C Precalculus Review. C.1 Real Numbers and the Real Number Line. Real Numbers and the Real Number Line

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1 C. Rel Numers nd the Rel Numer Line C C Preclculus Review C. Rel Numers nd the Rel Numer Line Represent nd clssif rel numers. Order rel numers nd use inequlities. Find the solute vlues of rel numers nd find the distnce etween two rel numers. Rel Numers nd the Rel Numer Line Rel numers cn e represented coordinte sstem clled the rel numer line or -is (see Figure C.). The rel numer corresponding to point on the rel numer line is the coordinte of the point. As Figure C. shows, it is customr to identif those points whose coordintes re integers The rel numer line Figure C Rtionl numers Figure C The point on the rel numer line corresponding to zero is the origin nd is denoted 0. The positive direction (to the right) is denoted n rrowhed nd is the direction of incresing vlues of. Numers to the right of the origin re positive. Numers to the left of the origin re negtive. The term nonnegtive descries numer tht is either positive or zero. The term nonpositive descries numer tht is either negtive or zero. Ech point on the rel numer line corresponds to one nd onl one rel numer, nd ech rel numer corresponds to one nd onl one point on the rel numer line. This tpe of reltionship is clled one-to-one correspondence. Ech of the four points in Figure C. corresponds to rtionl numer one tht cn e written s the rtio of two integers. (Note tht 4.5 = 9 nd.6 = 3.) 5 Rtionl numers cn e represented either terminting decimls such s 5 = 0.4 or repeting decimls such s 3 = = 0.3. Rel numers tht re not rtionl re irrtionl. Irrtionl numers cnnot e represented s terminting or repeting decimls. In computtions, irrtionl numers re represented deciml pproimtions. Here re three fmilir emples π e (See Figure C.3.) e π Irrtionl numers Figure C.3

2 C Appendi C Preclculus Review Order nd Inequlities One importnt propert of rel numers is tht the re ordered. For two rel numers nd, is less thn when is positive. This order is denoted the inequlit <. 0 < if nd onl if lies to the left of. Figure C.4 This reltionship cn lso e descried sing tht is greter thn nd writing >. If three rel numers,, nd c re ordered such tht < nd < c, then is etween nd c nd < < c. Geometricll, < if nd onl if lies to the left of on the rel numer line (see Figure C.4). For emple, < ecuse lies to the left of on the rel numer line. Severl properties used in working with inequlities re listed elow. Similr properties re otined when < is replced nd > is replced. (The smols nd men less thn or equl to nd greter thn or equl to, respectivel.) Properties of Inequlities Let,, c, d, nd k e rel numers.. If < nd < c, then < c. Trnsitive Propert. If < nd c < d, then + c < + d. Add inequlities. 3. If <, then + k < + k. Add constnt. 4. If < nd k > 0, then k < k. Multipl positive constnt. 5. If < nd k < 0, then k > k. Multipl negtive constnt. Note tht ou reverse the inequlit when ou multipl the inequlit negtive numer. For emple, if < 3, then 4 >. This lso pplies to division negtive numer. So, if > 4, then <. A set is collection of elements. Two common sets re the set of rel numers nd the set of points on the rel numer line. Mn prolems in clculus involve susets of one of these two sets. In such cses, it is convenient to use set nottion of the form {: condition on }, which is red s follows. The set of ll such tht certin condition is true. { : condition on } For emple, ou cn descrie the set of positive rel numers s {: > 0}. Set of positive rel numers Similrl, ou cn descrie the set of nonnegtive rel numers s {: 0}. Set of nonnegtive rel numers The union of two sets A nd B, denoted A B, is the set of elements tht re memers of A or B or oth. The intersection of two sets A nd B, denoted A B, is the set of elements tht re memers of A nd B. Two sets re disjoint when the hve no elements in common.

3 Intervls on the Rel Numer Line C. Rel Numers nd the Rel Numer Line C3 The most commonl used susets re intervls on the rel numer line. For emple, the open intervl (, ) = {: < < } Open intervl is the set of ll rel numers greter thn nd less thn, where nd re the endpoints of the intervl. Note tht the endpoints re not included in n open intervl. Intervls tht include their endpoints re closed nd re denoted [, ] = {: }. Closed intervl The nine sic tpes of intervls on the rel numer line re shown in the tle elow. The first four re ounded intervls nd the remining five re unounded intervls. Unounded intervls re lso clssified s open or closed. The intervls (, ) nd (, ) re open, the intervls (, ] nd [, ) re closed, nd the intervl (, ) is considered to e oth open nd closed. Intervl Nottion Set Nottion Grph Bounded open intervl (, ) {: < < } Bounded closed intervl [, ] {: } Bounded intervls (neither open nor closed) [, ) (, ] {: < } {: < } Unounded open intervls (, ) (, ) {: < } {: > } Unounded closed intervls (, ] [, ) {: } {: } Entire rel line (, ) {: is rel numer} Note tht the smols nd refer to positive nd negtive infinit, respectivel. These smols do not denote rel numers. The simpl enle ou to descrie unounded conditions more concisel. For instnce, the intervl [, ) is unounded to the right ecuse it includes ll rel numers tht re greter thn or equl to.

4 C4 Appendi C Preclculus Review Liquid nd Gseous Sttes of Wter Descrie the intervls on the rel numer line tht correspond to the tempertures (in degrees Celsius) of wter in. liquid stte.. gseous stte. Solution. Wter is in liquid stte t tempertures greter thn 0 C nd less thn 00 C, s shown in Figure C.5(). (0, 00) = {: 0 < < 00}. Wter is in gseous stte (stem) t tempertures greter thn or equl to 00 C, s shown in Figure C.5(). [00, ) = {: 00} () Temperture rnge of wter (in degrees Celsius) Figure C.5 () Temperture rnge of stem (in degrees Celsius) If rel numer is solution of n inequlit, then the inequlit is stisfied (is true) when is sustituted for. The set of ll solutions is the solution set of the inequlit. Solve 5 < 7. Solution Solving n Inequlit If = 0, then (0) 5 = 5 < 7. If = 5, then (5) 5 = 5 < If = 7, then (7) 5 = 9 > 7. Checking solutions of 5 < 7 Figure C.6 5 < 7 Write originl inequlit < Add 5 to ech side. < Simplif. < < 6 Simplif. The solution set is (, 6). Divide ech side. In Emple, ll five inequlities listed s steps in the solution re clled equivlent ecuse the hve the sme solution set. Once ou hve solved n inequlit, check some -vlues in our solution set to verif tht the stisf the originl inequlit. You should lso check some vlues outside our solution set to verif tht the do not stisf the inequlit. For emple, Figure C.6 shows tht when = 0 or = 5 the inequlit 5 < 7 is stisfied, ut when = 7 the inequlit 5 < 7 is not stisfied.

5 C. Rel Numers nd the Rel Numer Line C5 Solve 3 5. Solution Solving Doule Inequlit [, ] Solution set of 3 5 Figure C Write originl inequlit Simplif Simplif. The solution set is [, ], s shown in Figure C.7. Sutrct from ech prt. Divide ech prt 5 nd reverse oth inequlities. The inequlities in Emples nd 3 re liner inequlities tht is, the involve first-degree polnomils. To solve inequlities involving polnomils of higher degree, use the fct tht polnomil cn chnge signs onl t its rel zeros (the -vlues tht mke the polnomil equl to zero). Between two consecutive rel zeros, polnomil must e either entirel positive or entirel negtive. This mens tht when the rel zeros of polnomil re put in order, the divide the rel numer line into test intervls in which the polnomil hs no sign chnges. So, if polnomil hs the fctored form ( r )( r )... ( r n ), r < r < r 3 <... < r n then the test intervls re (, r ), (r, r ),..., (r n, r n ), nd (r n, ). To determine the sign of the polnomil in ech test intervl, ou need to test onl one vlue from the intervl. Solve < + 6. Solving Qudrtic Inequlit Choose = 3. ( 3)( + ) > 0 Choose = 4. ( 3)( + ) > 0 Solution < + 6 Write originl inequlit. 6 < 0 Write in generl form. ( 3)( + ) < 0 Fctor. 3 0 Choose = 0. ( 3)( + ) < 0 Testing n intervl Figure C The polnomil 6 hs = nd = 3 s its zeros. So, ou cn solve the inequlit testing the sign of 6 in ech of the test intervls (, ), (, 3), nd (3, ). To test n intervl, choose n numer in the intervl nd determine the sign of 6. After doing this, ou will find tht the polnomil is positive for ll rel numers in the first nd third intervls nd negtive for ll rel numers in the second intervl. The solution of the originl inequlit is therefore (, 3), s shown in Figure C.8.

6 C6 Appendi C Preclculus Review Asolute Vlue nd Distnce If is rel numer, then the solute vlue of is = {,, 0 < 0. The solute vlue of numer cnnot e negtive. For emple, let = 4. Then, ecuse 4 < 0, ou hve = 4 = ( 4) = 4. Rememer tht the smol does not necessril men tht is negtive. REMARK You re sked to prove these properties in Eercises 73, 75, 76, nd 77. Opertions with Asolute Vlue Let nd e rel numers nd let n e positive integer.. =. =, 0 3. = 4. n = n Properties of Inequlities nd Asolute Vlue Let nd e rel numers nd let k e positive rel numer... k if nd onl if k k. 3. k if nd onl if k or k. 4. Tringle Inequlit: + + Properties nd 3 re lso true when is replced < nd is replced >. Solving n Asolute Vlue Inequlit Solve 3. 0 units units 4 5 Solution set of 3 Figure C Solution Using the second propert of inequlities nd solute vlue, ou cn rewrite the originl inequlit s doule inequlit. 3 Write s doule inequlit Simplif. The solution set is [, 5], s shown in Figure C.9. Add 3 to ech prt.

7 C. Rel Numers nd the Rel Numer Line C7 (, 5) (, ) Solution set of + > 3 Figure C.0 d d Solution set of d d d Solution set of d Figure C. d d + d + d Solve + > 3. A Two-Intervl Solution Set Solution Using the third propert of inequlities nd solute vlue, ou cn rewrite the originl inequlit s two liner inequlities. + < 3 or + > 3 < 5 > The solution set is the union of the disjoint intervls (, 5) nd (, ), s shown in Figure C.0. Emples 5 nd 6 illustrte the generl results shown in Figure C.. Note tht for d > 0, the solution set for the inequlit d is single intervl, wheres the solution set for the inequlit d is the union of two disjoint intervls. The distnce etween two points nd on the rel numer line is given d = =. The directed distnce from to is nd the directed distnce from to is, s shown in Figure C.. Distnce etween nd Directed distnce from to Directed distnce from to Figure C. Distnce on the Rel Numer Line Distnce = Figure C The distnce etween 3 nd 4 is 4 ( 3) = 7 = 7 or 3 4 = 7 = 7. (See Figure C.3.). The directed distnce from 3 to 4 is 4 ( 3) = 7. c. The directed distnce from 4 to 3 is 3 4 = 7. The midpoint of n intervl with endpoints nd is the verge vlue of nd. Tht is, Midpoint of intervl (, ) = +. To show tht this is the midpoint, ou need onl show tht ( + ) is equidistnt from nd.

8 C8 Appendi C Preclculus Review C. Eercises Rtionl or Irrtionl? In Eercises 0, determine whether the rel numer is rtionl or irrtionl The interest rte r on lons is epected to e greter thn 3% nd no more thn 7%. 4. The temperture T is forecst to e ove 90 F tod. 3. 3π 4. 3 Solving n Inequlit In Eercises 5 44, solve the inequlit nd grph the solution on the rel numer line ( )3 Repeting Deciml In Eercises 4, write the repeting deciml s rtio of two integers using the following procedure. If = , then 00 = Sutrcting the first eqution from the second produces 99 = 63 or = = Using Properties of Inequlities Given <, determine which of the following re true. () + < + (c) 5 > 5 () 5 < 5 (d) < (e) ( )( ) > 0 (f) < 6. Intervls nd Grphs on the Rel Numer Line Complete the tle with the pproprite intervl nottion, set nottion, nd grph on the rel numer line. Intervl Nottion (, 4] (, 7) Set Nottion {: 3 } Grph < 3 < < > > 3. < 3. 3 > > <, > < < < < < 9 3 Distnce on the Rel Numer Line In Eercises 45 48, find the directed distnce from to, the directed distnce from to, nd the distnce etween nd = = = 5 = () = 6, = 75 () = 6, = () = 9.34, = 5.65 () = 6 5, = 75 Using Asolute Vlue Nottion In Eercises 49 54, use solute vlue nottion to define the intervl or pir of intervls on the rel numer line. Anlzing n Inequlit In Eercises 7 0, verll descrie the suset of rel numers represented the inequlit. Sketch the suset on the rel numer line, nd stte whether the intervl is ounded or unounded < < = = = 3 = < 8 5. = 0 = 4 Using Inequlit nd Intervl Nottion In Eercises 4, use inequlit nd intervl nottion to descrie the set = 0 = 4. is t lest 4.. q is nonnegtive

9 C. Rel Numers nd the Rel Numer Line C9 53. () All numers tht re t most 0 units from () All numers tht re t lest 0 units from 54. () is t most two units from. () is less thn δ units from c. Finding the Midpoint In Eercises 55 58, find the midpoint of the intervl = = = 5 = () [7, ] () [8.6,.4] 58. () [ 6.85, 9.35] () [ 4.6,.3] 59. Profit The revenue R from selling units of product is R = 5.95 nd the cost C of producing units is C = To mke (positive) profit, R must e greter thn C. For wht vlues of will the product return profit? 60. Fleet Costs A utilit compn hs fleet of vns. The nnul operting cost C (in dollrs) of ech vn is estimted to e C = 0.3m where m is mesured in miles. The compn wnts the nnul operting cost of ech vn to e less thn $0,000. To do this, m must e less thn wht vlue? 6. Fir Coin To determine whether coin is fir (hs n equl proilit of lnding tils up or heds up), ou toss the coin 00 times nd record the numer of heds. The coin is declred unfir when For wht vlues of will the coin e declred unfir? 6. Dil Production The estimted dil oil production p t refiner is p,50,000 < 5,000 where p is mesured in rrels. Determine the high nd low production levels. Which Numer Is Greter? In Eercises 63 nd 64, determine which of the two rel numers is greter. 63. () π or () π or () 5 or () 8 or Approimtion Powers of 0 Light trvels t the speed of meters per second. Which est estimtes the distnce in meters tht light trvels in er? () () (c) (d) Writing The ccurc of n pproimtion of numer is relted to how mn significnt digits there re in the pproimtion. Write definition of significnt digits nd illustrte the concept with emples. True or Flse? In Eercises 67 7, determine whether the sttement is true or flse. If it is flse, eplin wh or give n emple tht shows it is flse. 67. The reciprocl of nonzero integer is n integer. 68. The reciprocl of nonzero rtionl numer is rtionl numer. 69. Ech rel numer is either rtionl or irrtionl. 70. The solute vlue of ech rel numer is positive. 7. If < 0, then =. 7. If nd re n two distinct rel numers, then < or >. Proof In Eercises 73 80, prove the propert. 73. = 74. = [Hint: ( ) = ()( )] 75. =, = 77. n = n, n =,, 3, k if nd onl if k k, k > k if nd onl if k or k, k > Proof Find n emple for which >, nd n emple for which =. Then prove tht for ll,. 8. Mimum nd Minimum Show tht the mimum of two numers nd is given the formul m(, ) = ( + + ). Derive similr formul for min(, ).

10 C0 Appendi C Preclculus Review C. The Crtesin Plne Understnd the Crtesin plne. Use the Distnce Formul to find the distnce etween two points nd use the Midpoint Formul to find the midpoint of line segment. Find equtions of circles nd sketch the grphs of circles. The Crtesin Plne Just s ou cn represent rel numers points on rel numer line, ou cn represent ordered pirs of rel numers points in plne clled the rectngulr coordinte sstem, or the Crtesin plne, fter the French mthemticin René Descrtes. The Crtesin plne is formed using two rel numer lines intersecting t right ngles, s shown in Figure C.4. The horizontl rel numer line is usull clled the -is, nd the verticl rel numer line is usull clled the -is. The point of intersection of these two es is the origin. The two es divide the plne into four prts clled qudrnts. -is Qudrnt II Origin Qudrnt III Qudrnt I (, ) Qudrnt IV -is 4 (3, 4) 3 (, ) (0, 0) (3, 0) (, 3) 3 4 The Crtesin plne Points represented ordered pirs Figure C.4 Figure C.5 Ech point in the plne is identified n ordered pir (, ) of rel numers nd, clled the coordintes of the point. The numer represents the directed distnce from the -is to the point, nd the numer represents the directed distnce from the -is to the point (see Figure C.4). For the point (, ), the first coordinte is the -coordinte or sciss, nd the second coordinte is the -coordinte or ordinte. For emple, Figure C.5 shows the loctions of the points (, ), (3, 4), (0, 0), (3, 0), nd (, 3) in the Crtesin plne. The signs of the coordintes of point determine the qudrnt in which the point lies. For instnce, if > 0 nd < 0, then the point (, ) lies in Qudrnt IV. Note tht n ordered pir (, ) is used to denote either point in the plne or n open intervl on the rel numer line. This, however, should not e confusing the nture of the prolem should clrif whether point in the plne or n open intervl is eing discussed.

11 The Distnce nd Midpoint Formuls C. The Crtesin Plne C Recll from the Pthgoren Theorem tht, in right tringle, the hpotenuse c nd sides nd re relted + = c. Conversel, if + = c, then the tringle is right tringle (see Figure C.6). c (, ) (, ) (, ) The distnce etween two points Figure C.7 d The Pthgoren Theorem: + = c Figure C.6 Now, consider the prolem of determining the distnce d etween the two points (, ) nd (, ) in the plne. If the points lie on horizontl line, then = nd the distnce etween the points is. If the points lie on verticl line, then = nd the distnce etween the points is. When the two points do not lie on horizontl or verticl line, the cn e used to form right tringle, s shown in Figure C.7. The length of the verticl side of the tringle is, nd the length of. B the Pthgoren Theorem, it follows tht the horizontl side is d = + d = +. Replcing nd the equivlent epressions ( ) nd ( ) produces the Distnce Formul. Distnce Formul The distnce d etween the points (, ) nd (, ) in the plne is given d = ( ) + ( ). Finding the Distnce Between Two Points Find the distnce etween the points (, ) nd (3, 4). Solution d = ( ) + ( ) Distnce Formul = [3 ( )] + (4 ) Sustitute for,,, nd. = = =

12 C Appendi C Preclculus Review 6 4 (, ) d d d3 (4, 0) 4 6 Verifing right tringle Figure C.8 (5, 7) Verifing Right Tringle Verif tht the points (, ), (4, 0), nd (5, 7) form the vertices of right tringle. Solution Figure C.8 shows the tringle formed the three points. The lengths of the three sides re s follows. Becuse nd d = (5 ) + (7 ) = = 45 d = (4 ) + (0 ) = 4 + = 5 d 3 = (5 4) + (7 0) = + 49 = 50 d + d = = 50 d 3 = 50 Sum of squres of sides Squre of hpotenuse ou cn ppl the Pthgoren Theorem to conclude tht the tringle is right tringle. Ech point of the form (, 3) lies on this horizontl line. (, 3) (5, 3) Using the Distnce Formul Find such tht the distnce etween (, 3) nd (, ) is 5. Solution Using the Distnce Formul, ou cn write the following. d = 5 d = (, ) Figure C ( 5, 3) (, 0) Midpoint of line segment Figure C.0 (9, 3) 5 = ( ) + [3 ()] Distnce Formul 5 = ( 4 + 4) + 6 Squre ech side. 0 = 4 5 Write in generl form. 0 = ( 5)( + ) Fctor. So, = 5 or =, nd ou cn conclude tht there re two solutions. Tht is, ech of the points (5, 3) nd (, 3) lies five units from the point (, ), s shown in Figure C.9. The coordintes of the midpoint of the line segment joining two points cn e found verging the -coordintes of the two points nd verging the -coordintes of the two points. Tht is, the midpoint of the line segment joining the points (, ) nd (, ) in the plne is ( +, + ). Midpoint Formul For instnce, the midpoint of the line segment joining the points ( 5, 3) nd (9, 3) is ( 5 + 9, ) = (, 0) s shown in Figure C.0.

13 Center: (h, k) Rdius: r Point on circle: (, ) Equtions of Circles C. The Crtesin Plne C3 A circle cn e defined s the set of ll points in plne tht re equidistnt from fied point. The fied point is the center of the circle, nd the distnce etween the center nd point on the circle is the rdius (see Figure C.). You cn use the Distnce Formul to write n eqution for the circle with center (h, k) nd rdius r. Let (, ) e n point on the circle. Then the distnce etween (, ) nd the center (h, k) is given ( h) + ( k) = r. B squring ech side of this eqution, ou otin the stndrd form of the eqution of circle. Definition of circle Figure C. Stndrd Form of the Eqution of Circle The point (, ) lies on the circle of rdius r nd center (h, k) if nd onl if ( h) + ( k) = r. The stndrd form of the eqution of circle with center t the origin, (h, k) = (0, 0), is + = r. If r =, then the circle is clled the unit circle (3, 4) Writing the Eqution of Circle The point (3, 4) lies on circle whose center is t (, ), s shown in Figure C.. Write the stndrd form of the eqution of this circle. (, ) 6 4 Figure C. Solution The rdius of the circle is the distnce etween (, ) nd (3, 4). r = [3 ()] + (4 ) = = 0 You cn write the stndrd form of the eqution of this circle s [ ()] + ( ) = ( 0) ( + ) + ( ) = 0. Write in stndrd form. B squring nd simplifing, the eqution ( h) + ( k) = r cn e written in the following generl form of the eqution of circle. A + A + D + E + F = 0, A 0 To convert such n eqution to the stndrd form ( h) + ( k) = p ou cn use process clled completing the squre. If p > 0, then the grph of the eqution is circle. If p = 0, then the grph is the single point (h, k). If p < 0, then the eqution hs no grph.

14 C4 Appendi C Preclculus Review Completing the Squre Sketch the grph of the circle whose generl eqution is = 0. Solution To complete the squre, first divide 4 so tht the coefficients of nd re oth = 0 Write originl eqution = 0 Divide 4. 4 r = 5, ( ) ) 5 ( + + ( ) = A circle with rdius of nd center t ( 5, ) Figure C.3 ( ) + ( 4 + 0) = 37 4 ( ) + ( 4 + 4) = Group terms. Complete the squre dding ( 5 ) = 5 4 nd ( 4 ) = 4 to ech side. (hlf) (hlf) ( + 5 ) + ( ) = Write in stndrd form. Note tht ou complete the squre dding the squre of hlf the coefficient of nd the squre of hlf the coefficient of to ech side of the eqution. The circle is centered t ( 5, ) nd its rdius is, s shown in Figure C.3. You hve now reviewed some fundmentl concepts of nltic geometr. Becuse these concepts re in common use tod, it is es to overlook their revolutionr nture. At the time nltic geometr ws eing developed Pierre de Fermt nd René Descrtes, the two mjor rnches of mthemtics geometr nd lger were lrgel independent of ech other. Circles elonged to geometr, nd equtions elonged to lger. The coordintion of the points on circle nd the solutions of n eqution elongs to wht is now clled nltic geometr. It is importnt to ecome skilled in nltic geometr so tht ou cn move esil etween geometr nd lger. For instnce, in Emple 4, ou were given geometric description of circle nd were sked to find n lgeric eqution for the circle. So, ou were moving from geometr to lger. Similrl, in Emple 5, ou were given n lgeric eqution nd sked to sketch geometric picture. In this cse, ou were moving from lger to geometr. These two emples illustrte the two most common prolems in nltic geometr.. Given grph, find its eqution. Geometr Alger. Given n eqution, find its grph. Alger Geometr

15 C. The Crtesin Plne C5 C. Eercises See ClcCht.com for tutoril help nd worked-out solutions to odd-numered eercises. Using the Distnce nd Midpoint Formuls In Eercises 6, () plot the points, () find the distnce etween the points, nd (c) find the midpoint of the line segment joining the points.. (, ), (4, 5). ( 3, ), (3, ) 3. (, ), ( 3, 5) 4. ( 3, 3), ( 5 6, ) 5. (, 3), (, ) 6. (, 0), (0, ) Locting Point In Eercises 7 0, determine the qudrnt(s) in which (, ) is locted so tht the condition(s) is (re) stisfied. 7. = nd > 0 8. < 9. > 0 0. (, ) is in Qudrnt II. Vertices of Polgon In Eercises 4, show tht the points re the vertices of the polgon. (A rhomus is qudrilterl whose sides re ll the sme length.) Vertices Polgon. (4, 0), (, ), (, 5) Right tringle. (, 3), (3, ), (, 4) Isosceles tringle 3. (0, 0), (, ), (, ), (3, 3) Rhomus 4. (0, ), (3, 7), (4, 4), (, ) Prllelogrm 5. Numer of Stores The tle shows the numer of Trget stores for ech er from 006 through 05. Select resonle scles on the coordinte es nd plot the points (, ). (Source: Trget Corp.) Yer, Numer, Yer, Numer, Conjecture Plot the points (, ), ( 3, 5), nd (7, 3) in rectngulr coordinte sstem. Then chnge the sign of the -coordinte of ech point nd plot the three new points in the sme rectngulr coordinte sstem. Wht conjecture cn ou mke out the loction of point when the sign of the -coordinte is chnged? Repet the eercise for the cse in which the signs of the -coordintes re chnged. Colliner Points? In Eercises 7 0, use the Distnce Formul to determine whether the points lie on the sme line. 7. (0, 4), (, 0), (3, ) 8. (0, 4), (7, 6), ( 5, ) 9. (, ), (, 0), (, ) 0. (, ), (3, 3), (5, 5) Using the Distnce Formul In Eercises nd, find such tht the distnce etween the points is 5.. (0, 0), (, 4). (, ), (, ) Using the Distnce Formul In Eercises 3 nd 4, find such tht the distnce etween the points is (0, 0), (3, ) 4. (5, ), (5, ) 5. Using the Midpoint Formul Use the Midpoint Formul to find the three points tht divide the line segment joining (, ) nd (, ) into four equl prts. 6. Using the Midpoint Formul Use the result of Eercise 5 to find the points tht divide the line segment joining the given points into four equl prts. () (, ), (4, ) () (, 3), (0, 0) Mtching In Eercises 7 30, mtch the eqution with its grph. [The grphs re leled (), (), (c), nd (d).] () (c) (, 0) (0, 0) () (d) 7. + = 8. ( ) + ( 3) = 4 9. ( ) + = ( + ) + ( 3 4) = 4 6 (, 3) 4 (, ) Writing the Eqution of Circle In Eercises 3 38, write the stndrd form of the eqution of the circle. 3. Center: (0, 0) 3. Center: (0, 0) Rdius: 3 Rdius: Center: (, ) 34. Center: ( 4, 3) Rdius: 4 Rdius:

16 C6 Appendi C Preclculus Review 35. Center: (, ) Point on circle: (0, 0) 36. Center: (3, ) Point on circle: (, ) 37. Endpoints of dimeter: (, 5), (4, ) 38. Endpoints of dimeter: (, ), (, ) 39. Stellite Communiction Write the stndrd form of the eqution for the pth of communictions stellite in circulr orit,000 miles ove Erth. (Assume tht the rdius of Erth is 4000 miles.) 40. Building Design A circulr ir duct of dimeter D is fit firml into the right-ngle corner where sement wll meets the floor (see figure). Find the dimeter of the lrgest wter pipe tht cn e run in the right-ngle corner ehind the ir duct. 53. Proof Prove tht ( +, ) is one of the points of trisection of the line segment joining (, ) nd (, ). Find the midpoint of the line segment joining ( +, ) nd (, ) to find the second point of trisection. 54. Finding Points of Trisection Use the results of Eercise 53 to find the points of trisection of the line segment joining ech pir of points. () (, ), (4, ) () (, 3), (0, 0) True or Flse? In Eercises 55 58, determine whether the sttement is true or flse. If it is flse, eplin wh or give n emple tht shows it is flse. Writing the Eqution of Circle In Eercises 4 48, write the stndrd form of the eqution of the circle nd sketch its grph = = = = = = = = 0 Grphing Circle In Eercises 49 nd 50, use grphing utilit to grph the eqution. Use squre setting. (Hint: It m e necessr to solve the eqution for nd grph the resulting two equtions.) D 55. If < 0, then the point (, ) lies in either Qudrnt II or Qudrnt IV. 56. The distnce etween the points ( +, ) nd (, ) is. 57. If the distnce etween two points is zero, then the two points must coincide. 58. If = 0, then the point (, ) lies on the -is or on the -is. Proof In Eercises 59 6, prove the sttement. 59. The line segments joining the midpoints of the opposite sides of qudrilterl isect ech other. 60. The perpendiculr isector of chord of circle psses through the center of the circle. 6. An ngle inscried in semicircle is right ngle. 6. The midpoint of the line segment joining the points (, ) nd (, ) is ( +, + ) = = 0 Sketching Grph of n Inequlit In Eercises 5 nd 5, sketch the set of ll points stisfing the inequlit. Use grphing utilit to verif our result ( ) + ( ) >

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