Transcript Solution
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 1- 1
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
Whole Numbers
Standard Notation
Addition and Subtraction
Multiplication and Division; Rounding and Estimating
Solving Equations
Applications and Problem Solving
Exponential Notation and Order of Operations
Factorizations
Divisibility
Least Common Multiples
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1.1
STANDARD NOTATION
a Give the meaning of digits in standard
notation.
b Convert from standard notation to expanded
notation.
c
Convert between standard notation and word
names.
d Use < or > for to write a true sentence in a
situation like 6 10.
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Objective
a
Give the meaning of digits in
standard notation.
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Slide 1- 4
Place Value
A digit is a number 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 that names a
place-value location.
For large numbers, digits are separated by commas into groups
of three, called periods.
Each period has a name: ones, thousands, millions, billions,
trillions, and so on.
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Ones
Tens
Ones
Hundreds
Ones
Tens
Hundreds
Ones
Tens
Hundreds
Ones
Tens
Hundreds
Ones
Tens
Hundreds
Trillions
PLACE-VALUE CHART
Billions
Millions Thousands
Slide 1- 5
2
4
5
8
4
0
2
8
1
5
4
8
Tens
Ones
Hundreds
Tens
Ones
Hundreds
Tens
Ones
Hundreds
Tens
Ones
Ones
Hundreds
Ones
Tens
Hundreds
Trillions
PLACE-VALUE CHART
Billions
Millions Thousands
245 billions, 840 millions, 281 thousands, 548 ones
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Slide 1- 6
Example A
What does the digit 4 mean in each number?
Solution
1. 234,598
4 thousands
2. 456,901
4 hundred thousands
3. 24,355,567,222
4 billions
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Slide 1- 7
Objective
b
Convert from standard notation to
expanded notation.
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Slide 1- 8
Whole Numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, …
Natural Numbers: 1, 2, 3, 4, 5,…
Standard Notation: 34,123
Expanded Notation:
34,123 = 3 ten thousands + 4 thousands
+ 1 hundred + 2 tens + 3 ones
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Slide 1- 9
Example B
Write expanded notation for 5280 feet, the number of
feet in a mile.
Solution
5280 = 5 thousands + 2 hundreds + 8 tens + 0 ones
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Slide 1- 10
Example C
Write standard notation for 8 ten thousands +
4 thousands + 5 hundreds + 2 tens + 9 ones.
Solution
Standard notation is 84,529.
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Slide 1- 11
Objective
c
Convert between standard notation
and word names.
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Slide 1- 12
Example D
Write a word name for 123.
Solution
123 = one hundred twenty-three
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Slide 1- 13
Example E
Write a word name for 123,456,789.
Solution
123, 456, 789
One hundred twenty-three million,
four hundred fifty-six thousand,
seven hundred eighty-nine
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Slide 1- 14
Example F
Write standard notation.
Three hundred four million,
two hundred thirty-five thousand,
eight hundred eleven
Solution
Standard notation is 304,235,811
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Slide 1- 15
Objective
d
Use < or > for to write a
true sentence in a situation
like 6 10.
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Slide 1- 16
Order of Whole Numbers
For any whole numbers a and b:
1. a < b (read “a is less than b”) is true when a is to
the left of b on a number line.
2. a > b (“read a is greater than b”) is true when a is
to the right of b on a number line.
We call < and > inequality symbols.
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Slide 1- 17
Example G
Use < or > for
84
to write a true sentence: 84
88
94
94.
98
Solution
Since 84 is to the left of 94 on a number line, 84 < 94.
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Slide 1- 18
1.2
ADDITION and SUBTRACTION
a Add whole numbers.
b Use addition in finding perimeter.
c Convert between addition sentences and
subtraction sentences.
d Subtract whole numbers.
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Objective
a
Add whole numbers.
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Slide 1- 20
Addition of whole numbers corresponds to combining
or putting things together.
We combine two sets.
A set of 3
CD players
This is the resulting set.
A set of 5
CD players
A set of 8
CD players
The addition that corresponds to the figure above is
3 +
Addend
5
Addend
= 8
Sum
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Slide 1- 21
Example A
Add: 8456 + 2484.
Solution
1
1
8 4 5 6
+ 2 4 8 4
1 0 9 4 0
Add ones. We get 10 ones. Write the 0 in
the ones column and 1 above the tens.
This is called carrying, or regrouping.
Add tens. We get 14 tens. Write 4 in the
tens column and 1 above the hundreds.
Add hundreds. We get 9 hundreds. Write
9 in the hundreds column.
Add thousands. We get 10 thousands
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Slide 1- 22
Example B
Add: 3401 + 4387 + 9765 + 1356
Solution
1
3
4
9
+ 1
1 8
2
4
3
7
3
9
1
0
8
6
5
0
1
7
5
6
9
Add ones. We get 19. Write 9 in the ones
column and 1 above the tens.
Add tens. We get 20. Write 0 in the tens
column and 2 above the hundreds.
Add hundreds. We get 19. Write 9 in the
hundreds column and 1 above the
thousands.
Add thousands. We get 18.
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Slide 1- 23
Objective
b
Use addition in finding perimeter.
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Slide 1- 24
Perimeter
The distance around an object is its perimeter.
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Slide 1- 25
Example C
5m
5m
Find the perimeter of the
object shown.
The letter m denotes meters (a
meter is slightly more than 3 ft).
8m
8m
9m
Solution
Perimeter = 5 m + 8 m + 9 m + 8 m + 5 m
= 35 m
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Slide 1- 26
Example D
Find the perimeter of the figure shown.
18 yd
14 yd
Solution
24 yd
10 yd
10 yd
8 yd
Perimeter = 18 yd + 24 yd + 8 yd + 10 yd + 10 yd + 14 yd
= 84 yd
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Slide 1- 27
Objective
c
Convert between addition sentences
and subtraction sentences.
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Slide 1- 28
Subtraction of whole numbers applies to two kinds of
situations. The first is called “take away.”
A restaurant starts with 8 pies and sells 5 of them.
8
8–5=3
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Slide 1- 29
The minuend is the number from which another
number is being subtracted.
The subtrahend is the number being subtracted.
The difference is the result of subtracting the
subtrahend from the minuend.
8
Minuend
5
=
Subtrahend
3
Difference
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Slide 1- 30
Subtraction
The difference a b is that unique number c for
which a = c + b.
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Slide 1- 31
Example E
Write a related addition sentence: 9 – 4 = 5.
Solution
9–5=4
This number
gets added
9=4+5
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Slide 1- 32
Example F
Write two related subtraction sentences: 6 + 8 = 14.
Solution
6 + 8 = 14.
This addend gets
subtracted from
the sum.
6 = 14 8
6 + 8 = 14.
This addend gets
subtracted from
the sum.
8 = 14 6
The related subtraction sentences are 6 = 14 – 8
and 8 = 14 – 6.
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Slide 1- 33
The second kind of situation to which subtraction can
apply is called “how many more.”.
You have 2 notebooks, but you need 7. You can think
of this as “how many do I need to add to 2 to get 7?”
Need 7 notebooks
Have 2
notebooks
5 notebooks
What must be added to 2 to get 7? The answer is 5.
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Slide 1- 34
Objective
d
Subtract whole numbers.
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Slide 1- 35
Example G
Subtract: 8453 2311.
Solution
8453
2311
6142
Subtract ones.
Subtract tens.
Subtract hundreds.
Subtract thousands.
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Slide 1- 36
Sometimes we need to borrow.
Example H
Subtract: 5024 1956
Solution
11
4 9 1 14
5024
19 56
3 0 68
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Slide 1- 37
Example I
Subtract: 7000 – 2754.
Solution
6
–
9
9
10
7 0 0 0
2 7 5 4
4 2
4 6
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Slide 1- 38
1.3
MULTIPLICATION AND DIVISION ;
ROUNDING AND ESTIMATING
a Multiply whole numbers.
b Use multiplication in finding area.
between division sentences and
c Convert
multiplication sentences.
Divide whole numbers.
d
Round to the nearest ten, hundred, or
e thousand.
Estimate sums, differences and products by
f rounding.
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Objective a
Multiply whole numbers.
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Slide 1- 40
The multiplication 4 5 corresponds to this repeated
addition:
We combine 4 sets of 5 desks each.
5 desks
+
5 desks
+
5 desks
+
5 desks
Factor
Factor
Product
4 5 desks = 20 desks
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Slide 1- 41
Example A
Multiply: 5 652.
Solution
652
5
10
250
3000
3260
Multiply the 2 ones by 5: 5 2 = 10
Multiply the 5 tens by 5: 5 50 = 250
Multiply the 6 hundreds by 5: 5 600 = 3000
Add.
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Slide 1- 42
Instead of writing each product on a separate line, we
can use a shorter form.
Solution
2
1
652
5
3 2 60
Multiply the ones by 5: 5 (2 ones) = 10 ones = 1
ten + 0 ones. Write 0 in the ones column and 1
above the tens.
Multiply the 5 tens by 5: 5 (5 tens) = 25 tens, 25
tens + 1 ten = 26 tens = 2 hundreds + 6 tens.
Write 6 in the tens column and 2 above the
hundreds.
Multiply the 6 hundreds by 5 and add 2 hundred:
5 (6 hundreds) = 30 hundreds + 2 hundreds = 32
hundreds.
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Slide 1- 43
Example B
Multiply: 53 47.
Solution
1
2
53
47
371
2120
2491
Multiplying by 7
Multiplying by 40. (We write a 0 and multiply 53 by 4)
Adding
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Slide 1- 44
Example C
Multiply: 325 674.
Solution
3
6
1 3
2 2 7
1 9 5 0
2 1 9, 0
2
7
0
5
0
5
5
4
0
0
0
0
Multiplying 325 by 4
Multiplying 325 by 70
Multiplying 325 by 600
Adding
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Slide 1- 45
Example D
Multiply: 450 326.
Solution
326
450
1630 0
13040 0
1 4 6,7 0 0
Multiplying by 5 tens. (We write 0 and
then multiply 326 by 5.
Multiplying by 4 hundreds. (We write
00 and then multiply 326 by 4.
Adding
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Slide 1- 46
Objective b
Use multiplication in finding area.
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Slide 1- 47
The area of a rectangular region can be considered to
be the number of square units needed to fill it. Here is
a rectangle 4 cm (centimeters) long and 3 cm wide. It
takes 12 square centimeters (sq cm) to fill it.
1 cm
3 cm
1 cm
4 cm
In this case, we have a rectangular array of 3 rows,
each of which contains 4 squares. That is,
A = l w = 3 cm 4 cm = 12 sq cm.
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Slide 1- 48
Example E
An adult soccer goal is 24 feet wide by
8 feet high. Find the shooting area.
Solution
A=lw
= 24 ft 8 ft
= 192 sq ft.
8 ft
24 ft
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Slide 1- 49
Objective
c
Convert between division sentences
and multiplication sentences.
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Slide 1- 50
Division of whole numbers applies to two kinds of
situations. The first is repeated subtraction. Suppose
we have 20 notebooks in a pile, and we want to find
out how many sets of 5 there are. One way to do this is
to repeatedly subtract sets of 5.
1
How many
sets of 5
notebooks
each?
2
3
4
20
Dividend
5
=
Divisor
4
Quotient
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Slide 1- 51
We can also think of division in terms of rectangular
arrays. Consider again the pile of 20 notebooks and
division by 5. We can arrange the notebooks in a
rectangular array with 5 rows and ask, “How many are
in each row?”
We can also consider a rectangular array with 5
notebooks in each column and ask, “How many
columns are there?” The answer is still 4.
In each case, we are asking, “What do we multiply 5
by in order to get 20?”
5 ?? = 20
20 5 = ??
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Slide 1- 52
Division
The quotient a b, where b 0, is that unique number
c for which a = b c.
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Slide 1- 53
Example F
Write two related division sentences: 9 8 = 72.
Solution
9 8 = 72
This factor
becomes a divisor.
9 = 72 8
9 8 = 72
This factor
becomes a divisor.
8 = 72 9
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Slide 1- 54
Objective
d
Divide whole numbers.
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Slide 1- 55
Dividing by 1
Any number divided by 1 is that same
a
number: a 1 a.
1
Dividing a Number by Itself
Any nonzero number divided by itself
is 1: a 1, a 0.
a
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Slide 1- 56
Dividends of 0
Zero divided by any nonzero number
is 0: 0 0, a 0.
a
Excluding Division by 0
Division by 0 is not defined. (we agree
not to divide by 0.)
a
is not defined.
0
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Slide 1- 57
Example G
Divide by repeated subtraction: 24 7.
Solution
24
7
17
7
10
7
3
Subtracting 3 times
Remainder
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Slide 1- 58
Example H
Divide 4369 6.
Solution
728
6 4369
42
16
12
49
48
1
The answer is 728 R 1.
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Slide 1- 59
To do division of whole numbers:
a) Estimate.
b) Multiply.
c) Subtract.
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Slide 1- 60
Example I
Divide 9858 62.
Solution
159
62 9858
62
365
310
558
558
0
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Slide 1- 61
Example J
Divide 1223 4.
Solution
305
4 1223
12
023
20
3
The answer is 305 R 3.
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Slide 1- 62
Objective e
Round to the nearest ten,
hundred or thousand.
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Slide 1- 63
Example K
Round 57 to the nearest ten.
Solution
50
55
57
60
Since 57 is closer to 60, we round up to 60.
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Slide 1- 64
Example L
Round 52 to the nearest ten.
Solution
50
52
55
60
Since 52 is closer to 50, we round down to 50.
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Slide 1- 65
Example M
Round 55 to the nearest ten.
Solution
50
55
60
We agree to round up to 60.
When a number is halfway between rounding
numbers, round up.
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Slide 1- 66
Rounding Whole Numbers
To round to a certain place:
a) Locate the digit in that place.
b) Consider the next digit to the right.
c) If the digit to the right is 5 or higher, round up. If
the digit to the right is 4 or lower, round down.
d) Change all digits to the right of the rounding
location to zeros.
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Slide 1- 67
Example N
Round 7564 to the nearest hundred.
Solution
a) Locate the digit in the hundreds place, 5.
7564
b) Consider the next digit to the right, 6.
7564
c) Since that digit is 5 or higher, round 5 hundreds up
to 6 hundreds.
d) Change all digits to the right of the hundreds digit to
zeros.
7600
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Slide 1- 68
Example O
Round 88,696 to the nearest ten.
Solution
a) Locate the digit in the tens place, 9.
8 8, 6 9 6
b) Consider the next digit to the right, 6.
8 8, 6 9 6
c) Since that digit is 5 or higher, round 9 tens to 10
tens and carry the 1 over to the hundreds.
d) Change the digit to the right of the tens digit to
zeros.
8 8, 7 0 0
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Slide 1- 69
Objective
f
Estimate sums and
differences by rounding.
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Slide 1- 70
Example P
Mario and Greta are considering buying a new
computer. There are two models, and each has options
beyond the basic price, as shown below. Mario and
Greta have a budget of $1100. Make a quick estimate
to determine if the XS with a monitor, memory
upgrade to 80 gig and a printer is within their budget.
XS Model
Basic price: $595
Monitor: $220
Memory upgrade:
40 gig: $75
80 gig: $90
Printer: $120
LT Model
Basic price: $825
Monitor: $275
Memory upgrade:
80 gig: $110
Printer: included
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Slide 1- 71
Solution
First, we list the base price and then the cost of each
option. We then round each number to the nearest
hundred and add.
XS Model
Basic price: $595
Monitor: $220
Memory upgrade:
40 gig: $75
80 gig: $90
XS
$595
Monitor
$220
Memory
$90
Printer
+ $120
$600
$200
$100
+ $100
1000
Printer: $120
The price of the computer is within their budget.
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Slide 1- 72
Example Q
Estimate the difference by first rounding to the nearest
thousand: 8426 3840.
Solution
8426
3840
8000
4000
4000
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Slide 1- 73
1.4
SOLVING EQUATIONS
a Solve simple equations by trial.
b Solve equations like x + 28 = 54,
28 x = 168, and 98 ∙ 2 = y.
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Objective a
Solve simple equations by trial.
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Slide 1- 75
Solutions of an Equation
A solution is a replacement for the variable that
makes the equation true. When we find all the
solutions, we say that we have solved the
equation.
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Slide 1- 76
Example A
Solve x + 15 = 39 by trial.
Solution
We replace x with several numbers.
If we replace x with 22, we get a false equation: 22 + 15 = 39.
If we replace x with 23, we get a false equation: 23 + 15 = 39.
If we replace x with 24, we get a true equation: 24 + 15 = 39.
No other replacement makes the equation true, so the
solution is 24.
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Slide 1- 77
Example B
Solve: 1. 8 + n = 35
2. 54 9 = y
Solution
1. 8 + n = 35
(8 plus what number is 35?)
The solution is 27.
2. 54 9 = y
(54 times 9 is what?)
The solution is 486
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Slide 1- 78
Objective
b
Solve equations like
x + 28 = 54, 28 x = 168, and
98 ∙ 2 = y.
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Slide 1- 79
Solving x + a = b
To solve x + a = b, subtract a on both
sides.
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Slide 1- 80
Example C
Solve: t + 37 = 83.
Solution
t + 37 = 83
t + 37 37 = 83 – 37
t + 0 = 46
t = 46.
Subtracting 37 from both sides.
Check: t + 37 = 83
46 + 37 83
83
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true
Slide 1- 81
Example D
Solve: 8365 + x = 9301.
Solution
8365 + x = 9301
8365 + x 8365 = 9301 – 8365
x = 936
Subtracting 8365
from both sides.
Check: 8365 + x = 9301
8365 + 936 = 9301 True
The solution is 936.
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Slide 1- 82
Solving a x = b
To solve a x = b, divide by a on both sides.
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Slide 1- 83
Example E
Solve: 18 y = 1476
Solution
18 y = 1476
18 y = 1476
18
18
y = 82
Dividing both sides by 18
Check: 18 y = 1476
18 82 = 1476
1476 = 1476 True
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Slide 1- 84
Example F
Solve: 4032 = 56 y
Solution
4032 = 56 y
4032 = 56 y
56
56
y = 72
Dividing both sides by 56
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Slide 1- 85
1.5
APPLICATIONS AND PROBLEM SOLVING
a Solve applied problems involving addition,
subtraction, multiplication, or division of
whole numbers.
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Objective
a
Solve applied problems
involving addition, subtraction,
multiplication, or division of
whole numbers.
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Slide 1- 87
Five Steps for Problem Solving
1. Familiarize yourself with the situation.
a) Carefully read and reread until you understand
what you are being asked to find.
b) Draw a diagram or see if there is a formula that
applies to the situation.
c) Assign a letter, or variable, to the unknown.
2. Translate the problem to an equation using the letter or
variable.
3. Solve the equation.
4. Check the answer in the original wording of the problem.
5. State the answer to the problem clearly with appropriate
units.
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Slide 1- 88
Example A
The balance in Megan’s checking account is $712. She
uses her debit card to buy the power saw shown in the
ad. Find the new balance in her checking account.
Solution
1. Familarize. Visualize the situation.
We let M = the new balance in her
account.
Take away
$112
$712
New balance
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Now
$112.00
Slide 1- 89
continued example A
2. Translate.
Money
checking
Money
minus
712
spent
112
New
is
balance
=
M
3. Solve. This sentence tells us what to do. Subtract.
712 – 112 = M
712
600 = M
112
600
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Slide 1- 90
continued example A
4. Check. To check we can repeat our calculation.
We can also estimate:
712 – 112 700 – 100 = 600.
5. State. Megan has a new balance of $600 in her
checking account.
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Example B
The captain of a blimp is traveling between two cities
that are 745 miles apart. The first day he traveled 375
miles. How much further does he need to travel?
1. Familarize. Visualize the situation.
We let x = the remaining distance.
distance traveled
distance to go
375
x
started
2. Translate. Distance
Distance
Total
traveled
plus
to go
is
distance
x
745
375
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Slide 1- 92
example B continued
3. Solve. To solve the equation we subtract 375 from
both sides.
375 x 745
375 x 375 745 375
x 370
4. Check. Estimate: 750 – 380 = 370
5. State. The captain needs to travel 370 miles to the
next city.
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Slide 1- 93
Example C
What is the total cost of 6 calculators if each one
costs $36?
1. Familiarize. Make a drawing or visualize the
situation. Let T = total cost
$36
$36
$36
$36
$36
$36
2. Translate.
# of
calc.
6
cost of
times calc. is
Total
cost
36
T
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Slide 1- 94
continued example C
3. Solve. The sentence tells us what to do. Multiply.
36
6
216
4. Check. We can repeat the calculation or estimate.
40 6 = 240
5. State. The total cost of 6 calculators is $216.
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Key Words, Phrases, and Concepts
Addition (+)
Subtraction (–)
add
subtract
added to
subtracted from
sum
total
plus
difference
minus
less than
more than
increased by
decreased by
take away
how much more
missing addend
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Key Words, Phrases, and Concepts
Multiplication ()
multiply
multiplied by
Division ()
divide
divided by
product
times
of
quotient
repeated subtraction
missing factor
repeated addition
rectangular arrays
finding equal quantities
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1.6
EXPONENTIAL NOTATION AND ORDER
OF OPERATIONS
a Write exponential notation for products such
as 4 4 4.
b Evaluate exponential notation.
c
Simplify expressions using the rules for
order of operations.
d Remove parentheses within parentheses.
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Objective a
Write exponential notation for
products such as 4 4 4.
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Exponential Notation
4 4 4 4 4 we write as 4
5
5 factors
The 5 is called an exponent.
The 4 is the base.
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Example A
Write exponential notation for 777777.
Solution
Exponential notation is 76
6 is the
exponent.
7 is the base.
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Slide 1- 101
Objective b
Evaluate exponential notation.
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Example B
Evaluate: 84 and
104
Solution
84 = 8 8 8 8 = 4096
104 = 10 10 10 10 = 10,000
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Objective c
Simplify expressions using the
rules for order of operations.
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Rules for Order of Operations
1. Do all calculations within parentheses ( ),
brackets [ ], or braces { } before operations
outside.
2. Evaluate all exponential expressions.
3. Do all multiplications and divisions in order from
left to right.
4. Do all additions and subtractions in order from
left to right.
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Example C
Simplify: 24 3 4
Solution
There are no parentheses or exponents, so we start
with the third step.
all multiplications and divisions in
24 3 4 = 8 4 Doing
order from left to right
= 32
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Example D
Simplify: 20 4 3 (10 – 7)
Solution
20 4 3 (10 – 7) = 20 4 3 3
=533
= 15 3
=5
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Example E
2
2
6
(10
7)
2 8
Simplify:
Solution
62 (10 7)2 2 8 62 32 2 8
62 9 2 8
36 9 2 8
4 2 8
8 8
0
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Average
The average of a set of numbers is the sum
of the numbers divided by the number of
addends.
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Slide 1- 109
Example F
Mason has 4 math tests with scores of 76, 85, 92, and
91. Find the average of all four tests.
Solution
76 85 92 91
344
86
4
4
The average on Mason’s four tests is 86.
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Objective
d
Remove parentheses within
parentheses.
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Slide 1- 111
When parentheses occur within parentheses, we can
make them different shapes, such as [ ] (also called
“brackets”) and { } (also called “braces”). All of these
have the same meaning. When parentheses occur
within parentheses, computations in the innermost
ones are to be done first.
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Example G
Simplify: [32 (3 8) 2] (18 13).
Solution
[32 (3 8) 2] (18 13) [32 11 2] (18 13)
[32 22] (18 13)
10 5
2
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1.7 FACTORIZATIONS
a Determine whether one number is a factor of another,
and find the factors of a number.
b Find some multiples of a number, and determine
whether a number is divisible by another.
c Given a number from 1 to 100, tell whether it is prime,
composite, or neither.
d Find the prime factorization of a composite number.
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Objective a
Determine whether one number is a
factor of another, and find the factors
of a number.
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Slide 1- 115
Factor
In the product a b, a and b are factors.
If we divide Q by d and get a remainder of 0, then
the divisor d is a factor of the dividend Q.
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Example A
Determine by long division whether 12 is a factor of
3915.
Solution
326
12 3915
36
31
24
75
72
3
The remainder is not 0, so
12 is not a factor of 3915.
Not 0
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Example B
Find all the factors of 72.
Solution
Check sequentially the numbers 1, 2, 3, and so on, to
see if we can form any factorizations.
1 72
2 36
3 24
4 18
6 12
89
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Objective b
Find some multiples of a number, and
determine whether a number is divisible
by another.
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Multiples
A multiple of a natural number is a product of that
number and some natural number.
We find multiples of 2 by counting by twos:
2, 4, 6, 8, and so on. We can find multiples of 3 by
counting by threes: 3, 6, 9, 12, and so on.
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Example C
Multiply by 1, 2, 3,… and so on, to find 6 multiples of
seven.
Solution
17=7
3 7 = 21
5 7 = 35
2 7 = 14
4 7 = 28
6 7 = 42
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Divisibility
The number a is divisible by another number b if
there exists a number c such that a = b c. The
statements “a is divisible by b,” “a is a multiple of
b,” and “b is a factor of a” all have the same
meaning.
Thus,
15 is divisible by 5 because 15 is a multiple of 5 (15 = 3 5)
40 is divisible by 4 because 40 is a multiple of 4 (40 = 10 4)
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Example D
Determine whether 102 is divisible by 4.
Solution
25
4 102
8
22
20
2
Since the remainder is not
0 we know that 102 is not
divisible by 4.
Not 0
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Slide 1- 123
Objective c
Given a number from 1 to
100, tell whether it is prime,
composite, or neither.
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Slide 1- 124
Prime and Composite Numbers
A natural number that has exactly two
different factors only itself and one is called a
prime number.
The number 1 is not prime
A natural number, other than 1, that is not
prime is composite.
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Example E
Determine whether the numbers listed below are
prime, composite, or neither.
8
13
Has factors of 1, 2, 4 and
8, composite
Has only two factors 1 and
itself, prime
24
33
Has factors 1, 2, 3, 4, 6, 8,
12, 24, composite
Has factors 1, 3, 11, 33,
composite
85
97
Has 5 as a factor,
composite
Has only two factors 1 and itself,
prime
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A Table of Primes from 2 to 157
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47,
53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107,
109, 113, 127, 131, 137, 139, 149, 151, 157
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Slide 1- 127
Objective d
Find the prime factorization of a
composite number.
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Slide 1- 128
Example F
Find the prime factorization of 50.
Solution
25
a) Since 50 is even, it must have 2 as a factor. 2 50
5
b) Since 25 ends in 5, we know 5 is a factor. 5 25
Because 5 is prime, we can factor no further.
The prime factorization can be written
as 2 5 5 or 2 52
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Every number has just one (unique) prime
factorization.
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Example G
Find the prime factorization of 48 using a factor tree.
48
Solution
6
2
·
8
3 2
2
3
Had we begun with
different factors (2 ∙ 24, or
4 ∙ 12), the same prime
factorization would result.
·2
4
·2 · 2 = 48
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Example H
Find the prime factorization of 220.
220
Solution
22
2
11 5
10
2
220 2 2 5 11
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Example I
Find the prime factorization of 1424.
Solution We use a string of successive divisions.
89
2 178
178
2 356
356
2 712
1424 = 2 2 2 2 89
712
2 1424
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1.8 DIVISIBILITY
a Determine whether a number is divisible by
2, 3, 4, 5, 6, 8, 9, or 10.
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Objective a
Determine whether a number
is divisible by 2, 3, 4, 5, 6, 8,
9, or 10.
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By 2
A number is divisible by 2 (is even) if it has a
ones digit of 0, 2, 4, 6, or 8 (that is, it has an
even ones digit).
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Example A
Determine whether each of the following numbers is
divisible by 2.
1. 457
2. 3488
3. 3200
4. 7893
Solution
1. 457
is not divisible by 2; 7 is not even.
2. 3488
is divisible by 2; 8 is even.
2. 3200
is divisible by 2; 0 is even.
4. 7893
is not divisible by 2; 3 is not even.
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By 3
A number is divisible by 3 if the sum of its
digits is divisible by 3.
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Example B
Determine whether the number is divisible by 3.
1. 12
2. 96
3. 303
4. 374
Solution
1. 12
1+2=3
2. 96
9 + 6 = 15
3. 303
3+0+3=6
4. 374
3 + 7 + 4 = 14
Each is divisible by 3
because the sum of its
digits is divisible by 3.
The sum of the digits, 14, is not
divisible by 3, so 374 is not
divisible by 3.
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By 6
A number is divisible by 6 if its ones digit is
0, 2, 4, 6, or 8 (is even) and the sum of its
digits is divisible by3.
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Example C
Determine whether the number is divisible by 6.
1. 840
2. 90
3. 83
Solution
1. 840
is even, divisible by 2. Also 8 + 4 + 0 =
12, so 840 is divisible by 3, 840 is
divisible by 6.
2. 90
is even, divisible by 2. Also 9 + 0 = 9, so
90 is divisible by 3, 93 is divisible by 6.
3. 83
83 is not divisible by 6 because it is not
even.
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By 9
A number is divisible by 9, if the sum of its
digits is divisible by 9.
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Example D
Determine whether the number is divisible by 9.
1. 4824
2. 524
Solution
1. 4824
2. 524
4 + 8 + 2 + 4 = 18 and 18 is divisible by 9,
so 4824 is divisible by 9.
5 + 2 + 4 = 11 and 11 is not divisible by 9,
524 is not divisible by 9.
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By10
A number is divisible by 10, if its ones digit
is 0.
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Slide 1- 144
Example E
Determine whether the number is divisible by 10.
1. 4810
2. 1524
Solution
1. 4810
2. 1524
is divisible by 10 because the ones
digit is 0.
is not divisible by 10 because the ones
digit is not 0.
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By 5
A number is divisible by 5 if its ones digit is
0 or 5.
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Slide 1- 146
Example F
Determine whether each of the following numbers is
divisible by 5.
1. 340
2. 885
3. 6721
Solution
1. 340
is divisible by 5; because its one digit is 0.
2. 885
is divisible by 5; because its one digit is 5.
3. 6721
is not divisible by 5; because its one digit
is neither 0 nor 5.
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By 4
A number is divisible by 4 if the number
named by its last two digits is divisible by 4.
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Example G
Determine whether each of the following numbers is
divisible by 4.
1. 7732
2. 8453
Solution
1. 7732
2. 8453
is divisible by 4 because 32 is
divisible by 4.
is not divisible by 4 because 53 is
not divisible by 4.
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By 8
A number is divisible by 8 if the number
named by its last three digits is divisible by 8.
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Example H
Determine whether each of the following numbers is
divisible by 8.
1. 1264
2. 43,911
Solution
1. 1264
2. 43,911
is divisible by 8 because 264 is
divisible by 8.
is not divisible by 8 because 911
is not divisible by 8.
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1.9 Least Common Multiples
a Find the least common multiple, LCM, of
two or more numbers.
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Objective
a
Find the least common
multiple, LCM, of two or more
numbers.
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Least Common Multiple, LCM
The least common multiple, or LCM, of two
natural numbers is the smallest number that is
a multiple of both.
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Example A
Find the LCM of 40 and 60.
Solution
a) First list some multiples of 40 by multiplying 40 by
1, 2, 3, and so on:
40, 80, 120, 160, 200, 240, 280, …
b) Then list some multiples of 60 by multiplying 60 by
1, 2, 3, and so on:
60, 120, 180, 240, …
c) Now list the numbers common to both lists, the
common multiples: 120, 240, …
d) These are the common multiples of 40 and 60.
Which is the smallest? 120
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Method 1: To find the LCM of a set of numbers
using a list of multiples:
a) Determine whether the larger number is a multiple
of the others. If it is, it is the LCM. That is, if the
largest number has the others as factors, the LCM is
that number.
b) If it is not, check multiples of the largest until you
get one that is a multiple of each of the others.
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Example B
Find the LCM of 6 and 28.
Solution
1. 28 is the larger number, but it not a multiple of 6.
2. Check multiples of 28:
2 28 = 56
3 28 = 84
A multiple of 6
The LCM = 84.
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Example C
Find the LCM of 12 and 36.
Solution
1. 36 is the larger number and 36 is a multiple of 12,
so it is the LCM.
The LCM is 36.
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Method 2: To find the LCM of a set of
numbers using prime factorizations
a) Find the prime factorization of each number.
b) Create a product of factors, using each factor the
greatest number of times that it occurs in any one
factorization.
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Example D
Find the LCM of 35 and 90.
Solution
1. We write the prime factorization of each number.
35 = 5 7
90 = 2 3 3 5
2. a) We note that 2 3 3 5 does not contain the
other factorization 5 7.
b) To find the LCM of 35 and 90, we multiply
2 3 3 5 by the factor of 35 that it lacks, 7:
90 is a factor.
LCM = 2 3 3 5 7
35 is a factor.
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Example E
Find the LCM of 36 and 48.
Solution
1. Write the prime factorization of each number:
36 2 2 3 3
48 2 2 2 2 3
2. We multiply the factorization of 36, by any prime
factors of 48 that are lacking. In this case, 2 2.
48 is a factor.
LCM = 2 2 2 2 3 3
36 is a factor.
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Example F
Find the LCM of 15, 30 and 25.
Solution
1. Write the prime factorization of each number.
15 = 3 5
30 = 2 3 5
25 = 5 5
2. Consider 15 and 30, the factorization of 30
contains 15
235
Multiply by the factors of 25 that are still missing.
2 3 5 5 = 150
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