-Imaginary numbers: 1 ai la is a cal Ilumber Hnd i is the number whose square is -11; j 2 I: the sets of' Real numbers and Imaginary Ilumbers have no clements in COllllllon and arc.. 4.
Trang 1A COMPREHENSIVE OVERVIEW OF BASIC MATHEMATICAL CONCEPTS
NOTATION
{ } Brul"e.• indicate the beginning and end ofa set notation; when
listed elements or members mllst be separated by commas;
EXAMPLE: In A= {, 8, 16~ the 4, 8, and 16 arc called
elements or members 0 the set; sets are finite (ending, or
having a last element), unless olherwise indicated
In the middle ofa set indicates c(Jllfilluuti(J1I (JFa patte",;
EXAMPLE: 8 = {S, 10, 15, ,85, 90}
At the end of a set indicates an ill finite set that is, a set
with no last element;
EXAMPLE: C {3, 6 '1 12, }
Is a symbol which literally means '"such that
E Means j ' U member (~ror ;~
EXAMPLE: IfA={4, 8,12 1, then 12 E A because 12 i
E Means is 1101 u ",ember of or i.\' not all
EXAMPLE: If 8={, 4, 6, X}, then 3 8 because
in s
o Empty set or "ull 'te t : A sct containing no elements
members, but which is a subset of all sets;
as : }
e Means is a slibset of; also may be written as ~
(l Means is 11M a ."b •et ot; also may be written as
AeB Indicates that every element of set A is
(Jf s
EX1MPLE: If A=1) } and 8=: I, 3 S, 6,
because the 3 and 6 which are in set A are also in set 8
2" Is the IIl1l1/ber of slibsef.,· ".1" a set when II equals
number 01' elements in that s
E."<AMPLE: If A-{, 6:, then set A has 8
because A has 3 clements and 23=
OPERATIONS AuB Indicates the /llIi(J1I of
u UNION
set A with sct 8; every
clement of this set is EITH ER
an element of set A, OR an
' Iement 01' set B; that is, to lorm
the union of two sets, put all of
the clements of the two sets
together into one set, making
sure not to write any clemcnt more than oncc;
EXAMPLE: II' A={ ,4 6, 8, 10, 121 and B={3,6,9, 12,
15,18:' then: Au8={2, 3, , 6, 8, 9,10,12, 5 18
A ns~t I~d~~i i ::~ ~ e~h~:i~~':.~~i:'~~e~: · n INTERSECTION
01' this set is als
SCI A AND set 8 ; that is,
the intersection of two scts,
only
found in 80TH of the
EXAMPLE: If A=:2 4 6 8
_ 10, 12: and 8 ={3 6, 'J 12, 15, 18}, Ihen i\nB=16, 12}
A Indicates the ('oll/plell/em ofset
A; that is, all clements in the COMPLEMENT
universal set which arc NOT in SET
set A;
EXAMPLE: If the universal s
A ~ { O , 12 3 f,
A = :-I ,-2, -3, -4, f
PROPERTIES
A=B If all of the elements in set i\
elements in set 13 are also in set A,
have to be in the s
EXAMPLE: If A={, 10} and 8={ 10, j, then i\=
II(A) Indicates the IIl1mber ()f e1emellf it, .,et A, i.e.,
, EXA,~PLE: Ir A={2 4 6 then n(A)=
A-B Means is eqllivalellt to;
have
clcments thcmselves may
EXAMPLE: If A {2 , 6) and B=:6 12, 18 j, then A-8
because II(A) = 3 and 11(8) = 3
AnB=0 Indicates Ji •j()illf sets
EXAMPLE: IfA=: 3, 4,5 land 13; j7, 8 9), then An 8
because there are no common clements
CLOSURE
oa + h is a real number; when you add 2 real numbers, the result is also a real number
EXAMPLE: 3 and 5 are both real numbers; 3 + S= and the slim 8 is also a real number
oa - b is a rca I number; when you subtract 2 real numbers, the result is also a real number
EXAMPLE: 4 and II are both real numbers; 4- II =-7 and the diHerenee, -7, is also a real number
o(a)(b) is a real number; when you multiply 2 real numbers, the result is also a real number
EXAMPLE: 10 and -3 are both real numbers;
(10)(-3) = -30 and the product, ·-30, is also a real number
oalb is a real number when b>,O; when you divide 2 real numbers, the result is also a real number unless the denom
inator (divisor) is zero
EXAMPLE: -20 and 5 arc both real numoers; -20/5 =-4 and the quotient, -4, is also a real number
COMMUTATIVE
oa + b = b + a; you can add numbers in either order and get the same answer
EXAMPLE: 9+ 15=24 and 15 +9 24 so 9+ 15= 15 +9
o(a)(b)=(b)(a); you can multiply numbers in either order and get the same answer
EXAMPLE: (4)(26) = 104 and (26)(4) ~· 104, so (4)(26) = (26)(4)
oa-b b-a; you cannot subtract in any order and get the same answer
EXAMPLE: 8 2=6 but 2 - 8= 6 There is no commuta
tive property for subtraction
oa I b b I a; you cannot divide in any order and get the same answer
EXAMPLE: 8 12 = 4, but 2/8=.25, so there is no cOlllmu
tative property for division
ASSOCIATIVE o(a+b)+c=a+(b+c); you can group numbers in any arrangement wh':l1 adding and get the same ans\ver
EXAMPLE: (2+5)+9 =7+9= 16and2+(5+9) =2+ 14= 16
so (2 -15) 1 =2+ +9)
o(ab) e = 8 (be): you can group numbers in any arrangement when multiplying and get the same answer
EXAMPLE: (4x5) 8 =(20)8= 160 and 4(Sx 8) = 4(40) - 11i0,
so (4xS)8 = 4(5 x8 )
oThe assuciative property dues 1101 lI ' or k Cor subtraction or division
EXAMPLES: (10-4)-2=6-2=4but 10-(4-2)= 10-2-8;
fix division, (12 / 6) / 2 (2) / 2 = I, but 12 /(6/2) = 12/3 : 4
Notice that these answers are not the samc
IDENTITIES
oa+ 0= a; 7ero is the identity for addition because adding Zero does not change the original number
EXAMPLE: 9+0=9 and (H 9 9
oa(I)=8; one is the identity lor multiplication because multi
plying by one does not change the original number
EXAMPLE: 23 (I )=23 and (I )23 = 23
oldentities for suotraction and division become a problem It
is true that 45-0=45, but 0-45= 45, not 45 This is also the case for division because 4/1 =4 but 1 /4 = 25 , so the identities do not hold when the numbers are reversed
INVERSES
°a +(-a)=O; a number plus its additive inverse (the number with the opposite sign) will always equal zero
EXAMPLE: 5+ S) = O and (-5) r = 0 The exception is zero because 0 + 0 = 0 already
oa(l/a)= I; a number times its multiplicative inverse or reciprocal (the number written as a fraction and flipped) will always equal one
EXAMPLE: S(1 /5) = I The exception is zero because zero cannot be multiplied by any number and result in a product
of onc
DISTRIBUTIVE PROPERTY oa(b+c)=ab + ac or a(b -c) = ah -ac; each term in the parentheses must be multiplied by Ihe term in Cronl 01' the parentheses
EXAMPLE: 4( S + 7) 4( 5 ) + ( 7) = 0 + 28 = 8
This is a simple example and the distributive property is not required 1 0 Cind the answer When the problem involves
a variable owever, the distributive property is a nece ity
E."<AMPLE: 4(5a+7) 4(5al+4(7) =10a+28
PROPERTIES OF EQUALITY oReflexive: a = a; both sides of the equation are identical
EXAMPLE: 5+ k = 5 +k
oSymmetric: Ifa = h, then b = a This property a110\\ s ) OU to exchange the two sides of an equation
EX4MPLE: 4a-7=9- 7a+ 15 becomes 9 7a+ 15 =4a 7 oTransitive: If a=b and b=e, th~n a = e Thi:; pro erty allows y u to connect statemcnts whieh are each equal to the same comlllon statement
EXAMPLE: Sa 6 = 9k and 9k = a 12; you can eliminate the COllllllon term 9k and COllnect the 1()lIowing into olle equation: Sa (, = a + 2
oAddition Property of Equality: If a=b, then a +c=h+c This property allows you to add any number or algebraic term
to any equation as long as y u add it to hoth sides to keep the equation truc
EXAMPLE: 5= 5; if vou acid 3 to one sidc and nut the other the equation become; 8 = 5 which is "lise bill iI' you add 3
to botlt sides, you get a true equaliun X X Also 5a+4= 14 becomes 5a+4+( 4)= 14+( 4) ifYllU add 4 to bOlh sides This results in the cquation Sa 10
o Multiplication Pl'Operty of Equality: If a= b, then ac = be when c O This property allows you to multiply both sides
of an equation by any nonzero value
EXAMPLE: If 4a= 24, then (4a)( 2S ) = ( 24)(.25) and
a = Ii Notice that both ,ides of the = were multiplied by 25
SETS OF NUMBERS
- - - , DEFINITIONS
• Natural or Counting numhers: : 1 2 ], 4, , I I 12 :
• \Vholc numbers: {O, 1,2 3 , In, 11 12 13 : oIntegers: { , -4, -3, -2, -I , n, I 2 3 4, } oRational numbers: :p/ql p and q arc integers q>'O): the
sts of Natural numbers, Whole numbers, and Integers as well as numbers which can be ritten as proper or impro er fractions, arc all subsets 01" thc set of Ratonal numbcrs olrrational numbers: {x 1 x is a real lIumber but is not a Rational number}: thc s ts of R ~ l tio na l n mbers and Irrational numbers have no clements in common a d are
therefore, disjoint sets oReal numbers: :x1 x is Ihe coordinate oC p int on a n mber line }; the union of the set llf Rational numhers with the SCI
of Irrational numbers equals the SCI of Re l numbers
-Imaginary numbers: 1 ai la is a cal Ilumber Hnd i is the number whose square is -11; j 2 I: the sets of' Real numbers and Imaginary Ilumbers have no clements in COllllllon and arc therefore disjoint sets
oComplex numbers: :a+bil a and b arC Real numbers and
i is the number whose square is -I:; the SCI of Re l numbers and the set of Imaginar:v numbers arc both subsets of the set of Complex numbers
E XA MPLES: 4 + 7i and J-2i arc Comple, numbers
COMPLEX NUMBERS
Numbers Rational
~
Trang 2
OPERATIONS OF REAL
NUMBERS
• otal or sum is the answer to an addition problem The
numbers added are called addends
EXAMPLE : In S + 9 ~ 14, Sand 9 arc addends and 14 is the
total r slim
• Differencl' is the answer to a subtraction problem The number
subtracted is called the subtrabend The number t,'om which
the subtrahend is subtracted is called the minuend
EXAM P LE : In 2S - 8~ 17, 25 is the minuend, 8 is the
subtrahend, and 17 is the difference
• Product is the answer to a multiplication problem The
numbers multiplied are each called a factor
EXAMPLE: In 15 x 6 ~90, 15 and 6 arc factors and 90 is
the product
• Quotient is the answer to a division problem The number
being divided is called the dividend The number that you
arc dividing by is called the divisor If there is a number
remaining aller the division process has been completed,
that number is called the remainder
EXAMPLE : In 45 : 5~9, which may also be written as 5)45
or 4S 15, 45 is the dividend, 5 is the divisor and 9 is the quotient
• An exponent indicates the number of times the base is
multiplied by itsel f; that is, used as a factor
EX AM P LE : In S , S is the base and 3 is the exponent, or
power and 5) (5)(S)(5)~ 125; notice that the base, 5, was
multiplicd by itself' 3 times
• Prime numbers arc natural numbers greater than I having
exactly two factors, self and one
EXA M PLES: 7 is prime because the only two natural
numbers that multiply to equal 7 are and I; 3 is prime
because the only two natural numbers that multiply to equal
I 3 are I 3 and I
• Composite numbers are natural numbers that have more
than two factors
E XAM PL ES: 15 is a composite number because I, 3, 5
and 15 all multiply in some combination to equal 15; 9 is
composite because I, 3 and 9 all mUltiply in some combi
nation to equal 9
• The greatest common factor (GCF) or greatest common
divisor (GCD) of a set of numbers is the largest natural
number that is a tactor of each of the numbers in the set; that
is the largest natural number that will divide into all of the
numbers in the set without leaving a remainder
EXAMPLE: The greatest common factor (GCF) of 12, 30
and 42 is 6 because 6 divides evenly into 12, 0 and 42
without leaving remainders
• The least common multiple (LCM) of a set or numbers is
the smallest natural number that can be div ided (without
remainders) by each of the numbers in the set
EXAM P LE :The least common multiple of2, 3, and 4 is 12
because although 2, 3, and 4 divide evenly into Illany
numbers, including 48,36, 24 and 12, the smallest is 12
• The denominator of a fraction is the numbcr in the boltom;
that is, the divisor of the indicated division or the li·action
EX ' J MPLE : In 5 / 8, 8 is the denominator and also the divi
sor in the indicated division
• The numerator of a fraction is the number in the top; that
is, the dividend of indicated division of the fraction
EXAMPLE : In 3/4 3 is the numerator and also the dividcnd
in the indicated division
The Fundamental Theorem ofArithmetic states that every
composite number can be expressed as a unique product of
prime numbers
EXA MPLES: 15 ~ (3)(5), where 15 is composite and both
3 and 5 are prime; 72 ~ (2)(2)(2)(3)(3) where 72 is compos
ite and both 2 and 3 arc prime; otice that 72 also equals
(8)(9) but this does not demonstrate the theorem because
neither 8 nor 9 are prime numbers
ORDER OF OPERATIONS
• Description: The order in w ch additio , subtraction, multi
plication, and division arc performed determines the answer
• Order
I Parentheses: Any operations contained in parentheses
are done first if' there are an This also applies to these
cnclosure symbols, : : and [ ]
2 Exponents: Exp nent expressions are simpli fied
second, if there arc any
3 Multiplication and Division: These operations are d ne
next in the order in which they are found, goin left to
right; that is, if division comes rirst, goin lell to right,
then it is done first
4 Addition and Subtraction: These operations are done
next in the order in which they arc found goin left to
right; that is, if subtraction comes fir ·t, g ing left to
right, then it is done first
D ECI M AL NUMBERS
• The place value of each digit in a base 10 number is deter
mined by its position with respect to the decimal oin
Each position represents multiplication by a power or 10
(I 02~ 100) means 20 because it is 2 times 10 1(I 0 1~ 10)
and 4 means 4 times one because it is 4 times I011 (IO()~ I)
There is an invisible decimal point to the right of the 4 In
5.8 5 means 5 times one because it is 5 times
10°(1 OO~ I), X means 8 times one tenth because it is 8
ti mes 10-1(10-1~.1 ~ 1110) and 2 means 2 times one
hundredth because it is 2 times 10-2(10-2~ OI ~ 111 0 )
FRACTIONS
• Write the digits that are behind the decimal point as the numerator (top) of the li·action
• Write the place value of the last digit as the denominator (bottom) of the fa tion Any digits in front of the decimal
point are whole numbers
E)(AMPLE : In 4.068, the last digit behind the decimal
point is 8 and it is in the 1,000ths place; therefore 4.068
becomes 4 ~ '
t O O
• Notice the number ofzeros in the denominator is equal to the number of digits behind thc decimal point in the original
ADDITION
• Write the decimal umbers in a vertical lorm with the deci
mal points lined up one under the other, so digits of equal
place valu are under each other
' Add
EXAMPLE : 23.045 +7.5 + 143 + 034 would become
23.045 7.5
143,0 becabehind use the there 143 is an invisible decimal oint + 034
173.579
• Write the decimal numbers in a vertical [arm with the deci
mal points lined up one under the other
• Write additional zeros aller the la t digit behind the decimal
point in thc minuend (number on top) if needed (both the minuend and the subtrahend should havc an equal number
of digit behind the decimal point)
• EXAMPLE : In 340.06-27.3057, 40.06 only has 2 digits
behind the decimal point, so it needs 2 more zeros because
27.3057 has 4 digits behind the decimal point; therefo , the problem becomes:
340.0600
-27.3057
MULTIPLICATION
• Multiply
• Count the number of digits behind the decimal points in all factors
' Count the number of digits behind the decimal p int in the answer he answer must have the same number of digits
behind the decimal oint, as there are digits behind the deci
mal points in all thc factors It is n t necessary to line the decimal points up in multiplication
EXAMPLE :In (3.0 ) (.007), multiply the numbers and count the 5 digits behind the decimal points in the problem so you can pu 5 digits behind the decimal point in the product
(answer); therefore, ( 3 05 )(.0 07) ~ 02 1 35 This process works because 3 times 2 can be written as fractio s, )/10
times 2/10, which equals 6/tOO, which equals 06 as a decimal
number - two digits behind the decimal points in the prob
lem and two digits behind the decimal point in the answer
DIVISION
• Rule: Always divide by a whole number
'Ifthe divisor is a whole number simply divide and bring the
decimal point up into the quotient (answer)
2
70
EXAMPLE : 04
'1 f the divisor is a decimal number, move the decimal oint
behind the last digit and move the decimal point in the dividend the same number of places Divide and bring the
decimal point up into the quotient (answer)
•This process works because both the divisor and the
dividend arc actually multiplied by a po\\cr of 10; that
is, 10 100, 1,000, or 10,000 to move the decimal point
EXAMPLE : .1.5 x ~ = 35 0 ~ 7()
.05 tOO 5
ABSOLUTE VALUE
• Definition: Ixl-x ifx > II or x -() and Ix~ x if x<0; that is
the absolute value of a number is always the pOSitive \alue
of that number
both case~
ADDITION
·If the signs of the numbers arc the sUllie ADD The answer has the same sign as the Ilumbers
·If the signs of the numbers are diUel'(!III SUBTRACT
The HIlS\Vcr has the sign of the largl'r number
signs or taking the absolute value of the Ilumhers to dc
mine the larger
EXAMPLES: ( -4 ) + (9'1 ~ 5 and (4)+( 9)= 5
SUBTRACTION
• Change subtraction to addition of tbe opposite number:
a - b ~ a I ( b); that is, change the subtraction sign to addi tion and also change the sign~()f the number dircc11y behind
the subtraction sign to the opposite Theil to llo\\' the addi tion rules ahovC'
( 8) (12)-( X) -I ( 1 1 ) ~ 20 and
( 8) ( 1 2 ) ~ ( R) 1-(12) -4 Notice the sign or thl' number
in front of the subtraction sign never changes
Multiply or divide then fililow these rules to the sign of the
·Ifthe numbers have the same signs the answer is POSITIVE,
• If the numbers have different Signs th(, answer is NEGATI\'[,
-It makes no LlitTcrcncc which Ilumncr is lanler \\ hen you an~
trying to dL'h.:rmine the sign of the answer.'- ~
(- 2)('1) ~- 18
' -(-a )~a; that is the sign ill ti'on! or the parentheses changes the sign of the contents of the parentheses
EXAMPLES: ( 3)= +3 or (3) 3; also (5<1-6) 5a + 6
REDUCIN
• Divide nUl1lerator (op) and den minator (bottom) by the
same number, thereby renamin it to an cqui, alent fraction
in lower terms This process may be repeated
E XAM PLE: ~O 4 5
.12 4
ADDITION
• Change to equivalent factions with common d~nom i nator
EXAMPLE: To evaluate ~+ ~ + ~ lo llow these st
.1 .j
I Find the least common denominator by determining the
sma est umber which can be divided e\enly (no remain
ders) by all of the numbers in the denominators
(bottoms)
2 Multiply the numerator and denominator of each
fraction so the fradion value has n t changed but the COllllllon denominator has been obtained
EX AM PL E:
2 4 I 3 5 2 X > 10
.1 4 4 1 b 2 t2 t 2 1 2
Trang 3because the addition of fractions is counting equal
EXAMPLE: 8 3 10 21 '}
- +- +- = - = 1- = 1
12 12 12 12 12 4
SUBTRACTION
"-_"-=~ ,wherec;<O
• Change to equivalent fractions with a common
denominator
Find the least common denominator by determining
thc smallest number which can be divided evenly by all
of the numbers in the denominators (boltoms)
EXAMPLE: '2 _-'
<) 3
2 Multiply the numerator and denominator by the samc
number so the fraction value has not changed, but the
common denominator has been obtained
EXAMPLE: '2 _ -'- x ~=L~
9 3 3 <) <)
3 Subtract the numerators and keep the EXAMPLE:
same denominator because subtraction 3
of fractions is finding the dinerenee
between equal parts ~ 9
MULTIPLICATION
"- x "-= ~,wherc c;<O and d;<O
• Common denominators are ]'I;OT needed
I Multiply the numerators (tops) and mUltiply the denomi
nators (bottoms), then reduce the answer to lowest terms
EXAMPLE: ~x ~= 12
+~=-'-3 12 36 12 3 EXAMPLE:
2 OR - reduce any numerator (top)
any denominator (boltom)
multiply
DIVISION
"-+ "-= "-x "-= " x d , where c;< 0; d ;< 0; b ;< 0
d /, eX /,
'Common denominators are NOT needed
I Change division to multiplication by the reciprocal;
that is, Ilip the fraction in back of the division sign and
change the division sign to a multiplication sign
EXAMPLE: LI + ~ becomes LI x~ EXAMPLE:
, (2 ",I 2
2 Now, follow the steps for multiplication
uf fractions, as indicated above
PERCENT, RATIO
PERCENTS
• Definition: Percent means "out of I00" or "per 100."
• Percents and equivalent fractions
I Percents can be written as fractions by placing the
number over 100 and simplifying or reducing
EXAMPLES: 3IJ% = 30 =
100
4.5 45
4,5% = 100 = I.IX)(I =
2 Fractions can be changed to percents by writing them
with denominators of 100 The numerator is then the
percent number
EXAMPlE' 2~~x 20 =~=()O%
• Percents and decimal numbers
I To change a percent to a decimal number, move
" the decimal point 2 places to the left because percent
I{ means "out of 100" and decimal numbers with two
I ' digits behind the decimal point also mean "out of 100."
EXAMPLE: 45'%=~; 125'% = 18 ; 6%=~;
3.5% = ,035 bccause the 5 was already behind the decimal
point and is not counted as one of the digits in the "move
two places."
2 To change a decimal number to a perccnt, move the
EXAMPLES: .47=47'Yo; 3.2 = 320.%; .205= ~
• Definition: Comparison between two quantities
3
• Forms: 3 to 5, 3 : 5 3 / 5,5'
PROPORTION
• Definition: Statement of equality between two ratios or
tractions :J 9
Forms: 3 is to 5 as 9 is to 15,3:5::9:15, 5 15
SOLVING PROPORTIONS
•Change the fractions to equivalent fractions with common denominators, set numerators (tops) equal to eaeh other,
and solve the rcsulting statement
EXAMPLES: 2= -'-'-becomes -'2 = -'-'-, so n =
11+3 10 11+3
7 14 7
• Cross-multiply and solve the resulting equation NOTE:
Cross-multiplication is used to solve proportions illlh'
and mav ]'I;OT be used in fraction multiplication Cross
multiplication may be described as the product of the means being equal to the product of the extremes,
EXAMPLES:
- ~ - , 5n=21, n =21 +5, n= 4
7 5 5
MIXED NUMBERS
• Description of mixed numbers: Whole numbers followed
by fractions; that is, a whole number added to a fracton
EXAMPLE: 42 means 41 2 ; 4 /111t! 2
• Improper fractions are fractions that have a numerator (top number) larger than the denominator (bollom number)
• Conversions
I Mixed number to impmper fraction: Multiply the denom
inator (bottom) by the whole number
E XAM PI.E:
the numerator of the improper trac 2 3x 5+2 17
tion The denominator of the t 3 3 3 improper fi'3ction is the same as the
denominator in the mixed number
2 Improper fraction to mixed number:
Divide the denominator 3 ~
17 5 over the divisor (the divisor is the - nleans 5 }T7
thc improper fraction) 2
ADDITION
• Add thc whole numbers
• Add thc ,,'actions by following the stcps for addition of li'ac
tions in the fraction section of this study guide
• I f the answer has an improper fraction, change it to a
mixed number
EXAMPLE:
and add the
resulting whole
whole number in
the answer
SUBTRACTION
• Subtract the fractions first
I Ifthe fraction ufthe larger number is larger than the frac
tion of the smaller number, then follow the steps of subtracting fractions in the fraction section of this study guide and then subtract the whole numbers
EXAMPLE: 7 2-=5-=5
() 6 6 3
EXAMPLE:
2 If that is not the
borrow ONE from the 6-= 5 +-+ - = 5
whole number and add it to 7 7 7 7
5
the fraction (must have -3
common denominators)
2
7
3
•Short Cut For Borrowing:
6-=5 + - -= 5 one, replace the numerator by 7 7
?
Change each In ixcd number to an improper frae! io
follow the steps for multiplying and dividing fractio
GEOMETRIC FORMULAS
PER 1M ETER: The perimeter P of a two-dimensional shape is the SUIll of all side lengths
AREA: The area, A 01' a two-dimensional shape i, the number of square units that can be put in the region
cncloscd hy the sides NOTE: Area is obtaincd through
some combination of multiplying heights and bases which always form 90u angles with each other, except ill circles
VOLUME: Thc volullle, Y, of a threc-dimensional shapc is the number of cubic units that can be put ill the region enclosed by all the sides
Square Area: A = hb
If h=g, then b=H: also as all sides
arc equal in a square, the-n:
A = 64 sq uare: units
Rectangle rc : A = hb or A = Iw
If h=4 and b= 12, then: A =(4){ 121
A = 48 st.J Lmre unilS
If h= 8 and b= 12 th-en: A = '/,IKI( 12)
A = 48 square units
Il'h=6 and b=l) then: A=(6){YI
A = 54 square units
Cin.:le Area: A = rcr:!'
If1[= 3 14 and 1'=5, th~n: A=13.1 4)(:i)',
A;;; 78.5 square units Cin:ulllference: C = 2rcr C=(2)(3.14)(5)=31.4 units
If a ri ht triangle has hypotenuse c and
Rectangular Prism Ytllull1c V=lwh; ifl= 12 w=:1, h=4 then: hc:r7
V=( 1 31(4) V= 144 cubic units
Cubc Volume Y=
Each edge lenth c j:- equal to th
other edges in a c
If e =8 then: v=
V= 512 cubic units
Cylinder Volume V= 1[r'
If radius, 1'=9, h=8 the Y= 9'18i, Y=3.14(81\(
Y=2034.72 cubic unit
If r=6 and h= 8, th
Y= 'I; (3 1
V=30 1.44 cubic unit
TriangUI.<Ir Prism VOIUIl1.l'": V=(area 01"
If ~IHis an area equal (()
'/ 2(5)1121 then: V=30h alld if h=8 thell: \1
Y=(30j(XI, V=240 cubi~ units Rectangular
If I =5 and w = 4 the rCdanglc h a~
arca of 20, the V='/,(20Ih thcn:'V= '/,(20)191 V= nll cubic
Sphere VUIUIllI.!, V= 4/[,J Irradius, 1'= 5, 4(3.1.~ )(5)~ • then:
Trang 4PERCENT APPLICATIONS
_~ ( _!i_" c _r_t'l_ls_ C:' = amount oj'incI't:(lse or
IO!) origi/la/l',Jilie
FORMULAS:
through this subtraction:
SO,11 = 20 Uld the (!~ incrc:-L';C= 2fY~) becatlse 1Y;) means "out of 100:'
FOR~1ULAS: ';" diS( UIIII{ = WI1011llIo(dis("()f1ll1 or
100 originol \',dlle
for $250 on sale for $ISO Find the percent disCLlunt
II $llXI
so, II =40 and the (;0 discount ' 4U1%
SIMPLE INTEREST
r = rate: percent ratl'
t .":::: time: expressed in the same period as the rate,
~
3 months, find the total amount that she paid the bank
I Notice that the 3 momhs was changed to 25 of a year
l ~T _()Ia_I A ,()uI'tn _= I)' i ~ $'5 , _()()t)'$ 75 ~S:5 ,_O 7 5
EXAMPLE:
\/~) commission: ~:::::: '£C()!1I111is,~ or
100 origil/a!
each and sold them lor $44 each Find the percent markup
lOCI 20
so, n 120 and the f% markup = 120(%
I ~~ ~ -~ U S $ 4 95 free downloads &
"IS" & "OF"
Any problems that are or can be stated with percent and the words "is" and "of' can be solved using these formulas:
FORMULAS: Vi ; _ "is" numher
i7io - "of' number
I!~)
HXI
( 1.2S)(80) =II In either ease, the number ~ 100
FOR~,tULAS: % decrease = Oil/Olin! o(deCfCUSt' or
IfJf) originol mill e
II ~.(XX)
I ()() rOfU/ S illnJ/11('
5250,000 and $7,500 profit last month Find the percent
(% expenses:~:::::: 242.~
COMPOUND INTEREST
p=principal; money saved or invested
A ~ 137.49
ALGEBRA
VOCABULARY
oConstants arc specific numbers that arc not multiplied by anv variables
variables
• Like or similar ternlS arc terms that have the same vari
terms because although they have the same variable, x, it is to the power of 4 in one term and to the puwer uf 3 in the other
oAlgebraic expressions arc terms that are connected by
terms: 2s, 4a1 and 5
oAlgebraic equations arc statements of equality between at least two terms
that both statements have equal signs in them
< between at least two terms
'tles at
!\OTICI:: 1'0 STUDE'T Thi s Oui<.:kSllId y · gllid : j , a liall!!),
a!1n o hll<.' d ~.'U id~ to b aSIC m;r tli <,:Olll '>CS
\ ri ghl~ ~ rc li rH'd Nil p an oj' lhi~
duco.! d o r I r,lIls mi a d in
or ll :d la l1 i 1X'[(\lI~ ~ ((l!l 911~ 11 I)llII ~III! 1!IJ~ll ~llllr l llllillll l ll
4
o'lype 2: (a + b)(e+d)~a(c+d) + b(c+d) ~ae + ad + be + bd
using the FOIL Method j'.r Products of Binomials
(sec Algebra I chart) This is a pupular method ""
COMBINING LIKE TERMS
(adding or subtracting)
like terms and never change the c , \poncnt~ during addition
or subtraction: a + a = 2a
EX4MPLES: 4",3 and _7v3, arc like tcrl1l~ even thouch the
mantleI': 4~y ;l + 7y-'x = ] xv"; notice only th cn('trici~nt"
arc not likt ' terms bt 'l"[\usl.' the CXplllll.'nts of Ihl.' a art ' no! tht '
sallie in both terms ~o they Illay not he added or ~tlbtrat 'tl.'d
MULTIPLYING TERMS
am~(a)(a)(a) (a): that is, the a is multiplied by itselfm time,
the exponents
was only one h in the problem
or Equality (could be tric~y if not handled properly) ~
EXAMPl.E: 1/2(3 +S)~2!3(7a 5)+~ "ould be lTlulti
plied on both sides of the sign by thl? IO\\'l?st cornl11oll S'.
denominator of 1/2 [Ind 2/3• \vhil"h is 6: the result \\Quld
be 3(3a + 5) ~ 4(7a 5) + 54: notice that only 1,, 2'1and
• Comhine allY like h:Tll1S that arc 011 the sanll' side l.lfthe l'qual ; sign
side of the equals sign were the 20 and the + S 4
on both sides of the equals sign This may be done mOI"l: than
oncc The objective here is to get all terms with th ~arnc vari
'l,y (or divided by 10) til get" I in front of the a, so the
oCheck the answer by "ubstituting it Il)r the variable in the original equation to sec if it works
SOLVING A FIRST·DEGREE INEQUALITY WITH ONE VARIABLE
exception follows
o Exception: When applying the multiplication property the
negative number
sides by _ /5' Therefore, 5x( 1/,)<65( 1/5 ) This re>ldts in
x < 13 Notice the > did turn around and bC'L·0111C < bc lle