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### sol6p

Course: MA 320, Spring 2011
School: Kentucky
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ARITHMETIC Milo 1 DIGITAL D. Ercegovac and Toms Lang s a Morgan Kaufmann Publishers, an imprint of Elsevier Science, c 2004 Updated: September 23, 2003 Chapter 6: Solutions to Exercises With contributions by Elisardo Antelo and Fabrizio Lamberti Exercise 6.1 a) Radix-2, sj {1, 0, 1}, conventional (nonredundant) residual We have x = 144 28 = 0.10010000 and = 1. We choose s0 = 0. Therefore the...

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ARITHMETIC Milo 1 DIGITAL D. Ercegovac and Toms Lang s a Morgan Kaufmann Publishers, an imprint of Elsevier Science, c 2004 Updated: September 23, 2003 Chapter 6: Solutions to Exercises With contributions by Elisardo Antelo and Fabrizio Lamberti Exercise 6.1 a) Radix-2, sj {1, 0, 1}, conventional (nonredundant) residual We have x = 144 28 = 0.10010000 and = 1. We choose s0 = 0. Therefore the initialization is w [0] = x s0 = 0.10010000. We use the result-digit selection function for redundant residual but we consider only 2 integer bits since the range of the residual estimate is smaller than in the redundant case. 2w [0] = F1 [0] = w [1] = 001.00100000 11.10000000 00.10100000 y=1 F1 [0] = 11.10000000 s1 = 1 2w [1] = F1 [1] = w [2] = 001.01000000 10.11000000 00.00000000 y=1 F1 [1] = 00.11000000 s2 = 1 2w [2] = F1 [2] = w [3] = 000.00000000 10.01100000 10.01100000 y=0 F1 [2] = 01.01100000 s3 = 1 2w [3] = F1 [3] = w [4] = 100.11000000 01.10110000 10.01110000 y = 4 F1 [3] = 10.00110000 s 4 = 1 2w [4] = F1 [4] = w [5] = 100.11100000 01.10011000 10.01111000 y = 4 F1 [4] = 10.01011000 s 5 = 1 2w [5] = F1 [5] = w [6] = 100.11110000 01.10001100 1110.01111100 y = 4 F1 [5] = 01.10001100 s 6 = 1 Digital Arithmetic - Ercegovac & Lang 2004 Chapter 6: Solutions to Exercises 2 (continues on next page) 2w [6] = F1 [6] = w [7] = 100.11111000 01.10000110 10.01111110 y = 4 F1 [6] = 10.01110110 s7 = 1 2w [7] = F1 [7] = w [8] = 100.11111100 01.10000011 10.01111111 y = 4 F1 [7] = 10.01111011 s8 = 1 2w [8] = F1 [8] = w [9] = 100.11111110 01.10000001 10.01111111 y = 4 F1 [8] = 10.01111101 s9 = 1 We perform 9 iterations to compute the additional bit required for rounding. Since w [9] < 0 the correction step has to be performed. Thus s9 = 2. The result is s = 0.111 = (0.11000000)2 1 11 11 2 b) Radix-2, sj {1, 0, 1}, carry-save residual 2W S [0] = 2W C [0] = F1 [0] = W S [1] = W C [1] = 0001.00100000 0000.00000000 111.10000000 110.10100000 010.00000000 y=1 2W S [1] = 2W C [1] = F1 [1] = W S [2] = W C [2] = 1101.01000000 0100.00000000 110.11000000 111.10000000 000.10000000 y=1 2W S [2] = 2W C [2] = F1 [2] = W S [3] = W C [3] = 1111.00000000 0001.00000000 110.01100000 000.01100000 110.00000000 y=0 2W S [3] = 2W C [3] = F1 [3] = W S [4] = W C [4] = 0000.11000000 1100.00000000 001.10110000 101.01110000 001.00000000 s1 = 1 y = 4 F1 [0] = 111.10000000 s2 = 1 F1 [1] = 000.11000000 s3 = 1 F1 [2] = 001.01100000 s 4 = 1 F1 [3] = 110.00110000 (continues on next page) Digital Arithmetic - Ercegovac & Lang 2004 Chapter 6: Solutions to Exercises 3 2W S [4] = 2W C [4] = F1 [4] = W S [5] = W C [5] = 1010.11100000 0010.00000000 001.10011000 001.01111000 101.00000000 y = 4 2W S [5] = 2W C [5] = F1 [5] = W S [6] = W C [6] = 0010.11110000 1010.00000000 001.10001100 001.01111100 101.00000000 y = 4 2W S [6] = 2W C [6] = F1 [6] = W S [7] = W C [7] = 0010.11111000 1010.00000000 001.10000110 001.01111110 101.00000000 y = 4 2W S [7] = 2W C [7] = F1 [7] = W S [8] = W C [8] = 0010.11111100 1010.00000000 001.10000011 001.01111111 101.00000000 y = 4 2W S [8] = 2W C [8] = F1 [8] = W S [9] = W C [9] = 0010.11111110 1010.00000000 001.10000001 001.01111111 101.00000000 y = 4 s5 = 1 F1 [4] = 110.01011000 s6 = 1 F1 [5] = 001.10001100 s7 = 1 F1 [6] = 110.01110110 s8 = 1 F1 [7] = 110.01111011 s9 = 1 F1 [8] = 110.01111101 We perform 9 iterations to compute the additional bit required for rounding. Since w [9] < 0 the correction step has to be performed. Thus s9 = 2. The result is s = 0.111 = (0.11000000)2 1 11 11 2 c) Radix-4, sj {2, 1, 0, 1, 2}, carry-save residual a Since = r1 = 111.10010000. 2 3 < 1, s0 should be 1. Therefore w [0] = 1 s0 = Digital Arithmetic - Ercegovac & Lang 2004 Chapter 6: Solutions to Exercises 4 4W S [0] = 4W C [0] = F1 [0] = W S [1] = W C [1] = 1110.01000000 0000.00000000 001.11000000 11.10000000 00.10000000 S = 1.0000 y = 1110.010 S [0] = 1 s1 = 1 S [1] = 0.11 4W S [1] = 4W C [1] = W S [2] = W C [2] = 1110.00000000 0010.00000000 00.00000000 00.00000000 S = 0.1100 y = 0000.000 s2 = 0 S [2] = 0.1100 4W S [2] = 4W C [2] = 0000.00000000 0000.00000000 S = 0.1100 y = 0000.000 s3 = 0 S [3] = 0.110000 Since w = 0, the rest of the digits of S are 0. We perform 4 iterations to take into account the generation of the additional bit required for rounding. The radix-4 digits of the result are s0 = 1, s1 = 1 , s2 = 0, s3 = 0, s4 = 0 and s5 = 0. The result is s = (0.11000000)2 Digital Arithmetic - Ercegovac & Lang 2004 Chapter 6: Solutions to Exercises 5 Exercise 6.3 a) Use S [j ] in its original signed digit form In this case it is not necessary the on-the-y conversion of S [j ] for implementing the recurrence. Neverthless the register K [j ] is stil necessary. F [j ] is computed as Sj +1 2S [j ] + Sj +1 r(j +1) which requires a single concatenation of Sj +1 , and a digit multiplication by Sj +1 . Since F [j ] is represented in signed-digit form, the adder of the recurrence is more complex, that is, both operands are redundant. b) Convert S [j ] to twos complement representation The is conversion on-the-y, and since this conversion is already necessary, it does not introduce additional complexity. The adder is simpler that in a) since one operand is in nonredundant form. More specically the term Sj +1 2S [j ] + Sj +1 r(j +1) is generated in nonredundant form as follows: Sj +1 0 Concatenate Sj +1 to 2S [j ] in position j + 1 . Set the most signicant digit to one to have a negative operand (the weight of the most signicant digit is negative). Then perform digit multiplication. Sj +1 < 0 In this case 2S [j ] + Sj +1 r(j +1) = 2 S [j ] r j + (2r Sj +1 ) r(j +1) The term S [j ] r j is available from the on-the-y conversion module. The term 2r Sj +1 is precomputed for every digit and is concatenated to 2 S [j ] r j in postion j + 1 . Finally, the digit multiplication is performed. Digital Arithmetic - Ercegovac & Lang 2004 Chapter 6: Solutions to Exercises 6 Exercise 6.5 a) Network for digit selection Figure E6.5a shows the network for the selection of sj +1 and sj +2 in a radix-2 square root implementation using two radix-2 overlapped stages. 2 2 2 w[j] 2 w[j] 2F1 [j] 2F-1 [j] 2(2w[j]) CSA 4 4 4 SELSQR 2w[j] CSA 4 4 SELSQR 4 SELSQR 4 SELSQR 2 2 4 s j+1 2 2 3-1 MUX 2 s j+2 Figure E6.5a: Network for digit selection. b) Network to produce the next residual In Figure E6.5b the network producing the next residual is illustrated. c) Delay analysis Conventional implementation Computing the delay in the critical path we have tcycle = tSELSQRT (4) + tbuf f (1) + tmux (1) + tHA (1) + treg (2) = 9tg The latency of the conventional implementation (8 fractional bits) can be computed as 8 tcycle = 8 9tg = 72tg . Overlapped implementation Computing the delay in the critical path we have that the delay to produce W [j + 1] (that is, the delay from W [j ] to W [j + 1]) is tSELSQRT (4) + tbuf f (1) + tmux (1) + tHA (1) = 7tg Moreover, the delay to produce sj +2 (delay of CSA + delay of selection network + delay of 3-1 multiplexer) is tCSA (2) + tSELSQRT (4) + tmux (1) + tbuf f (1) = 8tg Digital Arithmetic - Ercegovac & Lang 2004 Chapter 6: Solutions to Exercises 7 w[0] initial values j=0 j=0 F1 , F-1 , K REGISTERS WS,WC REGISTERS K[j] F-1 [j] s j+1 F1 [j] F1 , F-1 , K MODULE MUX 2w[j] K[j+1] 3-2 CSA F-1 [j+2] s F1 [j+2] j+2 F1 , F-1 , K MODULE MUX 2w[j+1] K[j+2] F1 [j+2] 3-2 CSA F [j+2] -1 w[j+2] Figure E6.5b: Network to produce the next residual. Digital Arithmetic - Ercegovac & Lang 2004 Chapter 6: Solutions to Exercises 8 Finally, the delay to produce W [j + 2] (delay to produce sj +2 + delay of buer + delay of mux + delay of HA) can be computed as 8tg + 1tg + 1tg + 1tg = 12tg Adding the register delay we get tcycle = 11tg + 2tg = 13tg . Computing the latency of the overlapped implementation (8 fractional bits) we get 4 tcycle = 4 13tg = 52tg Exercise 6.8 We compute the radix-4 square root of x = (53)10 = (00110101)2 . Since n = 8, we perform a right-shift of m = 2 bits and produce x = .11010100. The number of bits of the integer result is 82 = 3. Consequently, two radix2 4 iterations are necessary. We have S [0] = 1 and w [0] = x 1 = 11.11010100. Note that no alignment to digit boundary is needed, since the square root algorithm does not require to compute a remainder. The iterations are as follows: 4W S [0] = 4W C [0] = F1 [0] = W S [1] = W C [1] = 1111.01010000 0000.00000000 001.11000000 10.10010000 10.10000000 4W S [1] = 4W C [1] = 1010.01000000 1010.00000000 y = 1111.0101 s1 = 1 S [1] = 0.11 y = 0100.0100 s2 = 2 S [2] = 0.1110 We do not need to compute w [2]. Therefore the result is s = 23 (0.111) = 111 = (7)10 Digital Arithmetic - Ercegovac & Lang 2004 Chapter 6: Solutions to Exercises 9 Exercise 6.13 We develop a radix-4 selection function for J = 3, t = 3 and = 4. k>0 min (Uk1 (Ii )) = 2 max (Lk (Ii )) = 2 1 + i 24 2 k 1 + (i + 1) 24 2 1 3 k 2 3 k0 min (Uk1 (Ii )) = 2 max (Lk (Ii )) = 2 1 + (i + 1) 24 2 1 + i 24 2 k 2 3 k + k Lk = max ( Lk (Ii ) 3 ) mk (i) min ( Uk1 (Ii )) 23 3 2 3 1 3 2 44 = Uk1 To improve the presentation of results, we use a bound for max (Lk (Ii )) . 2 More specically, we want an upper bound of the term k 2 44 . For 3 2 25 4 1 1 1 k = 0 we have 9 44 = 576 < 512 . For k = 1 we have 5 44 = 2304 < 64 . 3 The selection constants are presented in Table E6.13. Note that we give only half of the table (for S [j ] = 8, 9, 10, 11) since there is an interval U2 L1 that is negative. Consequently, there is no selection function for t = 3 and = 4. Digital Arithmetic - Ercegovac & Lang 2004 Chapter 6: Solutions to Exercises 10 S [j ] 8 9 10 11 L2 , U 1 12, 12 14, 14 15, 15 16, 17 m2 12 14 15 16 L1 , U 0 3, 4 4, 5 4, 5 4, 6 m1 4 4 4 4 L0 , U1 5, 4 5, 5 6, 5 7, 5 m0 4 5 6 6 L1 , U2 13, 13 14, 15 16, 16 18, 17 m1 13 X 16 18 Table E6.13: Selection interval and mk constants. S [j ]: real value= shown value/16. Lk , Uk1 and mk : real value = shown value/8. Digital Arithmetic - Ercegovac & Lang 2004 Chapter 6: Solutions to Exercises
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Kentucky - MA - 320
1DIGITAL ARITHMETICMilo D. Ercegovac and Toms LangsaMorgan Kaufmann Publishers, an imprint of Elsevier Science, c 2004 Updated: September 9, 2003 Chapter 5: Solutions to Selected Exercises With contributions by Elisardo Antelo and Fabrizio Lambert
Kentucky - MA - 320
1DIGITAL ARITHMETICMilo D. Ercegovac and Toms LangsaMorgan Kaufmann Publishers, an imprint of Elsevier Science, c 2004 Updated: February 13, 2004 Chapter 4: Solutions to Selected Exercises With contributions by Elisardo Antelo Exercise 4.1x= 30
Kentucky - MA - 320
1DIGITAL ARITHMETICMilo D. Ercegovac and Toms LangsaMorgan Kaufmann Publishers, an imprint of Elsevier, c 2004Chapter 3: Solutions to Selected Exercises with contributions by Elisardo Antelo Exercise 3.1As explained in the text, for twos compleme
Kentucky - MA - 320
1DIGITAL ARITHMETICMilo D. Ercegovac and Toms LangsaMorgan Kaufmann Publishers, an imprint of Elsevier, c 2004 Updated: February 13, 2004 Chapter 2: Solutions to Selected Exercises with contributions by Elisardo Antelo Exercise 2.1Assuming that
Kentucky - EE - 380
Kentucky - EE - 380
Kentucky - EE - 380
Kentucky - EE - 380
Purdue - MA - 26100
MA26100ASSIGNMENT SHEETSummer 2011Text: James Stewart Calculus, Early Transcendentals, Sixth EditionLesson1Section12.1-4Studyreview212.5, 12.6all313.113.213.313.414.114.214.314.414.514.614.714.714.815.115.215.315.415.415.5
Kentucky - EE - 380
Kentucky - EE - 380
Purdue - MA - 26100
MA26100CalendarSummer201113June12.112.41412.5,12.6151613.1,13.217Q13.32013.421Q 2214.12314.2,14.324Q14.4,14.52714.6,14.74JulyNoclass28Review,EXAM13014.7,14.81Q15.1,15.25Q 615.3,15.4715.4,15.58Q15.61115.71816.312
Kentucky - EE - 380
Kentucky - EE - 380
Purdue - MA - 26100
Ground Rules for MA 26100, Summer 2011Midterm Examinations: There will be two, one-hour, midterm exams.Final Examination: There will be a two-hour, comprehensive nal during exam week, August3 August 5. The time and place will be announced later.Grades
Kentucky - EE - 380
Kentucky - EE - 380
Purdue - MA - 26100
Kentucky - EE - 380
Purdue - MA - 26100
M A 26100: C OMMON OFFICE HOURS - S UMMER 2011MA 26100 is taught MTTHF 7:50-9:40 a.m., MTTHF 9:50-11:40 a.m. and MTTHF 1:00-2:50 p.m.C oordinator: D avid NorrisMONDAYTUESDAY11:00noon12:001:001:002:002:103:103:204:20* Denotes GraderWEDNESDAYTHU
Kentucky - EE - 280
Problem Solutions to Problems Marked With a * inLogic Computer Design Fundamentals, Ed. 2CHAPTER 3 2000 by Prentice-Hall, Inc.3-2.DT1 =T3 =X==Y==T3XCBYT4T1ABC,T2 = AD1T4 = D + BC,T3T4D + BCT2T4AD(D + BC) = A BCDT2YX11X0
Purdue - MA - 26600
MA26600ASSIGNMENT SHEETSUMMER 2011Text: Elementary Dierential Equations and Boundary Value Problems, by Boyce and DiPrima,9th Edition, WileyBoldface letters denote Supplemental Problems available on the course web page:www.math.purdue.edu./coursesL
Purdue - MA - 26600
MA26600CalendarSummer201113June1.1,dfield8141.2,1.315Q 162.12.2172.3(heavy)202.421Q 222.52.6232.7(heavy)24Q3.127Review4JulyNoclass28EXAM1293.2303.31Q3.45Q 63.53.67
Purdue - MA - 26600
Ground Rules for MA 26600, Summer 2011Midterm Examinations: There will be two, one-hour, midterm exams.Final Examination: There will be a two-hour, comprehensive nal during exam week, August3 August 5. The time and place will be announced later.Grades
Purdue - MA - 26600
Purdue - MA - 26600
M A 26600: C OMMON OFFICE HOURS - S UMMER 2011MA 26600 is taught M-F 8:40-9:40 a.m., M-F 9:50-10:50 a.m. and M-F 11:00 a.m.-12:00 p.m.C oordinator: Y oung Hwan You* Denotes graderMONDAYTUESDAYWEDNESDAYTHURSDAY12:001:00L ei Z hangM ath 1037B enj
Purdue - MA - 11100
MA 11100, Exam 1 Answers, Spring 2011Problem1)Form ADForm BCActual Answer472)CAEvery rational number is an integer.3)BA4)ADII and III only5)EBx=6)CB1057)BEx + 0.35 x + 2.50 = 19.358)DDa=9)BD15m8t 510)AC2 x 8y11
Purdue - MA - 11100
MA 11100, Exam 1 Grade Approximations, Spring 2011Exam Average: 68.0Exam Range: 33 100These cut-offs are intended for approximations only. If you arenear a cut-off, your score could be considered either grade. Forexample, if you received a 87, your g
Purdue - MA - 11100
MA 11100, Exam 2 Grade Approximations, Spring 2011Exam Average: 62.4Exam Range: 13 100These cut-offs are intended for approximations only. If you arenear a cut-off, your score could be considered either grade. Forexample, if you received an 80, your
Purdue - MA - 11100
MA 11100Final ExamSpring 2011NAME:_ Purdue ID:_-_1. Fill in your name and Purdue ID above.2. You must use a #2 pencil on the answer sheet.3. Make sure that the color of this page matches the color of your answer sheet.4. On the answer sheet, fill i
Purdue - MA - 11100
MA 11100Assignment SheetSpring 2011Text: Intermediate Algebra, Concepts and Applications by Bittinger and Ellenbogen, 8th edition, AddisonWesley, 2010. The textbook is shrink-wrapped with a 3-ring binder and the MyMathLab Student Access Kitthat is use
Purdue - MA - 11100
MA 11100 Exam 2 Answers, Spring 2011Problem1)Form ABForm BDActual Answer9 x + 5 y = 42)CAOne line has a negative slope, theother a zero slope.3)DCy=4)EC y = 3 x423 x 2 y = 105)DB( f + g )(10) = 2506)AEThe solution is in Quad
Purdue - MA - 11100
Purdue - MA - 11100
Purdue - MA - 11100
MA 11100CLASS PERIODGround RulesSpring 2011Students are expected to attend every class meeting and to read the appropriate sections of the text before coming toclass. Instructors frequently may not have time to cover every topic in class. Refer to yo
WPI - ECON - 1100
Final: ANSWER SHEETIntro Micro (ECON 1110)Name _Version ADate: _In each question there is only one correct answer. Select the answer you believe to be most correct andrecord it in the ANSWER SHEET.Name: _ECON 1110Date: _ID: AFinalMultiple Choi
WPI - ECON - 1100
Intro Micro (ECON 1110)Final: ANSWER SHEET, Version AName _Date: _1.abcde21.abcde2.abcde22.abcde3.abcde23.abcde4.abcde24.abcde5.abcde25.abcde6.abcde26.abcde7.abcde27.abc
Purdue - MA - 11100
MA 11100ScheduleSpring 2011Exam 1: Lessons 1-11; Exam 2: Lessons 12-22; Exam 3: Lessons 23-36JanuaryFebruaryMarchAprilMonday10Lesson 117 MLK day(no class)24Lesson 631Lesson 97Exam Review14Lesson 1421Lesson 1728Lesson 207Exam Revi
Purdue - MA - 11100
MA 11100 Calculator PolicyThe recommended calculator for MA 11100 students is a TI-30XA. It is the requiredcalculator for MA 15200 and MA 15300. You will need a calculator for some homeworkproblems. A calculator is never allowed on quizzes or exams.
Purdue - MA - 11100
These slides are to help thosestudents who have previously usedCourseCompass (previouslyregistered) for MA 15200 or MA11100. If you used MathXL; it isdifferent than MyMathLab. Youwould need to view thepresentation to register for the firsttime.Fo
Purdue - MA - 11100
Need your MyMathLab card with youraccess code (behind the pull off tab) Need a Valid E-Mail Address Need to know Purdues zip code is 47907and your course ID for your Class You may use an ITaP computer or on yourown computer (need Adobe FlashPlayer)
Purdue - MA - 11100
Purdue - MA - 11100
MA 11100Even AnswersSp11These even problems are from the bolded print problems on the assignment list. Anyeven problems that correspond to problems on coursecompass are not included in thisdocument. You will know the correct answers to those problems
Purdue - MA - 11100
MA 11100, Exam 3 Grade Approximations, Spring 2011Exam Average: 62.3Exam Range: 27 100These cut-offs are intended for approximations only. If you arenear a cut-off, your score could be considered either grade. Forexample, if you received a 87, your g
Purdue - MA - 11100
Exponent Rules and ExamplesAZero Exponent Rulea0 = 1Examples:1)12 0 = 12)( xyz ) 0 = 13)4 wrv 0 = 4 wr (1) = 4 wr4)60 + 4 = 1 + 4 = 56 x 0 6(1)5)==61h0BNegative Exponent Rulesna1=na1= annaabnb= anExamples:1)2)3)(3)
Purdue - MA - 11100
Lessons 23 Sections 4.2 and 4.33-part Inequality, Absolute Value Inequalities3-Part Inequality: 2 &lt; x &lt; 10 x &lt; 10 AND x &gt; 2The number must meet both conditions, therefore the conjunction and.Where are these numbers on the number line?-10-8-6-40-
Purdue - MA - 11100
Lesson 24, Section 5.1PolynomialsDefinition: A term is a number, a variable, a power of a variable, or the product of anyof these. A term is also commonly called a monomial. 3wExamples: 4x4 xy 2Definition: A Polynomial is a sum (or difference) of
Purdue - MA - 11100
Lesson 25Section 5.2Multiplication of PolynomialsTo multiply two monomials, use the rules of exponents.1)(8 x 2 y 3 z )(2 x5 y 2 z 2 ) =2)(5a 2b3 )(3a 5b 2 ) =To multiply a monomial and a polynomial with 2 or more terms, use the distributiveprope
Purdue - MA - 11100
Lesson 26, Sections 5.3 and 5.4 (part 1)Factoring out the Greatest Common Factor, Factoring by GroupingFactoring Trinomials (part 1)Factoring out the GCF is reversing the distributive property. It is putting thepolynomial back as a product (multiplied
Purdue - MA - 11100
Lessons 27Factoring Trinomials, Perfect Square Trinomials, Difference of SquaresTRINOMIALS (leading coefficient not a 1)Form: ax 2 + bx + c Always write terms in descending order!Notice: (3 x 5)(2 x + 3) = 6 x 2 + 9 x 10 x 15 = 6 x 2 x 156 x 2 is the
Purdue - MA - 11100
Lesson 28Factoring CompletelyFACTORING COMPLETELY1. Always factor out a GCF first, if possible.2. Count the number of terms. If there are 2 terms (binomial), look for a difference of squares pattern. If there are 3 terms (trinomial), look for a perf
Purdue - MA - 11100
Lesson 29Section 5.8Using Factoring to Solve Some Equations.Principle of Zero Products: If two factors have a product of 0, at least one of thefactors must be zero. ab = 0 a = 0 or b = 0Solve:( x 2)( x + 3) = 01)2)3 y (2 y + 1)( y 5) = 0Steps fo
Purdue - MA - 11100
Lesson 30Section 6.1Rational Expressions and FunctionsA Rational Expression is a polynomial divided by a non-zero polynomial.The following are examples of rational expressions.32w9+ xr2 ry2 2y + 5,,,,4x53r + 1y2 8A Rational Function is
Purdue - MA - 11100
Lesson 31Section 6.2Addition or Subtraction of Rational ExpressionsRemember: Fractions (rationals) can only be added or subtracted if they have a common325denominator. For example: + =888To add or subtract rational expressions with the same denomin
Purdue - MA - 11100
Lesson 32Section 6.4Rational EquationsRemember that a fraction cannot have a zero denominator. Because a rationalexpression cannot have a zero denominator, you must determine any values of x thatwould make a zero denominator when solving equations wi
Purdue - MA - 11100
Lesson 33, Section 6.5Application Problems Using Rational Equations1.2.3.4.1)Define a VariableDevelop A PlanWrite an EquationSolve and Answer the QuestionThe reciprocal of 5, plus the reciprocal of 7, is the reciprocal of what number?Let x = t
Purdue - MA - 11100
Lesson 34Section 6.8, VariationExamine this table:# hours workedPay1\$82\$163\$244\$326\$4910\$80When a relation between pairs of numbers is a constant ratio, such as above; it is called a8Direct Variation. The ratio above is and we say the p
Purdue - MA - 11100
Lesson 36Section 7.212What is a value for 9 ?Consider This.121211+229 9 = 9= 91 = 9 using the product rule of exponents.Now, Think!What other number times itself equals 9? 3 3 = 912Since both products equal 9, we can conclude that 9 = 3
Purdue - MA - 11100
Lesson 37Examine the following:4 9 = 23 = 6Sections 7.3 &amp; 7.4Since both equal 6, the expressions are equal.4 9 = 36 = 6Conclusion: 4 9 = 4 9Likewise:16=44=22Since both equal 2, the expressions are equal.16= 4=24Conclusion:164=164Th
Purdue - MA - 11100
Lesson 38Sections 7.5When two radicals have the same indices (plural of index) and same radicands, they aresaid to be 'like radicals'. They can be combined the same as 'like terms'.Like Radicals: 3 r , r,2 3 5m ,4r 3 x 3 5m , 12 3 5m5 4 5, 10 4 5
Purdue - MA - 11100
Lesson 39 Sections 7.6 and 8.1Solving Radical EquationsUsing the Principle of Square Roots to Solve an Equationx=3You know you can add, subtract, multiply, or divide (by nonnegative number) and get atrue equation. Let's see if both sides can be raise