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A COMPREHENSIVE OVERVIEW OF BASIC MATHEMATICAL CONCEPTS - SET THEORY NOTATION { } Brul"e.• indicate the beginning and end ofa set notation; when listed e lements or members mllst be separated by commas; EXAMPLE: In A= {4 , 8, 16~. the 4, 8, and 16 arc called e lement s or members 01' the set; set s are finite (ending, or having a last element), unless olherwi se indicated. In the middle of a set indi cates c(Jllfilluuti(J1I (JFa patte", ; EXAMPLE: 8= {S, 10, 15, ... ,85, 90}. At the end of a set indi cates an illfinite set. that is , a set with no last element; EXAMPLE: C={3, 6, '1, 12, ... }. Is a sy mbol which literally means '"such that." E Means j ,\' U member (~ror ;~. an elemellt oj; EXAMPLE: IfA= {4, 8,12 1, then 12 E A because 12 is in set A. E Means is 1101 u ",ember ofor i.\' not all element of; EXAMPLE: If 8 = {2, 4, 6, X} , then 3 Of 8 because 3 is not in set 8. o Empty set or "ull .'tet: A sct containing no elements or members, but which is a subset of all sets ; also written as : }. e Means is a slibset of; also may be written as ~ . (l Means is 11M a.."b.•et ot; also may be w ritten as g:; . AeB Indicates that every element of set A is also all elelllellf (Jf scI8; EX1MPLE: If A= 1). 6 } and 8 = : I, 3. S, 6, 7, 9]. then Ae8 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 equal s the number 01' e lements in that set; E."<AMPLE: If A- {4, 5. 6:, then set A has 8 subsets because A has 3 clements and 23=8. OPERATIONS AuB Indicates the /llIi(J1I of u UNIONset A with sct 8; eve ry 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= {2, 4. 6, 8, 10, 12 1 and B= {3, 6 , 9, 12, 15,18:' then: Au8= {2, 3, 4 , 6, 8, 9,10,12, 15, 18 }. Ans~t I~d~~ii::~~e~h~:i~~':.~~i:'~~e~:· n INTERSECTION 01' this set is al so an clement of SCI A AND set 8 ; that is, to form the intersection of two scts , list only those elements which arc found in 80TH of the two sets; 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 of set A; that is, all clements in the COMPLEMENT universal set which arc NOT in SET set A; EXAMPLE: If the universal set is the set of Integers and 0) . A~ { O, 1, 2, 3, ... f, then A=:-I , -2 , -3 , -4 , ... f. PROPERTIES A=B If all of the elements in set i\ are also in set 8 and all elements in set 13 are also in set A , although they do not have to be in the samc order. EXAMPLE: If A= {5, 10 } and 8 ={ 10, 5 j , then i\= B. II(A) Indicates the IIl1mber ()f e1emellf.. it, .,et A, i.e. , the cardinal number of the set. ,..... EXA,~PLE: Ir A={ 2, 4. 6 }. then n(A)=3. A-B Means is eqllivalellt to; that is, set A and set 8 have thc same number of elements, although the clcments thcmselves may or may not be the same. EXAMPLE: If A {2.. 4 , 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 which have no elements in common. EXAMPLE: IfA = : 3, 4 , 5 l and 13; j7, 8, 9), then An 8=0 because there are no common clements. PROPERTIES OF REAL NUMBERS CLOSURE oa + h is a real number; when you add 2 real numbers, the result is al so a real number. EXAMPLE: 3 and 5 are both real numbers; 3 + S = 8 and the slim. 8, is also a real number. oa - b is a rca I number; when you subtrac t 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 al so a real number unless the denom inator (di visor) 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 e ither 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: 812 = 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 -1 5) 1'1 = 2 + (5 + 9). o(ab) e = 8 (be) : you can group numbers in any arrangement when multiplying and ge t the same an swer. EXAMPLE: (4x 5) 8 = (20) 8 = 160 and 4(Sx8) = 4(40) - 11i0, so (4xS)8 = 4(5x8 ). oThe assuciative property dues 1101 lI 'ork Cor subtraction or division. EXAMPLES: (10 - 4) - 2 = 6 - 2 = 4, but 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 divi sion 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 5 = 0. The exception is zero because 0 + 0 = 0 already. oa(l/a)= I; a number times its multiplicati ve inverse or rec iprocal (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 th e parentheses mu st be multiplied by Ihe term in Cronl 01' the parentheses. EXAMPLE: 4( S+ 7) 4(5 )+ 4(7) = 20 + 28 = 48. This is a simple example and the distributive property is not required 10 Cind the answer. Whe n the problem invo lves a variable. however, the distributi ve pro pert y is a necessity. E."<AMPLE: 4(5a + 7) 4( 5al + 4(7) = 10a + 28. PROPERTIES OF EQUALITY oReflexive: a = a ; both s ides of the equation are identical. EXAMPLE: 5 + k = 5 +k. oSymmetric: If a = h, then b = a. Thi s pro pert y 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. T hi:; pro pert y a llows yo 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 int o o lle equation: Sa (, = a + 2 . oAddition Property of Equality: If a=b, then a +c=h+c. This property all ows you to add any number or algebraic term to anyequation as long as you add it to hoth s ides 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 "li se. bill iI' you add 3 to botlt sides, yo u get a true equaliun X X. Also. 5a + 4 = 14 becomes 5a + 4 + ( 4) = 14 + ( 4) ifYllU add 4 to bOlh s ides. Thi s results in the cquation Sa 10. oMultiplication 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 , th en (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 , 5, .... I I. 12. ... : • \V holc nu mbers: {O, 1,2. 3.... , In, 11. 12. 13. ... : oIntegers: {... , -4, -3, -2, -I , n, I. 2. 3. 4, ... } oRationa l num bers: : p/ql p and q arc integers. q>'O): the sets of Natu ral num bers, Who le numbers, and In tegers. as well as numbers which can be w ritten as proper or improper fractions, arc all subsets 01" thc set of Rationa l nu m bc rs. olrrational numbers: { x 1 x is a real lI umber but is not a Ration al num be r}: thc sets o f R~l tiona l num bers a nd Irrational numbers have no c lement s in common and are. there fo re, di sjoint sets . oR eal numbers: : x 1 x is Ihe coordinate o Ca po int on a number line }; the union of the set llf Rational numhers with the SCI of Irrationa l numbers equals the SCI o f Real num bers. -Imaginary numbers: 1ai l a is a Rca l Ilum ber Hnd i is the number wh ose square is - 11; j 2 I: the sets of' Rea l numbers and Imag inary Ilu mbers have no cl ement s in COllllllon and arc . therefore. disjoint sets. oComplex number s : :a +bil a and b arC Real numbers and i is the number whose square is - I:; the SCI of Rea l numbers and the se t of Imagina r :v numbers arc bot h subsets of the se t of Com plex numbers. EXA MPLES: 4 + 7i and J-2i arc Compl e, numbers. COMPLEX NUMBERS Numbers Rational Integers Whole ~ ~ ~/ ...----.. OPERATIONS OF REAL NUMBERS V OCABU LARY • Total 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 o r s lim . • 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. EXAMPLE: In 2S -8~ 17, 25 is the m inuend, 8 is the subtrahend, and 17 is the diffe rence. • Product is the answer to a multiplication problem. T he numbers multip lied are each called a facto r. EXAMPLE: In 15 x 6 ~ 90, 15 and 6 arc factors and 90 is the produc t. • Quotient is the answer to a di vision problem. The number being divided is call ed the d ividend. The number tha t you arc dividing by is called the divisor. If there is a number remaining aller the divi s ion process has been completed, tha t number is called the remainder. EXAMPLE: In 45 : 5~ 9, which may also be written as 5)45 or 4S15, 45 is the di vidend, 5 is the d ivisor and 9 is the quotient. • An exponent ind icates t he number of t imes the base is multiplied by itsel f; that is, used as a facto r. EXAMPLE: In S3, S is the base and 3 is the exponent, or power. and 5) (5 ) (S )(5)~ 125; notice tha t the base, 5, was multi plicd by itself' 3 ti mes. • Prime numbers arc natura l numbers g reate r than I having exactly two fact ors, it se lf and one . EXAMPLES: 7 is prime because the only two natura l numbers th at mu ltiply to equa l 7 are 7 and I ; 13 is pr ime because the only two natural numbers that multiply to equal I 3 are I 3 and I. • Composite numbers are nat ura l numbers that have more than two fac tors . EXAMPLES: 15 is a composite number because I, 3, 5. and 15 a ll multiply in some combination to eq ual 15; 9 is composite because I , 3. and 9 a ll mUltiply in some combi nation to equa l 9. • The greatest common factor (GC F) or greatest common divisor (GCD ) of a set o f 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 w ithout leav ing a remainder. EXAMPLE: The g rea test com mon fac tor (GCF ) of 12, 30 and 42 is 6 because 6 divides even ly into 12, 30, and 42 w ithout leaving remai nders. • The least comm o n multiple ( LC M ) of a set or numbers is the sma llest natu ra l number that can be d iv ided (w ithout remainders ) by each of the numbers in the set. EXAMPLE:T he least com mon multip le o f2, 3, and 4 is 12 because a lthough 2, 3, and 4 div ide evenly into Illany numbers, including 48,36, 24. and 12, the smallest is 12. • The denominato r o f a fract ion is the numbcr in the boltom ; that is, the divisor o f the in d icated di v is ion or the li·act io n. EX'JMPLE: In 5/8, 8 is the denominator and al so the div i sor in the indicated divisi on. • The numerator of a frac tion is the number in the top; that is, the div idend of indicated division of the fract ion. EXAMPLE: In 3/4 . 3 is the numerator and a lso the di vidc nd in the indicated d ivision. FUNDAMENTAL THEOREM OF ARITHMETIC The Fundamental Theorem of Arithmetic states that every compos ite number can be expressed as a un ique p roduct of prime numbers. EXAMPLES: 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; notice that 72 al so equal s (8)(9 ). but thi s does not demonstrate the theorem because nei ther 8 nor 9 are prime num bers. ORDER OF OPERATIONS • Description: T he order in whi ch addition, subtraction, multi plicat ion, and divi sion arc performed determines the answer. • Order I. Parentheses : Any operat ions contained in parentheses are done first. if' there are any. Th is also appl ies to these cnclosure symbols, : : and [ ]. 2. E xpo nents: Ex po nent expressions are simpli fie d second, if there arc any. 3. M ultiplication and Division : T hese operations are done next in the order in which they are fou nd, go ing le ft to right; that is, if divis ion comes ri rst, going lell to right, then it is done first. 4. Addi tion and Subtracti on : These operations are done next in the orde r in which they arc fo und. go ing left to r ight; that is, if subtraction comes fir ·t, go ing left to right, then it is do ne firs t. DECIMAL NUMBERS • The place value of each digit in a base 10 number is dete r mined by its positi on with respect to the decima l po int. Each position represents multiplication by a power or 10. EXAMPLE: In 324, 3 means 300 because it is 3 ti mes 102 (I 02 ~ 100 ). 2 means 20 because it is 2 times 101(I 0 1~ 10). and 4 means 4 times one because it is 4 times I 011 ( IO()~ I). The re is an invisible dec imal point to the r ight of the 4. In 5. 82 . 5 means 5 ti mes one because it is 5 ti mes 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 tim es one hundredth because it is 2 t imes 10-2 (10-2~ .OI ~ 111 00 ). WRITING DECIMAL NUMBERS AS FRACTIONS • W rite the digits that are beh ind the dec imal point as the numerato r (top ) o f the li·action . • Write the place val ue of the last digit as the denominator (bottom ) o f the fracti on . Any dig its in front of the decima l point are w hole numbers . E)(AMPLE: In 4.068 , the las t digi t behind the decima l poin t is 8 and it is in the 1,000ths place ; the refore. 4. 068 becomes 4~ ' t,OOO • Notice the number ofze ros in the denominator is equal to the number o f digits behind thc decimal point in the o riginalnumber. ADDITION • Write the decima l n umbers in a verti cal lorm w ith the deci mal points lined up one under the other, so digits o f eq ual p lace va lue are under each othe r. ' A dd EXAMPLE: 23 .045 + 7.5 + 143 + .034 would become 23 .045 7.5 143,0 because there is an invisib le deci ma l po int beh ind the 143. + .034 173.579 SUBTRACTION • Write the decimal numbers in a ver tica l [arm wit h the deci mal points lined up one under the other. • Write add itional zeros aller the last di git beh ind the dec imal point in thc minuend (number o n top ) if needed (both the minuend and the subtrahend shou ld havc an equal number of digits behind the decimal point). • EXAMPLE: In 340.06 - 27.3057, 340.06 on ly has 2 digits beh ind the decimal po int , so it needs 2 more zeros because 27 .3057 has 4 digi ts behind the decima l poin t; there fo re, the problem becomes: 340.0600 -27.3057 MULTIPLICATION • Multiply • Coun t the number of digits behind the decimal points in all factors . ' Count the number o f d igits behi nd the decim al po int in the answer. T he answer must have the same number of digits behind th e decima l po int, as there are digits behind the deci ma l poi nts in all thc factors . It is not necessary to line the deci ma l poin ts up in multipl ication. EXAMPLE: In (3.05) (.007), mult iply the numbers and count the 5 digits behind the decima l points in the problem so you can p ut 5 digits behind the decimal poi nt in the product (answer); therefore, ( 3 . 05 )(.007) ~ . 02 1 35. This process works because .3 ti mes .2 can be written as fract ions, )/10 ti mes 2/ 10, which equal s 6/ tOO, which equals .06 as a dec imal number - two dig its behi nd the dec imal po ints in the prob lem and two digits behind the dec imal point in the an swer. DIVISION • Rule : Always d ivide by a who le number. 'I fthe d ivisor is a whole number. simply di vide and bri ng the decimal point up into the quotien t (answer ). 2 70. EXAMPLE: .044)!16 OJ) 3.8 1 '1 f the divisor is a dec imal number, move the dec ima l po in t beh ind the last digit and move the deci ma l point in the dividend the same number of p laces. Div ide and bring the decima l point up into the quotient (a nswer). • Thi s process works because both the divisor and the di vidend arc actually multipli ed by a po\\ cr of 10; that is, 10. 100, 1,000, or 10,000 to move the dec imal point. EXAMPLE: .1.5 x~ = 350 ~ 7() .05 tOO 5 ABSOLUTE VALUE • Definition: Ixl- x ifx > II or x - () and Ixl ~ x if x < 0; that is. the absolute value of a number is always the pOSiti ve \alue of that number. EXAMPLES: 161 ~ 6 and I 61- 6; the '''lSWer is pos itive 6 in both case~. ADDITION ·If the signs of the numbers arc the sUllie. ADD. The answer has the same sign as the Ilumbers. EXAMPLES: (- 4)+( 9) - 13and 5 + 11 - lt). ·If the signs of the numbers are diUel'(!III. SUBTRACT. The HIlS\Vcr has the sign of the largl'r number (ignoring tlH.' signs or taking the absolute va lue of the Ilumhers to dcter mine the larger Ilumher). 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 dircc11 y behind the subtraction sign to the oppos ite. Theil . to ll o\\' the add i tion rules ahovC'. EXAMPI.ES: (X) ( 1 2 )~( R) + ( 12) ~ 4 and ( 8) (12) - ( X) -I ( 1 1 ) ~ 20 and (- 8) ( 1 2 ) ~ ( R) 1- (12) - 4. Notice the s ign or thl' number in front of the subtraction sign never changes. --------------------------------~MULTIPLICATION & DIVISION Multiply or divide. then fililow these rules to determine the s ig n of the answer. ·Ifthe numbers have the same signs. the answer is POSITIVE, • If the numbers have different Signs. th(, answer is NEGATI\'[, -It makes no LlitTcrc ncc which Ilumncr is lanler \\ hen you an~ try ing to dL'h.:rmine the sign of the answer.'- ~ EXAMPLES: ( 2)( 5) - 10 and ( 7)(3) ~ 2 1 and (-2)('1) ~- 18 DOUBLE NEGATIVE ' -(-a)~a; that is. the s ign ill ti'on! or the parentheses changes the s ign of the contents of the parentheses. EXAMPLES: ( 3 )= +3 o r (3) 3; also. (5<1-6) 5a + 6. FRACTIONS REDUCING • Divide nUl1lerator (top ) and de nom inato r (bottom) by the same number, there by renam ing it to an cqui , alent fraction in lower terms. This process may be repeated . EXAMPLE: ~O 4 5 .12 4 ADDITION !.!.. + ~= ~,wherec;tO c c • Change to equivalent fractions with common d~nomi nator . EXAMPLE: To eva luate ~ + ~ + ~. lo llow these steps: .1 .j 6 I. Find the least common d enominator by dete rmi ning the sm a ll es t number which can be di vided e \ en ly (no remain ders) by all of the num bers in the denominators (bottoms). EX4 MPLE: 3.4. and 6 divide evenly into 12. 2. Multiply the n umerator and denomin a tor of each fraction so the fradion value has not changed but the COllllllon deno minator has been obtai ned. EXAMPLE: 2 4 I 3 5 2 X .> 10 - x -+- x -+- X -~- I - I - .1 4 4 .1 b 2 t2 t2 12 3. Add the numerators and keep the same denominator because the addition of fractions is counting equal parts. EXAMPLE: 8 3 10 21 '} 3 -+ - + -=-= 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 nu merator and deno m inator 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 fraction s is finding the dinerenee between equal parts. ~9 MULTIPLICATION "- x"- = ~,wherc c;<O and d;<O d e x d • 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) with any denominator (boltom) and then multip ly thc numerators and multiply the denominators. DIVISION "- + "- = "- x "- = " x d , where c;< 0; d ;< 0; b ;< 0 d /, eX /, 'Common denominators are NOT needed. I. Change division to mul t iplication by the recip rocal; 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: 9 3 <) 2 ,..(2 ", I 2 2. Now, follow the steps for multiplication »;ZI= } uf fractions, as indicated above. PERCENT, RATIO & PROPORTION 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 = 2 100 10 4.5 45 9 4,5% = 100 = I.IX)(I = 21KI 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% - . 5 5 20 100 • 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{ mean s "out of 100" and decimal numbers with two I' digit s 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 inthe " move two places." 2. To change a decimal number to a perccnt, move the decimal point two places to the right. EXAMPLES: .47=47'Yo; 3.2 = 320.%; .205=~5'% • 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 = 15; 4 20 20 20 11+3 10 11+3 5 -- = - becomes ~- = - , so n + 3 = 5 and n = 2. 7 14 7 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 de scribed as the product of the means being equal to the product of the extremes, EXAMPLES: II 3 I -~- , 5n = 21, n = 21 +5, n = 4 755 3 7 I '4"""';:;:2' 3n + 6 =28, 3n=22, n= 7) MIXED NUMBERS & IMPROPER FRACTIONS GENERAL COMMENTS • Description of mixed numbers: W hole numbers followed by fracti ons; that is, a whole number added to a frac tion. I I I EXAMPLE: 42 means 41 2 ; 4 /111t! 2 • Improper fractions a re fracti ons that have a numerator (top number) larger tha n the denominator (bollom number) . • Conversions I. Mixed number to impmper fraction : Multiply the denom inator (bottom) by the whole number EXAMPI.E: and add the numerator (top) to fi nd .,.... the numerator of the improper trac 2 3x5+2 17 5- = - -=tion. T he 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 num ber: EXAMPLE: Divide the denominator into the 3~ numerator and write the remainder 17 5 over the divisor (the divi sor is the - nleans 5}T7 same number as the denominator in 5 15 thc improper fraction). 2 ADDITION • Add thc whole numbers. • Add thc ,,'actions by follow ing the stcps for addition of li'ac tions in the fract ion section o f th is study gu ide. • I f the answer has an improper fraction, change it to a mixed number EXAMPLE:and add the result ing whole number to the who le number in the answer. SUBTRACTION • Subtract the fractions first. I. If the 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 nu mbers. 5 I 4 2 EXAMPLE: 7--2-=5-=5() 6 6 3 EXAMPLE:2. If that is not the case, then 2 7 2 9 borrow ONE from the 6 -= 5 +-+ -=5 whole number and add it to 7 7 7 7 5the fracti on (must have -3 common denominators) 7 before subtracting. 42 7 3 • Short Cut For Borrowing: EXAMPL E: Reduce the whole nwnber by 2 ' 2+ 7, l) 6-= 5+-- = 5 one, replace the numerator by 7 7 7 the sum (add) of the numera -3~ = -3~ tor and denom inator of the 7 7 fraction, and keep the same 4 ? denominator. -7 MULTIPLICATION & DIVISION Change each In ixcd nu mber to an improper frae! io n and fo llow the steps for multip lyi ng and di vid ing frac tions. GEOMETRIC FORMULAS P E R 1M ETE R: The perimeter. P, o f a two-dimens io nal shape is the SUIll of all side le ngths. AREA: The area, A. 01' a two-di me nsional shape i, the number of square units that can be put in the region cncloscd hy the s ides. NOTE: Area is obtaincd through some combination of multiplying he ights and bases . which always form 90u angles with each othe r, except ill circl es . VOLUME: Thc volullle, Y, of a threc-d imens iona l shapc is the numbe r 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 Arca: A = hb. or A = Iw If h=4 and b = 12, then : A =( 4){ 121 A = 48 st.J Lmre un ilS Triangle A rea: A = II., bh If h= 8 and b= 12. th-en : A = '/,IK I( 12) A = 48 square unit s - ParallclognllTl Area: A = hh 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 Pythagorean Theorem If a right triangle has hypote nuse c. and :,ides a and b. then c 2 ~a2 +b2 Rectangular Prism Ytllull1c V=lwh ; ifl= 12. w=:1, h=4. then: hc:r7V=( 12 1( 31(4). V= 144 cubic units w Cubc Volume. Y= e'\ Each edge le ngth . c. j:-. equal to the other edges in a cube. If e =8. then: v= (XIIX)iS). V= 512 cubic units Cy linder Volume. V= 1[r' h If radius, 1' =9, h =8 . the n: Y= 1[(9 1' 18i, Y= 3.14(8 1\(SI. Y=2034.72 cubic unit s Cone Vo lume. V= l !3nr1h If r=6 and h = 8, thell: V= 1/1 1((6)1(8 ). Y= 'I; (3. 14)(36)(81 V=30 1.44 cu bic uni ts TriangUI.<Ir Prism VOIUIl1.l'": V=(area 0 1" Irianglc)hc2#_ If ~ IHis an area equal (() 12 S h '/2 (5)1121. then: V=30h alld if h= 8. thell: \1 Y=(30j(XI, V=240 cubi~ units Rectangular Pyramid V()lulllc: V= 1/.1, (:.trca of r~L'l a nglc)h I f I =5 and w = 4. the rCdanglc ha~ all arca of 20, then: V='/,(20Ih and irh=9. thcn: 'V= '/ ,(20)191. V= nll cubic units Sphere VUIUIllI.!, V= 4/[,J Irradius, 1'= 5, 4(3.1.~ )(5)~ • then: V= 1.:70. V= 523.J cuhic unih PERCENT APPLICATIONS % INCREASE _~(_!i_"c_r_t'l_ls_ = amount oj'incI't:(lseC:' or IO!) origi/la/l',Jilie FORMULAS: (original "alue) x (% increase) ~ amount ofincrease If the amount of increase is not given, it may be fOllnd through this subtraction: (new value) -(original value) ~ amount of increase. EXAMPLE: The Smyth Company had 10,000 employees in 1992 and 12.000 in 1993. Find the % increase. Amount of increase ~ 12,000 - 10,000 = 2,OO() '% increase: ~ 2,O()() I(X) 10,000 SO,11 = 20 .Uld the (!~. incrc:-L';C = 2fY~) becatlse 1Y;) means "out of 100:' % DISCOUNT FOR~1ULAS: ';" diS ( UIIII{ = WI1011llIo(dis("()f1ll1 or 100 originol \',dlle (original price) x (%1 discount) = $ discount If not given, ($ discount) = (original price) - (new price) EXAMPLE: The Smyth Company put suits thai usually sell for $250 on sale for $ISO. Find the percent disCLlunt. II $llXI % discount: 100 ~ $250 so, II = 40 and the (;0 discount '---- 4U1%. SIMPLE INTEREST FORMULAS: i~prt. or (total amount) ~ (principal) + interest Vv'hcrc i = interest p = principal: money borrowed or lent r = rate: percent ratl' t .":::: time: expressed in the same period as the rate, i.e., ifrak is per year. then time is in years or part ora year. Ii'rate is per month. then time is in months. EXAMPLE: Carolyn borrowed S5,000 Irom the bank at 6% ~ simple inkrest per year. I f she borrowed the money for only 3 months, find the total amount that she paid the bank. I S interest = prl ~ ($5,000) (6"1.,) (.2S) " $75 I Notice that the 3 momhs was changed to .25 of a year. l ~T__ I __n____ ____ ~ $'5 ,_______ ~ S:5 ,_7 5. ..._()Ia_A ,()u I't_= I) ' i____()( )t) ' $ 75____O____________ a: % COMMISSION FORl\lULAS: '~-(~ cOf1lmissiun _ S cOlI/missioll I(}I) - S .';alt.'s = ($ sales) x (% commission) ~ S commission EXAMPLE: Missy earned 4'1, on a hOLlse she suld for S125,000. Find her dollar commission. \/~) commission: ~:::::: '£ C()!1I111is,~ or ItXI $1 co. (X~I ($ 12S,()(XlI x (4%) = $ commission, so S cOl1lmission ~ $5,000. % MARKUP FORl\lULAS: ~'I/ Il/wJ.lljJ = S lIIurkllP or 100 origil/a! IJI"ic(' (original price)x('Yo markup)~ $ markup lfnot given, ($ markup) ~ (new price) - (original price) EXAMPLE: The Smyth Company bought blouses lor S20 each and sold them lor $44 each. Find the percent markup. S markup ~ $44-$20~$24 '1;' markup: ~~ 24, lOCI 20 so, n = 120 and the f% markup = 120(%. % PROFIT FORMULAS: " (, Cmlit _ S prulil J()() lolu/ S illCOlllt' or (total $ income) x (% profit) ~ $ profit if not givcn, S profit ~ (total $ income) - ($ expensesl. I: EXAMPLE: The Smyth Company had expenses of $ISO,OOO and a profit of $1 0,000. Find the 'Yr, profit. Z total S incomc - $150,000+ $1 UJ)OO = $160,()OO % profit: ~~ 10.000 or ~ 100 16lJ,(JIlO ($ I60,(JOO) x (n)=S I 0.000. In either case, the 0;., profit ~ 6.2S%. I ~~~--------------~ U S $ 4 95 free downloads & "" •• • t U" '~ "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 or "of' means multiply and "is" means equals. EXAMPLE I: What percent of 12S is 50" -"- ~ ~ or I!~) 125 II x 125 ~ 50: in either case, the percent ~ 40%. EXAMPLE 2: What number is 125% of 80" 125 ~ -"- or HXI KO ( 1.2S)(80) = II. In either ease, the number ~ 100. % DECREASE FOR~,tULAS: % decrease = Oil/Olin! o(deCfCUSt' or IfJf) originol mille (original value)x(% decrease) ~ amount of decrease Irnot given, amount ordecrease~(original value) · (new value) EXAMPLE: The Smyth Company had 12,OO() employees in 1993 and 9 ,000 in 19l)4. Find the percent decrease. Amount of decrease = 12,000 - 9,000 = 3,000 II ~.(XX) ~/o decrease: 10() = 12.()(IO' so, n=25 and the (% del'fcase 25(~/~1. % EXPENSES OR COSTS FORMULAS: % expellses , S C.\{'('llses7: I ()() rOfU/ S illnJ/11(' or (tolal $ income)x(% expenses) ~ $ expenses EXAMPLE: The Smyth Company had a total income or 5250,000 and $7,500 profit last month. Find the percent expenses. $ expenses: = $250,000 - $7 ,SOO = $242,5()O (% expenses:~:::::: 242.~ IIXI 250.t)(X) so. II = 97 and the (!/(I expenses = 971!-r(1. COMPOUND INTEREST FORMULA: A = 1'( I +~r Whcre: A = total amount p=principal; money saved or invested r=rate of interest: llsually a 1% per year t= time: expressed in years II =total number of periods fit EXAMPLE: John put SI00 A- P (I +.':.) .. II (4 8II1to a savings account a~ 4(% .. ( .(M, ) ~ x~ 1 compounded quarterly for R A -- 100 I+ 4 years. How much was in the A = I OO( 1.0 I ),1 account at the end of 8 years? A ~ 100( 1.3749) A ~ 137.49 ALGEBRA VOCABULARY oVariables are letters used ill represent numbers. oConstants arc specific numbers that arc not multiplied by anv variables. oC,iefficients are numbers that are multiplied by one or more variables. EXAMPLES: -4xy has a coefficient of-4: 9m' has a coef ficient of 9: x has -an invisible coefficient of I. • Terms are constants or variable expressions. EXAMPLES: .la: -Sc4d: 25 mpJ r,; 7 are all terms. • Like or similar ternlS arc terms that have the same vari ables to the same degree or exponent value. Coelficients do not,- matter, they n~ay be eq;lal or. nut. ExAMPLES: 3rn- and 7m~ are like terms because they bOlh have the same variable to the same power or cx~onent ·valuc. - ISa"b and 6a"b are like terms, but 2,4 and 6x are not like 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 either addition or subtraction. EXAMPLES: 2s , 4a1 - 5 is an algebraic expression with 3 terms: 2s, 4a1 and 5. oAlgebraic equations arc statements of equality between at least two terms. EXAMPLES: 4z ~ 28 is an algebraic equation. 3(a-4) 'c 6a= IIJ-a is also an algebraic equation. Notice that both statements have equal signs in them. oAlgebraic inequalities are statements that have either > or < between at least two terms. EXAMPLES: 50 <-2x is an algebraic inequality. 3(2n + 7) > - lOis an algebraic inequality. 'tles att !df . ISBN-13: 978-157222506-0~ Customer Hotline # 1,800,230,9522 qUlc 5 U y.com ISBN - l0: 157222506 - 8 !\OTICI:: 1'0 STUDE'T Thi s Oui<.:kSllId y · gllid..: j , a liall!!), a!1nohll<.'d ~.'U id~ to baSIC m;r tli <,:Olll'>CS \ 11 righl~~ rclit' rH'd. Nil pan oj' lhi~ duco.!d or Ir,lIls mi a ....d in 1"(l1"l11. ell or llh.:dla l1 ic:d 911~ 11I)llII ~III!1!IJ~ll~llllr l llllillll lll 1X'[(\lI~ ~ ((l!l 6 3 4 DISTRIBUTIVE PROPERTY FOR POLYNOMIALS oType I: a(c+d) ~ ae+ad EXAMPLE: 4x-'(2xy -t y' ) ~ 8x4y + 4x3y' o'lype 2: (a + b)(e+d)~ a(c+d) + b(c+d) ~ ae + ad + be + bd EX1MPLE: (2x + y)(3x - 5v) 2xt:lx -5y ) + y(3x-Sy ) ~ 6x1- I Oxy + 3xy - 5y1 ~ Iix2 - 7xy: 5y' . T hi s may als,) be clone b) using the FOIL Method j'.r Products of Binomials (sec Algebra I chart). This is a pupular method "". multiplying 2 terms by 2 terms only. FO IL means firs! term timcs first term. outer term times outer term. inn!.!!" term times inner tcnn. and last tcrm tirncs last term. COMBINING LIKE TERMS (adding or subtracting) o RU LE: Co mbine (acid or subtrael) only the eoellie ient ,; of 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 x and yl arc 1l9t iii th~' S~l11lt: or9cr. and may be comhined 111 Ihi" mantleI': 4~y;l + 7y-'x =. ] xv" ; notice only the cn('trici~nt" \vere combined and no CXpO~1Cl1t chan~ed: I)a2lx' and 3bca:' arc not likt..' terms bt..'l"[\usl.' the CXplllll.'nts o f Ihl.' a art..' no! tht..' sallie in both terms. ~o they Illay not he added or ~tlbtrat..'tl.'d. MULTIPLYING TERMS oDefinition: 35 (3)(3)(3)(3)(31: that is, 3 is called the bali<' and it is multiplied by itself 5 times because the exponenl is 5. am~(a)(a)(a) ... (a): that is, the a is multiplied by itselfm time,. o Product Rule for Exponents: (am)(II") ~ a""": that i,. when Illultiplying the sam ..... hase, a in this cas...... simply add the exponents. • Any terms may he Illultiplied. not just like terllls. oRULE: Multiply the coelri,'icnts and multiply the \1IriahJc, (this m~aJ1S you have tn add the cxpon ..... llt ~ ()ftlle same \ariable). EXAMPLE: (4a~c)(-12a2h.1c)=-4!1a·b·lc2: notice that 4 times - 12 became - 4X. a4 times a 2 bel·am.: at" c tim~~ .." became c2. and thL' bJ \vas written 10 intiiGltc Illultiplication by b, but the expunent did not cltange un the h heeause thcre was only one h in the problem. SOLVING A FIRST·DEGREE EQUATION WITH ONE VARIABLE oEliminale any fractions by l"inl' the Multiplicatiun PfIlpr:11) or Equality (could be tric~y if not handled properly) . ~ EXAMPl.E: 1/2( 3a + S) ~ 2!3 (7a 5) + ~ "ould be lTlulti plied on both sides of the sign by thl? IO\\'l?st cornl11oll S'. j 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,, 2'1. and 9 were multiplied by 6 and not the contcilts-ofti le par,'ntheses: the parentheses will be handled in Ihe next ,tep, di"tribution. • [)istribute to remove any J1arl'nthe~l" s . ifthcre arC' all )'. EXAMPLE: 3(3a + 5) 4(7a 5) + 5-1 becomes 9a + 15 ~ 28a - 2() + 54. • Comhine allY like h:Tll1S that arc 011 the sanll' s ide l.lfthe l'qual..; sign. EXAMPLE: 9a + 15 = na 10 + ~4 becomcs 9a + 15 ~ 28a + 3-1 because the only like terms on the same side of the equals sign were the 20 and the +S4. • Use the addition property (If equality to add the same terms 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 able on one side of the equals Sign and all constants \\ ithulI! the variable on the other side of the equals s ign. EXAMPLE:9a + 15 ~ na T 34 becollles 'Ia + 15 - 28a 15 =2~a 1-34-288-15. Notice that both - 2Sa and 15 wcre added to both ~ ides of the equal s " ign at the same time. This results in - llJa = 19 after lik..: terms arc added or subtracted. oUse the lIlultiplication property of "'qualit~ to make the coellicient or the variable a I. EXAMPLE: 19a ~ 19 would be multiplied (In both s ides b) 'l,y (or divided by 10) til get" I in front of the a, so the equation would become la 19( _ I/ ,y )or s implya- - I. 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 • Follow the same steps tor solving a first -degree equality as described above, except for one step in the process. This exception follows. oException: When applying the multiplication property. the inequality sign must turn artlund if ~'ou multiplied b~ a negative number. EXAMPLES: In 4m > -48, you need to multiply both ~ sides or the > symbol by II.. Therefore, J 4m( 1/4»-48( 1/4), This results in Ill > 12 . otiee the '> did not turn around hecause you multipli ed hy a posi tive 1/4. However, in 5x > 65. you need to multiply both sides by _ 1/5 ' Therefore, 5x( 1/, )< 65( 1/5 ). Thi s re>ldts in x < 13. Notice the > did turn around and bC'L·0111C < becall se you multiplied by a negative number, _ 1/5' ·Check the solution by substituting some Ilumerical \alu..: s of the variable in the original inequalit y.
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