# 1 ELEMENTARY ALGEBRA and GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE

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1 ELEMENTARY ALGEBRA nd GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE Directions: Study the exmples, work the prolems, then check your nswers t the end of ech topic. If you don t get the nswer given, check your work nd look for mistkes. If you hve troule, sk mth techer or someone else who understnds this topic. TOPIC : ARITHMETIC OPERATIONS A. Frctions: Simplifying frctions: exmple: Reduce 7 6 : (Note tht you must e le to find common fctor; in this cse 9; in oth the top nd the ottom in order to reduce.) Prolems -: Reduce: Equivlent frctions: exmple: is equivlent to how mny eighths? Prolems -: Complete: How to get the lowest common denomintor (LCD) y finding the lest common multiple (LCM) of ll denomintors: exmple: 6 nd 8. First find LCM of 6 nd : 6 LCM 0, so 6 0, nd Prolems 6-7: Find equivlent frctions with the LCD: 6. nd nd 7 8. Which is lrger, 7 or? (Hint: find the LCD frctions) Adding, sutrcting frctions: if the denomintors re the sme comine the numertors: exmple: Prolems 9-: Find the sum or difference (reduce if possile): If the denomintors re different, find equivlent frctions with common denomintors, then proceed s efore: exmple: exmple: Multiplying frctions: multiply the top numers, multiply the ottom numers, reduce if possile. exmple: ( ) Dividing frctions: mke compound frction, then multiply the top nd ottom (of the ig frction) y the LCD of oth: exmple: exmple: B. Decimls: Mening of plces: in.9, ech digit position hs vlue ten times the plce to its right. The prt to the left of the point is the whole numer prt. Right of the point, the plces hve vlues: tenths, hundredths, etc.,.9 ( 00)+ ( 0)+ ( ) So, + ( 0)+ ( 00)+ ( 9 000)..Which is lrger:.9 or.7? To dd or sutrct decimls, like plces must e comined (line up the points). exmple:... exmple: +.. exmple: 6.0 (.)

2 \$. \$.68 Multiplying decimls: exmple:... exmple:...06 exmple: (.0) (.) Dividing decimls: chnge the prolem to n equivlent whole numer prolem y multiplying oth y the sme power of ten. exmple:..0 Multiply oth y 00, to get 0 0 exmple:.0 Multiply oth y 000, get C. Positive integer exponents nd squre roots of perfect squres: Mening of exponents (powers): exmple: 8 exmple: 6 Prolems -: Find the vlue: (.). (.) is non-negtive rel numer if 0 mens, where 0. Thus 9 7, ecuse 7 9. Also, D. Frction-deciml conversion: Frction to deciml: divide the top y the ottom. exmple:.7 exmple: exmple: Prolems -: Write ech s deciml. If the deciml repets, show the repeting lock of digits: Non-repeting decimls to frctions: red the numer s frction, write it s frction, reduce if possile: exmple:. four tenths 0 exmple:.76 three nd seventy six hundredths Prolems 6-8: Write s frction: E. Percents: Mening of percent: trnslte percent s hundredths : exmple: 8% mens 8 hundredths or.08 or 8 00 To chnge deciml to percent form, multiply y 00: move the point plces right nd write the % symol. exmple:.07 7.% exmple:. % Prolems 9-60: Write s percent: To chnge percent to deciml form, move the point plces left nd drop the % symol. exmple: 8.76%.0876 exmple: 67%.67 Prolems 6-6: Write s deciml: 6. 0% 6..0% To solve percent prolem which cn e written in this form: % of is c First identify,,c : Prolems 6-6: If ech sttement were written (with the sme mening) in the form of % of is c, identify,, nd c : 6. % of 0 is is 0% of out of is % Given nd, chnge % to deciml form nd multiply (since of cn e trnslted multiply ). Given c nd one of the others, divide c y the other (first chnge percent to deciml, or if the nswer is, write it s percent).

3 exmple: Wht is 9.% of \$000? ( % of is c : 9.% of \$000 is? ) 9.% \$000 \$70 (nswer) exmple: 6 prolems correct out of 80 is wht percent? ( % of is c :? % of 80 is 6) % (nswer) exmple: 60 people vote in n election, which is 60% of the registered voters. How mny re registered? ( % of is c : 60 % of? is 60); 60%.6; (nswer) 66. % of 9 is wht? 67. Wht percent of 70 is 6? 68. % of wht is 60? F. Estimtion nd pproximtion: Rounding to one significnt digit: exmple:.67 rounds to exmple:.09 rounds to.0 exmple: 80 rounds to either 800 or 900 Answers: , , 8. (ecuse 0 8 < 8 ) TOPIC : POLYNOMIALS \$ not rel numer Prolems 69-7: Round to one significnt digit To estimte n nswer, it is often sufficient to round ech given numer to one significnt digit, then compute. exmple: Round nd compute: is the estimte Prolems 7-7: Select the est pproximtion of the nswer: (, 0, 00, 000, 0000) (.0,.,.,, 0, 0) (,, 98, 0, 00) 7. (.6890) (, 6, 60, 000, 000) % % c % A. Grouping to simplify polynomils: The distriutive property sys: ( + c) + c exmple: ( x y ) x y (, x, c y)

4 x + 7x ( + 7)x x exmple: ( x,, c 7) exmple: + 6x +x Prolems -: Rewrite, using the distriutive property:. 6( x ). ( ). x x Commuttive nd ssocitive properties re lso used in regrouping: exmple: x + 7 x x x + 7 x + 7 exmple: x + + x 0 x exmple: x + y x + y x x + y + y x + y Prolems -9: Simplify:. x + x 7. x ++ x x x y + y 8x 9. x y + y x B. Evlution y sustitution: exmple: If x, then 7 x 7 () 7 exmple: If 7 nd, then ( 7) ( ) 9( ) 9 exmple: If x, then x x ( ) ( ) Prolems 0-9: Given x, y, nd z. Find the vlue: 0. x. x + y. z 6. x x. xz 7. ( x + z). y + z 8. x + z. y + z 9. x z C. Adding, sutrcting polynomils: Comine like terms: exmple: ( x + x +) ( x ) x + x + x + x + exmple: ( x ) + ( x + x ) x + x + x x + x exmple: ( x + x ) ( 6x x +) x + x 6x + x x + x Prolems 0-: Simplify: x + 0. x + x. ( x )+ ( x). ( )+ ( + ). ( y y ) ( y y + ). ( 7 x) ( x 7). x ( x + x ) D. Monomil times polynomil: Use the distriutive property: exmple: ( x ) x + ( ) x + ( ) x exmple: ( x + ) x + exmple: x( x ) x + x 0. ( x ) 6. x 7 7. ( ). ( x ) ( ) 8. x( x + ). 8( + 7) 9. ( x )7 E. Multiplying polynomils: Use the distriutive property: ( + c) + c ( x ) is + c exmple: x + if: ( x +), x, nd c So ( + c) + c ( x +)x + ( x +) ( ) x + x 8x x 7x Short cut to multiply ove two inomils (see exmple ove): FOIL (do mentlly nd write the nswer) F: First times First: ( x) ( x) x O: multiply Outers : ( x) ( ) 8x I: multiply Inners : ( x) x L: Lst times Lst: ( ) Add, get x 7x exmple: ( x + ) ( x + ) x + x + 6 exmple: ( x ) ( x + ) x + x exmple: ( x ) ( x + ) x exmple: ( x ) x + exmple: ( x ) 9x x +6 exmple: ( x + ) ( ) x x + Prolems -: Multiply:. ( x + ) 8. 6x( x). ( x ) 9. ( x ). ( x + ) ( x ) 0. ( x ) ( x + ) 6. ( x + ) ( x ). ( x ) ( x + ) 7. ( x ) ( x )

5 F. Specil products: These product ptterns (exmples of FOIL) should e rememered nd recognized: I. ( + ) ( ) II. ( + ) + + III. + Prolems -: Mtch ech pttern with its exmple:. ( x ) 9x 6x +. ( x + ) x + 0x + c. ( x + 8) ( x 8) x 6. I:. II:. III: Prolems -: Write the nswer using the pproprite product pttern:. ( +) ( ) 9. ( ) ( ) 0. ( x y). ( x +y) ( ). ( x + y) x y 6. y G. Fctoring: Monomil fctors: + c ( + c) exmple: x x x( x ) exmple: x y + 6xy xy( x + ) Difference of two squres: + ( ) ( x ) exmple: 9x x + Trinomil squre: exmple: x 6x + 9 ( x ) Trinomil: exmple: x x ( x ) ( x +) exmple: 6x 7x x + Prolems -67: Fctor: ( x ) x x x x. 8x 6. 8x + 8x + x 6. x 0x x +x + 7. xy +0x 6. 6x y 9x y 8. x x 66. x x 9. x x x 0x x y y x Answers:. 6x 8. x. +. x. 6. x + y 7. x x 9. x y x. x.. y y 0. x. x + 6. x x + x 9. x 7 0. x. x x + 6x + 9. x 6x + 9. x 9 6. x 9 7. x 6x x + 6x 9. x x + 0. x + x. x + x. c y y x xy + y. 6x + xy + 9y. 9x y ( x ) ( x +) ( x + ) ( x + ) ( + x) ( x ). x + 6. x 7. x y x 8. x 9. x 60. xy x y 6. x 6. x x 6. x x + 6. x + 6. x y y x 66. x 67. x

6 6 TOPIC : LINEAR EQUATIONS nd INEQUALITIES A. Solving one liner eqution in one vrile: Add or sutrct the sme vlue on ech side of the eqution, or multiply or divide ech side y the sme vlue, with the gol of plcing the vrile lone on one side. If there re one or more frctions, it my e desirle to eliminte them y multiplying oth sides y the common denomintor. If the eqution is proportion, you my wish to cross-multiply. Prolems -: Solve:. x 9 7. x 6 x. 6x 8. x x +. x x x x. 0. 7x x +0. x 9. x + x 6. x x + To solve liner eqution for one vrile in terms of the other(s), do the sme s ove: exmple: Solve for F : C F 9 Multiply y 9 : 9 C F Add : 9 C + F Thus, F 9 C + exmple: Solve for : + 90 Sutrct : 90 exmple: Solve for x : x + c Sutrct : x c Divide y : x c Prolems -9: Solve for the indicted vrile in terms of the other(s): y x x y x + x. P + h 8. x + y 0 x. y x 9. y x 0 x y B. Solution of one-vrile eqution reducile to liner eqution: Some equtions which do not pper to e liner cn e solved y using relted liner eqution: exmple: x+ x Multiply y x : x + x Solve: x x (Be sure to check nswer in the originl eqution.) exmple: x+ x+ Think of s nd cross-multiply: x + x + x x But x does not mke the originl eqution true (thus it does not check), so there is no solution. Prolems 0-: Solve nd check: 0. x x x+ x. x x+. x x+8. x x+. x x exmple: x Since the solute vlue of oth nd is, x cn e either or. Write these two equtions nd solve ech: x or x x x x or x Prolems 6-0: Solve: 6. x 9. x 0 7. x 0. x + 8. x C. Solution of liner inequlities: Rules for inequlities: If >, then: + c > + c c > c c > c (if c > 0 ) c < c (if c < 0 ) c > c (if c > 0 ) c < c (if c < 0 ) If <, then: + c < + c c < c c < c (if c > 0 ) c > c (if c < 0 ) c < c (if c > 0 ) c > c (if c < 0 ) exmple: One vrile grph: solve nd grph on numer line: x 7 (This is n revition for { x : x 7}) Sutrct, get x 6 Divide y, x Grph: Prolems -8: Solve nd grph on numer line:. x >. x <

7 . x x > x. < x 7. x > +. x < x + x D. Solving pir of liner equtions in two vriles: The solution consists of n ordered pir, n infinite numer of ordered pirs, or no solution. Prolems 9-6: Solve for the common solution(s) y sustitution or liner comintions: 7 9. x + y 7. x y x y 8 x + y 0. x + y. x y x y x + y. x y 9. x + y x 8 x + y. x y 6. x y y x 6x 9 y Answers: ( P h). ( y +). 6. y ( y ) y 9. x y TOPIC : QUADRATIC EQUATIONS no solution 6., 7. no solution 8., 9. 0.,. x > 7. x < x 0. x > x > 6. x < 7. x > x - 9. ( 9, ) 0. (, ). (8, ). (, 9). 8 ( 9, 9 ). (,0) 0. no solution 6. ny ordered pir of the form (, ) where is ny numer. One exmple: (, ). Infinitely mny solutions. A. x + x + c 0: A qudrtic eqution cn lwys e written so it looks like x + x + c 0 where,, nd c re rel numers nd is not zero. exmple: x x Add x: x + x Sutrct : 0 x + x or x + x 0 So,, c exmple: x Rewrite: x 0 (Think of x + 0x 0) So, 0, c Prolems -: Write ech of the following in the form x + x + c 0 nd identify,, c :. x + x 0. x x. x 0. x x. 8x B. Fctoring: Monomil fctors: + c + c exmple: x x x x exmple: x y + 6xy xy x + Difference of two squres: + ( ) ( x ) exmple: 9x x +

8 Trinomil squre: exmple: x 6x + 9 ( x ) Trinomil: ( x +) ( x ) exmple: x x x exmple: 6x 7x x + Prolems 6-0: Fctor: x x x x 8. 8x 6. x + 8x + 8x 9. x 0x x +x + 0. xy +0x 8. 6x y 9x y. x x 9. x x. x x 6 0. x 0x +. x y y x C. Solving fctored qudrtic equtions: The following sttement is the centrl principle: If 0, then 0 or 0. First, identify nd in 0: exmple: ( x) ( x + ) 0 Compre this with 0 ( x); ( x + ) Prolems -: Identify nd in ech of the following:. x( x ) 0. ( x ) ( x ) 0. ( x )x 0. 0 ( x ) ( x +) Then, ecuse 0 mens 0 or 0, we cn use the fctors to mke two liner equtions to solve: exmple: If x( x ) 0 then( x) 0 or x 0 nd so x 0 or x ; x. Thus, there re two solutions: 0 nd exmple: If ( x) ( x + ) 0 then x or ( x + ) 0 nd thus x or x. 0 0 exmple: If x + 7 then x x 7 x 7 (one solution) 8 Note: there must e zero on one side of the eqution to solve y the fctoring method. Prolems -: Solve:. ( x +) ( x ) 0 9. ( x 6) ( x 6) 0 6. x( x + ) 0 0. ( x ) ( x )x. x( x + ) x 8.0 x + ( x ) 0 D. Solving qudrtic equtions y fctoring: Arrnge the eqution so zero is on one side (in the form x + x + c 0), fctor, set ech fctor equl to zero, nd solve the resulting liner equtions. exmple: Solve: 6x x Rewrite: 6x x 0 Fctor: x x So x 0 or x 0 0 Thus, x 0 or x exmple: 0 x x 0 ( x ) ( x + ) x 0 or x + 0 x or x Prolems -: Solve y fctoring: ( x + ) ( x ) ( x ) 0. x x. x x 0 9. x +. x x 0. 6x x +. x( x + ) x 6x 6. x x. x x 7. x + x 6. x x 6 0 Another prolem form: if prolem is stted in this form: One of the solutions of x + x + c 0 is d, solve the eqution s ove, then verify the sttement. exmple: Prolem: One of the solutions of 0x x 0 is A. B. C. D. E. Solve 0x x 0 y fctoring: x( x ) 0 so x 0 or (x ) 0 thus x 0 or x. Since x is one solution, nswer C is correct.

9 . One of the solutions of x A. B. C. 0 D. E. ( x + ) 0 is 9. One solution of x x 0 is A. B. C. D. E. Answers: c. x + x 0 -. x x x x x x Note: ll signs could e the opposite ( x -) ( x +) ( x + ) 8. x + 9. x 0. x y x. x. x. xy x y TOPIC : GRAPHING ( x + ) ( + x) ( x ). x. x x 6. x x + 7. x + 8. x y y x 9. x 0. x. x x. x x. x x. x x +., 6., , 8., , 0,. 0,. 0,. 0,., 0 6., 7., 8., 9., 0.,..,.,. B. B A. Grphing point on the numer line: Prolems -7: Select the letter of the point on the numer line with coordinte: A B C D E F G Prolems 8-0: Which letter est loctes the given numer: Prolems -: Solve ech eqution nd grph the solution on the numer line: exmple: x + x. x 6 0. x + x. x x + B. Grphing liner inequlity (in one vrile) on the numer line: Rules for inequlities: If >, then: + c > + c c > c c > c (if c > 0) c < c (if c < 0) (if c > 0) > c c < c c (if c < 0) If <, then: + c < + c c < c c < c (if c > 0) c > c (if c < 0) (if c > 0) < c c > c c (if c < 0) exmple: One vrile grph: solve nd grph on numer line: x 7 (This is n revition for { x : x 7}) Sutrct, get x 6 Divide y -, x Grph:

10 0 Prolems -0: Solve nd grph on numer line:. x > 8. x < 6. x < 9. x > x 6. x x >+ 7. < x exmple: x > nd x < The two numers - nd splits the numer line into three prts: x >, < x <, nd x <. Check ech prt to see if oth x > nd x < re true: prt x vlues x >? x <? oth true? x < < x < no yes yes yes no yes (solution) x > yes no no Thus the solution is < x < nd the grph is: exmple: x or x < ( or mens nd/or ) prt x vlues x? x <? x x < x > yes no no yes yes no t lest one true? yes (solution) yes (solution) no So x or x <; these cses re oth covered if x <. Thus the solution is x < nd the grph is: Prolems -: Solve nd grph:. x < or x >. x 0 nd x >. x > nd x C. Grphing point in the coordinte plne: If two numer lines intersect t right ngles so tht: ) one is horizontl with positive to the right nd negtive to the left, ) the other is verticl with positive up nd negtive down, nd ) the zero points coincide Then they form coordinte plne, nd ) the horizontl numer line is clled the x-xis, ) the verticl line is the y-xis, ) the common zero point is the origin, ) there re four y qudrnts, II I numered x s shown: III IV To locte point on the plne, n ordered pir of numers is used, written in the form (x, y). The x-coordinte is lwys given first. Prolems -7: Identify x nd y in ech ordered pir:. (, 0) 6. (, -). (-, ) 7. (0, ) To plot point, strt t the origin nd mke the moves, first in the x-direction (horizontl) nd the in the y-direction (verticl) indicted y the ordered pir. exmple: (-, ) Strt t the origin, move left (since x ), then (from there), up (since y ) Put dot there to indicte the point (-, ) 8. Join the following points in the given order: (-, -), (, -), (-, 0), (, ), (-, ), (, 0), (-, -), (-, ), (, -) 9. Two of the lines you drew cross ech other. Wht re the coordintes of this crossing point? 0. In wht qudrnt does the point (, ) lie, if > 0 nd < 0? Prolems -: For ech given point, which of its coordintes, x or y, is lrger? D. Grphing liner equtions on the coordinte plne: The grph of liner eqution is line, nd one wy to find the line is to join points of the line. Two points determine line, ut three re often plotted on grph to e sure they re colliner (ll in line). Cse I: If the eqution looks like x, then there is no restriction on y, so y cn e ny numer. Pick numers for vlues of y, nd mke ordered pirs so ech hs x. Plot nd join.

11 exmple: x Select three y s, sy -, 0, nd Ordered pirs: (-, -), (-, 0), (-, ) Plot nd join: Note the slope formul 0 gives ( ) 0, which is not defined: verticl line hs no slope. Cse II: If the eqution looks like y mx +, where either m or (or oth) cn e zero, select ny three numers for vlues of x, nd find the corresponding y vlues. Grph (plot) these ordered pirs nd join. exmple: y Select three x s, sy -, 0, nd Since y must e -, the pirs re: (-, -), (0, -), (, -) The slope is ( ) And the line is horizontl. exmple: y x Select x s, sy 0,, : If x 0, y 0 If x, y If x, y Ordered pirs: (0, -), (, ), (, ) Note the slope is ( ) 0, And the line is neither horizontl nor verticl. Prolems -: Grph ech line on the numer plne nd find its slope (refer to section E elow if necessry):. y x 9. x 6. x y 0. y x 7. x y 8. y. y x + E. Slope of line through two points: Prolems -7: Find the vlue of ech of the following: ( ) ( ) nd The line joining the points P x, y hs slope y y P x, y x x. exmple: A (, -), B (-, ) slope of AB ( ) Prolems 8-: Find the slope of the line joining the given points: 8. (-, ) nd (-, -) 9. (0, ) nd (-, -) 0. (, -) nd (, -).. Answers:. D. E. C. F. B 6. G 7. B 8. Q 9. T 0. S... ½. x > 7. x < 6. x ½ 0 7. x > 6 8. x > 9. x < 0. x >. x < or x >. x >. < x x y (0,-) 0. IV. x. y. y. x none - -

12 0. -. ½ -. ½ none (undefined) TOPIC 6: RATIONAL EXPRESSIONS A. Simplifying frctionl expressions: exmple: (note tht you must e le to find common fctor in this cse 9 in oth the top nd ottom in order to reduce frction.) exmple: (common fctor: ) Prolems -: Reduce: c x x ( x+) ( x ) ( x ) ( x ). 6xy y 0. x 9x x 9. 8( x ) 6 x x 7y x. x 7y x x+ exmple: y 0x 0 x y x y x y x x y y y y y Prolems -: Simplify: x. xy y 6 y. x x x B. Evlution of frctions: exmple: If nd, find the vlue of + Sustitute: + () x( x ) x 6 Prolems -: Find the vlue, given,, c 0, x, y, z : x y y x x c x 7.. z y 8. c. z C. Equivlent frctions: exmple: is equivlent to how mny eighths? 8, 6 8 exmple: exmple: x+ x+ ( x+) x+ x+ x+8 x+ x+ x+ exmple: x x+ ( x+) ( x ) x ( x ) ( x ) x+ ( x ) ( x+) x x+ ( x+) ( x ) Prolems -7: Complete: ( +) ( ). x 7 7y 7. x 6 6 x x+. x+ ( x ) ( x+) How to get the lowest common denomintor (LCD) y finding the lest common multiple (LCM) of ll denomintors: exmple: 6 nd 8. First find LCM of 6 nd : 6 LCM 0, so 6 0, nd exmple: nd 6 : 6 LCM, so 9, nd 6 exmple: x+ nd x LCM ( x + ) ( x ), so x+ x ( x+) ( x ), nd x x+ ( x+) ( x ) Prolems 8-: Find equivlent frctions with the lowest common denomintor: 8. nd 9. x nd x 9. x nd. x nd x+ 0. x nd x+. x nd x x+

13 D. Adding nd sutrcting frctions: If denomintors re the sme, comine the numers: exmple: x x x x x y y y y Prolems -8: Find the sum or difference s indicted (reduce if possile): x+ x +x y xy. x x x If denomintors re different, find equivlent frctions with common denomintors, then proceed s efore (comine numertors): exmple: exmple: + x x+ ( x+) ( x ) ( x+) + ( x ) ( x ) ( x+) ( x+6+x x ) ( x+) x+ ( x ) ( x+) Prolems 9-: Find the sum or difference: x 7. x + x x x. x 8. x x x x x x+ x. 0. x x x x. c. x x x. + E. Multiplying frctions: Multiply the tops, multiply the ottoms, reduce if possile: exmple: exmple: ( x+) x x x ( x+) ( x+) ( x ) ( x ) ( x+) ( x ) x+6 x ( ). c d ( x+). 9. x+ y y x 6 x x x+6 F. Dividing frctions: A nice wy to do this is to mke compound frction nd then multiply the top nd ottom (of the ig frction) y the LCD of oth: exmple: c d c d 7 exmple: d d c c d d x x exmple: x x y y x y y x y x x 9 x+7 x c c xy y Answers:... x.. 6. x y y c x ( x +) x. ( x ) ( x +). x+ x. x. x undefined xy ( x + ) ( ). x + x or x 6. + or , , x x x 0. x ( x + ) ( x +), ( x +)., x x

14 . ( x+) ( x ) ( x+), ( x ) ( x ) ( x +), x x( x +). x + x x x x x. x 0 x... c x +x 9. x ( x ) ( x+) ( x ) 0. x + x. x +x x.. c d y x 9. x x +7 x c 7. c TOPIC 7: EXPONENTS nd SQUARE ROOT A. Positive integer exponents: mens use s fctor times. ( is the exponent or power of.) exmple: mens, nd hs vlue. exmple: c c c c Prolems -: Find the vlue:. 8. (.) ( ) exmple: Simplify: Prolems -8: Simplify:. x 7. x y B. Integer exponents: I. c +c II. c III. IV. V. c ( ) c c ( ) c c c c c c ( x) x VI. 0 (if 0) VII. Prolems 9-8: Find x: 9. x. 8 x 0. x. x x x. x 7. c. x 8. y y Prolems 9-: Find the vlue: c x 9. 7x 0 x 6. c+ x c 0. x 7. 6x. 8. x x x x x. ( ). xy. x c+ x c C. Scientific nottion: exmple: if the zeros in the ten s nd one s plces re significnt. If the one s zero is not, write.80 0 ; if neither is significnt:.8 0. exmple:

15 exmple: 0 00 exmple: Note tht scientific form lwys looks like 0 n where <0, nd n is n integer power of 0. Prolems -: Write in scientific nottion:. 9,000, Prolems 6-8: Write in stndrd nottion: To compute with numers written in scientific form, seprte the prts, compute, then recomine. exmple: (. 0 ) (.) exmple: exmple: Prolems 9-6: Write nswer in scientific nottion: D. Simplifiction of squre roots: (.9 0 )(. 0 7 ) 8. 0 if nd re oth non-negtive ( 0 nd 0). exmple: 6 exmple: 7 exmple: If x 0, x 6 x Note: If x < 0, x 6 x mens (y definition) tht ), nd ) 0 Prolems 7-69: Simplify (ssume ll squre roots re rel numers): x x E. Adding nd sutrcting squre roots: exmple: + exmple: Prolems 70-7: Simplify: F. Multiplying squre roots: if 0 nd 0. exmple: 6 6 exmple: 6 exmple: ( ) ( ) 0 Prolems 7-79: Simplify: ( ) ( ) 79. Prolems 80-8: Find the vlue of x: x 8. x G. Dividing squre roots:, if 0 nd > 0. exmple: (or 8 ) Prolems 8-86: Simplify: If frction hs squre root on the ottom, it is sometimes desirle to find n equivlent frction with no root on the ottom. This is clled rtionlizing the denomintor. exmple: exmple: 9

16 Prolems 87-9: Simplify: Answers: x x 8. y y x c 6. x 6 7. x 8. x 9 9. x x 6. 8x y x x 67. x

17 7 TOPIC 8: GEOMETRY A. Formuls for perimeter P nd re A of rectngles, squres, prllelogrms, nd tringles: Rectngle with se nd ltitude (height) h: P + h A h h If wire is ent in this shpe, the perimeter P is the length of the wire, nd the re A is the numer of squre units enclosed y the wire. exmple: A rectngle with 7 nd h 8: P + h units A h squre units A squre is rectngle with ll sides equl, so the rectngle formuls pply (nd simplify). If the side length is s: P s A s s s exmple: A squre with side s cm hs P s! cm A s cm (sq. cm) A prllelogrm with se nd height h nd other side : A h h P + exmple: A prllelogrm hs sides nd 6; is the length of the ltitude perpendiculr to the side. P units A h 0 squre units In tringle with side lengths,, nd c, nd ltitude height h to side : P + + c c A h h h

18 exmple: P + + c units A h 0 (.8 ) squre units Prolems -8: Find P nd A:. Rectngle with sides nd 0.. Rectngle with sides. nd.. Squre with sides miles.. Squre with sides yrds.. Prllelogrm with sides 6 nd, nd height 0 (on side 6). 6. Prllelogrm, ll sides, ltitude Tringle with sides,, nd. Side is the ltitude on side. 8. Tringle shown: B. Formuls for circumference C nd re A of circle: A circle with rdius r (nd dimeter d r ) hs distnce round (circumference) C!d!r (If piece of wire is ent into circulr shpe, the circumference is the length of the wire.) exmple: A circle with rdius r 70 hs d r 0 nd exct circumference C!r! 70 0! units If! is pproximted y 7, C 0" #0( 7 ) # 0units (pprox.) If! is pproximted y., C! 0(.) units The re of circle is A!r exmple: If r 8, exct re is A!r! 8 6! squre units Prolems 9-: Find the exct C nd A for circle with: 9. rdius r units 0. r 0 feet. dimeter d km Prolems -: A circle hs re 9! :. Wht is its rdius length?. Wht is the dimeter?. Find its circumference. r 8 Prolems -6: A prllelogrm hs re 8 nd two sides ech of length :. How long is the ltitude to those sides? 6. How long re ech of the other two sides? 7. How mny times the P nd A of cm squre re the P nd A of squre with sides ll 6 cm? 8. A rectngle hs re nd one side 6. Find the perimeter. Prolems 9-0: A squre hs perimeter 0: 9. How long is ech side? 0. Wht is its re?. A tringle hs se nd height ech 7. Wht is its re? C. Pythgoren theorem: In ny tringle with 90 (right) ngle, the sum of the squres of the legs equls the squre of hypotenuse. (The legs re the two shorter sides; the hypotenuse is the longest side.) If the legs hve lengths nd, nd c is the hypotenuse length, then + c. In words: In right tringle, leg squred plus leg squred equls hypotenuse squred. exmple: A right tringle hs hypotenuse nd one leg. Find the other leg. Since leg + leg hyp, + x 9 + x x! 9 6 x 6 Prolems -: Find the length of the third side of the right tringle:. one leg:, hypotenuse: 7. hypotenuse: 0, one leg: 8. legs: nd Prolems -6: Find x:

19 8. In right!rst with right ngle R, SR nd TS 6. Find RT. (Drw nd lel tringle to solve.) 9. Would tringle with sides 7,, nd e right tringle? Why or why not? Similr tringles re tringles which re the sme shpe. If two ngles of one tringle re equl respectively to two ngles of nother tringle, then the tringles re similr. exmple:!abc nd!fed re similr: The pirs of sides which correspond re AB nd FE, BC nd ED, AC nd FD. D A 6 C Prolems 0-: Use this figure: B 0. Find nd nme two similr tringles.. Drw the tringles seprtely nd lel them.. List the three pirs of corresponding sides. If two tringles re similr, ny two corresponding sides hve the sme rtio (frction vlue): exmple: the rtio to x, or x, is the sme s y nd c z. Thus x y, c x z, nd c. Ech of y z these equtions is clled proportion.. Drw the similr tringles seprtely, lel them, 0 nd write 9 proportions for the corresponding sides. 8 Prolems -7: Solve for x: exmple: Find x y writing nd solving proportion: x, so cross multiply nd get x or x x 7. E AC 0; EC 7 BC ; DC x D A B C F 6 A E D A B C E x x x D. Grphing on the numer line: Prolems 8-: Nme the point with given coordinte: A B C D E F G " ". "... Prolems 6-: On the numer line ove, wht is the distnce etween the listed points? (Rememer tht distnce is lwys positive.) 6. D nd G 9. B nd C 7. A nd D 0. B nd E 8. A nd F. F nd G Prolems -: On the numer line, find the distnce from:. 7 to. to 7. 7 to. to 7 Prolems 6-9: Drw sketch to help find the coordinte of the point : 6. Hlfwy etween points with coordintes nd. 7. Midwy etween nd. 8. Which is the midpoint of the segment joining 8 nd. 9. On the numer line the sme distnce from 6 s it is from 0. E. Coordinte plne grphing: To locte point on the plne, n ordered pir of numers is used, written in the form ( x, y). Prolems 60-6: Identify coordintes x nd y in ech ordered pir: 60. (,0) 6. (," ) 6. ", 6. ( 0,) To plot point, strt t the origin nd mke the moves, first in the x-direction (horizontl) nd 0

20 then the y-direction (verticl) indicted y the ordered pir. exmple: (-, ) Strt t the origin, move left (since x!), then (from there), up (since y ), put dot there to indicte the point (-, ) 6. On grph pper, join these points in order: (, ), (, ), (, 0), (, ), (, ), (, 0), (, ), (, ), (, ) Two of the lines drwn in prolem 6 cross ech other. Wht re the coordintes of the crossing point? 66. In wht qudrnt is the point (,) if > 0 nd < 0? Prolems 67-69: ABCD is squre, with C(," ) nd D (","). Find: 67. the length A y B of ech side. 68. the coordintes of A. x 69. the coordintes of the midpoint of DC. D C Prolems 70-7: Given A( 0,), B(,0): 70. Sketch grph. Drw AB. Find its length. 7. Find the midpoint of AB nd lel it C. Find the coordintes of C. 7. Wht is the re of the tringle formed y A, B, nd the origin? Answers:. 0 units, 0 units (units mens squre units). units, 6 units. miles, 9 miles. yrds, 9 6 yrds. 0 un., 60 un un., 7 un un., 0 un. 8. un., 6 un. 9. 0" un., " un. 0. 0" ft., 00" ft.. " km, " km. 7.. ". 6. Cnnot tell 7. P is times, A is times No, ecuse 7 + " 0. "ABE ~ "ACD D. E A. AB, AC ; AE, AD ; BE, CD. 9. or. or D 9. E 0. C. F. B. G. B. F B A C x,y 0 6. x ",y 6. x,y " 6. x 0,y (0, ) 66. IV ", 69. (,") ( 6, ) 7. 0

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