver weigh
together 69 J.146
ELEMENTS OF ALGEBRA
11.000
is
partly invested at
6%. and the 5% investment brings $15 more interest than the 4 % investment.
in 8 years to $8500.
and
money and
17. A man invested $750. and 4 %. the annual interest would be $ 195. 5 %.
. and B's age is \ the sum of A's and C's ages. Ten years ago the sum of their ages was 90. 12.
13. and in 5 years to $1125.
much money
is
invested at
A sum
of
money
at simple interest
amounted
in 6 years
to $8000. Find their present ages. a part at 6 and the remainder bringing a total yearly interest of $260. and partly at 4 %. If the sum of
how
old
is
each
now ?
at
invested $ 5000. the rate of interest?
18.
partly at
5 %. the rate of interest ?
What was
the
sum
of
A sum
of
money
at simple interest
amounted
in 2 years
to $090. Three cubic centimeters of gold and two cubic centimeters of silver weigh together 78 grains. What was the amount of each investment ?
A man
%
5%.
and 5 years ago
their ages is 55.
the three sides of a triangle E. but if A would double his pace. what are the angles of the triangle ?
22. and $15 for each sheep.
In the annexed diagram angle a = angle b. and F.
1
NOTE. and e.
An C touch ing the sides in D. and CF?
is
a circle
inscribed in the
7<7. A farmer sold a number of horses. and angle BCA = 70. what is
that
=
OF. and CE If AB = G inches.
E.
It takes
A two hours
longer
24 miles.
. he would walk it in two hours less than
than
to travel
B
B. and F '(see diagram).
respectively. three
AD = AF. If one angle exceeds the sum of the other two by 20. ED = BE. c. If angle ABC = GO angle BAG = 50. are taken so
ABC. for $ 740. and their difference by GO .
is
the center of the circum-
scribed circle.
Find the parts of the
ABC touching the three sides if AB = 9. BC = 7 inches. andCL4 = 8.
24. and sheep.
23. and F. BD = HE. angle c = angle d.
triangle
Tf
AD.
On
/). and angle e angle/. The number of sheep was twice the number of horses and cows together.
BC=7.SIMULTANEOUS LINEAR EQUATIONS
147
20.
the length of
NOTE.
points. How many did he sell
of each if the total
number
of animals
was 24?
21. BE.
A
r
^
A
circle is inscribed in triangle
sides in D.
Find their
rates of walking. then AD = AF. cows. receiving $ 100 for each horse. B find angles a. and AC = 5 inches. The sum of the 3 angles of a triangle is 180. $ 50 for each cow. and GE = CF.
.
25.
Thus the points A.
is the abscissa.
The
of
Coordinates.
It'
Location of a point.
2). and whose ordinate is usually denoted by (X ?/). (7.
the ordinate of point P.
or its equal
OM. and point the origin. and ordinates abore the x-axis are considered positive .
186.
two fixed straight lines XX' and YY' meet in at right angles.
(3.
-3).. hence
The
coordinates lying in opposite directions are negative.
(2.
PN are given. and PN _L YY'. and
respectively represented
Dare
and
by
(3 7 4).
is
The point whose abscissa is a.
The
abscissa
is
usually denoted by
line XX' is called the jr-axis.
lines
PM
the
and P^V are
coordinates
called
point P.
first
3). jr. the ordinate by ?/. and PJ/_L XX'. Abscissas measured to the riyht of the origin. YY' they-axis. B.
. and r or its equal OA is
.
(2.CHAPTER
XII*
GRAPHIC REPRESENTATION OF FUNCTIONS AND
EQUATIONS
184.
PN. then
the position of point is determined if the lengths
of
P
P3f and
185.
?/. PM.
* This chapter
may
be omitted on a
148
reading.
and(l.
4. (4.
6.
Where do Where do
Where do
all
points
lie
whose ordinates
tfqual
4?
9. (0.
71
2). 0). -4).
(-3.)
EXERCISE
1. 3).
4)
and
(4.GRAPHIC REPRESENTATION OF FUNCTIONS
The
is
149
process of locating a point called plotting the point.2).
Plot the points
(6. the mutual dependence of the two quantities may be represented
either by a table or
by a diagram. paper ruled with two sets of equidistant and parallel linos intersecting at right angles.
-2). (See diagram on page 151. 0).
. -3).
(-5.
6.(!.3). (4. which of its coordinates
known ?
13.4). -2). (0.
all all
points
points
lie
lie
whose abscissas equal zero ?
whose ordinates equal zero?
y) if y
10.
=3?
is
If a point lies in the avaxis.
(4.
Plot the points: (-4.
What
are the coordinates of the origin ? If
187. 0).1). 11.
What
is
the locus of
(a?.
Plot the points
:
(0.
Draw
the triangle whose vertices are respectively
(-l.
4).
. (-4.e. Graphic constructions are greatly facilitated by the use of cross-section paper. (-2.
Plot the points:
(4.
1). (4.
2.
0).
(-1.
4)
from the
origin ?
7.
two variable quantities are so related that
changes of the one bring about definite changes of the other. i. (-4.
the quadrilateral whose vertices are respectively
(4.
whose coordinates are given
NOTE. 1).
3.
and measure
their
distance.
12. 3).
2J-).
What
Draw
is
the distance of the point
(3.
8.
Graphs. -!).
ically
each representing a temperature at a certain date. and the amount of gas subjected to pressures from
pound
The same data. and the corresponding number in the adjacent column the ordinate of a point. 10
. may be represented graphby making each number in one column the abscissa.
ure the ordinate of F.
representation does not allow the same accuracy of results as a numerical table.. C. or the curved line the temperature.
15.
may be found
on Jan. we obtain an uninterrupted sequence
etc. however.
188.
1.150
ELEMENTS OF ALGEBRA
tables represent the average temperature
Thus the following
of
New
volumes
1
Y'ork City of a certain
to 8 pounds.
. Thus the first table produces 12 points. B.
Thus the average temperature on May
on April 20.
By representing
of points. D. in like manner the average temperatures for every value of the time.
from January 1 to December 1.
ABCN
y
the so-called graph of
To
15
find
from the diagram the temperature on June
to be 15
. A.
we meas1
. but it indicates in a given space a great many more
facts than a table.
A graphic
and
it
impresses upon the eye
all
the peculiarities of
the changes better and quicker than any numerical compilations.
Whenever a clear.
:
72
find approximate answers to the following
Determine the average temperature of New York City on (a) May 1. uses them. (d) November 20. as the prices and production of commodities. The engineer.
. Daily papers represent ecpnoniical facts graphically. etc. and to deduce general laws therefrom.
physician. (b) July 15. the rise and fall of wages the
matics. the graph
is
applied. concise representation of a
number
of numerical data
is
required.
EXERCISE
From the diagram
questions
1. the merchant. (c) January 15.
(c)
the average temperature oi 1 C.152
2.
During what months
above 18 C.
When
What
is
the temperature equal to the yearly average of
the average temperature from Sept. At what date is the average temperature highest the highest average temperature?
?
What What
is
4. 1?
11
0. 1 to Oct.
From what
date to what date does the temperature
increase (on the average)?
8.
is
10.
When
the average temperature below
C.
During what month does the temperature change least?
14.. (d) 9 0. from what date to what date would it extend ?
If
. (1)
10
C.
Which month
is
is
the coldest of the year?
Which month
the hottest of the year?
16. At what date is the average temperature lowest? the lowest average temperature ?
5..
ELEMENTS OF ALGEKRA
At what date
(a) G
or dates
is
New York
is
C.
15.?
is
is
the average temperature of
New York
6.. ?
-
3.
June
July
During what month does the temperature increase most
?
rapidly
12.
is
ture
we would denote the time during which the temperaabove the yearly average of 11 as the warm season. ?
9.
During what month does the temperature decrease most
rapidly ?
13.
How much warmer
1 ?
on the average
is it
on July 1 than
on
May
17.
How
much. (freezing
point) ?
7.
1 ?
does
the
temperature
increase from
11.
on
1 to
the
average.
19. transformation of meters into yards.GRAPHIC REPRESENTATION OF FUNCTIONS
18. From the table on page 150 draw a graph representing the volumes of a certain body of gas under varying pressures. Construct a diagram containing the graphs of the mean temperatures of the following three cities (in degrees Fahren-
heit)
:
21.
153
1?
When is the average temperature the same as on April
Use the graphs of the following examples for the solution of concrete numerical examples.
Draw
.09 yards. in a similar manner as the temperature graph was applied in examples 1-18. One meter equals 1.
Hour
Temperature
.
a temperature chart of a patient.
20.
Draw
a
graph for the
23.
NOTE.
Represent graphically the populations
:
(in
hundred thou-
sands) of the following states
22.
A
10 wheels a day.
to
27. the value of a of this quantity will change. and $.) On the same diagram represent the selling price of the books.50.
2. if he sells 0. 2 . etc. represent his daily gain (or loss). if x assumes
successively the
tively
values
1.50.
28. gas.
x*
x
19.
If
dealer in bicycles gains $2 on every wheel he sells. the daily average expenses for rent.
+7
If
will
respec-
assume the values 7.inch. 3.
function
If the value of a quantity changes.
ELEMENTS OF ALGEBRA
If
C
2
is
the circumference of a circle whose radius
is J2.
.
29.50 per copy
(Let 100 copies = about \. 9.
books from
for printing.)
T
circumferences of
25. e.
from
R
Represent graphically the = to R = 8 inches. 2 8 y' + 3 y is a function of x and
y.
then
C
irJl.154
24.
to 20 Represent graphically the weight of iron from cubic centimeters.
3.5
grams. etc. binding.
2
is
called
x
2 xy
+ 7 is a function of x. Represent graphically the distances traveled by a train in 3 hours at a rate of 20 miles per hour.g.
(Assume ir~
all circles
>2
2
. if 1 cubic centimeter of iron weighs 7.
The
initial cost of
cost of manufacturing a certain book consists of the $800 for making the plates.
26. 2 x -f 7 gradually from 1 to 2. x
7 to 9. Represent graphically the cost of butter from 5 pounds if 1 pound cost $.. An expression involving one or several letters a function of these letters.
4. 1 to 1200 copies.
Show
graphically the cost of the
REPRESENTATION OF FUNCTIONS OF ONE VARIABLE
189.. amount to $8.
x increases will change gradually from
13. if each copy sells for $1.
190..
is
A
constant
a quantity whose value does not change in the
same discussion.
is
supposed to change. however. E. 3 50.
To
obtain the values of the functions for the various values of
the
following arrangement
be found convenient
:
. The values of func192. to
x = 4.
Ex.GRAPHIC REPRESENTATION OF FUNCTIONS
191. hence
various values of x
The values of a function for the be given in the form of a numerical table.
and join
the
points in order.
2
(-1.
etc. Thus the table on page 1G4 gives the values of the functions x 2 x3 and Vsr.g.
Q-. construct
'.
Draw the graph of x2 -f. 4).
values of x2
nates are the corresponding i.
-J).
. as
1.
a*.0).1).
9).
2).
(-
2.
may. (1.
155
-A
variable is a quantity
whose value changes in the
same
discussion. to con struct the graph x of x 2 construct a series of -3 points whose abscissas rep2 resent X) and whose ordi1
tions
.
(1^.2 x
may
4 from x
=
4.
1
the points (-3.
it is
In the example of the preceding article. 4). while 7 is a constant.
may
. plot points which
lie
between those constructed above.1). and (3.
3
(0. x a variable. If a more exact diagram
is
required.
Graph
of a function.e. 2. (2. for x=l.
9). be also represented by a graph.
.. -1). the function
is
frequently represented
by a single letter.
r
*/
+*
01
.156
ELEMENTS OF ALGEBRA
Locating the points(
4.4)..2 x
. 2 4 and if y = x -f.
Draw
y
z x
the graph of
= 2x-3.
7
. hence two points are sufficient for the construction
of these graphs.
rf
71
. 5).
If
If
Locating
ing
by a
3) and (4.
= 0..
194. Thus in the above example. or ax + b -f c are funclirst
tions of the first degree. and joining in order produces
the graph
ABC.-.
(To avoid
very large ordinatcs. = 4. straight line produces the required graph.)
For brevity. (-2. 4). and join(0. (4.
(-3.
It can be
proved that the
graph is a straight
of a function of the first degree
line. Thus 4x + 7.
Ex. y = 6.
if
/*
4
>
1i >
>
?/
=
193.20). etc.
A
Y'
function of the
first
degree is an integral
rational function
involving only
the
power of the variable.
4J. the scale unit of the ordinatcs is taken smaller than that of the x. as y.. j/=-3.
2.
e.. i. we have to measure the abscissas of the intersection of the
195that
graph with the o>axis.where x
c is
a constant. 14 F.
Show
any convenient number). it is evidently possible Thus to find to find graphically the real roots of an equation.
if c
Draw
the locus of this equation
= 12.
to Fahrenheit readings
:
Change
10
C.24. what values of x make the function x2 + 2x 4 = (see 192).
A body
moving with a uniform
t
velocity of 3 yards per
second moves in
this
seconds a distance d
=3
1.
that the graph of two variables that are directly proportional is a straight line passing through the origin (assume
for c
27. then
y = .. the abscissas of 3.
..) scale are
expressed in
degrees of the Centigrade (C.) scale by the formula
(a)
Draw
the graph of
C = f (F-32)
from
to
(b)
4 F F=l. Therefore x = 1.. If two variables x and y are directly proportional.
ELEMENTS OF ALGEBRA
Degrees of the Fahrenheit
(F.
y=
formula graphically.
From
grade equal to
(c)
the diagram find the number of degrees of centi-1 F.
1
C. Represent 26..24 or x =
P and
Q.
25.158
24. 32 F. then
cXj
where
c is a constant. 9 F.
?/
=4
AB.1.
Ex.
(f
.
and joining by a straight
line.160
ELEMENTS OF ALGEBRA
GRAPHIC SOLUTION OF EQUATIONS INVOLVING TWO UNKNOWN QUANTITIES
198. 2).
i. and join the required graph.
T
. locate points
(0.
fc
= 3.
1)
and
0).
If
x
=
0.
if
y
=
is
0.
represent graphically equations of the form y function of x ( 1D2). solve for
?/.2 y ~ 2.
3x
_
4
.2. that can be reduced
Thus
to represent
x
-
-
-L^-
\
x
=2
-
graphically.
199.
4) and
them by
straight line
AB
(3.
produces the
7*
required locus.
NOTE.
Ex. Equations of the first degree are called linear equations.
Hence. y = -l.
If the given equation is of the we can usually locate two
y.
Hence we may
join (0.
first
degree.
Hence
if
if
x
x
-
2.
== 2.
y=
A
and construct
x
(
-
graphically.
unknown
quantities. Represent graphically
Solving for
y ='-"JJ y. we can construct the graph or locus of any
Since
we can
=
equation involving two
to the above form.
y y
2.
Graph
of
equations involving two
unknown
quantities.
?/. Draw the locus of 4 x + 3 y = 12.
4)
and
(2.e. because their graphs are straight lines. = 0.
X'-2
Locating the points
(2.
Thus
If
in
points without solving the equation for the preceding example:
3x
s
.
. 0).
AB
but only one point
in
AB
also satisfies
(2).
Since two straight lines which are not coincident nor simultaneous
Ex. P.
Solve graphically the equations
:
(1)
\x-y-\.
parallel have only one point of intersection. viz.
Graphical solution of a linear system.
and CD.
To
find the roots of
the system.
3.1=0.GRAPHIC REPRESENTATION OF FUNCTIONS
161
200.
AB
y
= .
By
the
method
of
the preceding article construct the graphs
AB
and
and
CD
of
(1)
(2) respectively. linear equations have only one pair of roots. The roots of two simultaneous equations are represented by the coordinates of the point (or points) at which their
graphs intersect.15. and every set of real values of x and y satisfying the given equation is represented by a point in
the locus.
equation
x=
By measuring
3.
202.
(2)
.
201. we obtain the roots.
The
every
coordinates
of
point in satisfy the equation
(1). the point of intersection of the coordinate of P.
203. The coordinates of every point of the graph satisfy the given equation.57.
(1)
(2)
cannot be satisfied by the same values of x and y.
y equals
3. the graph
of
points
roots.0. and joining by a
straight line.
4. parallel graphs indicate inconsistent equations.0).
There can be no point of
and hence no
roots.
3.162
ELEMENTS OF ALGEBRA
graph.
-
4.e.
x2
.
4. 0.
intersection.5.
1. if x equals
respectively
0..
. This is clearly shown by the graphs of (1) arid (2). construct CD the locus of (2)
of intersection. (-2.y~ Therefore. the point
we
obtain
Ex.
(1)
(2)
-C.
2 equation x
3).g.5.
3.
4.
2. etc. 2.
3x
2 y = -6. and
.
Solve graphically the
:
fol-
lowing system
= =
25.
P
graphs meet in two and $. we of the
+
y*
= 25.
Locating two points of equation (2). 4.9. they are inconsistent.
4. 3). (4.
4. AB the locus of (1).
Inconsistent equations. there are two pairs of By measuring the coordinates of
:
P and Q we find
204. 0) and (0.
Using the method of the preceding para.
4. i. 0. which consist of a
pair of parallel lines.
The equations
2
4
= 0. (-4. e.
5.
Solving (1) for y.
In general. 1. 5.
and
+ 3). Since the two
-
we obtain DE.
Measuring the coordinates
of P.
obtain the graph (a circle)
AB C
joining.
4.
Locating the
points
(5. 3.
V25
5. = 0.
etc.
Thus
V^I is an imaginary number.
1.
or
-3
for
(usually written
3)
.
V
\/P
214.
27
=y
means
r'
=
27. and ( v/o* = a.
= x means
= 6-.
a)
4
= a4
. or y
~
3.
tity
.
numbers.
Evolution
it is
is the operation of finding a root of a quan the inverse of involution.
for (-f 3) 2
(
3)
equal
0.CHAPTER XIV
EVOLUTION
213.
or x
&4 . and
all
other numbers are.
It follows
from the law of signs
in evolution that
:
Any
even root of a positive.
4
4
.
\/a
=
x means x n
=
y
?>
a. called real
numbers. (_3) = -27. Since even powers can never be negative. it is evidently impossible to express an even root of a negative quantity by Such roots are called imaginary the usual system of numbers. for distinction.
V9 = +
3.
2. which can be simplified no further. quantity
may
the
be either 2wsitive
or negative. for (+ a) = a \/32 = 2.
\/"^27=-3.
215.
109
.
Every odd root of a quantity has
same sign as
and
2
the
quantity.
and b (2 a -f b). however. In order to find a general method for extracting the square root of a polynomial.
term a of the root
is
the square root of the
first
The second term
of the root can be obtained
a.
14.
+ 6 + 4a&.b 2 2 to its square. let us consider the relation of a -f.
a2
+ & + c + 2 a& .172
7.
The work may be arranged
2
:
a 2 + 2 ab
+ W \a + b
.
multiplied by b must give the last two terms of the
as follows
square.
12.
15.e.
and
b.
ELEMENTS OF ALGEBEA
4a2 -44a?> + 121V2 4a
s
.
8
.2 ab + b
. a -f.2 ac . it is not known whether the given
expression is a perfect square.>
13.
2
.2 &c.
2ab
.
2
49a 8 16 a 4
9.
the given expression is a perfect square. the that 2 ab -f b 2
=
we have then to consider sum of trial divisor 2 a.
a-\-b
is
the root
if
In most cases.
#2
a2
-
16.
mV-14m??2)-f 49.
The
term
a'
first
2
.72 aW + 81 &
4
. 11. i.
.
second term 2ab by the double of
by dividing the the so-called trial divisor.
10.
2 2
218.
and so
forth.
*/''
.
. We find the first two terms of the root by the method used in Ex. 8 /-.EVOLUTION
Ex. we obtain the next term of the root 3 y 3 which has to be added to 2 the trial divisor.
.
-
24 a
3
+
25 a 2
-
12 a
+4
Square of 4 a First remainder. the required root
(4
a'2
8a
+
2}.
\
24 a 3
4-f
a2
10 a 2
Second remainder.
219.24 afy* -f 9 tf.
of x.
2.
. 1.24 a + 4 -12 a + 25 a8
s
.
Arranging according to descending powers of
10 a
4
a.
Explanation. The process of the preceding article can be extended to polynomials of more than three terms.
173
x*
Extract the square root of 1G
16x4
10 x*
__
.
First complete divisor.
Ex. By doubling 4x'2 we obtain 8x2 the trial divisor.
.
. and consider Hence the their sum one term. the first term of the answer.
10 a 4
8
a.
is
As
there
is
no remainder. 8 a 2
-
12 a
+4
a
-f 2. As there is no remainder. 6 a. 8 a 2
2. Multiply the complete divisor Sx' 3y 3 by Sy 8 and subtract the product from the remainder.
by division we
term of the
root.
First trial divisor. 24# 2 y 3 by the trial divisor Dividing the first term of the remainder.
The square
. Second trial divisor.
Extract the square root of
16 a 4
. 2 Subtracting the square of 4x' from the trinomial gives the remainder '24 x'2 + y.
double of this term
find the next
is
the
new
trial divisor.
Arrange the expression according to descending powers root of 10 x 4 is 4 # 2 the lirst term of the root. 8 a 2 . 8 a 2 Second complete divisor.
1.
4 x2
3
?/
8 is
the required square foot.
1. Hence the root is 80 plus an unknown number. and we may apply the method used in algebraic process.
The
is
trial divisor
=
160. and the square root of the greatest square in
units. then the number of groups is equal to the number of digits in the square root. of 10. two figures. Thus the square root of 96'04' two digits.EVOLUTION
220.
square root of arithmetical numbers can be found to the one used for algebraic
Since the square root of 100 is 10.000.
a 2 = 6400.
As
8
x 168
=
1344.
7744 80 6400
1
+8
160
+ 8 = 168
1344
1344
Since a
2 a
Explanation.
Find the square root of 7744. and the complete divisor
168 the first of which is 8.
= 80. etc.
From
A
will
show the
comparison of the algebraical and arithmetical method given below identity of the methods.. the first of which is 4.
175
The
by a method very similar
expressions. the square root of 7744 equals 88.000 is 100. the integral part of the square root of a number less than 100 has one figure. Hence if we divide the digits of the number into groups. etc.000 is 1000.
2.
Find the square root of 524.
the preceding explanation it follows that the root has two digits.
Ex.000. of 1. of a number between 100 and 10.176. Therefore 6 = 8. which may contain one or two).1344. the first of which is 9 the square root of 21'06'81 has three digits. beginning at the
and each group contains two digits (except the last. and the first remainder is.
the
consists of
group is the first digit in the root.
Ex.
first
.
ELEMENTS OF ALGEKRA
In marking
off groups in a number which has decimal begin at the decimal point.
in .
12.1T6
221.7 to three decimal places.10. and if the righthand group contains only one digit.70
6.
Roots of common fractions are extracted either by divid-
ing the root of the numerator by the root of the denominator.
EXERCISE
Extract the square roots of
:
82
.
Find the square root of
6/.0961
are
'. or by transforming the common fraction into a decimal.
3.
The groups
of 16724. annex a cipher.1 are
Ex.688
4
45 2 70
2 25
508
4064
6168 41)600
41344
2256
222.
places.
we must
Thus the groups
1'67'24.GO'61.
The
two numbers
(See
is
2
:
3.
and their product
:
150.
9
&
-{-
c#
a
x
+a
and
c.
Find
is
the number.180
on
__!_:L
ELEMENTS OF ALGEBRA
a.
If a 2 4. 24.
84
is
Find a positive number which
equal to
its
reciprocal
(
144).
22
a.b 2 If s
If
=c
.
A
number multiplied by
ratio of
its fifth
part equals 45.
Find the side
of each field.
solve for d.
4.
solve for
r. If the hypotenuse
whose angles
a
units of length.
2a
-f-
1
23.
:
6.
2.
find a in terms of 6
.
If 2
-f 2 b*
= 4w
2
-f c
sol ve for
m. The sides of two square fields are as 7 2.
. then
Since such a triangle
tangle.
and they con-
tain together 30G square feet.
If s
= 4 Trr
'
2
.
228.
is
one of
_____
b
The side right angle. and the first exceeds the second by 405 square yards. and the two other sides respectively
c
2
contains
c
a and b units.
may
be considered one half of a
rec-
square units.
2
.
EXERCISE
1.
If 22
= ~^-.
r.
2
.
27.
A
right triangle is a triangle.
If
G=m m
g
.
2
:
3.
Find the numbers.
solve for v.
Three numbers are to each other as 1 Find the numbers.
=
a
2
2
(' 2
solve for solve for
= Trr
.
108. opposite the right angle is called the hypotenuse (c in the diagram).
3. 25.
28.
. its area contains
=a
2
-f-
b2
.
26.
and the sum
The
sides of
two square
fields are as
3
:
5. Find the side of each field.
'
4. is 5(5.)
of their squares
5.
29.
make x2
Evidently 7 takes the place 7x a complete square
to
to
which corresponds
m
2
.
Solve
Transposing. and the third side is 15 inches. (b) 44 square feet.)
13.
of a right triangle Find these sides.
2m.
Find the
sides.
and the two smaller
11. To find this term.
8.7 x -f 10 = 0. let us compare x 2
The
left
the perfect square x2
2
mx -f m
to
2
. Find the unknown sides and the area.
9.QUADRATIC EQUATIONS
7.
sides.2
7
.
-J-
=
12.
Method
of completing the
square. passes in t seconds 2 over a space s yt Assuming g 32 feet. in how many seconds will a body fall (a) G4 feet.
4.
24.
Find these
10. the radius of a sphere whose surface equals
If the radius of a sphere is r.
. and the
other two sides are as 3
4.
181
The hypotenuse
of a right triangle
:
is
35 inches.
The following
ex-
ample
illustrates the
method
or
of solving a complete quadratic
equation by completing the square.
The area $
/S
of a circle
2
.
The hypotenuse
of a right triangle is 2.
Find the
radii.
radii are as 3
14.
The area
:
sides are as 3
4. x* 7 x=
10. its surface
(Assume
ir
=
2 . we have
of
or
m = |.
is
and the other
two
sides are equal.
7r
(Assume
and their
=
2 7
2
.
.
add
(|)
Hence
2
. A body falling from a state of rest. The hypotenuse of a right triangle is to one side as 13:12.
. (b) 100 feet?
=
.)
COMPLETE QUADRATIC EQUATIONS
229.
member can be made a complete square by adding 7 x with another term.
8 = 4 wr2 Find 440 square yards.
the formula
= Trr
whose radius equals r is found by Find the radius of circle whose area S
equals (a) 154 square inches.
Two
circles together contain
:
3850 square
feet.
and c in the general answer.
article.
Solving this equation we obtain
by the method of the preceding
2a
The
roots of
substituting the values of a.
Solution
by formula.c
= 0.
.
2
Every quadratic equation can be
reduced to the general form.
ao.
=0.
49.
=8
r/io?.
any quadratic equation may be obtained by 6.
x
la
48.
231.
= 12.
o^
or
-}-
3 ax == 4 a9
7 wr
. -\-bx-\.
2x
3
4.184
ELEMENTS OF ALGEBRA
45
46.
-2.
56.
1.
54.1. and consequently many prob-
235.
The
difference of
|.
and whose sum
is
is 36.
2. -5. and equals 190 square inches.
Divide CO into two parts whose product
is 875.
88
its reciprocal
A
number increased by three times
equals
6J.
8.
PROBLEMS INVOLVING QUADRATICS
in general two answers.
3. The
11.
G.
1.
3.
area
A
a perimeter of 380
rectangular field has an area of 8400 square feet and Find the dimensions of the field.2. Find the number.
The sum
of the squares of
two consecutive numbers
85.
of their reciprocals is
4.
Find
the numbers.9.3.
5. feet.3.
52. -2.
is
Find two numbers whose product
288.
57.
its
sides of a rectangle differ by 9 inches.
:
3.
What
are the
numbers
of
?
is
The product
two consecutive numbers
210.
-2. Find the sides.
.
55.
189
the equations whose roots are
53.
-2.
0.
Find the number.QUADRATIC EQUATIONS
Form
51. and the difference Find the numbers.
2.
and whose
product
9.
Find a number which exceeds
its
square by
is
-|.
-4. but frequently the conditions of the problem exclude negative or fractional answers.0.
Problems involving quadratics have
lems of this type have only one solution.
7.3.
58.
two numbers is 4.
6.
EXERCISE
1.
Twenty-nine times a number exceeds the square of the 190.
Find two numbers whose difference
is 40.0.
number by 10.
and the slower reaches its destination one day
before the other.
13.
as the
16. What did he pay for each
apple ?
A man bought a certain number of horses for $1200. and the line BD joining
two opposite
vertices (called "diagonal")
feet. it would have needed two hours less to travel 120 miles.
15. he had paid 2 ^ more for each apple.
other. had paid $ 20 less for each horse.
of a rectangle is to the length of the recthe area of the figure is 96 square inches.
If he
each horse ?
. one of which sails two miles per hour faster than the other. he would have received 12 apples less for the same money.
ELEMENTS OF ALGEBRA
The length
1
B
AB of a rectangle.190
12. If a train had traveled 10 miles an hour faster. Find the rate
of the train. start together on voyages of 1152 and 720 miles respectively.
A man
cent as the horse cost dollars.10.
ply between the same two ports. and Find the sides of the rectangle.
and gained as many per Find the cost of the horse. a distance One steamer travels half a mile faster than the two hours less on the journey.
. What did he pay for
21. dollars.
A man
A man
sold a
as the watch cost dollars. ABCD. he would have received two horses more for the same money. A man bought a certain number of apples for $ 2. Two vessels.
vessel sail ?
How many
miles per hour did the faster
If 20.
c equals 221
Find
AB and AD. and lost as many per cent Find the cost of the watch.
19. and lost as many per cent Find the cost of the watch.
17.
The diagonal
:
tangle as 5 4.
watch for $ 24.
watch cost
sold a watch for $ 21.
Two steamers
and
is
of 420 miles. At what rates do
the steamers travel ?
18. exceeds its widtK AD by 119 feet.
14. sold a horse for $144.
Find
TT r (Area of a circle .
By formula. 23 inches long.
EQUATIONS IN THE QUADRATIC FORM An equation is said to be in the quadratic form
if it
contains only two unknown terms.
Ex.
A rectangular
A
circular basin is surrounded
is
-
by a path 5
feet wide.
1.
=9
Therefore
x
=
\/8
= 2. Equations in the quadratic form can be solved by the methods used for quadratics. and the unknown factor of one of these terms is the square of the unknown factor of the
other.
Find the side of an equilateral triangle whose altitude
equals 3 inches. as
0. how wide is the walk ?
23.
B
AB
AB
-2
191
grass plot.) 25. How many eggs can be bought for $ 1 ?
236.
. 30 feet long and 20 feet wide. In how many days can B do the work ?
=
26. the two men can do it in 3 days. If the area of the walk is equal to the area of the plot.
^-3^ = 7.
24.
of the area of the basin. is surrounded by a walk of uniform width. A needs 8 days more than B to do a certain piece of work.
Solve
^-9^ + 8 =
**
0.
237.
or x
= \/l = 1.
27.
and the area of the path
the radius of the basin. constructed with and CB as sides. contains B 78 square inches. so that the rectangle.QUADRATIC EQUATIONS
22. The number of eggs which can be bought for $ 1 is equal to the number of cents which 4 eggs cost. and working together. Find and CB. a point taken.
(tf.I) -4(aj*-l)
2
= 9.
is
On the prolongation of a line AC.
II. 4~ 3 have meaning according to the original definition of power. very important that all exponents should be governed by the same laws.
>
m therefore.
244.
for all values
1
of
m and n. and
.a" = a m n
mn . (a ) s=a m = aw bm
a
.
a m a" = a m+t1 .
The
first
of these laws
is
nition of power.
We assume. that a
an
= a m+n
.*
III.CHAPTER XVI
THE THEORY OF EXPONENTS
242."
means "is greater than"
195
similarly
means "is
. however.
= a""
<
. while the second of the first.
must be
*The symbol
smaller than.
no
Fractional and negative exponents.
It is.
provided
w > n. ~ a m -f.
m
IV. (a m ) w
. The following four fundamental laws for positive integral exponents have been developed in preceding chapters
:
I. we let these quantities be what they must be if the exponent law of multiplication is generally true.
the direct consequence of the defiand third are consequences
FRACTIONAL AND NEGATIVE EXPONENTS
243.
(ab)
.
Then the law
of involution.
we may choose
for such
symbols any definition that
is
con-
venient for other work. such as 2*. instead of giving a formal definition of fractional and negative exponents. hence.
23. fractional.
a?*.
Let
x
is
The operation which makes the fractional exponent disappear evidently the raising of both members to the third power.
Write the following expressions as radicals :
22.
we
try to discover the
let the
meaning of
In every case we
unknown quantity
and apply to both members of the equation that operation which makes the negative. (xy$.
'&M
A
27.
etc.
laws. a .g.
n 2 a. as. at.
28.
m$. ml.
24.
a*.
a\
26. 25.
30.
Hence
Or
Therefore
Similarly.
disappear.
0?=-^.
31.
29. 3*.
-
we
find
a?
Hence we
define a* to be the qth root of of.
= a.
.
^=(a^)
3*
3
. since the raising to a positive integral power is only a repeated multiplication. 4~ .
245.
e.196
ELEMENTS OF ALGEBRA
true for positive integral values of n.
To
find the
meaning
of
a fractional exponent. or zero exponent
equal
x.
(bed)*.
Assuming these two
8*.
in which
obtained from the preceding one by dividing both
members by
a.
each
is
The
fact that a
if
=
we
It loses its singularity
1 sometimes appears peculiar to beginners.
a
a
a
= =
a a a
a1
1
a.
a8 a
2
=
1
1
.
Or
a"#
= l.
vice versa.
by changing the sign of
NOTE.198
247.
etc.g.
Multiplying both members by
a".
an x = a. consider the following equations.
Let
x=
or".2
=
a2
. Factors
may
be transferred
from
the
numerator
to
the
denominator of a fraction. e.
248. or
the exponent.
cr n.
.
ELEMENTS OF ALGEBRA
To
find the
meaning
of a negative exponent.
Arrange in descending powers of
Check.202
ELEMENTS OF ALGEBRA
32.
powers of x arranged are
:
Ex.
1.
V ra
4/
3
-\/m
33.
we wish to arrange terms according to descending we have to remember that.
40.
1 Multiply 3 or
+x
5 by 2 x
x.
6
35.
If
powers of
a?.
34.2 d
.
lix
=
2x-l
=+1
Ex. the term which does not contain x may be considered as a term containing #. The
252.
Divide
by
^
2a
3 qfo
4.
1.
2.
46.
268.
Ex.
it
more convenient to multiply dividend and divisor by a factor which makes the divisor rational.
.
(3V3-2Vo)(2V3+V5).
47. a
VS
-f-
a?Vy
= -\/ -
x*y
this
Since surds of different orders can be reduced to surds of
the same order. Monomial surdn of the same order may be divided by multiplying the quotient of the coefficients by the quotient of the
surd factors.
all
monomial surds may be divided by
method.
43.
44.
(5V7-2V2)(2VT-7V2).
V3 .
53.
-v/a
-
DIVISION OF RADICALS
267.
a fraction.
Va
-v/a.
(5V2+V10)(2V5-1).
52.
ELEMENTS OF ALGEHRA
(3V5-5V3)
S
.
E.
(3V5-2V3)(2V3-V3).
(V50-f 3Vl2)-4-V2==
however.
(2
45. the quotient of the surds
is
If.
Ex.
49.
51.214
42.
48.
60.y.
is
1
2.V5) ( V3 + 2 VS).
the rationalizing factor
x
' g
\/2.
Divide 4 v^a by
is
rationalizing factor
evidently \/Tb
hence.57735. is illustrated
by
Ex.
+ 4\/5 _ 12v 3 + 4\/5 V8 V8
V2 V2
269.
4\/3~a'
36
Ex. e.73205.
Divide
VII by v7. the by 3 is much easier to perform than the division by
1. To show that expressions with rational denominators are simpler than those with irrational denominators.RADICALS
This method.
Hence
in arithmetical
work
it
is
always best to
rationalize the denominators before dividing. arithTo find. Evidently.
1.by the usual arithmetical method.
.
The
2.
VTL_Vll '
~~"
\/7_V77
.
we have
to multiply
In order to make the divisor (V?) rational.73205
we
simplify
JL-V^l
V3
*>
^>
division
Either quotient equals .
by V7.
.
3.g.
Divide 12 V5
+ 4V5 by V.
. called rationalizing the
the following examples
:
215
divisor.
/~
}
Ex. metical problems afford the best illustrations. however.. we have
V3
But
if
1.
is
Since \/8
12 Vil
=
2 V*2.
Ex.
:
importance. If n is a Theorem that
1. if n For ( y) n -f y n = 0. xn y n y n y n = 0.
2 8 (3 a )
+8=
+
288.
Factor
consider
m
m
6
n9
.
is
odd. if n is even.
The
difference of
two even powers should always be
considered as a difference of two squares.230
285.
We may
6
n 6 either a difference of two squares or a
dif-
* The symbol
means " and so forth to.
2
Ex.
It
y is
not divisible by
287.
1.
ELEMENTS OF ALGEBRA
positive integer.
By
we obtain the other
factors.
and have
for
any positive integral value of
If n
is
odd.
-
y
5
=
(x
-
can readily be seen that #n -f either x + y or x y.
it
follows from the Factoi
xn y n is always divisible by x y. if w is odd."
.
286.y n is divisible by x -f ?/.
2.
xn -f.
Two
special cases of the preceding propositions are of
viz.
x* -f-/
= (x +/)O .
Factor 27 a* -f
27 a 6
8.
ar
+p=
z6
e.g.
2.
actual division
n. For substituting y for x.xy +/).
x'2 2 x =
1.
The
~~f
fraction .
Or.
customary to represent this result
by the equation ~
The symbol
304.
I.
.
By making x
any * assigned
zero.
i.
be the numbers.increases
if
x
de-
x
creases.decreases
X
if
called infinity.000
a.e.
of the second exceeds the product of the first
Find three consecutive numbers such that the square and third by 1.
ELEMENTS OF ALGEBRA
Interpretation of ?
e. equation.
i.
Interpretation of
QO
The
fraction
if
x
x
inis
infinitely large.can be
If
It is
made
larger than
number.
306. or infinitesimal) This result is usually written
:
305.
as
+ l.
ToU"
^-100 a.
The
solution
x
=-
indicates that the problem
is indeter-
If all terms of an minate. it
is
an
Ex. cancel. without exception.
Hence any number will satisfy equation the given problem is indeterminate. or that x may equal any finite number.242
303. however
x approaches the value
be-
comes
infinitely large.
(a:
Then
Simplifying.
x
-f 2.
.
is satisfied
by any number.
great.
(1).
(1)
= 0.g.
and
.
and becomes infinitely small.
oo is
= QQ.
Hence such an equation
identity.
the
If in an equation
terms containing
unknown quantity
cancel.
+
I)
2
x2
'
-f
2x
+
1
-x(x + 2)= . while the
remaining terms do not
cancelj the root is infinity.
(1)
is
an
identity.
Let
2.
creases.
1.e.i
solving
a problem
the result
or oo indicates that the
all
problem has no solution.
TO^UU"
sufficiently small.
1. the answer is indeterminate.
= 10.
Find these sides.
13.)
53 yards.
8. To inclose a rectangular field 1225 square feet in area.
9.
The hypotenuse
is
the other two sides
7. Find the side of each square.
and the diago(Ex. the area becomes
-f%
of
the original area.
6.
146 yards. p. The sum of the areas of two squares is 208 square feet. and the edge of one.
Find the
sides of the rectangle. But if the length is increased by 10 inches and
12.
of a right triangle is 73. increased by the edge of the other. Two cubes together contain 30| cubic inches. Find the dimensions of the
field.
Find the
sides. Find the numbers. Find the edges. and
its
The diagonal
is
is
perimeter
11.
of a rectangular field
feet.
is
is
17 and the
sum
4.244
3.
is 6.
Find the other two
sides.
The area of a
nal 41 feet. 148 feet of fence are required. Find the edge of each cube. and the side of one increased by the side of the other e. 190. 12.
the
The mean proportional between two numbers sum of their squares is 328.
rectangle is 360 square Find the lengths of the sides.
14.
is
the breadth
diminished by 20 inches.
and
is
The area of a rectangle remains unaltered if its length increased by 20 inches while its breadth is diminished by 10 inches.
255 and the sum of
5. Find two numbers whose product whose squares is 514.
ELEMENTS OF ALGEBRA
The
difference between
is
of their squares
325.
two numbers Find the numbers.)
The area
of a right triangle is 210 square feet.
103.quals 20 feet. equals 4 inches. The volumes of two cubes differ by 98 cubic centimeters.
10.
.
and the sum of
(
228.
and the
hypotenuse
is 37. and the edge of one exceeds the edge of the other by 2 centimeters.
(Surface of sphere
If a
number
of
two
digits be divided
its digits.
.) (Area of circle
and
=
1
16.SIMULTANEOUS QUADRATIC EQUATIONS
15.
by the product of 27 be added to the number.
and
if
the digits will be interchanged. their areas are together equal to the area of a circle whose radius is 37 inches.
The
radii of
two spheres
is
difference of their surfaces
whose radius = 47T#2. Find the number.
245
The sum of the radii of two circles is equal to 47 inches.
the quotient
is 2. Find the radii. and the equal to the surface of a sphere Find the radii.
differ by 8 inches. irR *.)
17.
is
20 inches.
of a series are its successive numbers.
of the following series is
3.
Hence
/
= a + (n .
11..
To
find the
nth term
/
of an A. (n 1) d must be added to a. An arithmetic progression (A. to produce the nth term.
The common differences are respectively 4.
17. .CHAPTER XX
PROGRESSIONS
307. except the first.
. The first is an ascending.) is a series. each term of which.
The terms
ARITHMETIC PROGRESSION
308. and d.1) d..
:
7..
series 9. is derived from the preceding by the addition of a constant number..
to each
term produces the next term.. to produce the 4th term. P. 12.
. 3 d must be added to a. a + 2 d.
+
2 d.
to
A series
is
a succession of numbers formed according
some
fixed law. a -f d.
-4. 15 is 9 -f. 19.
-f
. a
11.
a. the first
term a and
the
common difference d being given.. to produce the 3d term. 16.
a
+
d. P. 2 d must be added to a.
The common
Thus each
difference is the
number which added
an A.
.. progression. a
3d.
309.7..
Since d
is
a
-f
3
d.
added to each term to obtain the next one.
10.11 246
(I)
Thus the 12th term of the
3
or 42.. The progression is a. P. 3.. the second a descending.
if
a = 5.
2
EXERCISE
1.
3.
first
2
Write down the
(a)
(6)
(c)
6 terms of an A.
1.. the
term
a.
Or
Hence
Thus
from
(I)
= (+/).
Find the nth term of the
series 2. 2.
the last term
and the common difference d being given.
.
.8.
9.
7. 5.
6
we have
Hence
.. -7...
247
first
To
find the
sum s
19
of the first
n terms of an A.. -4^.. 2J... ?
(a) 1..3 a = -l. 2
sum
of the first 60
I
(II)
to find the
' '
odd numbers.. P.
-3.. d
. 5. a = 2.. d = 3. P.. 6.
series
.
..
1. 3.
= a + (a
Reversing the order. 19.' cZ == .
1-J. 99) = 2600.-. -10.
Adding.
Find the 101th term of the
series 1.
2.16. -24.PROGRESSIONS
310.
..
Which
(6)
(c)
of the following series are in A.
5.
5. 8.
= I + 49 = *({ +
.
. 3. 8.
= -2.
= 99.
115.
6.
9. 5. 7. 21.
8.-.
(d) 1J.
of the series 10...
2*=(a + Z) + (a + l) + (a + l)
2s = n
*
.
Find the 5th term of the
4. . 3.
-|.
series 2.
Find the 7th term of the Find the 21st term
series
. 6.
Find the 12th term of the
-4.-
(a
+ + (a +
l)
l).
Find the 10th term of the
series 17. 4. P.4.
29.
to 15 terms. 31. How much does he receive (a) in the 21st year (6) during the first 21 years ?
j
311. the other two may be found by the solution of the simultaneous equations
. strike
for the first yard.
33. hence if any three of them are given.
rf.
to 20 terms.5
H + i-f
-f-
to 10 terms. 1|.
1+2+3+4H
Find the sum of the
first
n odd numbers.
1.
.
.
to 20 terms.
.
2. 7.
3. 7.
12. 2J. 11.
1J.
21.
>
2-f
2.
.
15.
8. P.
(i)
(ii)
.
4.
6.
19. 15.
to 7 terms.
1.
to 8 terms.
+ 3. and a yearly increase of $ 120.
16.
Sum
the following series
14.
11.
-.
+ 2-f-3 + 4 H
hlOO.
Q^) How many times
in 12 hours ?
(&fi)
does a clock.
to 10 terms.
.. 11.
20.
15.
17.248
Find the
10. 16. In most problems relating to A. striking hours only.1 -f 3.7 -f
to 12 terms.
(x +"l) 4.
7. 12. Jive quantities are involved.
$1
For boring a well 60 yards deep a contractor receives yard thereafter 10^ more How much does he receive all
together ?
^S5 A bookkeeper accepts a position at a yearly salary of $ 1000.
\-n.
to 16 terms. and for each than for the preceding one. 11.
:
3.
ELEMENTS OF ALGEBRA
last
term and the sum of the following series :
.
22.
to 20 terms.
.
18.
.(#
1
2) -f (x -f 3) H
to
a terms.
'.
13.
23.
n = 4. 78. Find w.
3. Given a = |. s == 440. s = 70.
6?
9. Find?.
7.
13. = 52. n = 16. Find d and Given a = 1700.
How many terms How many terms
Given d = 3.
produced.
I.
14. d = 5.250
ELEMENTS OF ALGEBRA
EXERCISE
116
:
Find the arithmetic means between
1. 74. = 17.
T?
^. P.
15. n = 20.
= 16. and all his savings in 5 years amounted to $ 6540.
a x
-f-
b
and a
b. Find n.
8.
. n = 17.
How much
. n.
m
and
n
2.
4. Find d. Given a = 4. n = 13.
12.
A
$300
is
divided
among 6 persons
in such a
way
that each
person receives $ 10 did each receive ?
more than the preceding
one.
Between 4 and 8
insert 3 terms (arithmetic
is
means)
so
that an A.
17. = 45.
16.
11. Find a and Given s = 44.
a+
and
b
a
b
5.
y and #-f-5y. Find d.
has the series 82.
f
J 1 1
/
. Find a Given a = 7.
Between 10 and 6
insert 7 arithmetic
means
. Given a = 1.
I
Find
I
in terms of a.
f?
. of 5 terms
6. = 1870. How much did he save the first month?
19. = 83.
and
s.3. = ^ 3 = 1.
10. Given a = .
man saved each month $2 more than in the pre 18. n
has the series
^
j
. ceding one.
To find the sum s of the first n terms term a and the ratio r being given. is
it
(G.
is
16(f)
4
..
NOTE.
.
Therefore
Thus the sum
= ^ZlD.
4-
(1)
. +1.. called the ratio.
(I)
of the series 16.
2
arn
(2)
Subtracting (1) from
(2)..
and
To
find the
nth term
/ of
a G.
ratios are respectively 3. 24. rs =
s
2
-.
or 81
315. 36. P.
(II)
of the
8 =s
first
6 terms of the series 16.g.
A geometric progression
first..
the following form 8
nf +
q(l-r")
1
r
. 36. 12.
g==
it is
convenient to write formula' (II) in
*.
of a G. or. ar.
.
Hence
Thus the 6th term
l
= ar
n~l
.arn ~ l ...PROGRESSIONS
251
GEOMETRIC PROGRESSION
313. except the
multiplying
derived from the preceding one by by a constant number.
2 a.
.
4..
The progression is a.
The
314.
E..
.
If
n
is less
:
than unity. the first
= a + ar -for ar -f ar Multiplying by r. the first term a and
the ratios r being given. <zr .
fl
lg[(i)
-l]
==
32(W -
1)
= 332 J. P. 24.)
is
a series each term of
which.
-2. P..
4. a?*2 To obtain the nth term a must evidently be multiplied by
.
s(r
1)
8
= ar"
7*
JL
a.
. 36. 108. -I..
|.
r
n~ l
.
ar8
r.
whose
... + 5.
4. Jive quantities are in.
Evidently the total
number
of terms is 5
+ 2.
first
term
4.
+-f%9 %
. P.
volved . P._!=!>..
. 9. 144.. whose and whose common ratio is 4.
2
term
3.
144.. whose and whose second term is 8. P.
r^2.. 36.
6. 9. 144.
.
0.
EXERCISE
1.
8. the other two be found by the solution of the simultaneous equations :
may
(I)
/=<!/-'.
(it.
.. 18.4. I
= 670.
(d) 5.288..
.
a
=
I. 1.
. In most problems relating to G.
Find the 5th term of a G. 72.
is 16.
To
insert 5 geometric
means between 9 and 576.
l.
Write down the first 6 terms of a G.
i 288.
series
Find the llth term of the Find the 7th term of the
ratio is
^.
.
10. 20.
...6. 676. hence.
-fa.18.
7.
series 6.18.
first
term
is
125 and
whose common
.5.l.
first
5. P.
Find the 6th term of the
series J.
..
288.
.
.
Write down the first 5 terms of a G.
is 3. if any three of them are given.
f. ?
(c)
2. 3.
Hence n
=
7.
Hence the
or
series is
0.54.
.
series 5.
676
t
Substituting in
= r6 = 64..
72..
.
Ex. P.72.
(b) 1. 36.
36.252
ELEMENTS OF ALGEBRA
316.
And the
required
means are
18.
117
Which
(a)
of the following series are in G.
.5.-.
\
t
series
.
|.
or
7.*. 4. 80.
9.
576. 25.
-fa.
series
. f.
Find the 7th term of the Find the 6th term of the
Find the 9th term of the
^.
7/
191.
father. the ana of the floor will be increased 48 square feet.
-ll?/-102. 10x 2 192. 15 m.
The length
is
of a floor exceeds its width
by 2
feet.
178. 4 a 2
y-y
-42.
younger than his Find the age of
the father.
sister
.
+
a.-36.
number divided by
3.
The age
of the elder of
it
three years ago of each.
ELEMENTS OF ALGEBRA
A A
number increased by
3.
+x-
2.
13 a + 3.
. x*
185.
dimension
182. + 11 ~ 6.
What
is
the distance?
if
square grass plot would contain 73 square feet more Find the side of the plot.
two boys is twice that of the younger. was three times that of the younger.
181.
186. How many are there in each window ?
. side were one foot longer. 12 m.
176.
188. z 2 + x . and the middle row has 4 panes in each window more than the upper row there are in all 168 panes of glass.
180.
is
What are their ages ? Two engines are together
more than the
of 80 horse
16 horse power
other.266
173.
and | as old as his Find the age of the
Resolve into prime factors
:
184. and 5 h.
same
result as the
number
diminished by
175.
A
boy
is
father.
.
train.
An
The two
express train runs 7 miles an hour faster than an ordinary trains run a certain distance in 4 h. 189.
3 gives the
same
result as the
numbet
multiplied by
Find the number.
respectively. A house has 3 rows of windows.
A
the
boy
is
as old as his father
and
3 years
sum
of the ages of the three is 57 years.
Find the number.
179. Find the age
5 years older than his sister
183. aW + llab-2&. the
sum
of the ages of all three is 51. and the father's present age is twice what the son will be 8 years
hence.56.
if
each
increased 2 feet.
2
2
+
a
_ no.
187. z 2
-92. 6 in each row the lowest row has 2 panes of glass in each window more than the middle row. Four years ago a father was three times as old as his son is now.
A
each
177.
.
190.
power one of the two Find the power of each. Find the dimensions of the floor.
3 gives the
174.
418 ~j-o.278
410. Find the number of miles an hour that A and B each walk.
421. Tn 6 hours
. Find the number. far did he walk all together ?
A
. and at the rate of 3^ miles an hour.
4x
a
a
2 c
6
Qx
3 x
c
419.
a
x
a
x
b
b
x
c
b
_a
b
-f
x
414. How long is each road ?
423.(c rt
a)(x
-
b)
=
0.
420.a)(x b
b)
(x
b
~
)
412.
mx ~
nx
(a
~
mx
nx
c
d
d
c)(:r
lfi:r
a
b)(x
A
in 9 hours
B walks
11 miles
number of two digits the first digit is twice the second.
hour. In a
if
and
422.c) . 411. he takes 7 minutes longer than in going.(5 I2x
~r
l
a)
.
down again
How
person walks up a hill at the rate of 2 miles an hour.
A man
drives to a certain place at the rate of 8 miles an
Returning by a road 3 miles longer at the rate of 9 miles an hour.
-
a)
-2
6 2a and was out 5 hours. 18 be subtracted from the number.
-f
a
x
-f
x
-f c
1
1
a-b
b
x
415. the order of the digits will be inverted.
(x
a
x
)
~
a
2 b
2
ar
a
IJ a.
Find the sum and the rate of
interest. Find the
fraction.
Find
the number. and 5 times the less exceeds the greater by 3.
if
the
sum of
the digits be multiplied by
the digits will be inverted.
487. age.
by 4.
A
spends \ of his.
479. and the other number least. and if each be increased by 5 the Find the fraction. If 31 years were added to the age of a father it would be also if one year were taken from the son's age
. How much money
less
484.
to
. Find the numbers.
latter
would then be twice the
son's
A
and B together have $6000. In a certain proper fraction the difference between the nu merator and the denominator is 12. Find the principal and the rate of
interest.
least
The sum
of three
numbers
is
is
21.
whose difference
is
4. There are two numbers the half of the greater of which exceeds the less by 2.
A
number
consists of
two
digits
4. Find their ages. A sum of money at simple interest amounted in 10 months to $2100. and a fifth part of one brother's age that of the other.
If 1
be added to the numerator of a fraction
it
if 1
be added to the denominator
it becomes equal becomes equal to ^. A sum of money at simple interest amounts in 8 months to $260. the Find their ages.
486.
477. Of the ages of two brothers one exceeds half the other by 4 is equal to an eighth of 482. and in 18 months to $2180. also a third of the greater exceeds half the less by 2.
485. Find two numbers such that twice the greater exceeds the by 30.282
ELEMENTS OF ALGEBRA
476.
483.
481.
half the
The greatest exceeds the sum of the greatest and
480.
years. and in 20 months to $275. What is that fraction which becomes f when its numerator is doubled and its denominator is increased by 1.
thrice that of his son
and added to the father's. and becomes when its denominator is doubled and its numerator increased by 4 ?
j|
478. Find the numbers.
. had each at first?
B
B
then has
J
as
much
spends } of his money and as A. fraction becomes equal to |.
532.
E
533. and 23 pounds of lead lose 2 pounds. (a) How many pounds of tin and lead are in a mixture weighing 120 pounds in air.
. his father is half as old again as his mother was c years ago. Two persons start to travel from two stations 24 miles apart.
and third equals \\ the sum third equals \. Find the present ages of his father and mother. A can do a piece of work in 12 days B and C together can do the same piece of work in 4 days A and C can do it in half the time in which B alone can do it.
527. and losing 14 pounds when weighed in water? (b) How many pounds of tin and lead are in an alloy weighing 220 pounds in air and 201 pounds in water ?
in 3 days. If they had walked toward each other. if and L.
sum of the reciprocals of of the reciprocals of the first of the reciprocals of the second and
the
sum
528. An (escribed) and the prolongations of BA and BC in Find AD. Tf and run together.
. 37 pounds of tin lose 5 pounds. and one overtakes the other in 6 hours. it is filled in 35 minutes. BC = 5.REVIEW EXERCISE
285
525. and BE. A boy is a years old his mother was I years old when he was born. M.
touches
and
F respectively. What are their rates of
travel?
. When weighed in water. Tu what time will it be filled if all run
M
N
N
t
together?
529. 90. Throe numbers are such that the
A
the
first
and second equals
. in 28 minutes. if L and Af in 20 minutes. if the number be increased by Find the number. L.
530.
it
separately
?
531.
and B together can do a piece of work in 2 days. A number of three digits whose first and last digits are the same has 7 for the sum of its digits.
and CA=7. How long will B and C take to do
. the first and second digits will change places. Find the numbers. B and C and C and A in 4 days. CD. In how many days can each alone do the same work?
526. A vessel can be filled by three pipes. they would have met in 2 hours. AB=6. In
circle
A ABC. N.
AC
in /).
545.
-
3 x.
+
3.
. GERMANY. if x = f 1.
-
3 x. 2 x
+
5. 2 541. x *-x
+
x
+
1. formation of dollars into marks. Represent the following table graphically
TABLE OF POPULATION (IN MILLIONS) OF UNITED STATES. x
2
+
x.3
Draw
down
the time of swing for a
pendulum
of length
8 feet.
a. The value of x that produces the greatest value of y. then / = 3 and write
=
3.
547. If
to
feet is the length of a
seconds.
to
do the work? pendulum.
How
is
t /
long will
I
take 11
men
2
t' .
540. 2
-
x
-
x2
.
c. i. FRANCE.
from x
=
2 to x
= 4.e. The values of x if y = 2. The greatest value of the function.
542. x 2 544 3 x
539. x*
-
2
x. 2|.
2. AND BRITISH ISLES
535.
Draw
the graphs of the following functions
:
538. the time of whose swing a graph for the formula from / =0
537.
x*.
546.
-
7.286
ELEMENTS OF ALGEBRA
:
534.
b. One dollar equals 4.10 marks. x 8
549.
543.
d. The roots of the equation 2 + 2 x x z = 1. 536. the function. Draw the graph of y 2 and from the diagram determine
:
+
2 x
x*.
550.
The values of y.
e. z 2
-
x x
-
5.
.
548.
Find four consecutive integers whose product is 7920.
723. what is the
price of the coffee per
pound ?
:
Find the numerical value of
728.
ELEMENTS OF ALGEBRA
+36 = 0.
in value.
of a rectangle is 221 square feet and its perimeter Find the dimensions of the rectangle. 725. 721.
729. Find the altitude of an equilateral triangle whose side equals a. paying $ 12 for the tea and $9 for the coffee.
714
2
*2
'
+
25
4
16
|
25 a2
711.44#2 + 121 = 0. 12
-4*+
-
8. **-13a: 2
710. Find two numbers whose 719. How shares did he buy ?
if
726.25 might have bought five more for the same money. 717.
716.
722. Find the price of an apple.
What two numbers
are those whose
sum
is
47 and product
A man
bought a certain number of pounds of tea and
10 pounds more of coffee. 724. 727.40 a 2* 2 + 9 a 4 = 0. A man bought a certain number of shares in a company for
$375. and working together they can build it in 18 days.l
+
8
-8
+
ft)'
(J)-*
(3|)*
+
(a
+
64-
+ i. he
many
312?
he had waited a few days until each share had fallen $6.
sum is a and whose product equals J.
217
.
A
equals CO feet. What number exceeds its reciprocal by {$.
___ _ 2* -5 3*2-7
715.
The
difference of the cubes of
two consecutive numbers
is
find them.
. 16 x* . If a pound of tea cost 30 J* more than a pound of coffee.292
709.
The area
the price of 100 apples by $1. if 1 more for 30/ would diminish
720. needs 15 days longer to build a wall than B. Find two consecutive numbers whose product equals 600.
a:
713. 2n n 2 2 -f-2aar + a -5 = 0. In how many days can A build the wall?
718. 3or
i
-16 .
two squares is 23 feet.
34
939. The perimeter of a rectangle is 92 Find the area of the rectangle. In the first heat B reaches the winning post 2 minutes before A.square inches. feet. y(x + y + 2) = 133.000 trees. there would have been 25 more trees in a row. (3 + *)(ar + y + z) = 96. and 10 feet broader.
2240.
is
3
. a second rec8 feet shorter. A and B run a race round a two-mile course.
and the sum of their cubes
is
tangle
certain rectangle contains 300 square feet.300
930.
rate each
man
ran in the
first
heat.
and the
difference of
936.
two numbers Find the numbers. s(y
932.
find
the radii of the two circles. = ar(a? -f y + 2) + a)(* + y
933. A plantation in rows consists of 10.
Assuming
= -y.
937.
942. In the second
heat
A
.
is 20. and the Find the sides of the and
its
is
squares. Find the side of each two
circles is
IT
square.
The sum of two numbers Find the numbers.
A
is
938. (y
(* + y)(y +*)= 50. Find the sides of the rectangle.102. How many rows are there?
941.
935.
the difference of their
The
is
difference of
their cubes
270. If each side was increased by 2 feet. The
sum
of the circumferences of
44 inches. (y + *) = .
943.
ELEMENTS OF ALGEBRA
(*+s)(* + y)=10.
y(
934. and B diminishes his as arrives at the winning post 2 minutes before B. The diagonal of a rectangle equals 17 feet.
931. + z) =108. and also contains 300 square feet. Find the length and breadth of the first rectangle.
.
the
The sum
of the perimeters of
sum
of their areas equals 617 square feet. Find the numbers.
is 3. *(* + #) =24. the area of the new rectangle would equal 170 square feet.
The sum
of the perimeters of
sum
of the areas of the squares is 16^f feet.
944. z(* + y + 2) = 76.
and the sum of
their areas 78$.
feet. 152. The difference of two numbers cubes is 513. Tf there had been 20 less rows.
much and A then
Find at what
increases his speed 2 miles per hour.
+ z)=18.
diagonal
940.
two squares equals 140 feet.
set out from two places. that
B
A
955. was 9 hours' journey distant from P.
. each block. Two men can perform a piece of work in a certain time one takes 4 days longer.
unaltered.
sum
Find an edge of
954.
953.
overtook
miles.
is
407 cubic feet. Find two numbers each of which
is
the square of the other. the area lengths of the sides of the rectangle.
952.
whose
946. distance between P and Q. and
its
perim-
948.
Two
starts
travelers.
Find the number. and travels in the same direction as A. The area of a certain rectangle is equal to the area of a square side is 3 inches longer than one of the sides of the rectangle.
.
. Find its length and breadth. The square described on the hypotenuse of a right triangle is 180 square inches. The diagonal of a rectangular is 476 yards. at
the same time
A
it
starts
and
B
from
Q
with the design to pass through Q. The sum of the contents of two cubic blocks
the
of the heights of the blocks is 11 feet. triangle is 6.
A and
B. A certain number exceeds the product of its two digits by 52 and exceeds twice the sum of its digits by 53. If the breadth of the rectangle be decreased by 1 inch and its
is
length increased by 2 inches. if its length is decreased 10 feet and its breadth increased 10 feet. and if 594 be added to the number. A rectangular lawn whose length is 30 yards and breadth 20 yards is surrounded by a path of uniform width. Find in what time both will do it.
951.REVIEW EXERCISE
301
945.
949.
Find the
eter
947. its area will be increased 100 square feet.
P and
Q.
950. What is its area?
field is
182 yards. The area of a certain rectangle is 2400 square feet. Find the number. the square of the middle digit is equal to the product of the extreme digits. When
from
P
A
was found that they had together traveled 80 had passed through Q 4 hours before. the digits are reversed. and that B. the difference in the lengths of the legs of the Find the legs of the triangle. at Find the his rate of traveling. Find the width of the path if its area is 216 square yards. A number consists of three digits whose sum is 14. and the other 9 days longer to perform the work than if both worked together.
.
"(.REVIEW EXERCISE
978. and the
common
difference.2
. P.---
:
+
9
-
-
V2
+
. The Arabian Araphad reports that chess was invented by amusement of an Indian rajah. P. The sum
982. Find the value of the infinite product 4
v'i
v7-!
v^5
.+ lY L V.
985.
980. Find the sum of the series
988. doubling the number for each successive square on the board. The
term.
If
of
2
of
integers + 2 1 + 2'2
by which
is
it
is
the
sum
of
the series
2 n is prime. such that the product of the and fourth may be 55.
990.
How many
sum
terms of 18
+
17
+
10
+
amount
.-.
1.
all
A
perfect number
is
a number which equals the sum
divisible.
first
984. Find the
first
term.
. named Sheran. and of the second and third 03.
v/2
1
+ +
+
1
4
+
+
3>/2
to oo
+
+
.
What
2 a
value must a have so that the
sum
of
+
av/2
+
a
+
V2
+
. 2 grains on the 2d.
of n terms of 7
+
9
+ 11+
is
is
40.
to infinity
may
be 8?
. and the sum of the first nine terms is equal to the square of the sum of the first two.3 '
Find the 8th
983.
of n terms of an A..
987. and so on. Find four numbers in A. The 21st term of an A.01
3. is 225.04
+
..
to oo.. who rewarded the inventor by promising to place 1 grain of wheat on
Sessa for the
the 1st square of a chess-board.
to 105?
981. then this
sum multiplied by
(Euclid.001
+
..
Find
n.
0. 5
11.-.
Find four perfect
numbers.)
the last term
the series
a perfect number. 4 grains on the 3d.
992.
to
n terms.
303
979.. Insert 22 arithmetic means between 8 and 54. Insert 8 arithmetic means between
1
and
-.1
+
2.
986.
989.001
4. P.
Find the number of grains which Sessa should have received.
In a circle whose radius is 1 a square is inscribed. (6) the sum of the infinity. (6) after n
What
strokes?
many
1002. P. Each stroke of the piston of an air
air contained in the receiver.
of squares of four
numbers
in
G.
pump removes
J
of the
of air
is
fractions of the original amount contained in the receiver. are unequal. areas of all triangles.
is 4.
994.
and
if
so forth
What
is
the
sum
of the areas of all circles. The other travels 8 miles the first day and After how increases this pace by \ mile a day each succeeding day.
The sum and sum
. inches.
AB =
1004. find the series. In an equilateral triangle second circle touches the first circle and the sides AB and AC. and so forth to infinity. (a) after 5 strokes. One of them travels uniformly 10 miles a day. and so forth to Find (a) the sum of all perimeters. are
45 and 765
find the
numbers. P. P. 995. in this square a circle. The sides of a second equilateral triangle equal the altitudes of the first. P. The side of an equilateral triangle equals 2. Two travelers start on the same road. in this circle a square. Under the conditions of the preceding example. ft.
512
996. the sides
of a third triangle equal the altitudes of the second.
1000. Insert 4 geometric means between 243 and 32. Insert 3 geometric means between 2 and 162.304
ELEMENTS OF ALGEBRA
993. are 28 and find the numbers.
1001.
third circle touches the second circle and the
to infinity.
997. Find (a) the sum of all circumferences. The sum and product of three numbers in G. prove that they cannot be in A.
and the
fifth
term
is
8 times
the second .
998.
. (I) the sum of the perimeters of
all
squares. The
fifth
term of a G. P. after how strokes would the density of the air be xJn ^ ^ ne original
density ?
a circle is inscribed. c.
999.
ABC
A A
n
same
sides. If a. at the same time. many days will the latter overtake the former?
. and G. 1003.
and the Summation of Series is here presented in a novel form. etc. All subjects now required for admission by the College Entrance Examination Board
have been omitted from the present volume.
$1.ELEMENTARY ALGEBRA
By ARTHUR SCHULTZE. so that the Logarithms. but the work in the latter subject
has been so arranged that teachers
who
wish a shorter course
may omit
it
ADVANCED ALGEBRA
By ARTHUR SCHULTZE.
xiv+563
pages.
$1.
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. Particular care has been bestowed upon those chapters which in the customary courses offer the greatest difficulties to the beginner. and commercial life.
A
examples are taken from geometry. Ph. than by the
. proportions and graphical methods are introduced into the first year's course. The author
has emphasized Graphical Methods more than is usual in text-books of this grade. physics.
THE MACMILLAN COMPANY
PUBLISHERS. but none of the introduced illustrations is so complex as to require the expenditure of
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not
The Advanced Algebra is an amplification of the Elementary.
i2mo.D. 64-66 FIFTH AVBNTC. especially
duction into Problem
Work
is
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much
Problems and Factoring. which has been retained to serve as a basis for higher work. comparatively few methods are heretofore.25
lamo.
which have been omitted from the body of the work Indeterminate Equahave been relegated to the Appendix. but these few are treated so thoroughly and are illustrated by so many varied examples that the student will be much better prepared for further
The Exercises are superficial study of a great many cases. book is a thoroughly practical and comprehensive text-book.10
The treatment of elementary algebra here is simple and practical.
xi 4-
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In Factoring. very numerous and well graded there is a sufficient number of easy examples of each kind to enable the weakest students to do some work. great many
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examples are taken from geometry. there is a sufficient number of easy examples of each kind to enable the weakest students to do some work. physics.
12010. All subjects now required for admission by the College Entrance Examination Board
have been omitted from the present volume. Logarithms. 64-66
7HTH
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. so that the tions. etc. than by the superficial study of a great many cases. The more important subjects
which have been omitted from the body of the work Indeterminate Equahave been relegated to the Appendix.10
The treatment
of elementary algebra here
is
simple and practical. without
Particular care has been the sacrifice of scientific accuracy and thoroughness.
THE MACMILLAN COMPANY
PUBLISHBSS.
$1. save Inequalities.
$1. book is a thoroughly practical and comprehensive text-book.25
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time for the teaching of physics or geometry. The Exercises are very numerous and well graded. proportions and graphical methods are introduced into the first year's course. The author
grade. and commercial life.
not
The Advanced Algebra is an amplification of the Elementary.
xi
-f-
373 pages. but the work in the latter subject
has been so arranged that teachers
who
wish a shorter course
may omit
it
ADVANCED ALGEBRA
By ARTHUR SCHULTZE.
In Factoring. especially
duction into Problem
Work
is
very
much
Problems and Factoring. The introsimpler and more natural than the
methods given heretofore.
xiv+56a
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By ARTHUR Sen ULTZE. bestowed upon those chapters which in the customary courses offer the greatest difficulties to the beginner.
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HatF leather. which has been retained to serve as a basis for higher work.
THE MACMILLAN COMPANY
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wor.
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xtt-t
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izmo.D.r and. Ph.10
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methods of teaching mathematics the first propositions in geometry the original exercise parallel lines methods of the circle attacking problems impossible constructions applied problems typical parts of algebra. and not from the information that it imparts. making mathematical teaching less informational and more disciplinary.The Teaching
of
Mathematics
in
Secondary Schools
ARTHUR SCHULTZE
Formerly Head of the Department of Mathematics in the High School Commerce.
. of these theoretical views.
12mo.
THE MACMILLAN COMPANY
64-66 Fifth Avenue. Most teachers admit that mathematical instruction derives its importance from the mental training that it But in affords. 370 pages. a great deal of mathematical spite
teaching
is
still
informational. $1.
.25
The
author's long
and successful experience as a teacher
of mathematics in secondary schools and his careful study of the subject from the pedagogical point of view.
.
New York
DALLAS
CHICAGO
BOSTON
SAN FRANCISCO
ATLANTA
. Typical topics the value and the aims of mathematical teach-
ing
.
.
."
The treatment
treated are
:
is concrete and practical.
.
. New York City.
causes of the inefficiency of mathematical teaching. enable him to
" The chief object of the speak with unusual authority.
Students
to
still
learn
demon-
strations instead of learning
how
demonstrate.
. and Assistant Professor of Mathematics in New York University
of
Cloth. " is to contribute towards book/ he says in the preface.
All
smaller movements and single events are clearly grouped under these general movements. diagrams.40
is distinguished from a large number of American text-books in that its main theme is the development of history the nation. supply the student with plenty of historical
narrative on which to base the general statements and other classifications made in the text. This book is up-to-date not only in its matter and method.
Cloth. The book deserves the attention
of history teachers/'
Journal of Pedagogy.
is
an excellent example of the newer type of
school histories.
but in being fully illustrated with
many excellent maps.AMERICAN HISTORY
For Use
fa
Secondary Schools
By ROSCOE LEWIS ASHLEY
Illustrated.
Topics.
photographs.
New York
SAN FRANCISCO
BOSTON
CHICAGO
ATLANTA
. which have been selected with great care and can be found in the average high school library. Maps.
THE MACMILLAN COMPANY
64-66 Fifth Avenue.
An exhaustive system of marginal references. Studies and Questions at the end of each chapter take the place of the individual teacher's lesson plans.
diagrams.
$1. and a full index are provided. The author's aim is to keep constantly before the
This book
pupil's mind the general movements in American history and their relative value in the development of our nation. which put the main stress upon national development rather than upon military campaigns.
i2mo. " This volume
etc. | 677.169 | 1 |
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Pages
Tuesday, 6 January 2015
Math Lesson on 7 Jan (Wed)
we will be starting on the chapter of 'Function" tomorrow.
"Function" is an A-level topic however, the concepts are very useful for understanding of future Math topics.
For a brief understanding, please access this link:
Read up on Function, Domain, Range and Codomain. These are very useful.
After which, we will be going into E.Math Graphs.
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