little black book-nelson
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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
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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
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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 2: Use the point-slope formula
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 3: Use the point-slope formula
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 3: Use the point-slope formula
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.
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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
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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.
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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
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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
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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
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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>.
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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
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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
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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:
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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:
1. Original Problem:
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. Repeat this process. 3 2 0 -10 -19 0 -45
6 18 24
2 6 8 5
7. Repeat this process. 3 2 0 -10 -19 0 -45
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
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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:
1. Original Problem:
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
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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
http://www.cliffsnotes.com/WileyCDA/CliffsReviewTopic/Radicals.topicArticleId-38949,articleId-38923.html
19
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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===
−
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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
• If the exponent divided by 2 equals
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:
• If the exponent divided by 2 equals
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
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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
−±−=
−±−=
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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
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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
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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.
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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.
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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
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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.
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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)
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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
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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..
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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
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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
(L. Nelson, Algebra II class notes, March 24, 2009)
Rules:
• Addition : Matrices must be the same size. If they are, add the corresponding elements.
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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.
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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.
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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 =++
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−=
−
•−21
43
23
32
46
405
243
(L. Nelson, Algebra II class notes, March 24, 2009)
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
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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
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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 =++
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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.
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ReferencesHornsby, J., & Lial, M. L. (2000). Algebra for College Students (4th Edition). New York: Addison Wesley Publish
Company.
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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)
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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
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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
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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
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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.
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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.
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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..
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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.
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