Math 101 Final Exam Review Sheet, Exams of Mathematics

Download Math 101 Final Exam Review Sheet and more Exams Mathematics in PDF only on Docsity! Math 101 Final Exam Review Sheet (Answers and some detailed solutions at the end) Exam Date: Thursday, March Exam Time: 9:00 AM - 10:00 AM Exam Coverage: • Section 5.1 (only interest compounded m times per year) • Sections 5.2, 5.3, 2.1, 2.2, 2.3, 2.4, and 2.5. Exam Locations: Room Location Instructor Sections BSONE AUD (Mitchell Auditorium) L. Brown, D. Falco 003, 008, 010, 011, 016 NSBITT 111 (Stein Auditorium) P. Bobo, J. Stueber 007, 012, 013, 017, 018 PISB 120 O. Odintsova, J. Pereira 002, 004, 006, 009, 019 Exam Instructions: 1. BRING YOUR NON-GRAPHING CALCULATOR. No sharing of calculators allowed. Using a graphing calculator/cell phone/smart watch, or sharing calculators, will be considered cheating, and at a minimum, result in a grade of 0 on the exam for all those involved!! 2. The exam will be given to you in paper. You may not use scrap paper - all work should be written on the exam paper. Cross off any work you do not want graded. You must show all work to receive credit. 3. The exam is closed-book and closed-notes. 4. Arrive 10 minutes early and be seated leaving an empty seat between you and your neighbor. 5. On the last page of this review sheet is the formula sheet that will be provided to you during the exam. 6. Note that studying from this review sheet alone is not enough. To be adequately prepared for the exam you should review your notes, the relevant chapters from the textbook, the MyLab assignments and quizzes, and the assigned practice problems from the syllabus.            3 3 1 7 −1 3 6 3 5. For the opening night of an opera house, a total of 1000 tickets were sold. Front orchestra seats cost $80 each, rear orchestra seats cost $60 each, and front balcony seats cost $50 each. The combined number of tickets sold for the front orchestra and rear orchestra exceeded twice the number of front balcony tickets sold by 400. The total receipts for the performance were $62,800. Determine how many tickets of each type were sold. 6. Perform each matrix operation, if possible. If the operation is not possible, explain why. (a) 6 3 8 − 3 −2 −1 4 5 6 0 −5 −7  1 1 −3   −2 −1 8  (b) 3 3 2 3  + 4  4 2 2 (c) −4 0 6 − 1 6 1 −3 5 10 8 (d) −1 2 2 4 (e) 2 1 −1 −1 2 0 0 0 1 −1 −2 2 3 0 −2 1 1 2 0 −1 (f) 1 −2 0 1 3 1 2 −1 1  1 4 0  7. Find the inverse of each matrix, if possible. (a) 2 6 −1 1 (b) 4 2 6 3 (c) 2 5 1 3 3 0 −1 0 1 −2 8. Write each system as a matrix equation, and solve using inverse matrices. (a) 2x + 3y = 5 3x + 5y = 8 (b) x + 2y = 14 2x − y = 5 5   35 35 5 Final Exam Review Solutions • Practice Problems on Sections 5.1-5.3 1. (a) (b) $879.48 538 weeks, or 10 years and 18 weeks 2. (a) $23,295.70 (b) $8,704.30 (c) $10,021.70 (d) $1,370.29 3. (a) $1,213.86 (b) $67,494.80 4. $133,117.54 5. (a) $216.43 (b) $4985.80 (c) $4211.88 • Practice Problems on Chapter 2 1. (a) (13, 7) (b) no solution (c) − 3 y + 9 , y (d) − 4 , 36 2. $2,300 in the 6% account and $9,700 in the 7% account 3. 200 children and 800 adults 4. (a) (1, 1, 2) (b) no solution (c) 1 − 7 z, 1 + 3 z, z 2 4 2 4 (d) (1, 2, 3) 5. 240 front orchestra, 560 rear orchestra, and 200 front balcony 6. (a) 3 5 9 4 10 13  −5 −1 23  (b) 25 14 17 33 21 30 (c) Not possible. Matrices can be added/subtracted only if they have the same number of rows and the same number of columns. The first matrix has 3 columns and the second has 2 columns. (d) 4 −2 9 13 (e) Not possible. Matrices can be multiplied only if the number of columns in the first matrix is equal to the number of rows in the second matrix. The first matrix has 3 columns and the second has 2 rows. − 7 Practice Problems on Chapter 2 • Problem 1a): Solve the system of linear equations x − 4y = −15 −2x + y = −19 using any of the following methods: – Substitution – Elimination – Echelon Method – Gauss-Jordan elimination method We will focus on substitution and elimination methods. In problem 4 a) below, the computations for the echelon and the Gauss-Jordan elimination method are shown. – Substitution: −2x + y = −19 y = 2x − 19 x − 4(2x − 19) = −15 x − 8x + 76 = −15 −7x = −91 −→ x = −91 = 13 y = 2(13) − 19 = 7. Solution: x = 7 and y = 13. – Elimination: Multiply the first equation by 2 and add both equations 2(x − 4y) = 2(−15) −2x + y = −19 2x − 8y = −30 − 2 x + y = − 19 −7y = −49 y = −49 = 7 −7 −2x + 7 = −19 −2x = −26 x = 13 Solution: x = 7 and y = 13.   0 2 −4   −6  • Problem 4a): Solve the system of linear equations using matrices via the echelon or Gauss-Jordan elimination method. Specify all the row operations you use. x + y + z = 4 2y − 4z = −6 3x + 2z = 7 – Echelon method: ∗ Write the augmented matrix  1 1 1 4  0 2 −4 −6 3 0 2 7 ∗ −3R1 + R3 → R3 −3R1 : −3 −3 −3 −12  1 1 1 4  R3 : 3 0 2 7 −3R1 + R3 : 0 −3 −1 −5 ∗ 3R2 + 2R3 → R3 0 2 −4 −6 0 −3 −1 −5 3R2 : 0 6 −12 2R3 : 0 −6 −2 3R2 + 2R3 : 0 0 −14 1 1 −18 −10 −28  1 1 1 0 0 −14 4 −28  ∗ 2 R2 → R2 and − 14 R3 → R3  1 1 1 4  0 1 −2 −3 0 0 1 2 ∗ Substitute and solve  1 1 1 4  0 1 −2 −3 0 0 1 2 x + y + z = 4 y − 2z = −3 z = 2 y − 2(2) = −3 y = 1 x + 1 + 2 = 4 x = 1 Solution: x = 1, y = 1 and z = 2.        0 2 −4   0 1 0 1  0 1 0 1  0 1 0 1 −6  R2 : 2R3 : R2 + 2R3 : 0 0 0 1 −2 −3 024 101 R1 : −R3 : R1 − R3 : 1 0 1 114 102 0 −1 −2 – Gauss-Jordan elimination method: ∗ Write the augmented matrix  1 1 1 4  0 2 −4 −6 3 0 2 7 ∗ −3R1 + R3 → R3 −3R1 : −3 −3 −3 −12  1 1 1 4  R3 : 3 0 2 7 −3R1 + R3 : 0 −3 −1 −5 ∗ 3R2 + 2R3 → R3 0 2 −4 −6 0 −3 −1 −5 3R2 : 0 6 −12 2R3 : 0 −6 −2 3R2 + 2R3 : 0 0 −14 1 1 −18 −10 −28  1 1 1 0 0 −14 4 −28  ∗ 2 R2 → R2 and − 14 R3 → R3  1 1 1 4  0 1 −2 −3 0 0 1 2 ∗ R2 + 2R3 → R2 ∗ R1 − R3 → R1 ∗ R1 − R2 → R1 R1 : 1 1 0 2 −R2 : 0 −1 0 −1 R1 − R2 : 1 0 0 1 Solution: x = 1, y = 1 and z = 2.  1 1 1 4  0 0 1 2  1 1 0 2  0 0 1 2  1 0 0 1  0 0 1 2    