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Honors Algebra 2
1.1 Real Numbers and Real Operations Objectives:1. Know the categories of numbers2. Know where to find real numbers on the
number line3. Know the properties and operations of real
numbers
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…., -4, -3, -2, -1, 0, 1, 2, 3, 4,…
counting numbers
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…., -4, -3, -2, -1, 0, 1, 2, 3, 4,…
counting numbers
whole numbers
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…., -4, -3, -2, -1, 0, 1, 2, 3, 4,…
counting numbers
whole numbers
integers
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rational numbers - numbers that can be written as a fraction or a decimal that repeats or terminates. Examples?
irrational numbers - numbers that can’t be written as a fraction or a decimal that repeats or terminates. Examples?
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Locate these numbers on a number line:
,31
,8,5
127.2,,33.4,2,
1. Approximate to decimal2. Determine range and mark line3. Plot original values
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property addition multiplication
closure a + b = real number ab = real number
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property addition multiplication
closure a + b = real number ab = real number
commutative
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property addition multiplication
closure a + b = real number ab = real number
commutative a + b = b + a
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property addition multiplication
closure a + b = real number ab = real number
commutative a + b = b + a ab = ba
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property addition multiplication
closure a + b = real number ab = real number
commutative a + b = b + a ab = ba
associative
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property addition multiplication
closure a + b = real number ab = real number
commutative a + b = b + a ab = ba
associative (a + b) + c = a + (b + c)
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property addition multiplication
closure a + b = real number ab = real number
commutative a + b = b + a ab = ba
associative (a + b) + c = a + (b + c)
(ab)c = a(bc)
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property addition multiplication
closure a + b = real number ab = real number
commutative a + b = b + a ab = ba
associative (a + b) + c = a + (b + c)
(ab)c = a(bc)
identity
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property addition multiplication
closure a + b = real number ab = real number
commutative a + b = b + a ab = ba
associative (a + b) + c = a + (b + c)
(ab)c = a(bc)
identity a + 0 = a
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property addition multiplication
closure a + b = real number ab = real number
commutative a + b = b + a ab = ba
associative (a + b) + c = a + (b + c)
(ab)c = a(bc)
identity a + 0 = a a x 1 = a
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property addition multiplication
closure a + b = real number ab = real number
commutative a + b = b + a ab = ba
associative (a + b) + c = a + (b + c)
(ab)c = a(bc)
identity a + 0 = a a x 1 = a
inverse
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property addition multiplication
closure a + b = real number ab = real number
commutative a + b = b + a ab = ba
associative (a + b) + c = a + (b + c)
(ab)c = a(bc)
identity a + 0 = a a x 1 = a
inverse a + -a = 0
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property addition multiplication
closure a + b = real number ab = real number
commutative a + b = b + a ab = ba
associative (a + b) + c = a + (b + c)
(ab)c = a(bc)
identity a + 0 = a a x 1 = a
inverse a + -a = 0 a x 1/a = 1
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property addition multiplication
closure a + b = real number ab = real number
commutative a + b = b + a ab = ba
associative (a + b) + c = a + (b + c)
(ab)c = a(bc)
identity a + 0 = a a x 1 = a
inverse a + -a = 0 a x 1/a = 1
distributive
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property addition multiplication
closure a + b = real number ab = real number
commutative a + b = b + a ab = ba
associative (a + b) + c = a + (b + c)
(ab)c = a(bc)
identity a + 0 = a a x 1 = a
inverse a + -a = 0 a x 1/a = 1
distributive a(b + c) = ab + acopposite of a = -a
Inverse of a = 1/a
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property addition multiplication
closure a + b = real number ab = real number
commutative a + b = b + a ab = ba
associative (a + b) + c = a + (b + c)
(ab)c = a(bc)
identity a + 0 = a a x 1 = a
inverse a + -a = 0 a x 1/a = 1
distributive a(b + c) = ab + acIdentify the property:
1. 5 + -5 = 0
2. 2(3●5) = (2●3)5
3. 4(3 + 7) = 4●3 + 4●7
4. 5 + 3 = 3 + 5
5. (x + 5) + 4 = x + (5 + 4)
6. 1x = x
7. 3●1/3 = 1
8. 2●3●4 = 3●2●4
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Honors Algebra 2
1.2 Algebraic Expressions and Models
Objectives:1. Evaluate algebraic expressions2. Simplify expressions3. Apply expressions to real world examples
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Vocabulary: power – a number and it’s exponent
Vocabulary: power
base
exponent
52
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Vocabulary: power – a number and it’s exponent
Vocabulary: power
52=5∙55 to the second power5 squared
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Vocabulary: power – a number and it’s exponent
Vocabulary: power
53=5∙5∙55 to the third power5 cubed
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34
25
93
16
3∙3∙3∙3
2∙2∙2∙2∙2
9∙9∙9
1∙1∙1∙1∙1∙1
3 to the fourth power
2 to the fifth power
9 cubed (to the 3rd power)
1 to the sixth power
to the 4th power
4
7
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Please Excuse My Dear Aunt Sally
P – parenthesisE – exponentsM – multiplicationD – divisionA – additionS – subtraction
Left to Right
1421 65
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P – parentheses and other grouping symbols from left to right
E – exponents from L to R
M – multiplication/division from L to R
A – addition/subtraction from L to R
Please Excuse My Aunt
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Evaluate these expressions:
25 )13 )(( 4
153)4(1 2
54 5 )( 32
2
2
2731)(4
22 22)(
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Evaluate 5x2x3 2
when x = 2
when x = 2/3
11
-7/3
**Calculator Tip : Decimal Fraction
Math, >Frac
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Like terms –
2x4x2x31x7x 232 2222 yx4x y2yx3x y7yx
8 )5 ( x2 )3 ( x
terms that have the same variables with the same powers
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Simplify these expressions:
2 )x44 ( x2 )3 ( x 2
8 )r2 ( 3r3
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Real World Applications
• See Note Sheet…
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You have 122 dollars from bagging at Dominick’s and you want to buy some DVD’s. If each DVD cost $13, write an expression to represent the amount of money you have left buying in DVD’s.
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You want to buy either scented lotion or bath soap for 8 people. The lotions are $6 for each and the soaps are $5 each. Write and expression for the total amount you must spend. Evaluate the expression when 5 people get the lotion.
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Write an expression for the total amount of juice in 15 cans if some hold 8 oz and some hold 12 oz. What is the total if 9 of the cans hold 8 oz?
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Practice…Try These on the Back
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HMWK!
• Worksheet 1.1-1.3: #s 7-15, 13-24
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Today’s Agenda!
• Collect Signed Syllabus
• Return Syllabus Scavenger Hunt
• HW Questions/Concerns
• Section 1.3 Notes– 1.3 Domino Worksheet
• Word Problems
• HA2 Pretest Tomorrow! (Bring #2 Pencil)
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Honors Algebra 2
1.3 Solving Linear Equations
Objectives:1. Solve simplified linear equations2. Solve linear equations that need simplifying3. Solve linear equation from real life
Vocabulary: solution, equation, identity equation, inconsistent equation
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PEMA – used to evaluate an expression
5- x w h e n3 ,2 x
4 x when,3
8-3)2(x
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Rules for Solving an Equation
3
2
1. Distribute2. Combine Like Terms3. Move variable to one side4. Isolate variable using inverse
operations5. Check
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Solving an equation – working backwards
2(x 3) - (3x + 2) - 4 -27
63
8-3)2(x
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(4/5)(x – 2) = 16
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2x + 5 = 7x – 16
3(x – 7) + 2x = -5(2x – 4)
2x + 8 = 5x – 2(x – 8)
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Infinite Solutions/No Solutions
• 4x – 3 = 2(2x – 9) + 4 • 7x + 5 – 3x = -8x + 5 + 12x
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Domino WKST!
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Real World Applicaiton
• See Note Sheet…
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Katie works at a restaurant. She earns $3 per hour base plus tips. She averages $12 in tip per hour. How many hours until she has earned $333?
3
2
22.2 hours
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A car salesman base salary is $21,000 plus 5% commission on sales. How much must he sell to earn $65,000.
3
2
$880,000 of cars
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The bill from your plumber is $134. The cost for labor was $32 per hour. The cost for material was $46. How many hours did the plumber work?
2.75 hours
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Honors Alg 2
The perimeter of a triangle is 35 feet. If the sides are 3x – 5, 2x – 3, and 15-x, what are its dimensions?
3
2
x = 7sides of 16, 13, and 8 units
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Assignment:• WS 1.1-3, #11-
23 on backside
Honors Alg 2