MPE Review Section I: Algebra

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1 MPE Review Section I: lger t Colordo Stte Universit, the College lger sequence etensivel uses the grphing fetures of the Tes Instruments TI-8 or TI-8 grphing clcultor. Whenever possile, the questions on the Mth Plcement Em nd in this review hve een designed to ccommodte the student who hs no knowledge of these grphing fetures. However, students m find it helpful to lern out these fetures. Tes Instruments mintins list of TI-8/8 tutoril wesites t POLYNOMIL MNIPULTION Polnomil epressions contin letters tht smolize numers nd oe the usul lws of rithmetic. In dding, sutrcting or multipling polnomils, we ppl these lws nd comine like terms. Emple: Simplif the polnomil epression ( )( ) ( ). Solution: ( )( ) ( ) ( ) ( ) ( ) Prolems:. Simplif: ( 7)( ) ( ). Simplif: (7 ) [ ( )]. Simplif: ( )( ). Simplif: ( ) ( )( ) SOLVING ONE LINER EQUTION IN ONE UNKNOWN n eqution ( sttement equting two epressions) cn e mnipulted dding (or sutrcting) the sme quntit from oth sides, or multipling (or dividing) oth sides the sme non-zero quntit. To solve simple liner eqution we mnipulte the eqution into the form, then divide to conclude tht Emple: Solve for in the following eqution: ( ) ( ). Solution: First perform ll indicted multiplictions: 8. Net simplif ech side seprtel: 8. Now isolte ll terms on one side of the eqution nd ll constnt terms on the other dding to oth sides nd dding to oth sides. Divide to otin the solution. - -

2 Prolems:. Solve for : ( ). Solve for u: [u (u )] u 7. Solve for R: R (R ) [R ( R)] MTHEMTICL MODELS mthemticl model is formul or lgeric eqution formed to model rel-world situtions. To model rel-world sitution: Red the prolem crefull. Wht informtion is given to ou? Wht is the prolem looking for? Let (or n other vrile) represent one of the quntities of the prolem. Write n epression with the informtion given. Evlute the epression. Rered the prolem. Does our nswer mke sense with the contet of the prolem? Emple: The length of the rectngulr tennis court is feet longer thn twice the width. If the court s perimeter is 8 feet, wht re the courts dimensions? Solution: Let the width of the court. The length must e. We re given tht the perimeter 8. The formul for perimeter is: W L P. ( ) ( ) 8 Since, 8 length () length 78 The dimensions of the tennis court re ft. 78 ft. Prolems: 8. For lrge prties, mn resturnts utomticll dd n 8% grtuit to the ill. If the totl ill, including grtuit is $., wht ws the ill efore dding the grtuit? 9. jo ps n nnul slr of $,0, which includes er-end onus of $70. If pchecks re issued twice month, wht is the gross mount of ech pcheck? INEQULITIES/SOLUTE VLUES Mnipulte inequlities in the sme w s equlities with one notle eception. The direction of the inequlit chnges when multipling or dividing the inequlit negtive numer. Such is the cse in the following: 8 > so ( )(8) < ( )(), since <. Emple: Simplif the following:. Solution: Simplif the following:. dd to oth sides:. Sutrct from oth sides,, so tht. Note tht we divided positive numer, therefore, no chnge in the inequlit sign occurs. The solution of n inequlit is set of numers nd cn e grphed on the numer line. In the ove emple, the solutions re grphed s the (closed) hlf-line to the right of / (inclusive). - -

3 Emple: Simplif the following: >. Solution: Sutrct from oth sides: >. Sutrct from oth sides: >. Then <. Note tht we divided negtive numer; therefore, sign chnge occurs in our inequlit. In this emple, ll < / is grphed s the (open) hlf-line to the left of / ecluding /. Prolems: 0. Find the solution set of ( ) > ( ).. Find the solution set of ( ).. Find the solution set of ( ) Find the solution set of ( )( ) < ( )( ). The solute vlue of numer is defined s if 0 if < 0 Hence, nd. The solute vlue of cn e visulized on numer line s the distnce the point lies from the origin. Grphicll we see tht is three units from the origin nd is five units from the origin. Emple: Find the set of -vlues stisfing <. Solution: When 0, nd the inequlit ecomes <. Thus, the non-negtive -vlues tht stisf < re given 0 <. When < 0, nd the inequlit ecomes <, or <. Thus, the negtive - vlues tht stisf < re given < < 0. It follows tht the set of ll -vlues stisfing < consists of ll numers strictl etween nd (ecluding nd ). The set of -vlues stisfing <, is the collection of numers within two units of the origin on the numer line. Emple: Find the solution set of <. Solution: n solution of < must stisf oth of the inequlities < nd - -

4 <. Write these two inequlities together s < < nd then solved t the sme time. dd to ll memers of the inequlit to get < <. Divide to find tht the solution set is the intervl < < of ll -vlues strictl etween nd. Prolems:. Find the solution set of.. Find the solution set of. Emple: Find the set of -vlues stisfing. Solution: When 0, nd the inequlit ecomes. Thus, the non-negtive - vlues tht stisf re given. When < 0, nd the inequlit ecomes, or. Thus, the negtive - vlues tht stisf re given. Thus, the set of ll -vlues stisfing is comprised of two disjoint sets. One set is the rel numers greter thn or equl to. The other set is the rel numers less thn or equl to. The set of -vlues stisfing consists of the numers two or more units from the origin on the numer line. Emple: Find the solution set of. Solution: n solution of must stisf either the inequlit or the inequlit. These two inequlities cnnot e written together nd should not e solved t the sme time. When >, When <, >, <, so > so <. The solution set of is comprised of two pieces. One piece is the rel numers greter thn or equl to. The other is the rel numers less thn or equl to. Notice tht the solution set for < is n intervl nd the solution set for is the complement of the sme intervl. Prolems:. Find the solution set of. 7. Find the solution set of 7 >. - -

5 GRPHING ON THE PLNE The grph of n eqution m, where m nd re specific numers, is stright line. Equtions of this form re clled liner equtions. Emple: Find some points on the grph of. Solution: First compute tle of vlues which stisf the given eqution Plotting these pirs gives the grph shown in Figure 8. Figure 8 The grph of is the stright line through these points. Prolems: 8. Sketch the grph of. 9. Sketch the grph of. (Consider.) 0. Sketch the grph of. (Consider.). Sketch the grph of.. Sketch the grph of. LINES IN THE PLNE In plne geometr, ou lern tht two distinct points lie on ectl one stright line. Its slope chrcterizes the direction of line. Given n two points on line, define the slope of the line is rise slope run chnge in chnge in ccording to this definition, the slope of horizontl line is zero (0) nd the slope of verticl line is undefined. Emple: Consider the line joining (, ) nd (7, ). The slope of this line is 7. Grphicll we hve Figure 9. Prolems: Figure 9. Wht is the slope of the line joining (, ) nd (7, )?. Wht is the slope of the line joining (, ) nd ( h, h h )? Through given point P (, ), there is ectl one line with slope m. If the coordintes of P nd the slope m re known, we cn esil find n eqution relting the - nd -coordintes of n point (, ) - -

6 on the line. In generl, the line through ( ) coordintes (, ) stisf P, 0 0 with slope m consists of ll points whose m which cn e rewritten s m ) ( 0 This is the point-slope form of n eqution of line. It displs the coordintes of one point on the line nd the slope m of the line. Emple: n eqution of line is ( ). The line then psses through the point ( 0, 0 ) (,) nd hs slope of m. Prolems:. Find n eqution of the line through (, ) with slope of.. Find the slope nd the coordintes of point on the line whose eqution is ( ). 7. Find n eqution of the line through (, ) with slope Find n eqution of the line through (, ) nd (, ). Now we consider nother useful form of n eqution of line, the slope-intercept form. Emple: If the point-slope form of n eqution of line is ( ), then we cn rewrite the eqution s 8 or. In this form one cn esil tell tht the point (0, ) is on the line nd tht the slope, given the coefficient of, is. More generll the slope-intercept form is m, where m is the slope nd (0, ) is the point where the line intercepts the -is. Prolems: 9. Write in slope-intercept form n eqution for the line through (, ) with slope. 0. Write 7 in the slope-intercept form.. Wht is the slope of the line 8? Wht is the -intercept? Emple: The following three points lie on the sme line: (,.8), (, ), nd (,-). the vlue of? Wht is Solution: First, find the slope of the line using the two points (,.8) nd (,-). (.8) m ( ) Find the eqution of the line using the slope nd either one of the given points. ( ) 0. ( ) Now we cn use this eqution to find the vlue of. 0.().8 - -

7 Prolems:. The following three points lie on the sme line: (, ), (, ), nd (,).. The following three points lie on the sme line: (,.), (,.7), nd (,-.). Wht is the vlue of? Wht is the vlue of? nother w to think of the slope of line is the verge rte of chnge. The verge rte of chnge over n intervl is the chnge in cross the intervl divided the chnge in cross the sme intervl. Emple: The tle elow could e mthemticl model for some sitution. Wht is the verge rte of chnge over the intervl from to? Solution: chnge in chnge in. 7. Prolems: Use the tle ove to nswer the following prolems.. Wht is the verge rte of chnge over the intervl from 0 to?. Wht is the verge rte of chnge over the intervl from to 9?. Wht is the verge rte of chnge over the intervl from to? SYSTEMS OF LINER EQUTIONS Review our reference tet for vriet of techniques for solving sstem of two liner equtions in two unknowns nd for solving three liner equtions in three unknowns. Emple: Consider the two equtions nd. The ssocited lines hve different slopes ( nd ) nd therefore intersect t single point. The coordintes of this point give the solution to this pir of equtions. Emple: Now consider the two equtions nd 0. The ssocited lines hve the sme slope, /, ut different -intercepts, nd /, respectivel. Hence, these lines re prllel nd do not intersect. Since the lines hve no point in common, this pir of equtions hs no solution. Emple: Finll consider the equtions nd 8. These two equtions not onl hve the sme slope, ut the represent the sme line. In this cse, there re infinitel mn solutions, one corresponding to ech point (, ) on the line. Prolems: 7. Solve for nd :

8 8. Solve for,, nd z: z 0 z 7 z. 9. In one week, Ferless Frnk drove 00 miles on the freew nd 0 miles in town. His cr consumed 8. gllons of gsoline. The net week his journes took him over 80 miles of freew nd 0 miles of locl driving, nd he used onl 0.9 gllons of gs. Wht were his highw m.p.g. nd his locl m.p.g.? DISTNCE FORMUL To compute the distnce, d, etween n ritrr pir of points, use the distnce formul. Consider nd point Q with coordintes (, ). point P with coordintes ( ), d ( ) ( ) Emple: Find the distnce etween (, ) nd (, 9). Let P (, ) nd Q (, 9). Solution: d ( ) ( ) [ ( ) ] () ( ) ( 9 ) 9 Prolems: 0. Determine the distnce etween (, ) nd (, ).. Determine the distnce etween (, ) nd (, ).. Find the length of the digonl of squre of re. EQUTION OF CIRCLE The stndrd form of the eqution of circle with center (h, k) nd rdius r is ( r h) ( k). Emple: Wht is the eqution for the circle in Figure 0? Solution: The center of the circle in Figure 0 is the point (, ), the rdius is. The eqution of this circle is: ( ) ( ). Prolems:. Wht is the eqution for the circle in Figure?. Grph: ( ) ( ) 9 Figure 0. Grph: ( ) ( ). Figure - 8 -

9 PIECEWISE FUNCTIONS piecewise function is function tht is defined two (or more) equtions over specific domin. The solute vlue of numer is defined s piecewise function: if 0 if < 0 Emple: Emm s cellulr phone pln costs $ ech month nd includes 00 free minutes of cll time. Emm hs to p n etr cents for ech minute she uses over 00. The following piecewise function represents how much her phone ill,, is in dollrs depending on how mn minutes,, of cll time she used..( 00) 00 > 00 Emple: The following tle represents Emm s phone usge for the pst months. Wht re the missing entries? Numer of minutes used, Phone ill,, in dollrs Solution: Since 79 < 00, the first eqution is used to find :. The phone ill is $. Since 8 > 00, use the second eqution to find :.(8 00) 79.. The phone ill is $79.0. In ddition, since 00 00, the first eqution is used to find :. The phone ill is $. Prolems: t t 8 t <. Consider the piecewise defined function: s. Wht re the missing entries t t in this tle? t s s s s SOLVING POLYNOMILS Y FCTORING Fctoring is the process of writing polnomil s the product of two or more polnomils. See our reference tet for review of the techniques for fctoring polnomils. If the polnomil eqution hs zero on one side nd fctored epression on the other side, it cn e solved using the zero-product propert (if 0, then either 0, 0, or oth). qudrtic eqution m e epressed in the form c 0 nd then m possil e solved fctoring. Emple: The eqution 0 cn e written in fctored form s ( )( ) 0. Since the product with is required to e 0, one or the other fctor must equl 0. Thus, or

10 Emple: The eqution 0 cn e fctored grouping, ( ) ( ) ( ) - ( ) ( )( - ) 0. Thus, or ±. Prolems: 7. Solve: 0 8. Solve: Solve: ( )( ) ( 9)( ) 0. Solve: 7 0. Solve: 0 COMPLEX NUMERS Comple numers involve the imginr unit i. Note tht i, i i, i. Thus, ever positive integrl power of i is equl to one of the four numers i,, i, or. comple numer is n numer of the form i, where nd re rel numers. Emple: i 8 [i ] i i i Emple: ( i)( 7i) i i i 9i 9 9i i ( i)( i) i i i Emple: i. i ( i)( i) i Prolems:. Simplif: ( i)( 7i) i 0 i. Write in the form i. i. Simplif: i 8 QUDRTIC FORMUL Comple numers re sometimes solutions to qudrtic equtions. When we cnnot fctor given qudrtic convenientl, (such s ) we refer to the qudrtic formul for finding the roots of qudrtic eqution. The formul pplies to qudrtic equtions written in the form c 0 The qudrtic formul ss tht the solutions to this eqution re ± c In the erlier emple - 0, the coefficients re, -, nd c. So, the qudrtic formul, solutions re nd ( ) ± ( ) ()() ± which gives us.

11 When the qudrtic formul is pplied to some qudrtic equtions, negtive numers occur under the squre root sign. Since negtive numers do not hve squre roots tht re rel numers, the solutions to these equtions involve the imginr numer i. Emple: Solve 0. Solution: Using the qudrtic formul, we find the solutions re ± ±, ± i. The quntit - c tht occurs under the squre root in the qudrtic formul is clled the discriminnt of the qudrtic. It determines the numer nd nture of the solutions of the qudrtic eqution c 0. i. When - c > 0, the eqution hs two distinct rel solutions. ii. When - c 0, the eqution hs onl one rel solution. iii. When - c < 0, the eqution hs two distinct, comple solutions. Prolems:. Compute the discriminnt nd decide wht tpe of solutions - 0 hs.. Compute the discriminnt nd decide wht tpe of solutions - 0 hs. 7. Compute the discriminnt nd decide wht tpe of solutions hs. 8. Find the solutions in Prolem. 9. Find the solutions in Prolem. 0. o is feet high nd feet longer thn it is wide. If its volume is 00 cuic feet, find its width.. o rides ike 8 miles. The trip would hve tken 0 minutes less if he hd ridden miles per hour fster. Find his ctul speed. COMPLETING THE SQURE Ever qudrtic function cn e epressed n eqution in verte form. The epression ( h) k 0 is clled the verte (or completed squre) form of the qudrtic function c 0. The grph of qudrtic function is prol whose verte is locted t some point with coordintes (h, k). The verte of prol is lso the mimum (if < 0) or minimum (if > 0) vlue of the function. n eqution written in verte form cn e solved for. See our reference tet for review of the techniques for completing the squre. Emple: Rewrite 0 0 in verte form, nd solve for. Solution: Fctor the coefficient of 0 : 0 dd nd sutrct the squre of hlf the coefficient of : Fctor nd comine like terms:

12 This qudrtic is in verte form. This tells us tht the verte of the prol is found t, nd this vlue for is the minimum vlue of the function. We cn continue the process nd solve for : ± ± Emple: Rewrite 0 c in verte form, nd solve for. Solution: 0 c 0 0 c c c 0 c Note: The verte is t c,. this look fmilir?) (Does c c c c c ± ±

13 Emple: The eqution Y models the epected crop ield on n eperimentl plot, where Y is the mount of crop the plot is epected to produce (in ushels) nd is the mount of fertilizer pplied in pounds. How much fertilizer should e pplied to produce the mimum ield? Wht is the mimum ield? Solution: The coefficients re: -0.07,.07, nd c.899. Since < 0, the verte of prol is lso the mimum vlue of this eqution. Use the verte formul c, to find the verte..07 Sustitute vlues for nd pounds of ( 0.07) fertilizer should e pplied to produce the mimum ield. Sustitute vlues for, nd c, c.07 ()( 0.07)(.899) 0.8. ( 0.07) The mimum crop ield is 0.8 ushels. nother w to find the mimum is to sustitute.7 into the originl eqution: 0.07(.7).07(.7) Prolems:. Rewrite 0 in verte form.. Find the verte nd solve for : completing the squre. Is the verte mimum or minimum vlue?. Find the verte nd solve for : 8 0 completing the squre. Is the verte mimum or minimum vlue?. The eqution 0 models the numer,, of inmtes (in thousnds) incrcerted in Federl nd Stte Correctionl Fcilities in the United Sttes over time. In this model, is the numer of ers since Jnur, 980. ccording to this model, how mn ers fter 980 were there 0,000 inmtes incrcerted in Federl nd Stte Correctionl Fcilities in the United Sttes?. theter hs 0 sets. The theter will sell tickets for ll sets if the tickets cost $0 ech. However, for ech $ increse in ticket price, 0 fewer tickets re sold. Wht ticket price will generte the most revenue nd wht is the mimum revenue? CONSTRUCTING POLYNOMILS You cn esil write polnomil with specified zeros ut there re mn solutions since ou could include other zeros or constnt multiplier. To reconstruct specific polnomil in stndrd form from the given zeros ou need to know the degree of the polnomil nd point on the grph of tht polnomil. See our reference tet for review of the techniques for constructing (finding) polnomils. Emple: Construct polnomil of degree tht hs the numers,, nd s its zeros nd hs -intercept. Solution: Since the polnomil is of degree, we know the given three zeros re the onl zeros of the polnomil. First, multipl the fctors together to get polnomil in stndrd form. - -

14 ( )( )( ) ( )( ) Sustitute in the point (0, ) to find out which constnt numer,, is fctor of the polnomil. f ( ) ( ) f (0) (0 (0) 0 ) Multipl the constnt, to find the polnomil. f ( ) ( ) Prolems: 7. Construct polnomil of degree tht hs the numers, nd s its zeros nd hs -intercept of. 8. Construct polnomil of degree tht hs the numers,, nd s its zeros nd f( ). 9. Construct polnomil of degree tht hs the numers,, nd s its zeros nd hs -intercept of. LGERIC FRCTIONS lgeric frctions re mnipulted ccording to the sme rules s numericl frctions. The m e simplified cnceling fctors common to the numertor nd denomintor. 8 ( 9) ( ) Emple: (8 ) ( ) Prolems: 70. Simplif: 9 ( ) Simplif: 7. Simplif: 7. Simplif: ( h ) ( ) ( ) ( ) ( )( ) [ ] ( ( h) ) h lgeric frctions re multiplied nd divided in the sme w s numericl frctions. Emple: Emple: C D C D C D D C D C fter operting with the frctions, lws tr to simplif the resulting frction. Emple: 8 7 ( )( )( )( ) ( )( )( )( ) Prolems: Multipl nd simplif: Multipl nd simplif: 7 - -

15 Divide nd simplif: Divide nd simplif: 0 To dd or sutrct lgeric frctions, first rewrite them s frctions hving the sme denomintor. Emple: Emple: S R S R R S RS SR RS ( )( ) ( )( ) ( )( ) ( )( ) ( )( )( ) ( )( )( ) ( ² ) - ( ² ) ( )( )( ) Prolems: 78. Simplif: Simplif: 8 9 ( )( )( ) ll our previous skills with frctions re required to simplif lgeric frctions whose numertors nd/or denomintors involve frctions. Emple: Emple: Prolems: 80. Simplif: ( ) ( ) ( ) ( ) ( ) 8. Simplif: ( ) ( ) 8 ( )( ) - -

16 8. Simplif: When solving equtions involving frctions, multipl the eqution the lest common multiple of the denomintors to otin n eqution involving no frctions, nd then solve the resulting eqution. The finl eqution m hve solutions, which do not stisf the originl frctionl eqution. Thus, ll potentil solutions found in this w must e checked in the originl eqution. Emple: Solve. ( ) Solution: Multipl ( - ) to eliminte frctions. ( ) ( ) ±. Check sustituting into the originl eqution. We find tht is not solution nd is solution. Prolems: 8. Solve: 8. Solve: Solve: 8. Solve: 7 0 INTEGER EXPONENTS The sic lws for working with eponents re i) N M N M, iv) 0 ( 0), M N, v) N N NM ii) ( ) iii) () N N N N, vi) N Emple: Simplif ( ) ( ) Solution: ( ) ( ) Emple: Eliminte negtive eponents nd simplif ( 8 - )(- - ). Solution: ( 8 - )(- - ) N nd ( ) Emple: Eliminte negtive eponents nd simplif. () ( ) - -

17 - 7 - Solution: 9 8 ) ( ) ( ) ( Emple: Eliminte negtive eponents nd simplif ( ). Solution: ( ) Prolems: Eliminte negtive eponents nd simplif. 87. ( ) ( )( ) S R S R S R S R S R S R 89. ( ) u v u v u u 90. ( ) 9. ( ) 0 9. ( ) ( ) ) ( RTIONL EXPONENTS ND RDICLS Recll tht for > 0 nd n positive integer, n n mens positive numer such tht n. It follows tht n m n m. Thus, rdicl epressions cn e converted to epressions involving frctionl eponents. These frctionl eponents oe the sme lws s integer eponents previousl reviewed. For instnce,. word of wrning!, not ±. Emple: Simplif epressing rdicls s non-negtive rtionl powers nd comining powers whenever possile: 9 8 Solution:

18 Prolems: Simplif epressing rdicls s non-negtive rtionl powers nd comining powers whenever possile z For tpogrphicl convenience nd historicl resons, it is customr to write epressions involving oth frctions nd rdicls so tht no frction ppers under rdicl sign nd no rdicl ppers in denomintor. This "rtionlizing the denomintor" is ccomplished in severl ws, s illustrted in the following emples. Emple: Emple: Emple: Emple: Prolems: Rtionlize the denomintor in C Rtionlize the denomintor in R S T. 00. Rtionlize the denomintor in. 0. Rtionlize the denomintor in SOLVING RDICL EQUTIONS ( ). Equtions involving rdicls re often solved rewriting the eqution, rising oth sides to some power to remove the rdicl, nd then solving the resulting eqution. s with frctionl equtions, this process m introduce etrneous roots. Potentil solutions must e checked in the originl eqution. Emple: Solve for : Solution: 8 Squring oth sides we get, so 0. Thus, 0 is possile solution.

19 Checking in the originl eqution we find 0 8. Thus 0 is the solution. Emple: Solve for : 0. Solution: - Squring oth sides we get - 9, so 0. Thus, is possile solution. Check in the originl eqution nd find 0 0. (Recll tht mens non-negtive squre root!) Thus, is not solution. This eqution hs no solution. Emple: Solve for : -. Solution: Squre oth sides nd find ( ) 9. Isolte the squre root term to get -, nd squre gin to get We find , or , so - / or. Checking sustituting into the originl eqution, we find tht is solution nd - / is not. The onl solution is. Prolems: 0. Solve for : 0. Solve for : - 0. Solve for : 0 0. Solve for : 7-0 POWER FUNCTIONS function, f ( ) where is rel numer is clled power function. You hve lred worked with mn power functions including functions like /. Emple: 8 Emple: Emple: Solve for : 7 Solution: Emple: Solve for : 0-9 -, /,,

20 Solution: 0 0 Prolems: So, 0 0 ± 0. Solve for : 07. Solve for : ± Solve for : Solve for : 0 or ± ± MORE RTIONL EXPONENTS Here re some more prolems involving eponents. Simplif nd rtionlize the denomintors. 0. ( C ).. u u v 8 v POLYNOMIL MNIPULTION I. LGER - NSWER SECTION SOLVING ONE LINER EQUTION IN ONE UNKNOWN. 0. u 7. R MTHEMTICL MODELS 8. $ $,8-0 -

21 INEQULITIES/SOLUTE VLUES 7 0. >... > or 7. < or > GRPHING IN THE PLNE

22 LINES IN THE PLNE.. h. ( ). m,( 0, 0 ) (,) 7. 0 or 8. slope undefined m, SYSTEMS OF LINER EQUTIONS , 8.,, z 0 9. pproimtel nd DISTNCE FORMUL EQUTION OF CIRCLE. ( ) ( 7) 9.. PIECEWISE FUNCTIONS. t s

23 SOLVING POLYNOMILS Y FCTORING 7., 8., 9., 8 0.,,.,, COMPLEX NUMERS. i. i. QUDRTIC FORMUL. distinct comple. rel, distinct 7. ectl one rel 8. ± i 9. COMPLETING THE SQURE ± 0. feet. miles per hour. 0. verte:,, ±. verte:,,, minimum..8 ers. Ticket prices of $ 70 generte mimum revenue of $,00. CONSTRUCTION POLYNOMILS 7. f ( ) 8. f ( ) 8 9. f ( ) LGERIC FRCTIONS ±, mimum ( ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( )( )( ) ( )( ) ( )( 9)( ) (7 )( 7) ( 7)( ) no solution 8. ± h 79. ( )( ) - -

24 INTEGER EXPONENTS S 87. R v (v ) u ( ) 7 R S ( ) RTIONL EXPONENTS ND RDICLS z TS RS SOLVING RDICL EQUTIONS 0. onl 0. onl 0. inspection, there is no solution 0. onl POWER FUNCTIONS 0. ± , 09. ± 8, ± 7 MORE RTIONL EXPONENTS 0. C u. v. - -

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac

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