math 20 mod 1 cover & lesson1

8/7/2019 Math 20 Mod 1 Cover & Lesson1

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Mathematics 20

Copyright Saskatchewan Ministry of Education

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Mathematics 20

Course Introduction

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Importance of Mathematics

The aim of the mathematics program as stated in the Ministry of EducationCurriculum Guide is to graduate numerate individuals who value mathematics andappreciate its role in society. For years mathematics has been thought of assomething to learn, but not very useful when being applied to everyday life. The

Saskatchewan curriculum aims at integrating mathematics into different aspects oflife so that students can cope confidently and competently with everydaysituations that require the use of mathematical concepts.

Mathematical concepts are still the focus of learning. It is important to be able tocompute, measure, estimate, and interpret mathematical data. It is also importantto be able to transfer this information to life in a technology based society.

Problem solving skills and abilities are also an integral part of the course. This isthe link between problem solving in the real-world and the mathematics that youwill be studying.

The course is designed to develop logical thinking skills, effective work habits andan appreciation of mathematics. Through the process of learning and applying thisknowledge to everyday living it is hoped that you will become a self-motivated,confident life-long learner.

Mathematics 20 has been developed to follow Mathematics 10 with the same goalsand objectives. Many of the concepts in Mathematics 10 are reviewed and thenexpanded further in the Mathematics 20 course.

Course Overview

The contents of the course follow the curriculum very closely. The structure of thecurriculum centres around seven mathematical strands.

Data Analysis and Consumer Mathematics Numbers and Operations Equations Algebra Functions

Geometry Trigonometry

Problem solving is an integral component of all of the strands and is incorporatedthroughout the course. Students will learn the steps necessary when approachinga problem and going through the problem solving process.

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Contents of the Course

The Mathematics 20 course takes many different aspects of mathematics andintegrates them in such a way as to give them a meaning and purpose in everydaylife.

The Mathematics 20 course is a combination of Algebra, Consumer Mathematicsand Geometry. All of these aspects of mathematics are a part of the content andeven though they are separate areas of study, the process of learning brings themtogether.

The course has been developed using a modular design. There will be threecontent areas or modules. Each will concentrate on one or two knowledge areas.While the modules are distinct in content from each other, certain concepts orcertain kinds of information presented in one may have some relationship orapplication to other modules. The modules or topic areas are:

Module 1 Review of Algebraic SkillsPolynomials and Rational ExpressionsIrrational Numbers

Module 2 Quadratic FunctionsQuadratic EquationsProbability

Module 3 Consumer MathematicsGeometry Angles and Polygons

Circles

References

The course has been developed using a variety of resources. This concurs with theMinistry of Educations initiative of resource-based learning. It is possible,however, to use mathematics-related references from public or school libraries asa supplement to the course material. A list of key resources is developed by theMinistry of Education every year. Many different sources of information were used

in the development of this course.

A pamphlet will be included for use in the consumer mathematics section of thecourse. This section will also require the use of a newspaper or other forms ofmedia coverage to complete the activities.

Any use of computer-based learning programs or various internet sites willcertainly enhance your learning and increase your knowledge of the link between

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mathematics and technology.

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Method of Study

Each person generally has, or develops, his or her own method of study.Therefore, you may consider the following as suggestions which you could follow,or modify to suit yourself.

Decide on the amount of time you wish, or feel you need, to cover thecourse. .

When you receive the course, place it in a three-ring binder. This shouldmake the handling of the modules easier and more orderly, with less chanceof losing pages.

At the beginning of each module is a table of contents for the lessons itcontains. When you begin a lesson, check the contents for that particularlesson to obtain some idea of the material it covers. You can then glanceover the objectives listed at the beginning of the lesson. Quickly read or

skim through the contents of the lesson and possibly the questions of theassignment which follows. Following this first quick reading, begin goingthrough the course material in more detail. As you do this, you may wish touse other sources of supplementary information.

Do the practice problems. This will help you understand the lesson.

Always do your own work. Be honest and fair to yourself.

Once the lesson material has been completed, begin the assignment. Do nothesitate to review course material or to seek out information from other

references when working on the individual questions or tasks. Aftercompleting a lesson and its assignment, you could review the objectiveslisted at the beginning of the lesson. This will give you some idea of howclose you have come to achieving them.

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Mathematics 20

Module 1

Lessons 1 7

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Module 1 Introduction

The first module reviews algebraic skills from past courses, and develops newconcepts. These new concepts include rational expressions and irrationalnumbers. These concepts are presented in such a way as to tie the information toreal world problems.

The material that you will learn here is the foundation to many new concepts thatwill be developed later in this course and in higher grade levels.

The process of developing the skills is very important. A method of solvingproblems is outlined in the first lesson. This will help you throughout the entirecourse when you are asked to solve problems and you need to work through theprocess.

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Mathematics 20

Module 1

Table of Contents

Page

Lesson 1 Review of Algebra

Introduction........................................................................................ 3

Objectives........................................................................................... 7

1.1 Whole Number Exponents......................................................... 71.2 Order of Operations and Solving Equations............................. 13

1.3 Polynomials................................................................................. 19

1.4 Calculating Percentages............................................................. 26

Answers to Exercises......................................................................... 33

Assignment 1...................................................................................... 43

Lesson 2 Irrational Numbers and Square Root Radicals

Introduction........................................................................................ 53Objectives........................................................................................... 55

2.1 Absolute Value............................................................................ 57

2.2 Irrational Numbers and Square Root Radicals........................ 62

2.3 Expressing Radicals in Simplest Form..................................... 69

2.4 Applications of Radicals............................................................. 75


Assignment 2...................................................................................... 89

Lesson 3 Operations with Radicals

Introduction........................................................................................ 101

Objectives........................................................................................... 103

3.1 Addition and Subtraction of Radicals....................................... 105

3.2 Multiplication of Radicals.......................................................... 109

3.3 Division of Radicals.................................................................... 115

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3.4 Radical Equations and Problem Solving................................... 120


Assignment 3...................................................................................... 137

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Lesson 4 Factoring Polynomials

Introduction........................................................................................ 151

Objectives........................................................................................... 153

4.1 Common Factors and Grouping................................................. 155

4.2 Factoring Special Polynomials................................................... 160

4.3 Factoring Trinomials of the Form ax2 + bx + c Where a = 1.... 167

4.4 Factoring Trinomials of the Form ax2 + bx + c Where a 1... 175


Assignment 4...................................................................................... 193

Lesson 5 Powers and Exponents

Introduction........................................................................................ 207

Objectives........................................................................................... 209

5.1 Dividing a Polynomial by a Binomial........................................ 211

5.2 Evaluating Powers...................................................................... 2165.3 Properties of Exponents............................................................. 224

5.4 Real World Exponential Problems............................................ 229


Assignment 5...................................................................................... 247

Lesson 6 Rational Expressions

Introduction........................................................................................ 259

Objectives........................................................................................... 2616.1 Non Permissible Values of Rational Expressions.................... 263

6.2 Simplifying Rational Expressions............................................. 269

6.3 Multiplying and Dividing Rational Expressions...................... 275

6.4 Adding and Subtracting Rational Numbers............................. 281


Assignment 6...................................................................................... 305

Lesson 7 Review of Lessons 1 6

Introduction........................................................................................ 319Objectives........................................................................................... 321

7.1 Review of Lesson 1...................................................................... 323

7.2 Review of Lesson 2...................................................................... 327

7.3 Review of Lesson 3...................................................................... 330

7.4 Review of Lesson 4...................................................................... 332

7.5 Review of Lesson 5...................................................................... 334

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7.6 Review of Lesson 6...................................................................... 337


Assignment 7...................................................................................... 353

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Mathematics 20

Module 1

Lesson 1 Review of Algebra

Mathematics 20 1 Lesson 1


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Review of Algebra

Introduction

This lesson reviews some of the main topics in algebra which were studied in

Mathematics 10 and which are needed as a background for further study in this course.

The topics in this lesson are a review and therefore each section will be in the form of a

summary of the important points, followed by a few examples.

An introduction to problem solving will follow on the next page. It will be important to

refer back to this throughout the course.

Since this is your first lesson, a few reminders are in order.

Do all the exercises at the end of each section. There are four

sections in this lesson. Answers are given at the end of the lesson.

You are encouraged to use your scientific calculator wherever possible.

Only the assignment at the end of the lesson is to be returned for grading.

Whole Number Math

By the Pythagoras Theorem, a2 + b2 = c2 is true for a right triangle with sides a, b, and

hypotenuse c.

There are many pairs of consecutive whole numbers such that the

sum of their squares is a perfect square. One such example is

32

+ 42

= 52 . Can you find other examples?



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Can you find four whole numbers a, b, c, d such that a3 + b 3 + c3 = d3 ?



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Problem solving is an integral part of the Mathematics 20 curriculum. With this in

mind it is important to develop a plan or strategy to use when you are faced with

solving real world problems.

Word problems are not to be feared! They are very valuable in showing us how the

mathematical knowledge that we are learning can be transferred into situations that are

happening around us daily.

The following are steps that can be used to make problem solving easier regardless of the type

or difficulty of the problem.

1. Read the problem

Identify what is given.

What is to be found?

Look for key words and phrases.

Put the problem in your own words.

2. Develop a plan

Choose a symbol for the variable.

Write down the information that you know and how it relates to what you have

to find.

Consider possible strategies for carrying out the plan.

drawing a diagram

looking for patterns

making a table or graph guessing and checking

working backwards

Write an equation using the information that you have attained.

3. Carry out the plan

This can be done by solving the equation.

The solution to the problem is the value that you find the variable to be.

4. Check the solution Check your solution by substituting the answer back into the original equation.

If necessary, estimate the answer to make sure that the answer is reasonable.

5. Write a concluding statement

State the answer to the problem in a clear and concise manner. Go back to the

original problem and make sure that you have answered the questions in this



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problem.



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Objectives

After completing this lesson, you will be able to

simplify and evaluate powers with whole number exponents by applying the laws of

exponents.

simplify numerical and algebraic expressions following the correct order of

operations.

solve linear equations by following the correct order of operations.

add and subtract polynomial expressions.

multiply binomial expressions.

apply the method of isolating a variable in an equation to solve percentage

problems.

understand and apply the steps necessary to solve real world problems.



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1.1 Whole Number Exponents

Symbols are frequently used in mathematics so that long expressions may be written in a

short and concise way. One of these symbols is the power.

For example, the long expression 2222222 is written using the symbol 27

indicating that there are seven factors of 2.

In the expression 27 :

2 is called the base.

7 is called the exponent.

the whole expression 27 is called the power.

In general, ifb is any number and n is a natural number, then the nth power

ofb is:

bn = b b b . . . b

There are n factors ofb. n is called the exponent.

b is called the base.

bn is called the power.

Calculator Use

A scientific calculator can be used to evaluate powers.

The y x key on the calculator evaluates powers.

Use the following key stroke pattern to evaluate the


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expression 27 :

clear 2 yx 7 enter

display: 128


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Example 1

Evaluate a) 23 b) 52 c) 1.0153 d) 1.110

Solution:

a) 23=8

clear 2 yx 3 enter display: 8

b) 52=25

clear yx 2 enter display: 25

c) 1.0153=1.045678375

clear 1.015 yx 3 enter display: 1.045678375

d) 1. 110=2.59374246

clear 1.1 yx1 10 enter display: 2.59374246

Often the base of a power is a variable. Observe carefully how a number replaces a

variable in a base.

Example

Evaluate x6 when: a) x = 3 b) x = 2 .

Solution:

a) Write the original expression. x6

Substitute x = 3 . = 36

Evaluate. = 729


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clear 3 yx 6 enter display: 729


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b) Write the original expression. = x6

Substitute x = 2 . = 2 6

Evaluate. = 64

clear ( 2 ) y x 6 enter display: 64

In case b, parentheses are needed to make it clear that there must be 6 factors of 2 .

Without parentheses, 26 would indicate there are 6 factors of 2 multiplied by 1 , or

1222222 = 64 , for b.

In general:

Without parentheses the exponent affects only the one position immediately to

the left.

ab2

= a b b

With parentheses the exponent affects all the elements enclosed in the parentheses.

ab 2

= ab ab = a2

b2

Example 2

Evaluate a) 7 + 22 b) 7 + 22 c) 32

3

Solution:

a) 7 + 22 = 92 = 81

b) 7 + 22 = 7 + 4 = 11

c) 32 3=6

3=6 6 6 =216


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Example 3

Evaluate a) 4 2 b) 4

2

Solution:

a) 4 2 b) 4

2

Parentheses No Parentheses

4 2 =4 4 4

2 = 44

= 16 = 16

When there are parentheses, the negative is included when the exponent is applied

to the base. ex) x 4=x x x x =x4

x 3=x x x =x3

When there are no parentheses, the negative is not included when the exponent is

applied to the base. ex) x 4=1 x x x x

The laws of exponents were developed to help in solving expressions that involve powers.

There are five laws that you should become familiar with and know how to apply todifferent situations.


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Properties of Exponents

Let x and y be any number and a, b, and c be any integer.

Product of Powers

To multiply powers with like bases, add the exponents.

yay

b= y

a + b

Quotient of Powers

To divide powers with like bases, subtract the exponents.

ya

yb

= ya - b

or

ya

yb = y

a - b

Power of a Power

To find the power of a power, multiply the exponents.

yab

= ya b

Power of a Product

To find the power of a product, find the power of each factor and multiply.

xy c=xc yc

Power of a Quotient

To find the power of a quotient, find the power of the numerator and the power of

the denominator and divide.

xy c

=x

c

yc


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Example 4

Simplify4a

32

2a2.

Solution:

Write the original expression. 4a32

2a2

Apply the power of a power rule to the numerator. =4

2a32

2a2

Simplify. =16a

6

2a2

Apply the quotient rule. = 162 a6- 2

Simplify. = 8a4

Exercises 1.1

1. Use your calculator to evaluate each power.

a. 1. 12, 1.1

4, 1. 1

8, 1. 1

16

b. 0 .92, 0 .9

4, 0 .9

8, 0. 9

16

c. 29

+ 35

d. 36

+ 632

2. Evaluate each expression for the given value of the variable.

a. x7 if x = 5

b. 2x 4 if x = 3

c. a9 if a = 2

d. b6 if b = 2

e. 2x2 x3 if x = 3


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3. Simplify each expression.

a.4

54

7

410

b. 2a53

c. 3a6 5a

2

d. 32363e. a

5b3a

2b6

f. xy 3 x

4y

9

g.a2b4

ab3

h. 3n3

i. x4

y2 y

2

x 3

j.2x

2y25

8x2

y3

xy 3

1.2 Order of Operations and Solving Equations

Order of Operations for Simplifying Expressions

For simplifying numerical or algebraic expressions a set of rules establishes the order in

which computation is to be done. This ensures that the final answers will be consistent.

First: Do the computations inside the brackets if there are any.

Second: Simplify the numbers with exponents.

Third: Divide or multiply in the order in which and appear from left to

right.

Fourth: Add or subtract in the order in which + and appear from left to

right.

An easy way to remember this is to use the letters that spell BEDMAS. Each letter stands


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for an operation and the order that the operations are to be performed.


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B Brackets

E Exponents

D Division

M Multiplication

A Addition

S Subtraction

Example 1

Simplify 122 22 + 2

2 5 1 .

Solution:

Write the original expression. 122 22 + 2

2 5 1

Brackets. = 122 22 + 4 5 1

= 122

26 5 1Exponents. = 144 26 5 1

Divide. = 726 5 1

Multiply. = 2160 1

Subtract. = 2159

The following key stroke pattern can be used.

clear 12 yx 2 2 ( 2 + 2 yx 2 ) 5 1 enter

display: 2159

** Some calculators may not perform order of operations for you. If you have a

calculator where this is the case, you will have to do the order of operations on your

own.


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Solving Linear Equations in One Variable

To solve an equation means to find a value of the variable that makes the equation true.

This value is called the root of the equation.

The procedure for solving an equation may also be called isolating the variable.

The order of operations for solving equations is the reverse of the order for simplifying

numerical expressions used above.

Whatever operation is performed to one side of the equation, the same operation must be

performed to the other side.

Example 2

Solve 5 x + 2 = 6 .

Solution:

Write the original equation. 5 x + 2 = 6

Subtract 2 from both sides of the equation. 5 x + 22=62

5 x = 4Divide both sides of the equation by 5.

5x

5=4

5

Simplify. x =4

5

There are two ways to solve an equation that has brackets.

Reverse the order of operations and do the brackets last.

Distribute the value in front of the brackets first and then continue to solve the

equation.


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Example 3

Solve 32 x + 5 + 7 = 1

Solution:

Write the original equation. 32 x + 5 + 7 = 1

Subtract 7 from both sides of the equation. 32 x + 5 + 77 = 17 32 x + 5 = 6

Divide both sides of the equation by 3. 32x + 5

3=6

3 2 x + 5 = 2

Remove the brackets and repeat the order of operations for the expression found inside thebrackets.

2 x + 5 = 22 x + 55 = 25

2 x = 7

2x

2=7

2

x =7

2

The solution may also be obtained the following way:

Write the original equation. 32x5 7=1

Distribute the value in front of the parenthesis. 6x157=1

Simplify. 6x22=1

Add or subtract equal values to both sidesof the equation. 6x=21

Multiply or divide both sides of the equation6x

6=21

6


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by the same value. x=7

2


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h =V

r2

Isolating a Variable in an Equation

Some equations have more than one variable and it is often convenient to isolate one of

the unknown variables when the values of the other variables are known. This occurs

most frequently when you are working with formulas. The usual procedure for solving an

equation is followed.

Example 1

The formula for the volume of a circular cylinder or can is

V = r2h, where h is the height and r is the radius.

What is the height of the can if its radius is 6 cm and its

volume is 546 cm3?

Solution:

V = 546cm3

r = 6 cm

h = ?

Write the formula. V = r2h

Isolate the variable h in the formula. Divide

both sides of the equation by r2 .

Substitute the known values. h =546

6 2

{=4 .8 3c

The value of is 3.14.


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Example 2

Isolate the variable x in the equation2

3x + 3 y = 1 .


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Solution:

Write the original equation. 2

3x + 3 y = 1

Subtract 3y from both sides.2

3x + 3 y 3 y = 13y

Simplify. 23 x = 13y

Multiply by the reciprocal 32 .3

2 2

3x = 32 13y

Simplify. x =3

29

2y

or x=39y

2

Exercises 1.2

1. Simplify, using the rules for order of operations.

a. 352644

b. 8 + 932

c. 72 84 + 3 + 122

23

d. 2 .45. 84 . 8 5 .44 .5

2. Determine if the given value is a root of the equation.

a. 3 x 7 = 5 x = 4

b. 2 x 7 = x + 1 x = 4

3. Solve each of the following equations and check your answer by substituting it into

the original equation.

a. 7 = 1 + 2x

b. 4 m 1 = 15

c. 6 + 2 m 1 = 0

d. 7 x + 1 3 = 11

e. 7 4 x + 2 + 15 = 1


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f. 326x 1 + 8 = 1


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4. Isolate the indicated variable.

a. 2 x + 3 y 1 = 0 y

b. 2 x 3y 1 = 0 x

c. A =1

2b

d. y = 5 x + 7 xe. y = 5 x + 7 x

5. The formula for the volume of a cone is V =1

3r2h . Isolate h and

evaluate h when r is 10 cm and Vis 1256 cm3.

6. The formula for the volume of a sphere is V =4

3r3 . Find the

a) volume (V) if the radius (r) is 3.6 cm.

b) radius (r) if the volume (V) is4

3 cm3 .

1.3 Polynomials

A term is a mathematical expression using numbers and variables combined in the formof a product.

The following are examples of terms:

5x 7 xy

x2 5x4 y

The degree of a term is the sum of the exponents of the variables. Each of these terms is

of degree 3.

2x3, 6 xy

2, 7 xyz

Single terms are also called monomials.

A polynomial is a sum of terms.


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A polynomial can be classified by the number of terms that it has.

Monomial A polynomial with one term 5x4 y

Binomial A polynomial with two terms 2x3 + 5

Trinomial A polynomial with three terms 2x2

+ 5 x + 6

The degree of a polynomial is the greatest degree of any of its terms after simplification

of the polynomial.

2x3 + 6x is a binomial of degree 3.

2 x + 1 is a binomial of degree 1.

x2 + 2 x + 1 is a trinomial of degree 2.

The formula for the area of a square is A = x2 .

The formula for the volume of a cube is V = x3 .

The formulas for the area of a square and the volume of a cube suggest the names for

these special polynomials.


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Polynomials of degree 2 are also called quadratic polynomials and those of degree 3 are

called cubic polynomials.


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Two terms are similar if they have the same variable factors.

7x2 and 3x

2 are similar terms.

8x2

y and 9 yx2 are similar terms.

Only similar terms can be combined by addition or subtraction.

Example 1

Simplify 7x23x2 .

Solution:

Write the original expression. 7x23x2

Write down the common base and combine the coefficients.

Simplify. = 4x2

Example 2

Simplify 8x2 y + 9x2 y .

Solution:

Write the original expression. 8x2

y + 9x2

y

Write down the common base and combine the coefficients.

Simplify. = 17 x2

y

Example 3

Simplify 10 x

2

+ 6x

3

7x

2

5x

3

.

Solution:

Write the original expression. 10 x2

+ 6x37x

25x

3


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Simplify by distributing the negative and

removing the brackets.

Combine like terms. = 107 x2 + 6 + 5 x3

Simplify. = 3x2 + 11x3


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Multiplying Two Monomials

The laws of exponents are used when multiplying two monomials.

Example 4

Simplify x2 x5 y2 .

Solution:

Write the original expression. x2 x5 y2 Apply the product rule. =x

2x

5y

2

=x25

y2

=x7

y2

Example 5

Simplify 2x5 y6 15 x4 y .

Solution

Write the original expression. 2x5 y6 15 x4 y Apply the product rule. =2 15 x

5x4

y6

y

=30 x54 y61

Simplify =30 x9

y7

Example 6

Simplify 2x5 y6 2

.

Solution:

Write the original expression. 2x5 y6 2

Apply the power of a power rule. = 22 x5 2 y6

2

Apply the power of a power rule. = 4x10

y12


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Multiplying a Polynomial by a Monomial

The distributive property is used to multiply a polynomial by a monomial.

The Distributive Property

The product ofa and b + c is given by:a b + c = ab + ac or b + c a = ba + ca .

The product ofa and b c is given by:a b c = ab ac or b c a = ba ca .

Example 7

Simplify 2x 5 xy x2

+ 1 .

Solution:

Write the original expression. 2x 5 xyx21

Apply the distributive property. =

Simplify. = 10 x2

y + 2x32x

2 ( 5x x y2


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Multiplying Two Binomials

The distributive property is used twice to multiply two binomials together.

Distributive Property

a + bc + d = a + bc + a + bd a + b is distributed.

= ac + bc + ad + bd c and d are distributed.

Example 8

Expand and multiply x3y 2 .

Solution:

Write the original expression. x3y 2

Expand = x3y x3y Use the distributive property = x3y x x3y 3y

=x23 xy3 xy9y2Simplify by combining like terms =x

26 xy9y

2

Another method used to multiply two binomials together is referred to as the FOIL

method.

Multiply the: F First terms of the binomials.

O Outside terms of the binomials.

I Inside terms of the binomials.

L Last terms of the binomials.


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Example 9

Multiply x + 3mm x .

Solution:

Write the original expression. x3m mx

Distribute or use FOIL. x3m mx

F O I L

=xmx2

3m2

3mxSimplify by combining like terms. = 2mxx

2+ 3m

2

Exercise 1.3

1. Simplify as much as possible by combining similar terms.

a. 4c + 4 + 2c = 3c 16

b. 53x2

1 + 12 = 4 2x

2

+ 3 + 9c. 12 = 2 x31 + 4 x

3+ 2

2. Multiply and combine similar terms.

a. 7 xy 3 x + 1 + 21x2

y

b. x22x x + 3x

2 + 6x

3

c. 2a 2a b a + 2ab

d. k + 1 k + 2

e. a 5 a + 5 f. x 3 2 + 6xg. 22 y y + 1

h. 3 x + y x + y 2x + y 2

i. 1 x y 2

j. 5 x + y x + y + 25 x2

F

O

I

L


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1.4 Calculating Percentages

The sentence "six percent of 250 is 15" can be translated into an equation.

6

100250 = 15

Formula for Percent

In general, the sentence "r percent ofA is p" is translated to the equation

r

100 A = p ,

where

r represents the percent.

A represents the amount from which a percentage is taken.

p represents the percent amount ofA.

To solve certain problems it is convenient to have A isolated on one side of the equation

and for others it is convenient to have r isolated.

A =100 p

r, r =

100 p

A

Example 1

How much money is saved by buying a computer

printer which is 15% off the regular price of $670?

Solution:

Read the problem.

The regular price (A) of the computer printer is $670.


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The percentage of savings (r) is 15%.

Find the amount of the savings (p).


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Develop a plan.

Substitute the known values into the formula for calculating percentages.

A = 670

r = 15

p = ?

Carry out the plan.

Write the formula. p =r

100 A

Substitute the known values. p =15

100670

Simplify. p=100.50

Check the solution.

p= r100

A

$ 100.50=?15

100

$ 100.50 = $ 100.50 ()

Write a concluding statement.

The money that is saved is $100.50.

Example 2

$27 is taken off the regular price of $599 for a set of golf clubs.

What percent discount is that?

Solution:

Read the problem.

The regular price (A) of the golf clubs is $599.

The amount of the percentage (p) is $27.Find the percentage (r).

Develop a plan.

Substitute the known values into the formula for calculating percentages.


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A = 599

p = 27

r = ?


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Carry out the plan.

Write the formula. r =100 p

A

Substitute the known values. r =10027

599

Simplify. r=4 .5

Check the solution.

r=100

Ap

4 .5=?10027

599

4 .5 = 4 . 5 ()


The discount is 4.5%.

Example 3

The bill for furnace repair parts including 7% GST is $543.79.

What is the cost of the parts before GST? What is the amount of GST?

Solution:

Read the problem.

The percentage (r) of the total amount (A) is 7%.

The total amount (A) is unknown.

The amount of the percentage (p) is unknown.

The total amount A + p is $543.79.

The final cost is COST BEFORE GST + GST = $543.79.

A + p = $543.79

Develop a plan.

Use the formula to compare A and p.


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p =7

100 A

Substitute the expression for p into the final cost equation.


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Carry out the plan.

Write the equation for final cost. A + p = $ 543.79

Substitute the known values. A +7

100

Simplify.100

100

A +7

100

A = $543.79

107

100

A =100

107$543.79

A=$508.21

Check the solution.

A7

100A=P

5 0 8.2 17

1 0 05 0 8.2 1=

?

5 4 3.7 9

543.79=543.79 ()


The cost before GST is $508.21.

The amount of GST is $ 543.79$508.21=$35.58 .

Exercise 1.4

1. a. 5% of 930 is ? .

b. 11% of ? is 100.

c. ? % of 900 is 800.

2. If 15% tax is paid on an item priced at $1250, find the final cost.


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3. The 1996 school enrolment is 751 and is 7% more than the previous years

enrolment. What was the previous years enrolment?

4. Holly paid $763 for a washing machine after 16% tax was added. What is the price

before tax?

5. The total bill, including a 7% tax, was $596.37. How much tax was paid?

6. The Goods and Services Tax is 7%. IfTrepresents the total bill including tax, show

that GST = TT

1.07=

7

107T .

math 20 mod 1 cover & lesson1

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