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Mrs. Nelson’s Little Black Book

Table of Contents

Linear Equations …………………………………………………………………………………………. 2

Absolute Value Equations ………………………………………………………………………………... 4

Piecewise Functions ………………………………………………………………………………………. 6

Simplifying Polynomials …………………………………………………………………………………. 8

Scientific Notation ………………………………………………………………………………………… 9

Factoring ………………………………………………………………………………………………….. 11

Rational Expressions ……………………………………………………………………………………… 12

Synthetic Division ………………………………………………………………………………………… 14

Radical Expressions ………………………………………………………………………………………. 17

Complex Numbers ………………………………………………………………………………………… 20

Deriving the Quadratic Formula …………………………………………………………………………… 22

Exponential/Logarithmic Equations ……………………………………………………………………….. 23

Systems of Equations -2variables …………………………………………………………………………. 27

Systems of Equations -3variables …………………………………………………………………………. 29

Matrices …………………………………………………………………………………………………… 32

Binomial Theorem ………………………………………………………………………………………. 36

Counting and Probabiltiy ………………………………………………………………………………… 38

Sig Figs …………………………………………………………………………………………………… 41



Linear Equations/Inequalities By Mrs. Lenora Nelson 1/05/09 From www.freewebs.com/mrs--nelson

1.

y-axis

Quadrant II Quadrant I

(−,+) (+,+)

origin x-axis

Quadrant III Quadrant IV

(−,−) (+,−)

Coordinate Grid

x-axis

y-axis

originQuadrant I

Quadrant II

Quadrant IIIQuadrant IV

2. Given the equation: 34 −=− y x Graphing with intercepts. You can graph and

equation if you find the points were the linecrosses the x-axis (x,0) and the y-axis (0,y). To

find these substitute zero in the place of the

variables.

3

3)0(4

34

=

−=−

−=−

y

y

y x

75.

3)0(4

34

−=−=−

−=−

x

x

y x

3.

y=5

x=5

Plot these two equations:5= y in green5= x in blue

4. Plot these two inequalities:5−< y in red

5≤ x in purple

5. These are the formulas for linear equations.

slope: x x

y ym

−−

=

standard form: fractionsno A

C By Ax

&0>

=+

slope-intercept form: bmx y +=

(y-intercept form)

point-slope form: )( 11 x xm y y −=−

parallel slope: has the same slope of given line

perpendicular slope: has negative reciprocal

slope of given line.

6. Solution Descriptions Problems generated by

2

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http://www.easyworksheet.comAll rights reserved. Copyright 1998 Triple Threat Inc.

Problem a: Write the equation of the line in standard form

Step 2: Find the slope

Step 3: Use the point-slope formula

Step 4: Write in standard form: x&y same side, x positive, no fractions.

Original problem: Line going through

(-3,4) and (-4,-2)

Step 2: 643

24=

+−+

=m

Step 3:1864

)3(64

+=−

+=−

x y

x y

Step 4: 226 −=− y x

Problem b: Write the equation of the line in standard form


Step 3: Write in standard form: x&y same side, x

positive, no fractions.

Original problem: slope=7 through (-7,6)

Step 2:4976

)7(76

+=−+=−

x y

x y

Step 3: 557 −=− y x

Problem c: Write the equation of the line in slope-intercept form

Step 2: Find slope of given line.


Step 4: Solve for y to get into slope-intercept form.

Original problem: Parallel to line:153 −=−− y x going through (-4,-5)

Step 2: in standard form B

Am

−= so

5

3−=m

Step 3:

5

12

5

3

5

)4(5

35

−

−

=+

+−=+

x y

x y

Step 4:5

37

5

3−−= x y

Problem d: Write the equation of the line in slope-intercept form

Step 2: Perpendicular slope is the negative reciprocal


Step 4: Solve for y to get into slope-intercept form.

Original problem: Perpendicular to line from problem c and going through same point.

Step 2:3

5

5

3=⊥−= m som

Step 3:320

355

)4(3

55

+=+

+=+

x y

x y

Step 4:3

5

3

5+= x y

Graphs courtesy of Mrs. Nelson's membership to http://webgraphing.com/index.jsp Copyright © 2004-2009

WebGraphing.com. All Rights Reserved.

3

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M. Ernst 5/4/09http://www.freewebs.com/julio4heisman

Absolute Value Equations

Definitions:

Absolute Value: the distance of “x” from zero

Absolute Value Equation: an equation that involves the absolute value of a variable expressionAbsolute Value Inequalities: an inequality that involves the absolute value of a variable expression

Examples:

Ex 1: Solve 712 =+ x

-1st step with any absolute value equation is to clear the absolute value bars1. 2x+1 = 7 or 2x+1= -7 -answer could be either 7 or -7 since variable

was

inside the absolute value bars

2. 2x= 6 or 2x= -8 -finish solving like a regular equation

3. x=3 or x= -4

4. {-4, 3}

Ex2: Solve 712 >+ x

1. 2x+1> 7or 2x+1< -7

- re-written because 2x+1 bust represent a number more than 7 units from 0 on

the

number line.

2. 2x> 6 or 2x< -8

3. x>3 or x< -4

4. (-∞, -4) U (3, ∞)

Ex 3: Solve 712 <+ x

1. -7<2x+1<7

-re-written because it must be less than 7 units from 0, so must be between -7

and 7

2. -8<2x<6

3. -4<x<3

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4. (-4, 3)

Ex 4: Solve 1253 =++ x

1. 73 =+ x

2. x+3= 7 or x+3= -7

3. x= 4 or x=-10

4. {-10, 4}

Ex 5: Solve 326 −=+ z z

1. z+6 = 2z-3 or z+6 = -(2z-3)

2. 6+3 = 2z-z or z+6 = -2z+3

3. z = 9 or z = -1

4. {-9, 1}

Special Cases:

1) Absolute value of an expression can never be negative. 0≥a for all real numbers.

a. −=− ,435r no solution

2) The Absolute value of an expression equals 0 only when the expression equals 0.

a. 7/3,037,037 ==−=− x x x

References:

Hornsby, J. (2000). Algebra for College Students. Reading, MA: Addison-Wesley.

Stapel, E. (2009). Absolute-Value Inequalities. In PurpleMath. Retrieved April 19, 2009, from

http://www.purplemath.com/modules/absineq.htm

5



Piecewise Linear FunctionsBy: Steven 08-09

Different linear equations for different parts

Graph of the Piecewise Function

y = -x + 3 on the interval (-3, 0)

and y = 3x + 1 on the interval (0, 3)

<<+

<<−+−=

30,13

03,3)(

xi f x

xi f x x f

To graph, you use the “if” portion, substitute it into the equation to find

the endpoints of each line.

>+

≤+−=

0,13

0,3)(

xi f x

xi f x x f

If only one endpoint is given in the “if” statement, then find it and then

use the slope to graph the remainder of the line in the direction of the

inequality. Remember that ≥≤ or means a solid dot.

Absolute Value Function-Graph with portions of two lines

Graph of the absolute value function: y = |x|

To graph put in vertex form: k h xa y +−=

Vertex = (h,k ) graph is symmetric to the line x = h

Opens up if a>0 and down if a<0

Wider if |a|<1 and narrower if |a| >1Slope: a can be graphed like the slope, in both direction from the vertex.

Greatest Integer Function-Greatest Integer = x

Step Function-Graph shaped like a staircase

On TI-84 go to Y=, then MATH, NUMSelect int( ) to enter the step function. You may need to switch to dot

mode.

6



Works Cited

Hornsby, John, and Margaret L. Lial. Algebra for College Students (4th Edition). New York: Addison Wesley

Publishing Company, 2000.

Kastberg, T. Barron & S.. "Piecewise Linear Functions." Jim Wilson's Home Page. 12

May2009<http://jwilson.coe.uga.edu/EMT668/EMAT6680.Folders/Barron/unit/Lesson%204/4.html>.

"Mathwords: Step Function." Mathwords. 12 May 2009 <http://www.mathwords.com/s/step_function.htm>.

"greatest integer, greatest integer function, step function." Welcome to www.mathnstuff.com. 12 May 2009<http://www.mathnstuff.com/math

7



Simplifying Polynomials By Mrs. Lenora Nelson 12/6/08

From www.freewebs.com/mrs--nelson

Exponential Rule Example

Definitions 32 x 2 is the coefficient

x is the base

3 is the exponent1. Addition/Subtraction Rule – If the base and exponent

are the same you add/subtract the coefficients.

3232 4623 x x x x +−+32 64 x x +−

2. Product Rule – Bases and exponents DO NOT need toMATCH. Multiply coefficients and add the exponents.

)10)(2)(3(47

yww

yw1160

3. Quotient Rule – Divide the coefficients and subtractthe exponents of the same base. 102

35

5

15

w x

w x=

7

33

w

x

4. Zero Exponent Rule – Any base raised to the zero power = 1 7177

17

0

0

=•=

=

x

5. Negative Exponent Rule – Polynomials must be

expressed with positive exponents. Bases with negativeexponents are moved to the other side of the fraction line.

2

2 3

3 x x =−

6. Power Rule – Exponents that appear on the outside of

parenthesis must be applied to all items within. Raise the

coefficient and multiply the exponent.

63216)4( x x =

7. Solution Description

Step 2- Apply outer exponents

Step 3-Use negative exponent rule to move all bases withnegative exponents.

Step 4: Expand or simplify exponents. This helps to see

factors to cancel.

Step 5: Reduce canceling out.. any top factor with any

bottom factor

Step 6: Use the product rule for final exponents.

Original problem:

3

2

13

2

3

15

8

2

5−

−

−−

x

a

a

x

Step 2:

−

−

−−

−−

63

33

63

93

15

8

2

5

x

a

a

x

Step 3:

3

63

3

93

63

815

52

xa

xa

Step 4:

••

••

6

3

9

6

888

151515

125

8

x

a

x

a

Step 5:

6

3

9

6

64

27

x

a

x

a

Step 6:

15

9

64

27

x

a

8





By: Brandon 08-09

Rules for Multiplication in Scientific Notation:1) Multiply the coefficients2) Add the exponents (base 10 remains) (Multiplication Using Scientific Notation)

3) Adjust the final power of 10 if the decimal needs to be re-placed behind the first non-zero digit.

Rules for Division in Scientific Notation:1) Divide the coefficients2) Subtract the exponents (base 10 remains) (Division Using Scientific Notation)

3) Adjust the final power of 10 if the decimal needs to be re-placed behind the first non-zero digit.

Problem A: convert to scientific notation

4760Step 1: place asterik to right of 4

Step 2: count spaces to decimal point

Step 3: exponent is positive because asterik is

to left of original decimal point

3

1076.4 ×Step 1: 760*4

Step 2: 4*760

3 spaces

Step 3: 31076.44760 ×=

Problem B: convert to scientific notation

.00091

Step 1: place asterik to right of 9

Step 2: count spaces to decimal point

Step 3: exponent is negative because asterik is

to right of original decimal point

4101.9 −

×

Step 1: .0009*1

Step 2: .0009*1

4 spaces

Step 3: .00091 = 4101.9

−×

Problem C: convert from scientific notation31,000,900

Step 1: Move decimal 7 points to the right of itscurrent position

3.10009×107

Step 1: 3.1000900.

Converting: When converting into or out of

scientific notation moving the decimal in the

correct direction can be tricky. Try this

Into: Large numbers produce + exponents

Small numers (.0125) produce – exponents

Out of: – exponents will make a small

decimal # (.08)+ exponents will make a large #

(15,000)

Problem D: solve by dividing 1×105

Step 1: multiply terms on the top, add

exponents

Step 2: divide coefficients

Step 3: subtract exponents

(1×104)(2.5×10-5)2.5×10-6

Step 1: (1*2.5)=2.5(4+-5) = (4-5) = -1

2.5×10-1

Step 2: 2.5/2.5 =1

Step 3: [-1-(-6)]=(-1+6)=5

1×105

9

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http://www.edinformatics.com/math_science/scinot_mult_divb.htm


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Problem E: solve by multiplying

9. x 10-6

Step 1: multiply the coefficients

Step 2: add the exponents, base ten that

remains

(3 x 10 -3) (3x 10-3)

Step 1: 3*3=9

Step 2: -3+-3=-6

9 x 10-6

References"Multiplication using Scientific Notation." Edinformatics -- Education for the Information Age. 04

May 2009 <http://www.edinformatics.com/math_science/scinot_mult_div.htm>.

“Scientific Notation Using Division." Edinformatics -- Education for the Information Age. 04 May2009 <http://www.edinformatics.com/math_science/scinot_mult_divb.htm>.

10



Rational Expressions By Mrs. Lenora Nelson 12/17/08

From www.freewebs.com/mrs--nelson

Exponential Rule Example

Definitions The quotient of two polynomials with the

denominator 0≠ . Also called an algebraic fraction.(You must always check the domain by setting the denom=0)

33

4

≠−

+ xwhere x

x

1. Simplifying to Lowest Terms – Factor the numerator

and the denominator. Cancel any common factors. Check the domain.

}2,3|{3

3

)3)(2(

)3)(2(

652

62

−−≠+−

++−+++

−−

x x x

x

x x

x x

x x

x x

2. Multiplying/Dividing – Factor all parts. Find the

domain. Cancel any common factors. Remember for

dividing you will need to flip the second fraction.

**Notice here that the (1- x) and the ( x-1) would be able to

cancel if you factor out a negative one.

}1,0,3|{2

1

)2(

1

)2)(1(

1

)2)(1(

4

)2)(3(

)1)(3(

)1(4

4

62

322

244

−≠+−

−−−

−−−−

−−

−+•−+

−

−+•

−+

−

x x x

x

x

x x

x

x x

x

x x

x x

x x

x

x x

x x

x x

3. Adding/Subtracting – Factor all the parts. Find the

domain. Find the LCD. Make common denominators andsimplify.

}3,0|{)3(212

301023

)3(212

233010

)3(212

)3()3)(2(5

)3(426

5 12

2

4

2

6

5

≠−

−+

−

+−

−

+−

−+

−

+

x x x x

x x

x x

x x

x x

x x x

x x

x

x

x x

x

x

12





4. Complex Fractions – Factor any parts that need it.

You may then find common denominators for the top and

bottom separately and then simplify. I prefer to “clear thefractions” by multiplying through by a complex fraction

made up of the LCD of the problem (this works because

the LCD complex fraction is equivalent to one).

**Notice the parts that cancel

}0,2|{43

2

42

2

)2(2

2

&

1

)2(1

)2(

2

2

12

2

)2(:2

2

12

2

−≠+

++

++

+

+

•

++

+

++

+

+

x x x

x

x x

x

x x

x

bottomtoponmultilpy x x

x x

x x

x

x x LCD

x x

x

5. Solution Description

Problem a:

Step 2- Factor and Find the domain.

Step 3-Find the LCD.

Step 4: Multiply through by the LCD to clear the

fractions.

Step 5: Cancel and simplify

Step 6: Compare answer to domain. Write answer.

Original problem:45

8

3

2

5

6 −=−

x x

Step 2: }0|{45

8

3

2

5

6≠

−=− x x

x x

Step 3: x

x x

45

45

8

3

2

5

6

−

=−

Step 4: x x

x

x

x

45*

45

845*

3

245*

5

6 −=−

Step 5: x83054 −=−

Step 6:}3{

3

248

−

−=

−=

x

x

Problem b:

Step 2- Factor and Find the domain.

Step 3-Find the LCD.

Step 4: Multiply through by the LCD to clear thefractions.

Step 5: Cancel and simplify. Compare answer to domain.

Write answer.

Original problem: 63

75

482 152w w w w+

+ −

−= −

− −

Step 2:

}5,3|{

)5)(3(

1522

48

5

7

3

6

−≠−+−−

−=

−−

++

x x

ww

wwww

Step 3:

)(3(:)5)(3(

48

5

7

3

6+

−+−

=−−

++

ww LCDwwww

Step 4: 48)3(7)5(6 −=+−− ww

Step 5:φ

3

48217306

−=

−=−−−

w

ww

Synthetic DivisionBy: M. Richardson (http://www.freewebs.com/malrich10) 08-09

13

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#1: Definitions & Rules

a. Identify the following parts of function notation:

• (x-k) = divisor• g(q) = quotient (answer)• r = remainder **always +, if the remainder is negative enclose it in ( )

Sometimes you will be asked for the answer only, not in function notation: r q g +)(

• g(q) = quotient (answer)• r = remainder (written over the divisor)

• This is sometimes called mixed number format.b. Synthetic Division: a shorthand, or shortcut, method of polynomial division in

the special case of dividing by a linear factor (and it only works in this case).Synthetic division is generally used, however, not for dividing out factors but forfinding zeroes (or roots) of polynomials. (Staple, 1998)

c. To find zeroes of polynomial equations: If you are given the polynomialequation y = x 2 + 5x + 6, you can factor the polynomial as y = ( x + 3)( x + 2).

Then you can find the zeroes of y by setting each factor equal to zero and solving. You will find that x = –2 and x = –3 are the two zeroes of y . (Staple, 1998)d. ***It is always true that, when you use synthetic division, your answer will be

raised to a power one less than what you'd started with.e. Step-by-Step Instructions:

Written Steps: Math Steps:

1. Original Problem:

2. First, carry down the “2” that indicates the

leading coefficient.

Write k on the outside, and the coefficientson the inside.

**Remember that in the form ( x-k ), that the –

is part of the form so adjust the signaccordingly.

3. Multiply by the number on the left and carry

the result into the next column.

4. Add down the next column. Obey integer

rules:

5. Multiply by the number on the left and carry

the result into the next column.

6. Add down the next column:

14

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7. Multiply by the number on the left and carrythe result into the next column.

8. Add down the next column:

9. Multiply by the number on the left and carrythe result into the next column.

10. Add down the column. This last number will

be the remainder.

11. Final Answer (in mixed number format):

(Staple, 1998) #2: Solution Descriptions

a) Written Steps: Math S teps:


2. First, write down all the coefficients. If

there is a gap of the exponent (like how

is missing in this problem), make sure to put a “0” wherever the absence of

exponent is. is already given to

you, so you do not need to solve for it. Put

the “3” on the left.

3 2 0 -10 -19 0 -45

3. Next, carry down the leading coefficient. 3 2 0 -10 -19 0 -45

2

4. Multiply by the potential zero, carry up tothe next column, and add down.

3 2 0 -10 -19 0 -456

2 6

5. Repeat this process. 3 2 0 -10 -19 0 -45

6 18

2 6 8


6 18 24

2 6 8 5


6 18 24 152 6 8 5 15

8. Repeat this process. 3 2 0 -10 -19 0 -456 18 24 15 45

2 6 8 5 15 0

15

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9. Because you started with a polynomialraised to the third power, you are left with

a polynomial raised to the second power.

*Note: There is no remainder, so you

wouldn’t put anything after .

Final answer (in proper function notation):

)155862)(3()( 234 ++++−= x x x x x x f

b) Written Steps: Math Steps:


2. First, write down all thecoefficients, and put the zero

from (so x =3) at theleft.

3. Next, carry down the leadingcoefficient.

4. Multiply by the potential zero,carry up to the next column, andadd down.

5. Repeat this process.

6. Repeat this process.

7. Because you started with apolynomial raised to the thirdpower, you are left with apolynomial raised to the secondpower.

Final Answer (in mixed number format):

(Staple, 1998)

References

Stapel, E. (1998). PurpleMath.com. Synthetic division: The process. Retrieved May 1, 2009from http://www.purplemath.com/modules/synthdiv2.htm.

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Radical Expressions By Mrs. Lenora Nelson 1/27/09 From www.freewebs.com/mrs--nelson

#1: Definitions & Rulesa. Identify the following parts of the radical expression 3 27

Index = 3

Radical =

Radicand = 27 b. Define:

Principal root: The positive root of a square root. If asked to find “the” root, principal

root is implied. If asked to find “all” roots then you would want the positive andnegative roots.

Following are true statements regarding radical expressions.

(CliffNotes, 2009)

c. Rules for Radicals

Simplifying : To simplify numbers that aren’t perfect squares you must factor into

perfect squares and work from there.

Adding/Subtracting: The index and radicand must be the same. If you have numbers in

front of the radical, treat as coefficients and obey integer rules. The radicand acts as the

variable and remains unchanged.

Multiplying/Dividing: The index must be the same. If you have numbers in front of theradical, treat as coefficients and mult/div obeying integer rules. The radicands are also

mult/div within the radical and then simplified if possible.

Radicals in Denominator: These are not allowed. You must rationalize thedenominator by multiplying by something equivalent to 1. See example

Ex: 22

22

2

2

2

2

2

2==•=

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For Algebra 2 Students:

#2: Solution Descriptions Simplify Problem a:

Step 1: Factor.

Step 2: Take the square root of 25 and leave the 3 inside

Step 3: For the variables divide the exponent by the index

and answer goes on outside remainder if any stays inside.

Original problem: 6375 y x

Step 1: 63*3*25 y x

Step 2: 6335 y x

Step 3: x xy 35 3

Problem b:

Step 1: Because these are all factors I can move thenegative exponent to the top.

Step 2: Make common denominators for your exponents.

Step 3: Multiply the top by keeping the base and addingthe exponents.

Step 4: Divide by keeping the base and subtracting theexponents.

Optional: You may leave in that form unless directionssay write in radical form.

Original problem:13/2

3/1

* − x x

x

Step 1:3/2

13/1 *

x

x x

Step 2:3/2

3/33/1 *

x

x x

Step 3:3/2

3/4

x

x

Step 4: 3/23/23/4 x x =−

Optional: 3 2 x

Problem c:

Step 1: Indices are the same bur radicands are not so

simplify to see if they become the same.

Step 2: Now that the radicands are the same combine the

integer coefficients.

Original problem: 27123 +−

Step 1: 3*93*43 +− 3336 +−

Step 2: 33−

Problem d:

Step 1: Change to exponential form. Index becomes thedenominator.

Step 2: Reduce fractional exponent if possible.

Original problem: 8 4 x (leave in exponential form)

Step 1: 8/4 x

Step 2: 2/1 x

18



Problem e:

Step 1: Multiply by the conjugate of the denominator.

This gives us the difference of two squares and no radicals

left in denominator.

Step 2: FOIL or distribute as needed.

Step 3: Simplify if possible.

Original problem:34

2

+ (rationalize)

Step 1:34

34

34

2

−−

•+

Step 2: 316

)34(2

−−

Step 3:13

)34(2 −

Problem f:

Step 1: With the radical on one side you can square both

sides to get rid of the radical part. Don’t forget to FOIL.

Step 2: Move everything to one side. Try to get thesquared term to be positive.

Step 3: Try to Factor to solve for x.

Step 4: Because the original problem contained a variable

under the radical you must check to see if both answersare solutions.

Only the principal roots are considered here.

Step 5: Write final answer in solution set.

Original problem: x x −=+ 843 (solve for x)

Step 1: 22 )8()43( x x −=+

2166443 x x x +−=+Step 2: 060192 =+− x x

Step 3: 0)15)(4( =−− x x

15,4= x

Step 4: sub x = 4 sub x = 15

7744

749416

)15(84)15(3)4(84)4(3

843843

−≠=

−==

−=+−=+

−=+−=+ x x x x

Step 5: }4{

References

CliffNotes.com (n.d.). CliffNotes.com. Retrieved January 15, 2009, from


19

http://www.cliffnotes.com/



http://www.cliffnotes.com/





Complex Numbers

H. Norris 08-09

Definitions:i: any real number whose square is -1.

For any positive real number b: bib =−2i = - 1

Complex Numbers: if a and b are real numbers, and an number of the formula a+bi is called a complex

number.

a: and real number

bi: imaginary number

Conjugate: Used in division of complex numbers. It is the opposite operation on the bottom of the

division problem. For example, if you have 1+1 over 2+2, then the conjugate would be 2-2.

Simplifying Square Roots of Negative Numbers:

Ex. 1

It is impossible to take the square root of a

negative number, so first you have to get rid of thenegative and then you can find the square root.

Since you know that i is any number whose square

is -1, you know that you can take out i and be leftwith 100. Next, find the square root of 100, and

don't forget the i.

Ex. 2This problem doesn't have a perfect square, but we

can still do this. It helps if you write out -27 in

simpler terms.

Next, you do the same thing that you did in theearlier problem when you eliminated the negative

by taking out i. Once this is done, you can see thatwhile 27 itself is not a perfect square, there is a

perfect square within it(9). With the square of 9

(3) taken out, you should only have one left withinthe radical.

Ex. 1

ii

i

10100

100100

=

=−

Ex. 2

33

9*3)3*3*3(

)3*3*3(27

i

i=−

−=−

Dividing Square Roots of Negative Numbers

Ex. 1

First, begin by taking out the negative, which

turns into ani. This leaves you with regular squareroots. You can now divide 75 by 3. And then

simplify.

Ex. 2In this example there is only one negative number.

This automatically tells you that there will be an i

in your answer. Start by removing the negative,divide your numbers, and then simplify. Don't

forget to include the i in the final answer!

Ex. 1

5253

75

3

75

3

75====

−

−

i

i

Ex. 2

iii

248

32

8

32===

−

20



Adding/Subtracting Complex Numbers

Ex. 1Begin by identifying the complex numbers and

those without the i. Deal with any parentheses and

combine like terms.

Ex. 2Here is an example with a subtraction part. Notice

the negative was distributed over the parentheses.

Ex. 1

i

ii

ii

78

4632

)46()32(

+

+++

+++

Ex. 2

i

i

i

34

395

)39(5

+−

+−

−−

Multiplying Complex Numbers

Ex. 1

When you multiply complex numbers, you have

to factor them. Remember FOIL, so that way youget all the right numbers. Once you've FOIL'd,

you can combine like terms. Remember from the

definitions earlier, that i^2 is equal to a -1 and

since -1 * -10=10, you add 10 instead of subtract10.

Ex. 1

i

i

iii

ii

1422

101412

1020612

)24)(53(

2

+

++

−+−

−+

Dividing Complex Numbers

Ex. 1When you divide, you have to remember the

conjugate, which is defined above. You multiply

both parts of the problem by the conjugate andthen foil. You put those answers over each other,

and then divide like normal.

Ex. 1

i

ii

i

i

i

i

+

+=

+−−

++

2

29

)2(29

29

2958

25

25*

25

98

Calculator HelpMany calculations with i can be performed using the i button on your TI-84 calculator. The i button is locatedon the decimal key. Don’t forget to use parentheses.

Simplifying Powers of i by Hannah 08-09 Pattern even exponents of i :

• If the exponent divided by 2 equals

an even number, then it is +1


an odd number, then it is -1i2= -1 i4=1

i6= -1 i8=1

i10= -1 i12=1

i14= -1 i16=1

i18= -1 i20=1

Pattern odd exponents of i:


an even number, then it is +i

• If the exponent divided by 2 equalsan odd number, then it is – i

i1

= i i3

= -ii5= i i7= -i

i9= i i11= -i

i13= i i15= -i

i17= i i19= -i

21



Deriving the Quadratic Formula By Mrs. Lenora Nelson 5/13/09 From www.freewebs.com/mrs--nelson

Original Problem: 02 =++ cbxax

Step 1: Move c to the other side cbxax −=+2

Step 2: Divide everything by a

a

c x

a

b x

−=+2

Step 3: Complete the Square by takinghalf of b (which is now b/a) and square it;

add it to both sides

222

22

+

−=

++

a

b

a

c

a

b x

a

b x

Step 4: Factor the left side. On the right,

apply the power and then make commondenominators.

2

22

2

22

4

4

2

42

a

acb

a

b x

a

b

a

c

a

b x

−=

+

+

−

=

+

Step 5: Take the square root

a

acb

a

b x

2

4

2

2 −±=+

Step 6: Solve for x/ simplify

a

acbb x

a

acb

a

b x

2

4

2

4

2

2

2

−±−=

−±−=





Exponential and Logarithmic Functions By Mrs. Lenora Nelson 3/2/09 From www.freewebs.com/mrs--nelson

#1: Definitions & Rulesd. Identify the following parts of this logarithm 125log5= y

Exponent = y

Yielded Number = 125

Base = 5e. Define:

One-to-One Function: a function in which each x-value corresponds to only one y-value and

vice versa (Lial & Hornsby, 2000, p. 634). None of the x’s or y’s repeat.

Exponential Function: a function with a variable in the exponent. F( x) = a x where a>0 and

1

Logarithm: the inverse of an exponential function. It describes the exponent needed to

produce a given answer. x y alog= (a ≠ 1, a>0, x>0 ). 125log5= y With the base of 5 a 3

needed to yield a 125.

The red line shows the exponential function.The green line is the line of reflection.The blue line shows the logarithmic function.

Common Logarithm: Logarithms to the base10. Calculators evaluate Logs base 10.

Natural Logarithms: “The logarithm base e of a number. That is, the power of e necessary t

equal a given number. The natural logarithm of x is written ln x. For example, ln 8 is2.0794415... since e2.0794415... = 8” (Simmons, 2006). They are called natural because they occ

in biology and the social sciences in natural situations that involve growth or decay (Lial &

Hornsby, 2000, p. 672). All of the Logarithm Rules below apply to ln as well.

e : e ≈ 2.7182818284.... is a transcendental number commonly encountered when workingwith exponential models of growth, decay,and logistic models, and continuously compound

interest (Simmons, 2006).

e is the unique number with the property that the area of the region boundethe hyperbola , the x-axis, and the vertical lines and is 1.

other words,

With the possible exception of , is the most important constant in

mathematics since it appears in myriad mathematical contexts involving lim

and derivatives (Sondow & Weisstein, n.d.)

f. Rules


http://www.mathwords.com/l/logarithm.htm


http://www.mathwords.com/b/base_of_a_logarithm.htm

http://www.mathwords.com/p/power.htm


http://www.mathwords.com/e/e.htm


http://www.mathwords.com/n/natural_logarithm.htm

http://www.mathwords.com/t/transcendental_numbers.htm

http://www.mathwords.com/e/exponential_growth.htm

http://www.mathwords.com/e/exponential_decay.htm

http://www.mathwords.com/l/logistic_growth.htm

http://www.mathwords.com/c/continuously_compounded_interest.htm



http://mathworld.wolfram.com/e.html



http://www.mathwords.com/b/base_of_a_logarithm.htm




http://www.mathwords.com/t/transcendental_numbers.htm

http://www.mathwords.com/e/exponential_growth.htm

http://www.mathwords.com/e/exponential_decay.htm

http://www.mathwords.com/l/logistic_growth.htm




http://mathworld.wolfram.com/e.html



How to determine if One-to-One Function: If the graph of the function passes the horizonta

line test (no 2 points touch the function) then it is one-to-one. If you are given an equation

format you recognize apply the horizontal line test otherwise graph first.

How to find the inverse of a one-to-one function:

1. First verify that it is one-

to-one.2. Swap the x and the y.

3. Solve for y

4. Write in the format

...)(1 =− x f

3 1)( += x x f

3 1+= y x

1

13

3

−=

+=

x y

y x

1)(1 +=− y x f

Product Rule: baab logloglog +=

Quotient Rule:

bab

alogloglog −=

Power Rule: aa log2log

2=

Change of Base Rule:2log

9log8log 2 = This example shows taking a log base 2 and converti

to log base 10. This is most commonly done to find a numerical value using a calculator orgraph logs other than base 10 in a calculator. The Change of Base Rule can be used to conv

to any base if needed.

#2: Sample Problems: Solution DescriptionsProblem a: Rewrite in exponential form

Step 1: Write the base of 49 raised to the ½. Place the

yield number on the other side of the = sign.

Original problem:2

17log49 =

Step 1: 749 2/1 =

Problem b: Rewrite in exponential form

Step 1: ln is the natural logarithm to the base of e so raise

t to the 2nd. Place the yield number x on the other side of

he = sign.

Original problem: 2ln = x

Step 1: xe =2

Problem c: Rewrite in logarithmic form.

Step 1: Write the base as a log to base 81. Place the yieldnumber next. Then place the exponent number ½ on the

other side of the = sign.

Original problem: 981 2/1 =

Step 1: 219lo g81 =

Problem d: Rewrite in logarithmic form.

Step 1: Write the base e as ln. Place the yield number

next. Then place the exponent x on the other side of the =ign.

Original problem: 5= xe

Step 1: x=5ln



Problem e: Expand.

Step 1: For the numerator apply the power rule to move

he exponent in front. For the denominator you will

ubtract it. As you also apply the power rule you willminus a minus which gives you a plus.

Original problem:4

4

4log−c

a

Step 1: ca 44 log4log4 +

Problem f: Condense.

Step 1: First convert the 3 to log base 8 by raising 8 to therd power.

Step 2: Because the 512 is positive place it in thenumerator. The log of m is negative so place it in the

denominator.

Original problem: m8log23−

Step 1: m88 log2512log −

Step 2: 28

512log

m

Problem g: Solve for x.

Step 1: First see if you can convert to the same bases.

Step 2: If the bases are equal then the exponents must be

qual. Set the exponents equal to each other.

Step 3: Solve for x and write the solution set.

Original problem: 162 =− x

Step 1: 422 =− x

Step 2: 4=− x

Step 3:}4{

4

−

−= x

Problem h: Solve for x.

Step 1: Determine what power 3 would need to equal a 27.

OR you can use the Change of Base Rule to type into

alculator.Step 2: Write solution set.

Original problem: x=27log3

Step 1: 273[] = OR 3log

27log

Step 2: }3{

Problem i: Solve for x.

Step 1: Because you can’t get the same base you canither take the log or ln of both sides.

Step 2: This allows you to now apply the Power Rule tomove the x into a position that is solvable.

Step 3: Solve for x.Step 4: Use calculator to get the decimal value.

Original problem: 102 = x

Step 1: 10log2log=

x

Step 2: 10log2log = x

Step 3:2log

10log= x

Step 4:}322.3{

322.3≈ x

Problem j: Solve for x.

Step 1: Apply the Power Rule.

Step 2: If you have a single log on both sides of the = signo the same base then the yield numbers are equivalent.

Set them equal.

Step 3: Convert to positive exponents.

Step 4: Take the cube root of both sides.

Original problem: 27loglog3 22 −= x

Step 1: 12

32 27loglog −= x

Step 2: 1327

−= x

Step 3:27

13 = x

Step 4:}3/1{

3

1= x

References

Lial, M. L. & Hornsby, J. (2000). Algebra for College Students (4th Edition). New York: Addison Wesley Publishin

Company.



Simmons, B. (2006). Definition of e. Mathwords: Terms and Formulas from Algebra I to Calculus. Retrieved Marc

6, 2009, from http://www.mathwords.com/e/e.htm.

Simmons, B. (2006). Natural Logarithm. Mathwords: Terms and Formulas from Algebra I to Calculus. RetrievedMarch 6, 2009, from http://www.mathwords.com/n/natural_logarithm.htm.

Sondow, J. & Weisstein, E. W. (n.d.). “e.” MathWorld --A Wolfram Web Resource. Retrieved March 6, 2009, from

http://mathworld.wolfram.com/e.html.

http://www.mathwords.com/



http://mathworld.wolfram.com/








System of Two EquationsBy: David & Gabby 08-09

) Defs/Rules:

1. System of Equations: a collection of linear equations involving the same set of variables

a. System of equations : a set of equations that finds numbers that make two or more equations true at the same ti

b. Linear system : two or more linear equations

c. Solution set of a linear system : contains all ordered pairs that satisfy all the equations of the system at the sam

time.

d. Inconsistent : when a linear system is graphed, if the lines are parallel; the problem cannot be solved.

e. Dependent : when a linear system is graphed, if the lines are the same line (as in both lines are in the exact sam

position); the answer is infinite. (Hornsby & Lial, 2000)2. Solve by:

i. Graphing: Graph by hand or on graphing calculator. Find where the lines intersect.

ii. Elimination: Method of solving a set of equations by combining the two equations of the system so th

one variable is eliminated.

iii. Substitution: Method of solving a set of equations that works best when a variable has the coefficient

(see example of substitution further down). Get x or y alone, then plug into other equation

3. If both variables are eliminated when a system of equations is solved:i. There is no solution if the resulting statement is false (4≠3)

ii. There are infinitely many solutions if the resulting statement is true (4=4)

4. Solutions:

) Solve by elimination

.362

43

=−−

=+

y x

y x

----------------------------------------------

2.362

862

=−−

=+

y x

y x

----------------------------------------------

. 110 ≠

1. Multiply the top by 2

2. Add like terms. Cancel out the x and y variables

3. Answer is undefined) Solve by elimination

.1332

425

=−

=−

y x

y x

----------------------------------------------

2.2664

12615

−=+−

=−

y x

y x

----------------------------------------------

.264

1215

−=−=

x

x

----------------------------------------------

4. 1411 −= x----------------------------------------------

. 14

11

−= x----------------------------------------------

6. 42)14

11(5 =−− y

28

1−= y

----------------------------------------------

.

−−

28

1,

14

11

1. Multiply the top by 3 and the bottom by -2

2. Cancel out the y variable

3. Add down

4. Solve for x

5. Plug x into original equation

6. Solve for y

7. Get final answer

). Solve by elimination

.936

32

=−

=−

y x

y x

----------------------------------------------

1. Multiply the top by -3

2. Add down

http://en.wikipedia.org/wiki/Linear_equation



http://en.wikipedia.org/wiki/Variable_(math)






2.936

936

=−

−=+−

y x

y x

----------------------------------------------

. 00 = 3. Answer is all real numbers

4). Solve by substitution

.42

5

=−

=+

y x

y x

----------------------------------------------

2. y x −= 5----------------------------------------------

. 4)5(2 =−− y y----------------------------------------------

4. 2= y----------------------------------------------

. )2(5−= x----------------------------------------------

6. 3= x. )2,3(

1. Solve the top equation for x.

2. Replace x in the bottom equation with (5-y).

3. Solve for y

4. Substitute 2 and solve for x.

5. Solve.

6. Write the ordered pair.

5) Solve by Graphing.

34

132

−=+−

=+−

y x

y x

----------------------------------------------

2.34

3

1

3

2

−=

+=

x y

x y

----------------------------------------------

. (1,1)

1. Solve both equations for y.2. Press “y =” on the top of the calculator

** Plug in both equations

** Press “graph” on the top of the calculator ** Find where the lines intersect. You can use the

INEQUALZ APP or …

2nd TRACE, intersect, verify the first curve/enter, verif

the next curve/enter, guess/enter.3. Write the ordered pair.

f your equations are in standard form you may use your TI-84 to solve. Choose APPS, select POLYSMT. This is t

polynomial root finder and the simultaneous equation solver. Follow the onscreen instructions.

References:

Hornsby, J., & Lial, M. L. (2000). Algebra for College Students (4th Edition). New York: Addison Wesley Publishi

Company.



Systems of Equations: 3 Variables (Ch. 11.2)By: Hannah 08-09

A. Definitions:

1. ordered triple- a solution of an equation in three variables, such as 2x + 3y –z = 4, is called an ordered tripl

and is written (x,y,z)

2. system of equations- a set of equations in which finding the numbers makes 2 or more equations true at the

same time3. linear system- 2 or more linear equations form a linear system

4. solution set of a linear system- contains all ordered pairs that satisfy all the equations of the system at the s

time

Lial & Hornsby, 2000)

B. The graph of a linear equation with 3 variables is a plane not a line.

Graphs of Linear Systems in Three Variables

1. The 3 planes may meet at a single, common point that is the solution of

the system.

2. The 3 planes may have the points of a line in common so that the set of

points that satisfy the equation of the line is the solution of the system.

3. The planes may have no points common to all 3 so that there is nosolution for the system.

4. The 3 planes may coincide so that the solution of the system is the set of all points on a plane.

Lial & Hornsby, 2000)C. Steps

Solving Linear Systems in 3 Variables by Elimination

Step 1: Eliminate a variable. Use the elimination method to eliminate any variable

from any 2 of the given equations. The result is an equation in 2 variables.

Step 2: Eliminate the same variable again. Eliminate the same variablefrom any other two equations. The result is an equation in the same twovariables as in Step 1.

Step 3: Eliminate a different variable and solve. Use the elimination method to

eliminate a second variable from the two equations in two variables that result from

Steps 1 and 2. The result is an equation in one variable that gives the value of thatvariable.

Step 4: Find a second value. Substitute the value of the variable found in Step 3

into either of the equations in two variables to find the value of the second variable.

Step 5: Find a third value. Use the values of the two variables from Steps 3 and 4

to find the value of the third variable by substituting into any of the originalequations.

Step 6: Find the solution set. Check the solution in all of the original equations.Then write the solution set.

Lial & Hornsby, 2000)



D. Worked Examples

Original Problem:

Step 1: eliminate one variable from any two equations.To do that, I notice that equation A can be multiplied

by 3 to eliminate variable z.

Step 2: eliminate the same variable from any twodifferent equations. This time, I’m choosing to

multiply equation A by -2. This will eliminate

variable z again.

Step 3: Take equation D & E and eliminate another

variable. The LCM of 13 & 6 is 78; so I multiply D by 6 and E by 13. This will eliminate variable x.

Step 4: solve for the remaining variable

Step 5: Now plug the y = 1 into either equation D or E

to solve for x.

Step 6: Choose one of the original equations A, B, or

C. I choose A. Plug in both found values into theequation with 3 variables to solve for the final variable

Step 7: Check your answers and write in solution set if correct

Original Problem: A: 4x + 8y + z = 2

B: x + 7y – 3z = -14C: 2x – 3y + 2z = 3

Step 1: 12x + 24y + 3z = 6x + 7y – 3z = -14

D: 13x + 31y = -8

Step 2: -8x – 16y -2z = -42x – 3y + 2z = 3

E: -6x – 19y = -1

Step 3: D: 13x + 31y = -8

E: -6x – 19y = -1

78x + 186y = -48

-78x – 247y = -13-61y = -61

Step 4: -61y = -61

y = 1

Step 5: -6x – 19(1) = -1

-6x – 19 = -1-6x = 18

x = -3

Step 6: 4(-3) + 8(1) + z = 2-12 + 8 + z = 2

z = 6

Step 7: {(-3, 1, 6)}

Original Problem:

Step 1: eliminate one variable from two equations

Step 2: eliminate the same variable from two different

equations

Step 3: eliminate another variable from the two newequations; since zero is not equal to 13, there is no

solution. If the statement were true (0 = 0) then the

answer would be all real numbers or infinite solutions.

Original Problem: 2x + 2y – 6z = 5

-3x + y - z = -2

-x – y + 3z = 4Step 1: 2x + 2y – 6z = 5

18x – 6y +6x = 12

20x – 4y = 17

Step 2: -9x + 3y – 3z = -6

-x – y + 3z = 4

-10x + 2y = -2

Step 3: 20x – 4y = 17

-20x + 4y = -40 ≠ 13

no solution



Original Problem:

Step 1: eliminate a variable from two equations

Step 2: use the new equation and the equation with the

same variables to eliminate another variable; solve for

the variable

Step 3: plug in the new found variable to another

equation to find another variable

Step 4: plug in either of the found variables and solve

for the final variable

Step 5: check your answer; if correct, write in solution

set

Original Problem: 2x + y = 6

3y – 2z = -4

3x – 5z = -7Step 1: -6x – 3y = -18

3y – 2z = -4

-6x -2z = -22

Step 2: -6x – 2z = -22

6x – 10z = -14

-12z = -36z = 3

Step 3: -6x – 2(3) = -22-6x – 6 = -22

-6x = -16

x = 8/3

Step 4: 3y – 2(3) = -4

3y – 6 = -4

3y = 2y = 2/3

Step 5: {(8/3, 2/3, 3)}

f your equations are in standard form you may use your TI-84 to solve. Choose APPS, select POLYSMT. This is t

polynomial root finder and the simultaneous equation solver. Follow the onscreen instructions.

References

Lial, M., & Hornsby, J. (2000). Algebra for College Students. New York: Addison Wesley Longman, Inc..



Matrices

By: C.Meaux 08-09 http://cmeaux.webs.com/

Rows

↑↑↑

→ 217

532

Columns

• Each number or variable inside of a matrix is called an elemento 2, 3, 5, 7, 1, and 2 are the elements in this matrix

To name a matrix: (number of rows) x (number of columns)o So, this is a 2x3 matrix

o Read as “2 by 3 matrix”

(L. Nelson, Algebra II class notes, March 24, 2009)

Definitions and Examples:

• Square matrix : A matrix with the same number of row and columns

Example:

47

52 This is a 2 x 2

• Row matrix : A matrix with just one row

Example: [ ]652 This is a 1 x 3

• Column matrix : A matrix with just one column

http://cmeaux.webs.com/

http://cmeaux.webs.com/



Example:

4

7

2

This is a 3 x 1

• Equal matrices : Two matrices with equal corresponding elements

Example:

=

y x

q p 13

42

so: 4,2,1,3 ==== y xq p

• Zero matrix : All elements are zero

Example:

00

00


Rules:

• Addition : Matrices must be the same size. If they are, add the corresponding elements.



Example:

61

01

38

64

98

65

• Subtraction : Matrices must be the same size. If they are, flip the integers and add the corresponding element

Example:

13

42

85

23

42

65

• Multiplication with a coefficient : Multiply the coefficient by each element in the matrix.



Example:

−=

−

20

11

40

32

5

• Multiplication with two matrices : The first matrix’s number of columns must match the second matrix’s numof rows. The numbers that are left will tell you the size of the answer.

Example:

−

•

−

23

32

46

405

243

• The first matrix is 2x3

• The second matrix is 3x2

o 2 x 3 3 x 2

Because the numbers on the inside (the threes) match, this can be multiplied. Because the numbers on the outside are both twos, the answer will be a 2x2 matrix

Once you have determined that the matrices can be multiplied, you multiply each element in the first row of first matrix by the elements in the first column of the second matrix. You add these answers together to get p

of the final answer.



Example:

−

•

−

23

32

46

405

243

You repeat this procedure with the first row of the first matrix and the second column of the second matrix.

You repeat this procedure again with the second row of the first matrix and the first column of the secondmatrix, then once more with the second row of the first matrix and the second column of the second matrix.

So the final answer is:

632

824

1863

=⋅

=⋅

=−⋅−

326818 =++

422

1234

1243

=⋅

=⋅

−=⋅−

441212 =++−

1234

020

3065

=⋅

=⋅

−=−⋅ 1812030 −=++−

824

030

2045

=⋅

=⋅

=⋅

288020 =++



−=

−

•−21

43

23

32

46

405

243


Sample Problems

Problem a:

−+

−

−

34

61

53

42

Step 1: Check to see if the matrices are the same

size

Step 2: Add the corresponding elements

Step 1: 2x2 = 2x2

Step 2:

835

143

264

1)1(2

=+

=+−

=+−

=−+

81

21



Problem b:

−

−−

−

26

35

52

43

Step 1: Check to see if the matrices are the samesize

Step 2: Flip the integers and add the corresponding

elements

Step 1: 2x2 = 2x2

Step 2:

725

462

134

253

=+

−=−

=−

=+−

− 74

12

Problem c:

−

06

233

Step 1: Multiply the coefficient by each of theelements

Step 1:

003

1863

623

933

=⋅

=⋅

−=−⋅

=⋅

−01 8

69



Problem d:

−•

−

12

45

01

353

264

Step 1: Check to see if the matrices can bemultiplied

Step 2: Multiply the first row of the first matrix by

the first column of the second matrix.

Step 3: Multiply the first row of the first matrix by

the second column of the second matrix.Step 4: Multiply the second row of the first matrix

by the first column of the second matrix.

Step 5: Multiply the second row of the first matrix

by the second column of the second matrix.

Step 6: Plug the elements into the final answer.

Step 1: 2x3 3x2

Step 2:

Step 3:

Step 4:623

2555

313

=⋅

=⋅

=⋅

Step 5:

Step 6:

13 4

23 0

By: Morgan 08-09 http://morgan097.webs.com/

422

3056

414

−=⋅−=⋅=⋅

304304 =−+

212

2446

004

=−⋅−=⋅=⋅

262240 =++

173200 =−+

313

2045003

−=−⋅

=⋅

=⋅

346253 =++

http://morgan097.webs.com/

http://morgan097.webs.com/



1. IDENTITIY MATRIX

A=

−

026

954

731

1. (See definition for Identity Matrix

above)

To find the identity matrix for A, find

numerals, like in multiplying these two

matrices, wherein the answer of themultiplication problem would be the

original A.

I=

100

010

001

2. DETERMINANTS

86

43−

= A

-24 – 24 =

-48

2.

We start off by multiplying the top left

numeral with the bottom left numeral andget a product of negative 24.

Then, we multiply the bottom left numeralwith the top right numeral and get a

product of 24.

After, we subtract the second product from

the first and find that the determinant of

this matrix is -48.

While reading these steps, you may be asking yourself “But when will I ever need this inreal life?” Well, believe it or not, but many career path involve matrices. In small

businesses, matrices are used to compute revenues. Because of the fact that, in a store,

different products have different prices, matrices are great for figuring out costs, revenues,and profits of many different sales. In the same sector, matrices are also used to determine

production costs for similar products. Matrices are also used in cryptography as well as

economic mobility.



ReferencesHornsby, J., & Lial, M. L. (2000). Algebra for College Students (4th Edition). New York: Addison Wesley Publish

Company.



The Binomial TheoremBy Lauren 5/6/09

Pascal’s Triangle

1. To complete Pascal’s triangle start with a

pyramid consisting of two rows, all spaces shouldcontain a one.

2. When starting the next row add the two

numbers surrounding a space and put that sum

between those two numbers but in the next row.Then place ones on the outside.

3. The row containing “1 2 1” is row number two.

This means that if the exponent is 2, you should

refer to this row. The numbers count down by onefrom there as you advance in the triangle. The

same method used in #2 is what you should do to

get any row after that.

11 1

1

1 1

1 2 1

1

1 1

1 2 11 3 3 1

1 4 6 4 1

1 5 10 10 5 11 6 15 20 15 6 1

*These are the coefficients of each part of a binomial, expanded using the binomial theorem

The Binomial Theorem -Shortcut used to expand a binomial that is raised to a power.

1. Start with simple (x + y)2 2)( y x+

2. Match the exponent with the correspondingrow in Pascal’s triangle. 1 2 1

3. Create three blanks for the three coefficientsin the row of Pascal’s triangle, and place your

coefficients in them.

1 ___ +2 ___ +1 ___

4. Starting on the far left side, place x next to the

first coefficient and give the x the same exponent

on the outside of the original problem. **As you

go from left to right place an x with an exponent

one less than the one

before it until you have an exponent of zero. The

furthest blank to the right should not have an x at

all.

1x2 +2x +1___

5. Now starting on the far right side, place y next to

the last coefficient and give the y the same

exponent on the outside of the original problem.This will be multiplied by anything you’ve already

written for that blank . As you go from right to left

place a y with an exponent one less than the one

before it until you have an exponent of zero. The

furthest blank to the left should not have a y at all

1x2 +2xy +1y2

Final Answer:x2 +2xy + y2

(L. Nelson, Algebra II class notes, April 8, 2009)



Now for one a little bit more difficult

*The steps for this are the exact same as listed above (refer to them if needed).

1. (x + y)6

2. 1 6 15 20 15 6 1

3. 1___ + 6___ + 15___ + 20___ + 15___ + 6___ + 1___

4. 1x6 __ + 6x5 __ + 15x4 __ + 20x3 __ + 15x2 __ + 6x__ + 1___

5. 1x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6

Binomial Expansion: x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + y6

What if the letters in the parenthesis already have coefficients or there aren’t letters, but numbers?

t’s simple really. When you make your spaces with the Pascal’s Triangle coefficients already attached to them, you

are ready to make the next step. Now do exactly as you did before but place anything with a coefficient (watch out fohe sign, if it’s a negative include that too) inside of parenthesis next to its correct Pascal Coefficient. Leave the

xponent at the end of the parenthesis. Once you’ve expanded all of that with both parts of the binomial then you mdistribute your exponents, coefficients and signs. This will give you the Binomial expansion.

Example 1:

1. (2x + 4)4

2. 1__ + 4__ + 6__ + 4__ + 1__

3. 1(2x)4 _ + 4(2x)3 __ + 6(2x)2 __ + 4(2x)__ + 1__

4. 1(2x)4 _ + 4(2x)3(4)_ + 6(2x)2(4)2 _ + 4(2x)(4)3 _ + 1(4)4 _

5. 16x4 + 128x3 + 384x2 + 512x + 256

Binomial Expansion: 16x4 + 128x3 + 384x2 + 512x + 256



Counting & ProbabilityBy: Jacqueline -- May 4th, 2009

Definitions:

• Fundamental principle of counting: If one event can occur in m ways and a second ev

can occur in n ways, then both events can occur in mn ways, provided the outcomethe first even does not influence the outcome of the second. (Hornsby, 2000)

• Factoral: n! = n(n-1)(n-2)…(1) (Hornsby, 2000)o Ex: 5! = 5 * 4 * 3 * 2 * 1

o 0! = 1

• Permutations:o P (n, r)

P = permutation

n = number of elements

r = at a time

o key words: order, list, arrangement, positiono nPr = !n (multiply by only as many numbers as r; not a complete factorial)

5P3= 5(4)(3) = 60

On TI-84 first enter the number of elements, find MATH button, then selPRB, choose nPr, enter the r-number.

• Combinations:

o

r

nOR nCr

n = number of elements

r = at a timeo key words: group, team, class

o nCr =

5C3 =)1)(2(3

)3)(4(5= 10

On TI-84 first enter the number of elements, find MATH button, then selPRB, choose nCr, enter the r-number.

• Probability: the ratio where desired outcome is over the possible outcome (Horns2000)

o Probability = possible

desired

o If the problem states wants you to find the probability of x and y, you shocalculate the probability of each separate,

• Odds: ratio of favorable outcomes to unfavorable outcomeso Odds = favorable : unfavorable



Problem 1: 181,440

1. First determine that the ordermatters, therefore this is apermutation.2. Because r is 7, you must

multiply 9 by its descendingnumbers 7 digits.3. Solve.

Problem 1: How many ways can 7people be lined up in a row from aclass of 9 people?

1. 9P7

2. 9(8)(7)(6)(5)(4)(3)

3. 181,440Problem 2: 15

1. Because the students will be ona committee with no assignedpositions, this is a combination.2. Set up the problem. Because r

is 4, you must multiply 6 by itsdescending numbers 4 digits.3. Cancel any repeating numbers.4. Solve.

Problem 2: Six students want tobe on the Senior Prom Committee.If only four students are permittedon the committee, how manydifferent ways can they bechosen?

1.

4

6

2.)1)(2)(3(4

)3)(4)(5(6

3.)1)(2(

)5(6

4. 15

Problem 3: 26

1

1. Because there are only 2 blackkings in a standard deck of cards(52), place 2 over 52.

2. Reduce for answer.

Problem 3: A card is drawn from astandard deck of 52 cards. Findthe probability of a black king.

1.52

2

2. 26

1

Problem 4:13

4

1. Find the probability of drawinga jack or a spade. Because thereare 4 jacks in a deck, place 4 over52. Because there are 13 spadesin a deck, place 13 over 52.2. Find the probability of arepeating card: a jack of spades.Because there’s only one jackthat’s also a spade, place 1 over52.

3. Add the probabilities of the

Problem 4: A card is drawn from astandard deck of 52 cards. Findthe probability of drawing a jack ora spade. (Hornsby, 2000)

1. Jack =52

4Spade =

52

13

2. Repeating =52

1

3.52

4+

52

13-52

1=

52

16



jacks and spades together andsubtract the repeating probability.4. Reduce for answer.

4.13

4

Problem 5: 3:10

1. There are 12 face cards in adeck of 52 cards.2. Reduce for answer.

Problem 5: What are the odds of drawing a face card from astandard deck of cards? (Hornsby,2000)

1. 12:52

2. 3: 10

Works Cited

Hornsby, John (2000). Algebra for College Students. Reading, MA: Addison-Wesley.



By: S. Roberts 4/2/09

Significant Figures are the minimum amount of digits required to report a valuewithout loss of accuracy. “It is important to use significant figures when recording a

measurement so that it does not appear to be more accurate than the equipment is capaof determining.”(Fetterman, 2007) For example, when using a ruler that only measuresinches (it doesn’t have little marks for tenths of inches) you can’t measure somethingaccurately to the thousandth’s place because the ruler doesn’t go up that high.

Significant Digits

Rule for DeterminingSig Figs

Example# of SigFigs

Nonzero digits are always significant 257 33.5 2

Digits to the left of a decimal point arealways significant

92. 2

360. 3

Zeroes between significant figures aresignificant

1,003 4

6.0001 5

Placeholders (0’s that indicate positionof a decimal point) are NOT significant

210 2

0.003 1

Zeroes at the end of a decimal pointare significant

0.320 3

0.0045000 5(Louisiana iLEAP)

Problem Solution1.) How many significant figures are in the number

34.0900?

1.) Using the rules for determining sig figs, the answer to

the problem is 6. We know this because all nonzero digits

are sig figs, and 0’s between sig figs are significant, and

also numbers that come after the decimal after a sig fig aresignificant as well, so all of the numbers in the problem are

significant.

2.) How many significant figures are in the number

0.0000000030?

2.) The answer to this problem is 2 because when you use

the rules for determining whether figures are significant or

not, only 2 follow the criteria. Because there are no sig figs

in front of the decimal point, all of the zeroes in front of the

three are NOT significant because they don’t have sig figs

on both sides of them. And as for the sig fig at the end, it is

significant because it comes after a sig fig.



3.) Round the measurement of the image to the correct

sig figs.

3.) The answer to the problem is about 6.7. When

measuring things using sig figs, you include the numbers

you know for sure along with one estimated digit. We

know for sure that the image is somewhere between 6 and 7

inches. Because there are no tick marks between the two

measures, we can guess ONE digit. Some people may see

6.7, others, 6.8 or even 6.6. Sig figs are the number of

digits believed to be correct by the person doing the

measuring (2008). So, if you estimated more digits, your reading would become less precise. You can’t just

randomly guess numbers, saying that the measure of the

image was 6.7485297635 because there’s no way to justify

those numbers without the tick marks.

4.) Round the measurement of the image to the nearest

sig figs.

4. Rounded to the nearest sig figs, the measurement of the

image is 6.73. Although it is the exact same image as the

one in problem #3, the ruler has more tick marks, and the person doing the measuring can get a more precise reading.

Sig figs consist of all of the known digits and one estimated

digit. The 6 and the 7 are known because the ruler tells us.

But the 3 is unknown because the ruler doesn’t have a mark

for the hundredths place. The image could have well been

measured as 6.74 or 6.72 and would not have been wrong.

Adding/subtracting numbers

and determining Sig Figs5.) You have 4.7832 grams of salt and 1.234 grams of

sugar and 2.02 grams of flour. If you combine these,

how many grams, in significant figures, will you get

total?

5.) First line up the decimals.If you add these numbers together, you will get:

To make sure your answer has the correct amount of

significant figures, for adding/subtracting you go with the

least number of decimal places present. Here the last

number (2.02) has 2 decimal places; therefore, you roundyour answer to 2 decimal places..



Multiplying/dividing numbers

and determining Sig Figs6.) Multiply 2.8723 by 1.6. Give your answer to the

correct number of sig figs.

6.) First you multiply regularly.

When multiplying and dividing you round to the least

number of sig figs. Now count sig figs in each factor. The

first number has 5 sig figs. The second (1.6) has 2 sig figs.

Now round your answer in such a way that there will only be

2 sig figs. It helps to start at the left of your answer then

round. Sometimes it will be necessary to write your answer

in scientific notation to get the right amount of sig figs.

References

Fetterman, Lewis M. (2007). Significant Figures. Retrieved April 2, 2009, fromhttp://www.campbell.edu/faculty/fetterman/Significant%20Figures.htm

Significant figures. (2008). Math skills review: significant figures. Retrieved August 13, 2008, fromhttp://www.chem.tamu.edu/class/fyp/mathrev/mr-sigfg.html

Louisiana iLEAP test preparation and intervention. Illinois: McDougal Littel.

http://www.campbell.edu/faculty/fetterman/Significant%20Figures.htm

http://www.chem.tamu.edu/class/fyp/mathrev/mathrev.html

http://www.chem.tamu.edu/class/fyp/mathrev/mr-sigfg.html


http://www.campbell.edu/faculty/fetterman/Significant%20Figures.htm

http://www.chem.tamu.edu/class/fyp/mathrev/mathrev.html


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