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AAlphabetized Topics Pages Pages
Area 32 Place Value 9, 10 Circumference 32 Properties 12 Comparing 23, 26
Proportions 22, 23, 24, 25, 26
Congruent Figures 35 Pythagorean Theorem
36 Converting 14, 15, 23 R.A.C.E. 41 Divisibility Rules 8 Range 11 Equations 37, 38, 39 Rates 25 Flow Charts Ratios 25, 26 Formulas 32, 33, 34 Rounding 10 Fractions
18, 19, 20, 21, 22, 23, 24
Scale Factor 35 Geometric Figures 30, 31 Similar Figures 35 Greatest Common
Factor (GF or GCD) 22
Slide Method 22 Inequalities 40 Substitution 29 Integers 18, 19 Surface Area 33 Ladder Method 22 Symbols 5 Least Common
Multiple (LCM or LCD)
22 Triangles 30, 36
Mean 11 Variables 29 Median 11 Vocabulary Words 43 Mode 11 Volume 34 Multiplication Table 6 Word Problems 41, 42 Order of
Operations 16, 17
Percent 23, 24, 25, 26 Perimeter 32
3
TTable of Contents
Pages Cheat Sheets 5 – 42 Math Symbols 5 Multiplication Table 6 Types of Numbers 7 Divisibility Rules 8 Place Value 9 Rounding & Comparing 10 Measures of Central Tendency 11 Properties 12 Coordinate Graphing 13 Measurement Conversions 14 Metric Conversions 15 Order of Operations 16 – 17 Integers 18 – 19 Fraction Operations 20 – 21 Ladder/Slide Method 22 Converting Fractions, Decimals, &
Percents 23
Cross Products 24 Ratios, Rates, & Proportions 25 Comparing with Ratios, Percents,
and Proportions 26
Solving Percent Problems 27 – 28 Substitution & Variables 29 Geometric Figures 30 – 31 Area, Perimeter, Circumference 32 Surface Area 33 Volume 34 Congruent & Similar Figures 35 Pythagorean Theorem 36 Hands-On-Equation 37 Understanding Flow Charts 38
Solving Equations Mathematically 39
Inequalities 40 R.A.C.E. – Answering Questions 41 Word Problem Cheat Sheet 42
Math Vocabulary 43 - End
4
5
MMathematic Symbols Cheat Sheet
+ Plus or Positive Line AS
– Minus or Negative Line segment AS
• * x Multiplied by Ray AS
÷ / Divided by Triangle ABC
= Equal to Angle ABC
≠ NOT equal
Angle B
Approximately equal to
Right angle
Congruent to
Perpendicular to
< Is less than
Perpendicular to
> Is Greater than ˚
%
Degree
Percent
≥ Is greater than or equal to Σ
Sum
Square root of x
≤ Is less than or equal to π Pi (3.14.159)
a/b a:b Ratio of a to be or a divided by b or the fraction a/b
!
xn
Factorial
Nth power of x
(a, b) Ordered pair Infinity
6
7
TTypes of Numbers – Cheat Sheet Prime Number – A number that has exactly two (2) factors
Zero (0) and One (1) are neither prime nor composite because they only have one factor (itself)
Composite Number – A number that has three (3) or more factors
Even Numbers ending in 0, 2, 4, 6, 8
Odd Numbers ending in 1, 3, 5, 7, or 9
8
Divisibility Rules
- Divisible by 2 – All even numbers are divisible by 2. Even numbers end in 0, 2, 4, 6, or 8 and all are divisible by 2.
- Divisible by 3 – If the sum of the digits is divisible by 3 so is the number. Add up the digits in the number, if the answer is divisible by 3 so is the number.
- Divisible by 4 – Odd numbers are NEVER divisible by 4. Odd numbers end in 1, 3, 5, 7, or 9, so any number ending with one of this will NOT be divisible by 4. Even numbers MAY be divisible by 4. To check, look at the last 2 digits of the number. If the number formed by the last 2 digits is divisible by 4, then the number is divisible by 4.
- Divisible by 5 – If a number ends in a 5 or a zero then it is divisible by 5
- Divisible by 6 – If a number is divisible by 2 AND 3, it is divisible by 6.
- Divisible by 9 – – If the sum of the digits is divisible by 9 so is the number. Add up the digits in the number, if the answer is divisible by 9 so is the number.
- Divisible by 10 – Numbers that are divisible by 10 end in with a zero.
9
PPlace Value Cheat Sheet
From Billions to Ten-millionths
10
PPlace Value & Rounding Comparing & Ordering Decimals
Rounding Rules Example Example 1. Underline the determined value 42.3 576.8 2. Draw an arrow to number to the right of underlined
number 42.3 576.8
3. 0 – 4 = Round Down (Keep the underline number the same)
a. All numbers to the left of underlined number stay the same
b. Underlined number stays the same c. All numbers to the right of underlined number
go to zero 4. 5 – 9 = Round Up (Underline number goes up 1)
a. All numbers to the left of the underline number stay the same
b. Underline number goes up 1 c. All numbers to the right of underlined number
go to zero
Round Down 42.3 ≈ 42.0
Round Up 576.8 ≈ 580.0
Comparing Decimal Rules 1. Line up the decimals using their decimal point ** If you do not see a decimal
point, it is at the end of the number Example = 423 = 423.0
2. Fill in zeros so that all numbers have the same place value
3. Compare each number in their “lanes” (from left to right)
4. Determine greatest to least or least to greatest
Billions Millions Thousand Ones . Decimals
Hundred B
illion
Ten-B
illions
Billions
Hundred-M
illions
Ten-M
illions
Millions
Hundred-
Thousand
Ten-T
housands
Thousands
Hundreds
Tens
Ones
.
Tenths
Hundredths
Thousandths
Ten-Thousandths
Hundred-
Thousandths
Millionths
.
.
11
MMeasures of Central Tendency: The Mean, Median, Mode, and Range
When finding the measures of central tendency the first step is to place the numbers in order from least to greatest.
Mean (Average): Add up a list of values in a set of data and divide by the number of values you have.
6, 4, 4, 3, 8 Step 1 Put in order from least to greatest 3, 4, 4, 6, 8 Step 2 Add up all the numbers 3 + 4 + 4 + 6 + 8 = 25 Step 3 Divide by the number of values you have 25 ÷ 5 = 5
Answer The mean is 5
Median (Middle): The middle value in a set of data when the values are written in order. If there are 2 values in the middle, find the mean of the two.
6, 4, 4, 3, 8 Step 1 Put in order from least to greatest 3, 4, 4, 6, 8 Step 2 Find the middle number
**If there are an odd number of data values 3, 4, 4, 6, 8
Answer The median is 4
6, 4, 4, 3, 8, 5 Step 1 Put in order from least to greatest 3, 4, 4, 5, 6, 8 Step 2 Find the middle number
**If there are an even number of data values then there will be two middle numbers
3, 4, 4, 5, 6, 8
Step 3 Find the mean of the two middle numbers 4 + 5 = 9 9 ÷ 2 = 4.5
Answer Median = 4.5 Mode (MOST): The value in a set of data that is repeated most often. A set of data could have no mode, one mode, or more than one mode.
6, 4, 4, 3, 8 Step 1 Put in order from least to greatest 3, 4, 4, 6, 8 Step 2 Find the number that occurs most often 3, 4, 4, 6, 8
Answer The mode is 4
Range: The largest number minus the smallest number
6, 4, 4, 3, 8 Step 1 Put in order from least to greatest 3, 4, 4, 6, 8 Step 2 Subtract the largest number minus the smallest number 8 - 3
Answer The Range = 5
12
PProperties 1. Commutative Property
Numbers can be added or multiplied in any order and the answer is still the same.
Examples: Commutative Property of Addition: 3 + 2 = 2 + 3 a + b = b + a
Commutative Property of Multiplication: 5(4) = 4(5) ab = ba 2. Associative Property
When adding OR multiplying 3 or more numbers, they can be grouped in any way and the answer remains the same.
Examples: Associative Property of Addition: (2 + 4) + 9 = 2 + (4 + 9) a + (b + c) = (a + b) + c
Associative Property of Multiplication: (5x4)x2 = 5x(4x2) (cd)e = c(de)
3. Identity Property of Addition When you add 0 to any number your answer is that number.
Examples: 5 + 0 = 5 0 + 1,253 = 1,253 a + 0 = a 0 + b = b 4. Identity Property of Multiplication
When you multiply any number by 1 your answer is that number.
Examples: 4 ∙ 1 = 4 1 x 746 = 746 1 x a = a b x 1 =b
5. Property of Zero
Any number multiplied by zero is zero.
Examples: 0 x 8 = 0 52 ∙ 0 = 0 a ∙ 0 = 0 0 x b = 0
6. Distributive Property
Multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products.
Examples: 2(3 + 4) = 2∙3 + 2∙4 a x (b + c) = (a x b) + (a x c)
13
Coordinate Plane Cheat Sheet This is a coordinate plane. Sometimes it is referred to as a coordinate graph. It has two axes and four quadrants. The two number lines form the axes. The horizontal number line is called the x-axis ( ) and the vertical number line is called the y-axis ( ). The coordinate plane is divided into 4 part called quadrants. See the figure to the right to see the location and name of each quadrant. You can describe points on this graph by using a coordinate pair. A coordinate pair has an x-coordinate and a y-coordinate and looks like this: (x, y). The center of the coordinate plane is called the origin. The origin has coordinates of (0, 0). Locating Points on a Coordinate Graph Locating points on a coordinate graph is very similar to playing the game Battle Ships. The coordinates tell you exactly where the point will be located. The x- and y-coordinates in the coordinate pair tell you which way to go and how far to go.
Follow the steps below:
It takes 2 moves to plot a point. 1.) Start at the origin 2.) The x-coordinate comes first and it
moves to the right or left. Right for positive numbers and left for negative. Example: (-3, 5) For the 1st move, the x-coordinate is -3 so starting at the origin, move 3 places to the left.
3.) The y-coordinate comes last & it moves up or down. Up for positive numbers and down for negative. Example: (-3, 5) You have already moved to the left 3 places, and for the 2nd move go up 5.
14
MMeasurement Conversion
15
Met
ric
Con
vers
ion
KKin
g H
en
ry D
ied
By
Dri
nki
ng
Ch
oc
ola
te M
ilk
Kin
g H
en
ry D
oe
sn’t
Usu
ally
Dri
nk
Ch
oc
ola
te M
ilk
E
xam
ple
Conv
ert
Compa
re
16
Ord
er o
f Ope
ration
s Ch
eat
Shee
t Th
ere
is a
spe
cific
orde
r in w
hich
mat
h pr
oblems
shou
ld b
e wo
rked
out
. I
t is c
alled
the
“ord
er o
f op
erat
ions
.” If
you
do
not
wor
k mat
h pr
oblems
in t
he c
orre
ct o
rder
, yo
u pr
obab
ly w
ill g
et t
he w
rong
ans
wer.
It
is
like
a st
ep-b
y-st
ep
recipe
to
work
out
a m
ath
prob
lem t
hat
will
lead
you
to
the
corr
ect
answ
er.
1st Pa
rent
hesis
& Gr
ouping
Sym
bols –
2nd E
xpon
ents
– 3r
d Multiply
or D
ivide
– 4t
h Add
or
Subt
ract
H
int:
Plea
se g
uys,
exc
use
my
dear
Aun
t Sa
llly
Exam
ples
::
P Pa
rent
hesis
1st
Do
the
pare
nthe
sis
and
all o
ther
gro
uping
symbo
ls.
P
aren
thes
is:
(6
+ 7)
B
rack
ets:
[(3
+ 2)
– (2-
1)]
Bra
cket
s us
ually
go
arou
nd a
set
of
pare
nthe
sis.
Wor
k inside
the
bra
cket
s firs
t un
til th
ere
is n
othing
lef
t to
do.
F
ract
ion
Bars
: =
Do
ever
ything
abo
ve t
he f
ract
ion
bar,
the
n ev
eryt
hing
below
the
Fra
ction
bar,
and
the
n divide
.
G Gr
ouping
sym
bols
such
as
brac
kets
or
a f
ract
ion
bar.
E Ex
pone
nts
2nd
Do
all e
xpon
ents
. 2
3 =
2·2·
2 =
8
4
2 =
4(4)
= 1
6
M
Multiply
3rd
Multiply
or d
ivide
from
LEF
T TO
RI
GHT
Somet
imes
you
multiply
firs
t, b
ut s
omet
imes
you
divide
firs
t. Yo
u de
cide
by
going
left
to
righ
t.
6·2
÷ 4
18
÷ 3·
5 M
ultiplying
com
es
3
÷ 4
Dividing
comes
6 ·
5 f
irst
12
f
irst
30
D
Divid
e
A
Add
4t
h
Add
or
subt
ract
fro
m
LEFT
to
RIGH
T
Somet
imes
you
add
first
, bu
t so
met
imes
you
sub
trac
t firs
t.
You
decide
by
going
left
to
righ
t.
4 +
2 –
5
7 –
3 +
3
Add
ing
comes
6 –
5
S
ubtr
acting
com
es
4 +
3
f
irst
1
f
irst
7
S Su
btra
ct
17
Exam
ples
of
using
the
prop
er o
rder
of
oper
ations
: Ex
ample
1:
(1
7 +
3) +
23 ÷
4 ·
2 (1
7 +
3) +
23 ÷
4 ·
2
20
+ 2
3 ÷
4 ·
2
20
+ 8
÷ 4
· 2
20 +
2 ·
2
20
+ 4
24
1st –
pare
nthe
sis
2n
d –
expo
nent
s 3r
d –
divide
4t
h -
multiply
5th –
add
Ans
wer
Exam
ple
2:
2[
6 +
(4 –
3)] –
5
2[
6 +
(4 –
3)] –
5
2
[6 +
1] –
5
2
[7] –
5
14 –
5
9
1st –
inne
r pa
rent
hesis
2n
d –
brac
kets
3r
d –
multiply
4th -
subt
ract
Ans
wer
Exam
ple
3:
– 3
+ 1
0
- 3
+ 1
0
– 3
+ 1
0
3
– 3
+ 10
0 +
10
0
1st –
grou
ping
sym
bols
(abo
ve &
below
fra
ction
bar)
2nd –
divide
3r
d –
subt
ract
4t
h –
add
Ans
wer
Exam
ple
4:
33 –
+ 2
33 –
+ 2
33
–
+ 2
33
- 5 +
2
27
– 5
+ 2
22 +
2
2
4
1st –
grou
ping
sym
bols
(ab
ove
& be
low
frac
tion
bar
) 2n
d –
divide
(finish
the
grou
ping
sym
bol )
3rd –
expo
nent
4t
h –
subt
ract
(b
ecau
se it
comes
first
) 5t
h -
add
Ans
wer
18
Inte
gers
ADDIN
G IN
TEGE
RS
Chip B
oard
Num
ber
Line
Ru
les
1. S
et u
p ch
ipbo
ard
by p
utti
ng c
hips
on
the
chip
boa
rd f
or t
he f
irst
par
t of
the
pro
blem
–
Rem
embe
r bl
ack
chip
s ar
e po
siti
ve a
nd r
ed
are
nega
tive
. 2.
Add
mor
e ch
ips
to t
he c
hip
boar
d fr
om t
he
seco
nd p
art
of t
he p
robl
em
3. C
alcu
late
the
val
ue o
f th
e ch
ip b
oard
REM
EMBE
R:
- Pa
ir u
p th
e bl
ack
and
red
chip
s.
- O
ne b
lack
chi
p &
one
red
chip
equ
al
zero
. -
Rem
ove
each
pai
r fr
om t
he b
oard
-
The
fina
l val
ue is
rep
rese
nted
by
what
is
left
on
the
boar
d.
1. F
ind
star
ting
poi
nt
2. A
DD
ING
mea
n yo
u’ll M
OVE
to
the
RIGH
T.
3. I
f yo
u co
me
to a
NEG
ATI
VE S
IGN
in
the
pro
blem
, you
mus
t CH
AN
GE
DIR
ECTI
ON
S.
Mov
e an
d se
e wh
ere
you
land
, tha
t is
yo
ur a
nswe
r.
1. P
ositive
+ Po
sitive
= P
ositive
Ju
st a
dd
A
nswe
r is
pos
itiv
e
2. N
egat
ive
+ Neg
ative
= Neg
ative
Ig
nore
the
sig
ns &
just
add
A
nswe
r is
neg
ativ
e
3. N
egat
ive
+ Po
sitive
= N
eg.
or P
os.
Positive
+ N
egat
ive
= Neg
. or
Pos
.
Igno
re s
igns
& s
ubtr
act
If
you
hav
e m
ore
nega
tive
s, t
he
answ
er is
neg
ativ
e
If y
ou h
ave
mor
e po
siti
ves,
the
an
swer
is p
osit
ive.
19
SUBT
RACT
ING
INTE
GERS
Ru
les
Easy
Met
hod
Num
ber
Line
#1
Num
ber
Line
#2
1. R
ewri
te t
he s
ubtr
acti
on p
robl
em a
s an
ad
diti
on p
robl
em.
Su
btra
cting
a nu
mbe
r is t
he s
ame
as a
dding
it’s o
ppos
ite.
2. N
ow ju
st f
ollo
w th
e ru
les
for
addi
ng
inte
gers
Ex
amples
: 7
– 5
=
is t
he s
ame
as 7
+ (-
5) =
Subt
ract
ing
5 is
the
sam
e as
add
ing
its
oppo
site
(-5)
. N
ow ju
st a
dd.
**
****
****
****
**
-6 –
(-3)
is
the
sam
e as
-6
+ 3
= Su
btra
ctin
g -3
is t
he s
ame
as a
ddin
g it
s op
posi
te (3
). N
ow ju
st a
dd.
****
****
****
****
-2 –
9 =
is t
he s
ame
as -
2 +
(-9)
=
Subt
ract
ing
9 is
the
sam
e as
add
ing
its
oppo
site
(-9)
. N
ow ju
st a
dd.
1. C
ross
the
line
the
n ch
ange
th
e si
gn.
2. T
hen
just
fol
low
the
rule
s fo
r ad
ding
inte
gers
.
Exam
ples
: 6
– 2
= Cr
oss
the
line
and
chan
ge t
he s
ign.
You
get
:
6
+ - 2
=
Now
fol
low
the
rule
for
add
ing,
**
****
****
****
****
****
**
-8
– (-
5) =
Cro
ss t
he li
ne &
ch
ange
the
sig
n. Y
ou g
et:
-8
+ (+ 5)
=
Now
fol
low
the
rule
for
add
ing.
**
****
****
****
****
****
**
-
4 –
7 =
Cros
s th
e lin
e an
d ch
ange
the
sig
n. Y
ou g
et:
-4 +
- 7 =
Fol
low
rule
s; a
dd.
1. F
ind
star
ting
poi
nt
2. S
UBT
RACT
ING
mea
n yo
u’ll M
OVE
to
the
LEFT
. 3.
If
you
com
e to
a
NEG
ATI
VE S
IGN
in
the
prob
lem
, you
m
ust CH
ANGE
DIR
ECTI
ONS.
4.
Mov
e an
d se
e wh
ere
you
land
, tha
t is
you
r an
swer
.
1. S
ubtr
acti
on m
eans
you
are
fi
ndin
g a
“dif
fere
nce”
. “D
iffe
renc
e” b
asic
ally
mea
ns
that
you
nee
d to
fin
d ou
t ho
w fa
r ap
art
the
num
bers
ar
e fr
om e
ach
othe
r.
2. P
ut b
oth
num
bers
on
the
num
ber
line
and
see
how
man
y fa
r ap
art
they
are
. 3.
Now
you
mus
t de
term
ine
whet
her
you
answ
er is
po
siti
ve o
r ne
gati
ve.
A
larg
e nu
mbe
r m
inus
a
smal
ler
num
ber
has
a po
siti
ve a
nswe
r.
A
sm
all n
umbe
r m
inus
a
larg
er n
umbe
r ha
s a
nega
tive
ans
wer.
Larg
e –
Small =
Pos
itive
Sm
all –
Lar
ge =
Neg
ative
Multiplying
Int
eger
s
Posi
tive
x P
osit
ive
= P
osit
ive
N
egat
ive
x N
egat
ive
= Po
siti
ve
Po
siti
ve x
Neg
ativ
e =
Neg
ativ
e
Neg
ativ
e x
Posi
tive
= N
egat
ive
Dividing
Inte
gers
Posi
tive
÷ P
osit
ive
= Po
siti
ve
N
egat
ive
÷ N
egat
ive
= Po
siti
ve
Po
siti
ve ÷
Neg
ativ
e =
Neg
ativ
e
Neg
ativ
e ÷
Posi
tive
= N
egat
ive
20
Fraction Operations
Adding & Subtracting Fractions 1. Make sure the denominators are the same. 2. If needed, we have to build each fraction so
that the denominators are the same. 3. Then, we add or subtract the numerators. 4. The denominator of your answer will be the
same denominator of the built-up fractions. 5. Reduce or simplify the answer, if required.
Examples: To add or subtract fractions with a common denominator, you simply omit Step#1.
1/3 + 1/3 = 2/3
Note: DO NOT add or subtract denominators!
When adding fractions with different denominators, we do all the steps.
1/2 + 1/3
3/6 + 2/6 = 5/6
Multiplying Fractions Here are the Rules for multiplying fractions...
1. You do not have to worry about a common denominator!
2. If possible, simplify before you multiply. 3. Multiply the numerators. 4. Multiply the denominators. 5. Simplify or reduce the resulting fraction, if
possible.
Examples:
32 x
54 =
158
Remember: You do not have to worry about a common denominator! Just multiply the numerators & then multiply the denominators!!
Multiplying Mixed Numbers 1. Change the mixed numbers into improper
fraction 2. If possible, simplify first. 3. Multiply the numerators. 4. Multiply the denominators. 5. If necessary, rewrite your answer as a mixed
number and check to be sure it is in simplest form.
Examples: 311 X 2
43 =
Change mixed numbers to improper fractions then solve.
34 X
411 =
1244 =
311 = 3
32
21
Dividing Fractions A Key Word to Understand
Reciprocal A reciprocal of a number is when the numerator and denominator switch places.
If the fraction is a mixed number, change it to an improper fraction first, then write its reciprocal. The product of any number and its reciprocal is always one.
Example:
The reciprocal of 43 is
34 .
The reciprocal of 51 is
15 .
Example of reciprocal with mixed numbers:
211 equals
23 and it’s reciprocal is
32
Steps for Dividing Fractions 1. Rewrite the division problem as a
multiplication problem, but multiply by the reciprocal of the number you were dividing by.
2. Simplify before you multiply. 3. Multiply the numerators. 4. Multiply the denominators. 5. Be sure your answer in its simplified or
reduced form. Change improper fraction to whole numbers or mixed numbers.
Example:
21 ÷
31
Rewrite as a multiplication using the reciprocal.
21 X
13 Now solve.
21 X
13 =
23 Simplified = 1½
Hints for Dividing Mixed Numbers 1. Change the mixed numbers into improper
fraction 2. Rewrite the division problem as a
multiplication problem, but multiply by the reciprocal of the number you were dividing by.
3. Simplify before you multiply. 4. Multiply the numerators. 5. Multiply the denominators. 6. Be sure your answer in its simplified or
reduced form. Change improper fraction to whole numbers or mixed numbers.
Example:
211 ÷ 2
32
Rewrite division problem with improper fractions.
23 ÷
38
Now rewrite as a multiplication using the reciprocal, and solve.
23 X
83 =
169
22
LLadder / Slide Method Greatest Common Factor or Divisor (GCF/GCD):
Highest number that divides exactly into two or more numbers Least Common Denominator or Multiple (LCM or LCD):
Smallest number that is a multiple of two or more numbers Smallest Number that is a multiple of two or more denominators
Simplified Fractions: Reduce a number to make as simple as possible. (Numbers only have a factor of one that is the same)
Step 1: Write the two numbers in a box Step 2: Find a factor that goes into both numbers Step 3: Divide both numbers
Step 4: Continue this process until both numbers only have a factor of 1 that is similar
GCF/GCD Multiply the left side
LCM/LCD Multiply the left side and the bottom numbers Simplified Fractions Bottom numbers become you simplified fraction
23
Frac
tion
s, D
ecim
als,
& P
erce
nts
C
hang
e a
. . .
T
o a
. . .
To
a . .
.
FFra
ctio
n D
ecim
al
Perc
ent
Div
ide
the
num
erat
or b
y th
e de
nom
inat
or.
Exa
mpl
e: ¾
wou
ld b
e 3
÷ 4
= 0
.75
Cha
nge
the
frac
tion
to
a de
cim
al t
hen
mul
tipl
y th
e de
cim
al b
y 10
0.
Exa
mpl
e: ¾
= 0
.75
The
n 0.
75 x
100
= 7
5%
Cha
nge
a . .
.
To
a . .
. T
o a
. . .
DDec
imal
Perc
ent
Frac
tion
Mul
tipl
y th
e de
cim
al b
y 10
0.
Exa
mpl
e: T
o ch
ange
0.3
82 to
a p
erce
nt ju
st
mul
tiply
by
100.
0.
382
x 10
0 =
38.2
%
If y
ou c
an r
ead
the
deci
mal
pro
perl
y yo
u ca
n w
rite
it
as a
fra
ctio
n. S
impl
ify
the
frac
tion
.
Exa
mpl
e: 0
.875
read
s 875
thou
sand
ths –
as a
frac
tion
that
wou
ld b
e 10
0087
5 -
whi
ch re
ads e
xact
ly th
e sa
me.
Now
sim
plify
you
r ans
wer
and
you
are
fini
shed
1000
875
=87
.
Cha
nge
a . .
.
To
a . .
. T
o a
. . .
PPer
cen
t
Dec
imal
Fr
action
D
ivid
e th
e pe
rcen
t by
100
. Ex
ampl
e: 7
5% w
ould
be
75 ÷
100
= 0
.75
So
75%
= 0
.75
Wri
te t
he p
erce
nt a
s a
frac
tion
ove
r 10
0 th
en s
impl
ify
the
frac
tion
.
Exa
mpl
e: 7
5% w
ould
be 10
075
. Si
mpl
ified
100
75=
¾
Find
ing
the
Perc
ent
of a
Num
ber
Find
ing
the
Frac
tion
of
a Num
ber
To fi
nd th
e pe
rcen
t of a
num
ber –
Mul
tiply
the
num
ber b
y th
e pe
rcen
t w
ritte
n as
a d
ecim
al o
r a fr
actio
n.
EExam
ple:
75%
of 4
0 .
75%
= 0
.75
so th
is w
ould
be
0.75
x 4
0 =
30 O
R s
ince
75%
=10
075
= ¾
then
¾ x
40
= 30
.
Mul
tiply
the
num
ber b
y th
e fr
actio
n or
if th
e fr
actio
n ca
n be
writ
ten
as
a te
rmin
atin
g de
cim
al th
en y
ou c
an a
lso
mul
tiply
by
the
frac
tion
writ
ten
as a
dec
imal
.
EEx
ampl
e:: ¾
of 2
8 w
ould
be
¾ x
28
= 21
O
R
0.
75 x
28
= 21
24
Cross Products
The Rule of Cross Products states that when you multiply the diagonals of 2 fractions they are equal.
You can see in the example that 15 x 3 = 45 and 5 x 9 = 45 or we could say 15 x 3 = 5 x 9
The Rule of Cross Products has truths that are helpful in solving for a missing part of 2 equivalent fractions, ratios or proportions.
EXAMPLE:
Because of the Rule of Cross Product we know that 15n = 18 x 10 or 15n = 180.
This can be solved algebraically but most prefer the quick and easy way below.
QUICK AND EASY SOLUTION
25
Ratios Rates & Proportions Ratio: A comparison between two different amounts. There are 3 ways to write ratios
8 to 3 8:3 A ratio is usually a part-to-part comparison, but it can be a part to whole comparison.
Example: The score was 15 to 4. There are two parts being compared - the score of one team being compared to the score of the other team.
Proportion: Two ratios that are equal to each other. Example:
Proportions are used when two things are being compared and one of the parts is missing. Example: Margaret knows that she can serve 7 people with 2 cans of green beans. She will be feeling 84 people at the luncheon. How many cans of green beans will she need to buy?
N = 24 cans
Rate: A ratio comparing 2 amounts measured in 2 different units. Example: The ratio below is comparing minutes to kilometers. These are two different units of measurement so this ratio is a rate.
Unit Rate: A unit rate is the amount for 1 item Example:
The car gets 32 miles per gallon of gasoline. This is a unit rate because we are talking about 1 gallon of gasoline
A proportion can be used to find a unit rate. Example: A bottle of shampoo cost $3.99 for 13.5 ounces. Find the unit rate.
N = about $0.30 per ounce
26
Com
pari
ng w
ith
Frac
tion
s, P
erce
nts,
Rat
ios,
and
Pro
port
ions
W
hat
is b
eing
com
pare
d?
Frac
tion
s:
Alwa
ys a
par
t to
who
le c
ompa
riso
ns.
Num
erat
or
pa
rt
Den
omin
ator
wh
ole
Perc
ents
: Al
ways
a par
t to
who
le c
ompa
riso
n.
The
perc
ent i
s the
par
t out
of 1
00.
E
xam
ple :
53%
5
3 is
the
part
out
of 1
00.
The
100
repr
esen
ts th
e w
hole
.
Ratios
: Us
ually
a par
t to
par
t co
mpa
riso
ns, b
ut m
ay b
e
Part
to
whole
com
pari
sons
.
- M
ost o
f the
tim
e 1
part
is b
eing
com
pare
d to
ano
ther
par
t -
Som
etim
es 1
par
t is b
eing
com
pare
d to
the
who
le
- Y
ou n
eed
to lo
ok a
t wha
t the
num
ber r
epre
sent
then
thin
k….
Are
thes
e se
para
te p
arts
or i
s one
a w
hole
?
Prop
ortion
s:
Alwa
ys c
ompa
ring
2 e
qual r
atios.
Use
d to
hel
p fin
d a
mis
sing
par
t whe
n th
ings
are
bei
ng c
ompa
red.
Ex
ampl
e:
Key
Wor
ds
“to”
A
ratio
usu
ally
use
s “to
”.
Look
for 2
thin
gs b
eing
com
pare
d.
“altog
ethe
r”
“Alto
geth
er”
usua
lly re
fers
to a
who
le.
“all”
“A
ll” u
sual
ly re
fers
to a
who
le.
“t
otal”
“Tot
al”
usua
lly re
fers
to a
who
le.
Ther
e ar
e 8
girls
and
12 b
oys
in M
rs. Gr
een’s
4th
hour
class
.
Find
the
ratio
of b
oys t
o gi
rls.
Thi
nk:
A r
atio
is a
par
t to
part
com
pari
son.
Ask
you
rsel
f: W
hat p
art a
re b
oys?
12
boys
Ask
you
rsel
f: W
hat p
art a
re g
irls?
8 g
irls
N
ow w
rite
your
ratio
with
the
boys
firs
t and
then
the
girl.
1
2 :8
or 1
2 to
8 o
r 12/
8
Find
the
frac
tion
of th
e st
uden
ts th
at a
re g
irls
.
Thi
nk:
A fr
actio
n is
a p
art t
o w
hole
com
pari
son.
Ask
you
rsel
f: W
hat p
art a
re th
e gi
rls?
8 g
irls
A
sk y
ours
elf:
Wha
t num
ber r
epre
sent
s the
who
le c
lass
? 20
stud
ents
Find
the
perc
ent o
f stu
dent
s th
at a
re g
irls
.
Thi
nk:
A p
erce
nt is
a p
art t
o w
hole
com
pari
son.
Ask
you
rsel
f: W
hat p
art o
f the
cla
ss a
re g
irls?
8 b
oys
A
sk y
ours
elf:
Wha
t num
ber r
epre
sent
s the
who
le c
lass
? 2
0 st
uden
ts.
Thi
nk:
You
just
foun
d th
e fr
actio
n of
the
stud
ents
.
C
hang
e th
e fr
actio
n to
a d
ecim
al to
a p
erce
nt.
= 0
.4 =
40%
\
27
So
lvin
g P
erc
en
t P
rob
lem
s Fi
ndin
g Pe
rcen
t of a
Num
ber -
- The
re a
re 2
com
mon
way
s –
usin
g a
prop
ortio
n or
usi
ng a
n eq
uatio
n.
Find
ing
the
Perc
ent o
f a N
umbe
r U
sing
a P
ropo
rtio
n U
sing
an
Equ
atio
n Th
ings
you
nee
d to
kno
w:
-
Rem
embe
r: A
per
cent
is a
par
t to
who
le c
ompa
rison
. Th
e pa
rt is
the
perc
ent a
nd th
e w
hole
is 1
00.
- A
per
cent
can
be
writ
ten
as a
frac
tion
out o
f 100
.
-
Thin
gs y
ou n
eed
to k
now
: -
Rem
embe
r: A
per
cent
is a
par
t to
who
le c
ompa
rison
. Th
e pa
rt is
the
perc
ent a
nd th
e w
hole
is 1
00.
- A
per
cent
can
be
writ
ten
as a
dec
imal
by
divi
ding
the
perc
ent
by
100.
-
72%
= 7
2 ÷
100
= 0.
72
How
it w
orks
: 1.
Fi
nd 2
5% o
f 68
2.
Writ
e a
part
to w
hole
pro
porti
on.
3.
Solv
e th
e pr
opor
tion
by m
ultip
lyin
g di
agon
als a
nd d
ivid
ing
by le
ft-ov
er. S
o, n
= 1
7.
4.
Ther
efor
e, 2
5% o
f 68
is 1
7.
5.
Hin
t: T
he “
of”
in th
e pr
oble
m “
25%
of 6
8” w
ill u
sual
ly b
e ho
oked
to
the
num
ber t
hat r
epre
sent
s the
who
le.
How
it w
orks
: 1.
Fi
nd 2
5% o
f 68
2.
In m
ath
“of”
usu
ally
alw
ays m
eans
mul
tiply
.
3.
So 2
5% o
f 68
wou
ld m
ean
to m
ultip
ly 2
5% b
y 68
.
4.
Firs
t, ch
ange
25%
to a
dec
imal
. 25
% =
25
÷ 10
0 =
0.25
5.
Rew
rite
the
orig
inal
pro
blem
as a
mul
tiplic
atio
n pr
oble
m, b
ut
mul
tiply
by
the
perc
ent w
ritte
n as
a d
ecim
al.
25%
of 6
8 0.
25 x
68
= 17
6.
Ther
efor
e, 2
5% o
f 68
is 1
7 O
ther
exa
mpl
es:
1.
11%
of 8
40
Sol
ve a
nd n
= 9
2.4
So
11%
of 8
40 =
92.
4
2.
32%
of 9
12
S
olve
and
n =
29
1.84
S
o, 3
2% o
f 912
is 2
91.8
4
Oth
er e
xam
ples
: 1.
11
% o
f 840
Rem
embe
r: 1
1% =
0.1
1 0.
11 x
840
= 9
2.4
So 1
1% o
f 840
= 9
2.4
2.
32
% o
f 912
Rem
embe
r: 3
2% =
0.1
1
0.32
x 9
12 =
291
.84
So, 3
2% o
f 912
is 2
91.8
4
28
Oth
er
Ty
pe
s o
f P
erc
en
t P
rob
lem
s -
So fa
r you
hav
e le
arne
d to
find
the
perc
ent o
f a n
umbe
r. Y
ou a
re fi
ndin
g th
e pa
rt w
hen
give
n th
e w
hole
. -
Som
etim
es y
ou a
re g
iven
the
part
aske
d to
find
the
who
le, o
r you
mig
ht b
e gi
ven
the
part
and
the
who
le a
nd a
sked
to fi
nd th
e pe
rcen
t. -
It is
impo
rtant
that
you
und
erst
and
the
wor
d us
ed in
per
cent
pro
blem
s.
Hin
ts:
a.
) “IS
” us
ually
repr
esen
ts th
e pa
rt.
b.
) “O
F” u
sual
ly re
pres
ents
the
who
le
c.)
Prop
ortio
ns a
re th
e ea
sies
t way
to so
lve
thes
e pr
oble
ms.
EX
AM
PLE
S
1.) 2
4 is
wha
t per
cent
of 3
2?
Fi
ll in
you
r pr
opor
tion:
So o
ur p
ropo
rtion
is
Sol
ve:
p =
75
A
nsw
er is
75%
2.) W
hat n
umbe
r is 6
2% o
f 50?
Fill
in y
our
prop
ortio
n:
So o
ur p
ropo
rtion
is
So
lve:
n =
32
Ans
wer
is 3
2.
3.) 2
8 is
35%
of w
hat n
umbe
r?
Fi
ll in
you
r pr
opor
tion:
So
our
pro
porti
on is
So
lve:
n =
80
A
nsw
er is
80
4.) 8
is w
hat p
erce
nt o
f 400
?
Fill
in y
our
prop
ortio
n:
So
our
pro
porti
on is
So
lve:
p =
2
A
nsw
er is
2%
29
Substitution & Variable Cheat Sheet Substitution is used to replace a value for a variable in an expression, equation, or formula.
Things you need to know: - What is a variable? A variable is a letter that represents a number in an expression or equation.
Examples: 5 + n = 2 ‘n’ is the variable f – g ‘f’ and “g” are variables
- What does it mean when a number is right next a variable? When a number is right next to a variable it means multiply.
Example: 3t =15 Because the ‘t’ is right next to the 3, this means ‘t’ multiplied by 3.
- What does it mean when 2 variables are right next to each other? When a 2 variables are right next to each other it means multiply.
Example: xy Because the ‘x’ and ‘y’ are right next to each other it means the value ‘x’ Is multiplied by the value of ‘y’.
EXAMPLES: a. Solve the problem if a + b if a = 3 and b = 5 1st Write out the problem a + b
3 + 5
8
2nd Show the substitutions - Take out the “a” and put in a 3. - Take out the “b” and put in a 5.
3rd Solve the problem b. Solve the problem 6n + 4 if n = 0 1st Write out the problem 6n + 4
6(0) + 4
0 + 4
4
2nd Show the substitutions - Take out the ‘n’ and put in a 0. - Be sure to show some type of multiplication
sign between the 6 and the 0. 3rd Solve the problem c. Solve the problem 10 – tu if t = 2 and u = 4 1st Write out the problem 10 – tu
10 – 2(4)
10 + 8
18
2nd Show the substitutions - Take out the ‘t’ and put in a 2. - Take out the ‘u’ and put in a 4. - Be sure to show some type of multiplication
sign between the 2 and the 4. 3rd Solve the problem
30
GGeometric Figures Polygons are two-dimensional closed geometric figures formed by line segments.
Two-Dimensional Figures Triangles have 3 sides and 3 angles.
- The sum of the measure of the inside angles of any triangles is always 180º. - Angle + Angle + Angle = 180º
Scalene Triangle Isosceles Triangle Equilateral Triangle
No congruent sides or congruent angles
At least 2 congruent sides and at least 2 congruent angles
3 congruent sides and 2 congruent angles
Right Triangle Acute Triangle Obtuse Triangle
Has a right angle (measure 90º)
All angles measure less than 90º
Has an angle that measures more than 90º
Quadrilaterals have 4 sides and 4 angles.
- The sum of the measure of the inside angles of any triangles is always 360º. - Angle + Angle + Angle + Angle = 360º
Quadrilateral Parallelogram Trapezoid Any closed figure with 4 sides
Opposite sides are congruent and parallesl
Exacly 1 pair of parallel sides
Rectangle Rhombus Square A parallelogram with 4 right angles
A parallelogram with 4 congruent sides
A parallelogram with 4 right angles and 4 congruent sides. (A rhombus with 4 right angles) (A rectangle with 4 equal sides.)
31
Other Common Two-Dimensional Figures
Pentagon Hexagon Octagon
A polygon with 5 sides and 5 angles
A polygon with 6 sides and 6
angles
A polygon with 8 sides and 8 angles
Three Dimensional Figures A 3-dimensional figure has length, width, and height. The surfaces may be flat or curved. A 3-dimensional figure with flat surfaces is called a polyhedron.
Prisms Triangular Prisms: - 5 faces (2 bases) - 9 edges - 6 vertices
Rectangular Prisms: - 6 faces (2 bases) - 12 edges - 8 vertices
Cubes: - 6 faces (2 bases) - 12 edges - 8 vertices
Pyramids
Triangular Pyramid: - 4 faces (1 base – it’s a triangle) - 6 edges - 4 vertices
Rectangular Prisms: - 5 faces (1 base – it’s a rectangle) - 8 edges - 5 vertices
32
AREA (Covering) - The number of square units it takes to cover a figure or an object.
PERIMETER (Distance Around)- The sum of the sides of straight sided figures.
Shape Example Area Equation/Formula
Perimeter Equation/Formula
Rectangle
A = l w P = S1 + S2 + S3 + S4
(P = 2l + 2w) Triangle
A = 2
bh OR
A = ½ bh P = S1 + S2 + S3
Parallelogram
A = bh P = S1 + S2 + S3 + S4
Trapezoid
A = ½ h(b + b) or
A = h(b + b) 2
P = S1 + S2 + S3 + S4
Circle
A = π r2 Circumference
C = πd or C = 2πr
The Circle
Circumference The distance around a circle.
Radius The distance between the center of the circle and any point on the circle
Diameter The distance across the circle through the center
Pi π 3.14 or 722
Key b = base h = height l = length w = width d = diameter
r = radius A = Area π 3.14 or 722 C = Circumference
33
Surface Area - Covering Total area of a three-dimensional object (Sum)
** Find the area of every side and add them together**
Shape Example Equation/Formula
Rectangular Prism
SA = 2 (lw + wh + hl)
Triangular Prism
SA = bh + (S1 + S2 + S3)H
Cylinder
SA = 2πr2 + 2πrh
Cone
SA = πr2 + πrl
Rectangular Pyramid
SA = s2 + 2sl
Sphere
SA = 4π r2
Key
b = base h = height r = radius A = Area C = Circumference
V = Volume B = area of base π ≈ 3.14 or 722 SA = Surface Area
34
Volume - Filling
The number of cubic units needed to fill the space inside the figure
Cubic Unit: A cube with edges of one unit long.
Shape Example Equation/Formula
Rectangular Prism
V = lwh Volume = length x width x height
Triangular Prism
V = Bh Volume = area of the triangle x height
Cylinder
V = Bh Volume = area of base x height
V = π r2 • h
Cone
V = ⅓B x h Volume = ⅓ x Area of Base x Height
V = ⅓ π r2 • h
Rectangular Pyramid
V = ⅓B x h Volume = ⅓ x Area of Base x Height
V = ⅓ l • w • h
Sphere
V = π r3
Volume = 4/3 x Pi x radius cubed
Key
b = base h = height r = radius A = Area
C = Circumference V = Volume B = area of base π ≈ 3.14 or 722
35
Congruent and Similar Figures Understanding Congruent Figures
The symbol for congruent
Congruent Figures Must Have -Same Shape -Same Angles -Same Size -Same Side Lengths
EXAMPLE: Triangles ABC DEF
Therefore, they have the…. - Same Shape - Same Angles - Same Size - Same Side Lengths
Understanding Similar Figures The symbol for similar
Similar Figures Must Have -Same Shape -Same Angles -A Scale Factor* -Same Side-to-Side Ratios**
EXAMPLE: Rectangles ABCD EFGH
Therefore, they have the…. - Same Shape - Same Angles - A Scale Factor* - Same Side-to-Side Ratios**
*So, what does Scale Factor mean?
The Scale Factor is the magic number that all of the side lengths of one figure are multiplied by to get all of the side lengths of new figure.
Because all of the side lengths of the smaller figure are all multiplied by 3, the scale factor is 3 or SF = 3.
In similar figures the sides that are in the same position are called corresponding sides. We call the angles that are the same in similar figures, corresponding angles.
**Then what are Side-to-Side Ratios?
In Rectangle ABCD, if you compare the ratio of the long side to the short side, it should be equal to the ratio of Rectangle EFGH’s long side to its short side.
Rectangle ABCD: Rectangle EFGH: Therefore, these rectangles have the same side-to-side ratios.
Corresponding Sides and Corresponding Angles In congruent and similar figures the sides that are in the same position in both figures are called corresponding sides. The angles that are the same in both congruent figures and similar figures are called corresponding angles.
EXAMPLES: In the rectangles above the short sides in rectangle ABCD corresponds with the short sides in EFGH. In the triangles above, angle A corresponds with angle D because they are both 50º.
36
Pythagorean Theorem
Pythagoras was a Greek philosopher and mathematician, born in Samos in the sixth century B.C. He and his followers tried to explain everything with numbers. One of Pythagoras’s most popular ideas is known as The Pythagorean Theorem. Things you need to know:
1. Right triangles have 2 legs and a hypotenuse. - The legs are the short side. - The hypotenuse is the long side that is opposite the right angle.
2. What is the Pythagorean Theorem
- The Pythagorean Theorem says that the sum of the legs squares of a RIGHT triangle equal the square of the hypotenuse. a2 + b2 = c2.
3. You can find the missing parts of a right triangle. Examples
A. Find the hypotenuse. a2 + b2 = c2 1. Write formula.
32 + 52 = c2 2. Show substitutions. 9 + 25 = c2 3. Solve.
36 = c2
4. Find the square root of c2.
c = 6 cm 5. The hypotenuse equals 6 cm.
B. Find the missing side. a2 + b2 = c2 1. Write formula.
a2 + 72 = 252 2. Look closely & then show
substitutions. a2 + 49 = 625 3. Solve.
– 49 – 49 4. Subtract 49 from each side.
a2 = 576 fdd 5. Find the square root
of a2. a = 24 m 6. The missing side is 24 m.
37
Solving Equations with Hands-On-Algebra Solving equations is all based on maintaining balance. A scale is used to represent that balance.
Example 1 Example 2 1. Set up your balance scale.
4x + 5 = 2x + 13
1. Set up your balance scale. Hint: The 2 outside the parenthesis means you must do the inside of the parenthesis twice. 2(x +3) = x + 8
2. There are pawns on both sides so to maintain balance, remove 2 pawns from each side.
2. When you lay it all out it looks like this.
3. Now you are left with 2x + 5 = 13. 3. There are pawns on both sides so to maintain balance, remove 1 pawn from each side.
4. There are cubes on both side. Now remove 5 from the cubes on each side.
4. Now you are left with x + 3 = 8
5. You are now left with 2x = 8 5. There are cubes on both sides. Now remove 6 from the cubes on each side.
6. If 2 pawns equals 8, then each pawn must equal 4. So, x = 4 (Hint: 8÷2)
6. Because you have all your pawns on one side and all of your cubes on the other you are finished. You are now left with x = 2.
7. Finally check your answer if x = 4.
4x + 5 = 2x + 13 Substitute: 4(4) + 5 = 2(4) + 13 Solve: 16 + 5 = 8 + 13 21 = 21 It checks.
7. Finally check your answer if x = 2.
2(x +3) = x + 8 Substitute: 2(2 + 3) = 2 + 8
Solve: 2(5) = 10 10 = 10 It checks.
38
Understanding Flow Charts
A flow chart is a visual diagram that shows each step in evaluating an algebraic expression or equation.
EXAMPLES: I. Just follow the rules and arrows.
a. b.
II. Flow charts can be created from expressions. HINT: ORDER OF OPERATIONS IS VERY IMPORTANT. Start with the variable. What do you do first? Next? Notice the difference in these two flow charts. AGAIN, ORDER OF OPERATIONS IS VERY IMPORTANT!!
a. 6n + 1 b. 6(n + 1)
Solve if n = 4.
Solve if n = 4.
Your answer is the same when using substitution with the original expression:
Solve if n = 4 6n + 1 6(4) + 1
25
Your answer is the same when using substitution with the original expression:
Solve if n = 4 6n + 1 6(4 + 1)
6(5) 30
III. Flow charts can be used to solve equations. 5n + 4 = 79 1. Create a flow chart for the equation. Since 79 is
what comes “OUT” put it in the last oval.
2. Work backwards. - Start at the “OUT”, the 79. - Undo adding 4 by subtracting 4 from 79. - Finally, undo multiplying by 5 by dividing 75 by 5. - So n = 15
3. Substitute your answer in the original equation to check your answer.
n = 15 5n + 4 = 79 5(15) + 4 = 79 75 + 4 = 79 79 = 79 It checks.
39
Solving Equations Mathematically A few hints to solve equations mathematically: - Remember the importance of keeping the equation “balanced” like with Hands-On-Algebra. - Think of “undoing” like with the flow charts. “UNDO” adding by subtracting. “UNDO” subtracting by adding. “UNDO” multiplying or dividing. “UNDO” dividing by multiply.
Examples: 1) 5 + 3g = 23 5 + 3g = 23
Think about the flow chart
What would you “Undo” first? - Undo adding 5 by subtracting 5. Remember to keep thing balanced by subtracting 5 from both sides. 5 + 3g = 23
-5 -5 3g = 18
What do you “Undo” next? - Undo multiplying by 3 by dividing by 3. Keep things balanced by dividing both sides by 3. 3g = 18
3 3 So, g = 6
2) 2w – 4 = 8 +4 +4 Add 4 to both sides 2w = 12 2w = 12 2 2 Divide both sides by 2 w = 6
3) -3 -3 Subtract 3 from both sides. = -2
5 = (-2)5 Multiply both sides by 5 n = -10
4) 22 + 3n = 6n + 4 -4 -4 Take 4 from each side. 18 + 3n = 6n -3n -3n Take 3n’s from each side. 18 = 3n 18 = 3n Divide both sides by 3. 3 3 6 = n
5) 6 - 5p = p + 30 -6 - 6 Take 6 from each side. -5p = p + 24 -p = - p Take 1p from each side. -6p = 24 -6p = 24 Divide both sides by -6. -6 = -6 p = -4
Remember to substitute and check your answers!!
40
IInequalities
Inequality Two values that are not equal (less than, greater than)
< Greater than > Less than
≤ Greater than or equal to ≥ Less than or equal to
≠ Not equal
Graphing Inequalities x < 4 y ≥ - 3
1. Locate the value for the variable
2. Mark the point with one of the following a. Closed Circle if symbol is ≥ or ≤ b. Open Circle if symbol is < or >
3. Determine which direction you will draw the arrow
a. Left If variable is smaller than the value
b. Right If variable is larger than the value
Solving Inequalities by Adding & Subtracting Addition & Subtraction Properties of Inequality: You can add or subtract the number to each side of an inequality and the problem stays balanced. n + 3 ≤ -4 n – 14 > 10 – 3 – 3 - Undo adding by subtracting + 14 + 14 - Undo subtraction by adding n ≤ -7 n > 24
Solving Inequalities by Multiplying & Dividing Multiplication & Division Properties of Inequality: You can multiply and divide each side of the inequality by the same number, BUT you must be careful about the directions of the inequality sign.
- IF you multiply or divide by a positive number the sign stays exactly how it was. - IF you multiply or divide by a negative number, the sign flips the opposite way.
– 1 ≤ 7 +1 +1 ≤ 8
≥ (8)2
n ≥ 16
1) Add 1 to each side. 2) Multiply both sides by 2.
Since you are multiplying each side by a positive number, the sign stays the same.
-3n + 4 > 16 – 4 – 4 -3n > 12
-3n < 12 -3 -3
n < -4
1) Subtract 4 from each side. 2) Divide both sides by -3.
Since you are dividing each side by a negative number you must switch the sign from > to <.
41
Correctly Answering a Question:
R Restate the question
You need to restate the question so that the person reading your answer knows what the question was asked.
A Answer all parts of the question.
Many questions have multiple parts, be sure to read, and re-read and answer all parts of the question
C Cite Evidence How do you know that this is the correct answer. Many times this can be shown in your work.
E Explain Explain the process you used to get the correct answer.
42
WWor
d Pr
oble
m C
heat
She
et
If y
ou s
ee t
hese
wor
ds in
a w
ord
prob
lem
the
n us
e...
Ad
dit
ion
(S
um
) S
ub
tra
ctio
n
(Dif
fere
nce
)
Add
Alto
geth
er
A
nd
B
oth
H
ow m
any
H
ow m
uch
M
ore
than
In
all
In
crea
sed
by
Pl
us
Su
m
To
geth
er
To
tal
A
re n
ot
C
hang
e
Dec
reas
ed b
y
How
man
y di
d no
t hav
e
Less
than
H
ave
left
Le
ft ov
er
H
ow m
any
mor
e
How
muc
h m
ore
D
iffer
ence
Few
er
Mu
ltip
lica
tio
n
(Pro
du
ct)
Div
isio
n
(Qu
oti
en
t)
B
y (d
imen
sion
s)
D
oubl
e (ti
mes
tw
o)
Tr
iple
(tim
es
thre
e)
Ea
ch g
roup
Gro
up
M
ultip
lied
by
O
f
Prod
uct o
f
Tim
es
Tw
ice
(tim
es tw
o)
Ea
ch g
roup
has
Hal
f (d
ivid
e by
2)
H
ow m
any
in e
ach
Sh
are
som
ethi
ng e
qual
Frac
tions
– d
ivid
e by
de
nom
inat
or
Pa
rts
Q
uotie
nt o
f
Sepa
rate
d
Split
Div
ided
by
43
VVoca
bula
ry C
heat
She
et
Ter
m
Def
init
ion
E
xam
ple
Abs
olut
e Va
lue
Dis
tanc
e fro
m z
ero
– al
way
s po
sitiv
e R
ead
– Th
e ab
solu
te v
alue
of a
# is
#.
|5| =
5
Acu
te (A
ngle
) A
ngle
less
than
90˚
A
dden
d N
umbe
rs b
eing
add
ed to
geth
er
Add
end
+ A
dden
d =
Sum
5
+ 4
= 9
Adj
acen
t (an
gles
) A
ngle
s ha
ving
com
mon
sid
es a
nd c
omm
on v
erte
x (c
ente
r poi
nt)
A
lgeb
raic
A
pro
blem
, tab
le, e
quat
ion
that
invo
lves
a v
aria
ble
4m
+ 7
= 2
4
Ana
lyze
Lo
ok a
t dat
a an
d in
terp
ret t
he re
sults
Ang
le
The
amou
nt o
f tur
n be
twee
n tw
o st
raig
ht li
nes.
Mee
t at
a ve
rtex
App
roxi
mat
ion
See
Est
imat
ion
See
Estim
atio
n
Arc
P
art o
f the
circ
umfe
renc
e of
a c
ircle
Are
a C
over
s (s
quar
e un
its)
Fo
r spe
cific
form
ulas
: S
ee F
orm
ula
Che
at S
heet
44
Asc
endi
ng
Goi
ng u
p fro
m s
mal
lest
to la
rges
t
Ass
ess
Eva
luat
e or
est
imat
e if
som
ethi
ng m
ay b
e tru
e or
fals
e gi
ven
cond
ition
s 5
+ 3
= 8
??
Tru
e
Ass
ocia
tive
Prop
erty
of
Add
ition
&
Mul
tiplic
atio
n G
roup
ing
sym
bols
can
be
mov
ed w
ithou
t the
ans
wer
ch
angi
ng
(4 x
3) x
2 =
4 x
(3 x
2)
(4 +
3) +
2 =
4 +
(3 +
2)
Ave
rage
S
ee m
ean
Bar
Gra
ph
Gra
ph u
sing
rect
angu
lar b
ars
Box
-and
-Whi
sker
S
how
s ou
tlier
s an
d m
edia
ns
Div
ides
dat
a in
to 4
par
ts
Biv
aria
te
Two
varia
ble
equa
tion
yy =
4xx
+ 3
Cal
cula
te
Sol
ve b
y ap
plyi
ng th
e fo
ur o
pera
tions
Cen
ti-
Circ
umfe
renc
e D
ista
nce
arou
nd a
circ
le
45
Coe
ffici
ent
A n
umbe
r use
d to
mul
tiply
a v
aria
ble
C
omm
utat
ive
Prop
erty
of
Add
ition
&
Mul
tiplic
atio
n
Mul
tiply
or a
dd in
any
ord
er w
ithou
t cha
ngin
g th
e an
swer
3
x 6
= 6
x 3
5 +
2 =
2 +
5
Com
plim
enta
ry
Ang
les
Two
angl
es th
at a
dd u
p to
90˚
Com
posi
te �
um�e
rs
Num
bers
that
has
mor
e th
an tw
o fa
ctor
s E
xam
ple
: 4, 6
, 8, 9
, 12
Com
pute
To
sol
ve
Con
eA
3-d
imen
sion
al o
bjec
t tha
t has
a c
ircul
ar b
ase
and
it co
mes
to a
poi
nt
Con
grue
nt
Sam
e m
easu
res
(ang
les,
leng
th, s
hape
, or s
ize)
Con
secu
tive
Num
bers
that
follo
w e
ach
othe
r in
orde
r with
out g
aps
20
, 21
, 22
, 23
…
Con
vert
To
cha
nge
from
one
mea
sure
men
t to
a di
ffere
nt
mea
sure
men
t 6
mm
= _
____
km
Coo
rdin
ate
Gra
ph
Gra
ph th
at c
onta
ins
an x
-axi
s an
d y-
axis
that
inte
rsec
t
Crit
erio
n �C
riter
ia�
Sta
ndar
ds o
r rul
es th
at m
ake
som
ethi
ng tr
ue o
r fal
se
�f a
clos
ed fi
gure
has
� s
trai
ght s
ides
it is
a
pent
agon
�
46
Cu�
e �
oot
The
num
ber m
ultip
lied
by it
self
3 tim
es th
at g
ives
the
perfe
ct c
ube
(See
Per
fect
Cub
e)
Cyl
inde
rA
3-d
imen
sion
al (3
-D) s
hape
that
has
two
cong
ruen
t an
d pa
ralle
l rou
nd fa
ces
�ec
a-
Pre
fix fo
r ten
s - 1
0 D
ecad
e –
10 y
ears
D
ecag
one
– 10
side
d fig
ure
Dec
i -
Pre
fix fo
r Ten
ths
- 0.1
0.
1
Dec
imal
A
ny n
umbe
r inc
ludi
ng w
hole
num
bers
and
num
bers
w
ith a
dec
imal
poi
nt.
9 or
17.5
Den
omin
ator
B
otto
m n
umbe
r in
a fra
ctio
n
Des
cend
ing
Ord
erin
g fro
m b
igge
st to
sm
alle
st
Dia
met
er
Dis
tanc
e ac
ross
a c
ircle
goi
ng th
roug
h th
e ce
nter
Diff
eren
ce
Ans
wer
to a
sub
tract
ion
prob
lem
M
inue
nd –
Sub
trahe
nd =
Diff
eren
ce
8 –
5 =
3
Dila
tion
P
olyg
on g
row
s or
shr
inks
but
kee
ps e
xact
ly th
e sa
me
shap
e (S
imila
r Fig
ure
– m
ust h
ave
a sc
ale
fact
or)
47
Dis
tri�
�tio
n �D
ata�
D
ata
and
how
ofte
n (fr
eque
ncy)
it o
ccur
s
Dis
tri�
�ti�
e �r
o�er
t�
The
num
ber o
n th
e ou
tsid
e of
the
pare
nthe
ses
is
dist
ribut
ed (m
ultip
lied)
to th
e nu
mbe
rs o
n th
e in
side
of
the
pare
nthe
ses
Di�
iden
d N
umbe
r bei
ng d
ivid
ed
Div
iden
d ÷
Div
isor
= Q
uotie
nt
24 ÷
8 =
3
Di�
isor
N
umbe
r div
idin
g
Div
iden
d ÷
Div
isor
= Q
uotie
nt
24 ÷
8 =
3
���a
tion
Pro
blem
with
an
equa
l sig
n
���i
�ale
nt
Equ
al
�s
timat
e ��
stim
atio
n�
App
roxi
mat
e an
swer
(Aro
und
the
sam
e nu
mbe
r)
3.4 ≈
3
Eval
uate
S
olve
the
prob
lem
!!!!!!
Even
N
umbe
rs e
ndin
g in
0, 2
, 4, 6
, and
8
Exam
ple:
2, 1
2, 1
4, 1
02
Even
t A
sin
gle
inci
dent
(occ
urre
nce)
Expo
nent
S
how
s ho
w m
any
times
you
mul
tiply
a n
umbe
r
Expr
essi
on
Pro
blem
with
out a
n eq
ual s
ign
4 · 5
48
Exte
rior �
n�le
A
ngle
mea
sure
men
ts o
utsi
de o
f a p
olyg
on w
hen
the
lines
are
ext
ende
d ou
tsid
e th
e sh
ape.
�a�t
or
Num
ber b
eing
mul
tiplie
d Fa
ctor
x F
acto
r =
Prod
uct
6 x
5 =
30
�lo�
��a
rt
Vis
ual d
iagr
am th
at s
how
s ea
ch s
tep
in e
valu
atin
g an
al
gebr
aic
expr
essi
on o
r equ
atio
n
�or�
ula
Rec
ipe
for s
olvi
ng a
spe
cific
type
of p
robl
em
Exa
mpl
e: A
= l
• w
�ra�
tion
Par
t of a
who
le
�re�
uen�
� H
ow o
ften
som
ethi
ng o
ccur
s (u
sual
ly in
a s
peci
fic ti
me
perio
d
�un�
tion
A re
latio
nshi
p be
twee
n in
puts
and
out
puts
of a
spe
cific
ru
le.
Eve
ry in
put w
ill pr
ovid
e an
out
put.
�re
ater
��a
n B
igge
r >>
�re
ates
t �o�
�on
�a
�tor
��iv
isor
� ��
���
��
��
Hig
hest
num
ber t
hat d
ivid
es e
xact
ly in
to tw
o or
mor
e nu
mbe
rs
�ex
a�on
6
side
d fig
ure
�
ori�
onta
l R
uns
from
left
to ri
ght
49
��p
oten
use
Th
e si
de o
f a ri
ght t
riang
le th
at is
opp
osite
the
right
an
gle
��
enti�
� pr
oper
t� o
� �
��iti
on
Add
ing
zero
to a
ny n
umbe
r kee
ps th
e nu
mbe
r the
sa
me
5 +
0 =
5
��en
tit�
�rop
ert�
o�
�ul
tipli�
atio
n
Mul
tiply
ing
by 1
to a
ny n
umbe
r kee
ps th
e nu
mbe
r the
sa
me
1 x
10 =
10
��pr
oper
�ra
�tio
n Fr
actio
n th
at h
as a
larg
er n
umbe
r in
the
num
erat
or th
an
in th
e de
nom
inat
or
�ne�
ualit
� Tw
o va
lues
that
are
not
equ
al (l
ess
than
, gre
ater
than
)
�n�e
ren�
e ��n
�er�
U
sing
dat
a an
d in
form
atio
n to
com
e to
a c
oncl
usio
n.
You
can
infe
r th
at C
oke
is th
e fa
vori
te d
rink
�n�in
ite
Goe
s on
fore
ver w
ith n
o en
d. N
ot fi
nite
�nte
�er
All
coun
ting
num
bers
, inc
ludi
ng z
ero
and
it’s
oppo
site
s Ex
ampl
e: -1
, 0, -
5, 7
, 250
�nte
rpre
t D
escr
ibin
g th
e m
eani
ng b
ehin
d th
e da
ta.
Of t
he 6
2 vo
tes,
11 p
eopl
e lik
e Pe
psi.
Inte
rsec
t W
hen
lines
, sha
pes,
or d
ata
over
lap
or c
ross
ove
r eac
h ot
her.
(Lin
es in
ters
ect o
r mee
t at 1
poi
nt.)
Inve
rse
Opp
osite
ope
ratio
n
Mul
tiplic
atio
n D
ivid
e D
ivis
ion
Mul
tiply
A
dditi
on
Sub
tract
Su
btra
ctio
n A
dd
50
Irr�t
��n�
� ���
�er
A d
ecim
al th
at c
anno
t be
writ
ten
as a
frac
tion
– It
goes
on
fore
ver w
ithou
t rep
eatin
g.
π ≈
3.1
4159
…
Is�s
ce�e
s �r
��n�
�e
Tria
ngle
with
two
equa
l sid
es a
nd tw
o eq
ual a
ngle
s
��te
Q
uadr
ilate
ral w
ith tw
o pa
irs o
f con
grue
nt s
ides
adj
acen
t to
eac
h ot
her
�e�s
t ���
��n
�
��t��
�e
��en
���n
�t�r
� ��
��
���
��
Sm
alle
st n
umbe
r tha
t is
a m
ultip
le o
f tw
o or
mor
e nu
mbe
rs
Sm
alle
st N
umbe
r tha
t is
a m
ultip
le o
f tw
o or
mor
e de
nom
inat
ors
�ess
���
n S
mal
ler
<<
Line
ar
Mak
es a
line
y
= m
x +
b
Low
est T
erm
s S
ee S
impl
ify
Mea
n A
vera
ge (a
dd a
ll nu
mbe
rs to
geth
er a
nd d
ivid
e by
how
m
any
item
s th
ere
are
in a
set
of d
ata)
Ex
ampl
e: 5
+ 5
+ 8
+ 1
2
4
Med
ian
Mid
dle
num
ber i
n a
set o
f dat
a w
hen
the
num
bers
are
pu
t in
orde
r fro
m le
ast t
o gr
eate
st.
**If
ther
e ar
e tw
o m
iddl
e nu
mbe
rs m
ust f
ind
the
mea
n of
the
two
num
bers
**
51
Mi��
i�
Mi�
ed �
�m�e
r Fr
actio
n w
ith a
who
le n
umbe
r and
a p
rope
r fra
ctio
n
Mod
e N
umbe
r tha
t occ
urs
the
mos
t ofte
n in
a s
et o
f dat
a
3,
3, 5
, 6, 6
, 6, 9
, 9
The
mod
e =
6
M��
ti��e
R
esul
t of m
ultip
lyin
g by
a w
hole
num
ber
Mul
tiple
s of 3
: 3, 6
, 9, 1
2…
�on
�Lin
ear
Not
a s
traig
ht li
ne
�on
�Ter
min
atin
� �
e�im
a�
A d
ecim
al th
at d
oes
not e
nd, a
nd m
ay o
r may
not
re
peat
4.
2596
3911
4286
9281
…
�e�
ati�
e N
umbe
r les
s th
an z
ero
�
ot �
��a�
V
alue
s ar
e no
t the
sam
e am
ount
≠≠
Num
erat
or
Top
num
ber i
n a
fract
ion
Obt
use
(Ang
le)
Ang
le g
reat
er th
an 9
0˚ b
ut le
ss th
an 1
80˚
52
O�t
agon
8-
side
d fig
ure
O��
Num
bers
end
ing
in 1
, 3, 5
, 7 a
nd 9
O�e
rat�o
n A
dd, S
ubtra
ct, M
ultip
ly, D
ivid
e +
–
�
÷
O��
os�te
S
ame
dist
ance
from
zer
o bu
t in
the
othe
r dire
ctio
n N
egat
ive →
Opp
osite
= P
ositi
ve
Posi
tve →
Opp
osite
= N
egat
ive
Or�
er o
� O�e
rat�o
ns
The
rule
s of
whi
ch c
alcu
latio
ns c
ome
first
in a
n ex
pres
sion
or e
quat
ion
(The
ord
er w
e so
lve
a pr
oble
m)
�l
ease
�u�
s ��
�use
��
�ea
r Aun
t �al
l�
Or�
ere�
�a�
rs
Two
num
bers
writ
ten
in p
aren
thes
es s
how
ing
the
x an
d y
coor
dina
tes
O
r�g�n
W
here
the
x-ax
is a
nd y
-axi
s in
ters
ect
Poi
nt =
(0,0
) A
lway
s st
art a
t the
orig
in w
hen
plot
ting
poin
ts
Out
lier
Val
ue th
at “l
ies”
out
side
the
othe
r set
of d
ata
**
Eith
er m
uch
larg
er o
r sm
alle
r tha
n th
e re
st o
f the
dat
a
Para
llel
Line
s th
at a
re a
lway
s th
e sa
me
dist
ance
apa
rt an
d ne
ver t
ouch
53
Para
llel�
�ra�
Q
uadr
ilate
ral t
hat h
ave
oppo
site
sid
es p
aral
lel a
nd
equa
l in
leng
th. O
ppos
ite a
ngle
s ar
e al
so e
qual
Pe�t
a���
Fi
ve-s
ided
pol
ygon
Per
= 1
Mile
s PE
R H
our
Per�
e�t
Par
t out
of 1
00
/100
10
0%
Per�
e�t �
e�re
a�e
The
amou
nt th
e pr
ice
of a
n ite
m w
ent d
own
from
the
orig
inal
�� �
eter
�i�
e t�
e �e
�rea
�e�
a��u
�t
��
t� �
� �
�� �
e�re
a�e
�� �
i�i�
e ��
t�e
�l�
�alu
e
��
���
� ��
� ��
���
�ert
t� a
�er
�e�t
a�e
���
� ��
� �
���
�e�
rea�
e
Per�
e�t �
rr�r
Th
e ap
prox
imat
e er
ror i
n da
ta
Per�
e�t �
��re
a�e
Th
e am
ount
the
pric
e of
an
item
wen
t up
from
the
orig
inal
�� �
eter
�i�
e t�
e i�
�rea
�e�
a��u
�t
��
t� �
� �
�� i�
�rea
�e
�� �
i�i�
e ��
t�e
�l�
�alu
e
��
���
� ��
� ��
���
�ert
t� a
�er
�e�t
a�e
���
� ��
� �
���
i��r
ea�e
54
Per�
e�t �
u�e
A w
hole
num
ber c
reat
ed b
y m
ultip
lyin
g it
by it
self
thre
e tim
es -
cubi
ng (n
3 ) a w
hole
num
ber
(Per
fect
cub
es: 1
, 8, 2
7, 6
4))
�� �
� ��
� ��
� ��
� �
����
��
� �
�� �
���
���
� ��
��
�� �
��
�� �
���
���
� ��
��
�� �
��
�� �
���
���
� ��
��
�� �
���
���
� ��
��
���
� ��
��
Per�
e�t �
�uar
e A
who
le n
umbe
r cre
ated
by
mul
tiply
ing
it by
itse
lf -
squa
ring
(n2 ) a
who
le n
umbe
r (P
erfe
ct s
quar
es:
1, 4
, 9, 1
6)
Peri�
eter
D
ista
nce
arou
nd a
n ob
ject
Per�
e��i
�ula
r Li
nes
that
form
a ri
ght a
ngle
Pi
����
or
π Po
lygo
n
Mul
ti-S
ided
clo
sed
figur
e
Mus
t Con
tain
all
stra
ight
sid
es
Popu
latio
n W
hole
gro
up fr
om w
hich
a s
ampl
e is
take
n
55
Po�i
ti��
Num
bers
to th
e rig
ht o
f zer
o on
the
num
ber l
ine
P���
i�t
Bas
ed o
n da
ta m
ake
an e
stim
atio
n of
som
ethi
ng th
at
mig
ht h
appe
n in
the
futu
re o
r will
be a
con
sequ
ence
of
the
curre
nt d
ata
P�i�
� A
num
ber t
hat c
an b
e di
vide
d ev
enly
by
only
one
and
its
elf
Exam
ple:
2, 3
, 5, 7
, 11,
13,
17…
P�i�
�
A s
olid
figu
re th
at h
as tw
o fa
ces
that
are
con
grue
nt (t
he
sam
e or
equ
al)
P�o�
a�ili
ty
The
chan
ce s
omet
hing
will
hap
pen
(the
likel
ihoo
d of
an
even
t tak
ing
plac
e
P�o�
u�t
Ans
wer
to a
mul
tiplic
atio
n pr
oble
m
Fact
or x
Fac
tor =
Pro
duct
5
x 4
= 20
P�op
o�tio
n Tw
o ra
tios
set e
qual
to e
ach
othe
r
Py�a
�i�
A
sol
id o
bjec
t whe
re:
B
ase
is a
pol
ygon
Sid
es a
re tr
iang
les
whi
ch m
eet a
t the
top
(Ape
x)
Pyt�
ago�
�an
���o
���
R
ight
Ang
le T
riang
le –
The
long
sid
e (h
ypot
enus
e)
squa
red
equa
ls th
e su
m o
f the
squ
ares
of t
he o
ther
two
side
s a
� � �
� � �
�
56
�ua
��ila
t��a
l Fo
ur s
ided
figu
re
�ua
litat
i��
Info
rmat
ion
(Dat
a) th
at d
escr
ibes
som
ethi
ng
�ua
ntita
ti��
Info
rmat
ion
(Dat
a) th
at c
an b
e �o
unt�
� or
��a
�u��
�
�ua
ntity
H
ow m
uch
ther
e is
of s
omet
hing
�
uoti�
nt
Ans
wer
to a
div
isio
n pr
oble
m
Div
iden
d ÷
Div
isor
= Q
uotie
nt
45 ÷
9 =
5
�a�
iu�
Dis
tanc
e fro
m th
e ce
nter
to th
e ed
ge o
f a c
ircle
�an
�o�
�a�
pl�
A s
elec
tion
that
is c
hose
n ra
ndom
ly (b
y ch
ance
– n
o pr
edic
tion)
�an
g�
The
diffe
renc
e be
twee
n th
e lo
wes
t and
hig
hest
val
ue
5, 1
2, 1
3, 1
5, 2
4 R
ange
= 2
4 –
5 =
19
�at
� R
atio
that
com
pare
s tw
o di
ffere
nt q
uant
ities
usi
ng
diffe
rent
uni
ts
Mile
s per
hou
r $
per g
allo
n
�at
ioA
com
paris
on o
f tw
o qu
antit
ies
by d
ivis
ion
Writ
ten
in 3
diff
eren
t way
s
Mile
s : H
our
Mile
s to
Hou
r M
iles /
Hou
r
57
Rat
iona
l Num
ber
Num
ber t
hat c
an b
e m
ade
by d
ivid
ing
one
inte
ger b
y an
othe
r Ex
ampl
e: 0
.5, 1
.73,
-15.
23, 5
/3
Rec
ipro
cal
Num
ber y
ou m
ultip
ly a
noth
er n
umbe
r to
get o
ne (1
)
Rec
tang
le
4 si
ded
figur
e w
ith ri
ght a
ngle
s an
d tw
o se
ts o
f equ
al
side
s
Rec
tang
ular
Pris
m
Sol
id o
bjec
t tha
t has
six
(6) s
ides
that
are
all
rect
angl
es
Rec
tang
ular
Pyr
amid
A
sol
id o
bjec
t whe
re:
B
ase
is a
rect
angl
e or
squ
are
S
ides
are
tria
ngle
s w
hich
mee
t at t
he to
p (A
pex)
Ref
lect
ion
An
imag
e or
sha
pe a
s it
wou
ld b
e se
en in
a m
irror
(re
flect
s ov
er a
n ar
ea)
Reg
ular
Pol
ygon
A
ll si
des
and
angl
es a
re e
qual
Rep
eatin
g D
ecim
al
A fr
actio
n th
at w
hen
writ
ten
as a
dec
imal
repe
ats
in a
pa
ttern
that
goe
s on
fore
ver
Exam
ple:
1/3
= 0
.333
3333
…
Rig
ht (A
ngle
) A
ngle
that
is e
xact
ly 9
0˚
58
Rig
ht P
rism
A p
rism
that
has
the
base
s th
at li
ne u
p on
e on
top
of
the
othe
r. (L
ater
al fa
ces
are
rect
angl
es)
Pris
ms
that
can
be
stac
ked
stra
ight
up
on to
p of
eac
h ot
her
Rot
atio
n A
circ
ular
mov
emen
t
Rou
nd
(� –
�) F
our o
r Les
s L
et it
rest
(5
– 9
) 5 o
r Mor
e R
aise
the
Sco
re
45.2
3 4
5
Scal
e Th
e ra
tio o
f the
leng
th o
f a m
odel
to th
e re
al th
ing
Scal
e D
raw
ing
A d
raw
ing
that
sho
ws
a re
al o
bjec
t with
acc
urat
e si
zes
but t
hey
have
bee
n re
duce
d or
enl
arge
d us
ing
a sc
ale
Scal
e Fa
ctor
Th
e m
agic
num
ber t
hat a
ll of
the
side
leng
ths
of o
ne
figur
e ar
e m
ultip
lied
by to
get
all
of th
e si
de le
ngth
s of
ne
w fi
gure
Scal
ene
Tria
ngle
Tr
iang
le w
ith a
ll th
ree
side
s ha
ving
diff
eren
t len
gths
Scat
ter P
lot
A g
raph
of p
lotte
d po
ints
that
sho
ws
the
rela
tions
hip
betw
een
two
sets
of d
ata
Pos
itive
Cor
rela
tion:
Up
to th
e rig
ht
Neg
ativ
e C
orre
latio
n: D
own
to th
e rig
ht
No
Cor
rela
tion:
Ran
dom
dot
s th
roug
hout
59
Se��
ence
Li
st o
f num
bers
or o
bjec
ts in
spe
cial
ord
er
Si�
ilar
A s
hape
is s
imila
r if:
Sam
e Sh
ape
S
ame
Ang
les
S
ame
Sid
e to
Sid
e R
atio
s
Sca
le F
acto
r �
Si�
�li��
R
educ
e a
num
ber t
o m
ake
as s
impl
e as
pos
sibl
e. (N
o ot
her n
umbe
r oth
er th
an 1
can
go
into
bot
h nu
mbe
rs.
Slo�
e H
ow s
teep
a s
traig
ht li
ne is
y =
mx
+ b
Sol�
tion
Ans
wer
to a
pro
blem
4 +
3 =
�
S��e
re
Circ
ular
3-D
sha
pe –
Lik
e a
ball
Squa
re
4-si
ded
poly
gon
that
has
all
four
sid
es o
f equ
al le
ngth
an
d eq
ual 9
0˚ a
ngle
s
60
Squa
re �
���
The
num
ber t
hat i
s m
ultip
lied
by it
self
that
giv
es y
ou
the
perfe
ct s
quar
e.
(See
Per
fect
Squ
are)
S�e�
a��
�ea
� A
plo
t whe
re a
ch d
ata
valu
e is
spl
it in
to a
“lea
f” (u
sual
ly
the
last
dig
it) a
nd a
“ste
m” (
the
othe
r dig
it)
Exam
ple:
32
= 3
(ste
m) a
nd 2
(lea
f)
S�ra
����
����
�e�
Line
- 18
0˚
Su��
���u�
���
R
epla
cing
a v
aria
ble
with
a n
umbe
r
Su�
A
nsw
er to
add
ition
pro
blem
A
dden
d +
Add
end
= Su
m
4 +
3 =
7
Su��
�e�
e��a
r�
Two
angl
es th
at a
dd u
p to
180
deg
rees
61
Sur�
a�e
�rea
To
tal a
rea
of a
thre
e-di
men
sion
al o
bjec
t
�
ee �
he��
�he
e� ��
r ��r
mu�
��
�a��
e N
umbe
rs o
r qua
ntiti
es a
rrang
ed in
row
s an
d co
lum
ns
�a
� P
erce
ntag
e of
the
cost
of a
n ite
m a
dded
to th
e to
tal
cost
�er�
��a�
���
�e�
��a�
D
ecim
al n
umbe
r tha
t has
dig
its th
at s
top
0.5
�ra�
���r
�a�
���
Mov
ing
a sh
ape
in a
diff
eren
t pos
ition
, but
it w
ill n
��
chan
ge s
hape
, siz
e, a
rea,
ang
les
or le
ngth
s.
(See
Rot
atio
n &
Ref
lect
ion)
�ra�
��a�
���
Mov
ing
a sh
ape,
with
out r
otat
ing
or fl
ippi
ng it
(Slid
ing)
�ra�
��er
�a�
A li
ne th
at c
ross
es a
t lea
st tw
o ot
her l
ines
�ra�
e���
� Fo
ur s
ided
figu
re w
ith o
ne p
air o
f par
alle
l sid
es
62
�ree
��a
�ra�
A d
iagr
am to
hel
p yo
u de
term
ine
the
prob
abilit
y of
an
even
t M
ultip
ly a
long
bra
nche
s
Add
alo
ng c
olum
ns
���
que
Lead
ing
to o
nly
one
resu
lt 4
+ 5
= 9
���
� O
ne –
sin
gle
item
One
Oun
ce
���
� �a�
e A
mou
nt p
er it
em (O
ne It
em)
Varia
ble
A le
tter t
hat r
epre
sent
s a
num
ber i
n an
equ
atio
n or
ex
pres
sion
5
+ x
= 15
x
is th
e va
riab
le
Varia
bilit
y H
ow c
lose
or f
ar a
part
a se
t of d
ata
is
Vert
ical
R
uns
up a
nd d
own
Vert
ical
Ang
les
Ver
tical
ang
les
are
angl
es th
at a
re o
ppos
ite e
ach
othe
r w
hen
two
lines
cro
ss
V
ertic
al a
ngle
s ar
e al
way
s co
ngru
ent
63
V�l�
�e
The
amou
nt o
f spa
ce a
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64
