AK Algebra I RQ 2021-22
Transcript of AK Algebra I RQ 2021-22
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Table of Contents Chapter 1. Equations and Inequalities ............................................................. 4
Chapter 2. Verbal Problems ............................................................................. 8
Chapter 3. Linear Graphs................................................................................ 12
Chapter 4. Linear Systems .............................................................................. 16
Chapter 5. Polynomials .................................................................................. 29
Chapter 6. Irrational Numbers ....................................................................... 31
Chapter 7. Univariate Data ............................................................................. 33
Chapter 8. Bivariate Data ............................................................................... 35
Chapter 9. Introduction to Functions ............................................................. 39
Chapter 10. Functions as Models ..................................................................... 43
Chapter 11. Exponential Functions .................................................................. 49
Chapter 12. Sequences ..................................................................................... 53
Chapter 13. Factoring ....................................................................................... 55
Chapter 14. Quadratic Functions ..................................................................... 57
Chapter 15. Parabolas ...................................................................................... 65
Chapter 16. Quadratic-Linear Systems ............................................................. 71
Chapter 17. Cubic and Radical Functions ......................................................... 73
Chapter 18. Transformations of Functions ....................................................... 76
Chapter 19. Discontinuous Functions .............................................................. 78
Notation A code next to each Regents Question number states from which Algebra I Common Core Regents exam the question came. For example, AUG β18 [25] means the question appeared on the August 2018 exam as question 25.
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Chapter 1. Equations and Inequalities
1.1 Properties of Real Numbers 1. JUN '14 [1] Ans: 1 2. AUG β17 [1] Ans: 4 3. JAN β18 [1] Ans: 4 4. JAN β19 [8] Ans: 4 5. JUN β19 [9] Ans: 4 6. AUG β19 [26] Commutative; the property is correct because β5π + 7 = 7 β 5π. 7. JAN β20 [29] Distributive Property; Addition Property 1.2 Solve Linear Equations in One Variable 1. JUN β17 [19] Ans: 1 2. AUG β18 [4] Ans: 2 1.3 Solve Linear Inequalities in One Variable 1. JAN '15 [7] Ans: 1 2. JUN β16 [9] Ans: 4 3. AUG β16 [7] Ans: 1 4. JUN β17 [13] Ans: 4 5. AUG β17 [11] Ans: 1 6. JAN β18 [17] Ans: 2 7. JUN β18 [1] Ans: 4 8. JAN β20 [3] Ans: 1 9. JUN '14 [27] 2 β1 + π β1 β 7 > β12 β9 β π > β12 βπ > β3 π < 3 Ans: 2
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10. AUG '14 [30] 3π₯ + 9 β€ 5π₯ β 3 12 β€ 2π₯ 6 β€ π₯ Ans: 6 11. JUN '15 [30] β8π₯ + 7 < 15 β8π₯ < 8 π₯ > β1 Ans: 0 12. AUG '15 [34] 7π₯ β 12π₯ + 24 β€ 6π₯ + 12 β 9π₯ β5π₯ + 24 β€ β3π₯ + 12 12 β€ 2π₯ 6 β€ π₯ or π₯ β₯ 6 {6,7,8} is the set of integers that are greater than or equal to 6 in the interval 13. JAN β17 [27] 1.8 β 0.4π¦ β₯ 2.2 β 2π¦ 1.6π¦ β₯ 0.4 π¦ β₯ 0.25 14. JAN β19 [25] 3600 + 1.02π₯ < 2000 + 1.04π₯ 1600 < 0.02π₯ 80,000 < π₯ or π₯ > 80,000 1.4 Compound Inequalities [CC] There are no Regents exam questions on this topic. 1.5 Solve Equations with Fractions 1. JUN '14 [5] Ans: 1 2. AUG '14 [20] Ans: 1 3. AUG β17 [13] Ans: 2 4. JAN β18 [22] Ans: 4 5. JAN β19 [5] Ans: 2 6. AUG β19 [4] Ans: 3
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7. JAN β20 [5] Ans: 2 8. JUN β18 [30] 6 β π₯ + 5 = 4π₯ 3 6 β 3 π₯ + 5 = 3 4π₯ 18 β 2 π₯ + 5 = 12π₯ 18 β 2π₯ β 10 = 12π₯ 8 β 2π₯ = 12π₯ 8 = 14π₯ π₯ = = 9. JUN β19 [25] 12 β π₯ + 12 + 12 π₯ = 12 β π₯ + 12[2] β8 π₯ + 12 + 8π₯ = β15π₯ + 24 β8π₯ β 96 + 8π₯ = β15π₯ + 24 β96 = β15π₯ + 24 β120 = β15π₯ π₯ = 8 10. AUG β19 [29] 15 π₯ + 15 < 15 π₯ β 15 9π₯ + 5 < 12 β 5 10 < 3π₯ < π₯ 1.6 Solve Literal Equations and Inequalities 1. JAN '16 [6] Ans: 3 2. JAN β17 [4] Ans: 3 3. JUN β17 [2] Ans: 2 4. JUN β17 [23] Ans: 3 5. JUN β18 [23] Ans: 4 6. JAN β19 [20] Ans: 2 7. JUN β19 [13] Ans: 4
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8. AUG '14 [34] 2π΄ = β π + π = π + π β π = π π = β β 12 = 8 ft. 9. JAN '16 [31] ππ₯ β 3π β₯ ππ₯ + 7π β10π β₯ π β π π₯ β β₯ π₯ 10. JUN β16 [31] = π β 2 + 2 = π 11. AUG β16 [32] 4ππ₯ + 12 β 3ππ₯ = 25 + 3π ππ₯ + 12 = 25 + 3π ππ₯ = 13 + 3π π₯ = 12. AUG β18 [29] πΎ = πΉ + 459.67 πΉ = πΎ β 459.67 13. JUN β19 [30] π = ππ β 3π = ππ β β = 14. AUG β19 [28] ππ‘ = π£ β π£ ππ‘ + π£ = π£ 15. JAN β20 [32] 2π = π(π + π) = π + π β π = π
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Chapter 2. Verbal Problems
2.1 Translate Expressions 1. AUG '15 [3] Ans: 4 2. JAN β17 [18] Ans: 4 3. AUG β17 [12] Ans: 2 4. AUG β19 [1] Ans: 3 2.2 Translate Equations 1. MAY '13 [4] Ans: 4 2. JUN '14 [16] Ans: 2 3. JAN '16 [11] Ans: 2 4. AUG β16 [16] Ans: 3 5. JUN β19 [15] Ans: 2 6. AUG β19 [20] Ans: 2 2.3 Translate Inequalities 1. AUG '15 [5] Ans: 4 2. JUN β16 [7] Ans: 2 3. JUN β18 [6] Ans: 1 4. JUN β19 [10] Ans: 1 2.4 Linear Model in Two Variables 1. JUN '14 [7] Ans: 3 2. JUN '14 [22] Ans: 4 3. AUG '14 [2] Ans: 2 4. JAN '15 [1] Ans: 2 5. JAN '15 [23] Ans: 4 6. JUN '15 [1] Ans: 3 7. AUG '15 [8] Ans: 4 8. AUG β16 [14] Ans: 3 9. JAN β17 [9] Ans: 2 10. AUG β17 [9] Ans: 4 11. JAN β18 [7] Ans: 4
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12. JUN β18 [17] Ans: 3 13. AUG β18 [17] Ans: 2 14. JUN '15 [26] π(π₯) = 6.5π₯ + 4(12) π(π₯) = 6.5π₯ + 48 15. JAN β17 [30] πΆ = 0.99(π β 1) + 1.29 No. πΆ = 0.99(52 β 1) + 1.29 = 51.78 16. JAN β18 [33] π(π₯) = 0.035π₯ + 300; 0.035(8250) + 300 = 588.75 2.5 Word Problems β Linear Equations 1. AUG '15 [10] Ans: 2 2. MAY β13 [5] 12π₯ + 9(2π₯) + 5(3π₯) = 15 45π₯ = 15 π₯ = π₯ + 2π₯ + 3π₯ = 6π₯ = 6 = 2 lbs. 2.6 Word Problems β Inequalities 1. JAN '15 [13] Ans: 3 2. JUN '15 [24] Ans: 4 3. JAN β19 [4] Ans: 1 4. SEP '13 [9] 8π₯ + 11π¦ β₯ 200; 8π₯ + 11(15) β₯ 200 8π₯ + 165 β₯ 200 8π₯ β₯ 35 π₯ β₯ 4.375 5 hours
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5. AUG β18 [33] 135 + 72π₯ β₯ 580 72π₯ β₯ 445 π₯ β₯ 6.2 Minimum of 7 weeks 6. JUN β19 [33] π΄ = 5π₯ + 50 π΅ = 6π₯ + 25 5π₯ + 50 < 6π₯ + 25 25 < π₯ Minimum of 26 shirts 7. JAN β20 [26] 6.25π + 4.50π¦ β€ 550 6.25π + 4.50(45) β€ 550 6.25π + 202.50 β€ 550 6.25π β€ 377.50 π β€ 55.6 Maximum of 55 adult-sized T-shirts 2.7 Conversions 1. JAN '15 [2] Ans: 2 2. JUN β16 [8] Ans: 1 3. AUG β16 [9] Ans: 3 4. JUN β17 [20] Ans: 4 5. JUN β18 [15] Ans: 1 6. AUG β18 [12] Ans: 3 7. JAN β19 [24] Ans: 4 8. JUN β19 [24] Ans: 1 9. AUG β19 [9] Ans: 4 10. JAN β17 [26] Γ . = 7.44 ππβ . = . π₯ β 3.5 βππ
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11. AUG β17 [30] The grasshopper jumps 20 times its height. 5β²9" = 69 inches. Therefore, the athlete jumps 69 Γ 20 = 1380 in per jump. 1380 ππ β = 115 ft per jump. 1 mile = 5280 ft would take β 46 jumps. 12. JAN β18 [27] The rate of speed is measuring distance over time, so it would be expressed in feet per minute. 13. JAN β20 [27] Γ Γ Γ =
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Chapter 3. Linear Graphs
3.1 Determine Whether a Point is on a Line 1. AUG β16 [2] Ans: 3 2. AUG β17 [20] Ans: 3 3. JAN β20 [11] Ans: 1 3.2 Lines Parallel to Axes There are no Regents exam questions on this topic. 3.3 Find Intercepts 1. AUG '14 [8] Ans: 1 2. JAN '15 [9] Ans: 4 3.4 Find Slope Given Two Points There are no Regents exam questions on this topic. 3.5 Find Slope Given an Equation There are no Regents exam questions on this topic. 3.6 Graph Linear Equations 1. AUG '14 [13] Ans: 2 2. JUN '14 [29]
No, the point is not on the line.
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3. AUG β19 [27]
π = β4 3.7 Write an Equation Given a Point and Slope There are no Regents exam questions on this topic. 3.8 Write an Equation Given Two Points 1. JAN '15 [11] Ans: 4 2. JUN β16 [29] Sue used point-slope form and Kathy used slope-intercept form. Both are correct as shown: π = = β π¦ β π¦ = π(π₯ β π₯ ) π¦ β 4 = β (π₯ + 3) π¦ = ππ₯ + π 4 = β (β3) + π 4 = 1 + π 3 = π π¦ = β π₯ + 3 3.9 Graph Inequalities 1. JUN '15 [5] Ans: 1 2. JAN '16 [5] Ans: 2
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3. AUG '15 [26] π¦ > β2π₯ + 1
4. AUG β16 [34] The graph should be shaded below the line. Shawn may have forgotten to flip the sign when dividing by a negative while solving for y.
5. JAN β17 [29]
Any point in the shaded area but not on the dashed line, such as (0,0).
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6. JUN β17 [30]
7. JAN β19 [35] 7.5π₯ + 12.5π¦ β€ 100
7.5π₯ β€ 100 π₯ β€ 13. 3, so the maximum is 13
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Chapter 4. Linear Systems
4.1 Solve Linear Systems Algebraically 1. JUN '14 [14] Ans: 2 2. JAN '16 [21] Ans: 4 3. AUG β16 [22] Ans: 4 4. AUG β17 [24] Ans: 4 5. JAN β18 [15] Ans: 2 6. AUG β18 [22] Ans: 3 7. JAN β19 [22] Ans: 3 8. AUG β19 [22] Ans: 1 9. JAN β20 [20] Ans: 2 10. JUN '15 [33] 3(8π₯ + 9π¦ = 48) β 24π₯ + 27π¦ = 144 β2(12π₯ + 5π¦ = 21) β β24π₯ β 10π¦ = β42 17π¦ = 102 π¦ = 6 π¦ = . = 6, 8π₯ + 9(6) = 48 8π₯ = β6 π₯ = β , Yes, π₯ = β and π¦ = 6 for both 4.2 Solve Linear Systems Graphically 1. JAN '15 [18] Ans: 3 2. JUN β16 [18] Ans: 4 3. JUN β17 [8] Ans: 1 4. JAN β17 [25] No. The two lines coincide, so there are infinitely many solutions.
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5. JAN β19 [37] π‘ + π = 15 3π‘ + 2π = 38
No, the point of intersection on the graph shows that 8 tricycles were ordered. 6. JAN β20 [37] 3π₯ + 2π¦ = 170 4π₯ + 6π¦ = 360
(30,40); child tickets cost $30 and adult tickets cost $40. 4.3 Solutions to Systems of Inequalities 1. SEP '13 [1] Ans: 2 2. JAN β17 [16] Ans: 4 4.4 Solve Systems of Inequalities Graphically 1. JUN '14 [4] Ans: 2 2. AUG '14 [7] Ans: 1 3. AUG '15 [6] Ans: 3 4. JAN β18 [20] Ans: 3
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5. JAN '15 [34] π¦ β₯ 2π₯ β 3,
, Disagree since (2,1) is not a solution of π₯ + 2π¦ < 4 6. JUN β16 [34]
(6,2) is not a solution as its falls on the edge of each inequality. 7. AUG β17 [35]
(3,7) is not in the solution set because it is on a dashed boundary. 8. JAN β18 [28] (0,4) is located at the intersection of the two lines. However, since the line for the inequality π¦ < π₯ + 4 is dashed, this point is not a solution to the system.
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9. JUN β18 [35]
(1,2) is not in the solution set because it lies on the dashed line. 10. AUG β18 [35]
(6,1) is on a solid line; (β6,7) is on a dashed line. 11. JUN β19 [36] π¦ < β3π₯ + 3 and π¦ β€ 2π₯ β 2 Region A represents the solution set of the system. The gray region represents the solution set of π¦ β€ 2π₯ β 2.
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12. AUG β19 [33]
No, (1,8) falls on the boundary line of π¦ β 5 < 3π₯, which is a dashed line. The points on a dashed line are not included in the solution. 13. JAN β20 [34] π¦ < π₯ + 2 and π¦ β₯ β π₯ β 1
Stephen is correct because (0,0) lies in the double-shaded region S, so it is in the solution set.
4.5 Word Problems β Linear Systems 1. AUG '14 [19] Ans: 4 2. JUN '15 [6] Ans: 3 3. JUN β16 [5] Ans: 1 4. JAN β18 [3] Ans: 1 5. AUG β18 [9] Ans: 2
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6. JUN '14 [36] 2.35π + 5.50π = 89.50, No because 2.35(8) + 5.50(14) = 95.80, π + π = 22 π = 22 β π 2.35π + 5.50(22 β π) = 89.50 121 β 3.15π = 89.50 β3.15π = β31.50 π = 10 10 cats 7. JAN '15 [33] 2π + 3π = 18.25 and 4π + 2π = 27.50, β2(2π + 3π = 18.25) β β4π β 6π = β36.50 4π + 2π = 27.50 β4π = β9 π = 2.25 2π + 3(2.25) = 18.25 2π + 6.75 = 18.25 2π = 11.50 π = 5.75 popcorn $5.75, drink $2.25
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8. JUN β16 [37] 3π₯ + 2π¦ = 19 2π₯ + 4π¦ = 24
2(3π₯ + 2π¦ = 19) β 6π₯ + 4π¦ = 38 β1(2π₯ + 4π¦ = 24) β β2π₯ β 4π¦ = β24 4π₯ = 14 π₯ = 3.5 2(3.5) + 4π¦ = 24 7 + 4π¦ = 24 4π¦ = 17 π¦ = 4.25 Cupcakes cost $3.50 and brownies cost $4.25 per package.
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9. AUG β16 [37] 18π + 32π€ = 19.92 14π + 26π€ = 15.76 14(0.52) + 26(0.33) = 15.86 β7(18π + 32π€ = 19.92) β β126π β 224π€ = β139.44 9(14π + 26π€ = 15.76) β 126π + 234π€ = 141.84 10π€ = 2.4 π€ = 0.24 18π + 32(0.24) = 19.92 18π + 7.68 = 19.92 18π = 12.24 π = 0.68 A juice box is 68 cents and a water bottle is 24 cents. 10. JAN β17 [34] β5(π + 2π = 15.95) β β5π β 10π = β79.75 2(3π + 5π = 45.90) β 6π + 10π = 91.80 π = 12.05 11. JAN β17 [37] 1000 β 60π₯ = 600 β 20π₯ 400 = 40π₯ 10 = π₯ 10 months 1000 β 60(10) = $400 Ian is incorrect because 1000 β 60(16) = 40, so he would still owe $40.
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12. JUN β17 [37] π¦ = 10π₯ + 5 and π¦ = 5π₯ + 35.
(6,65) It took 6 years for the two clubs to have the same number of members, at which point they had 65 members each. 13. AUG β17 [28] For $50, Dylan can buy 14 games in Plan B but only 12 games in Plan A. Bobby can buy 20 games for $65 under both plans, so he can choose either plan. 14. JAN β18 [37] π = 2π β 5 = No, because it doesnβt make each equation true: eg, 20 β 2(15) β 5; ( ) = 4(π + 3) = 3(2π β 2) 4π + 12 = 6π β 6 18 = 2π π = 9 π = 2(9) β 5 = 13 She had 9 cats and 13 dogs
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15. JUN β18 [34] π΄(π₯) = 7 + 3(π₯ β 2) π΅(π₯) = 3.25π₯ 7 + 3(π₯ β 2) = 3.25π₯ 7 + 3π₯ β 6 = 3.25π₯ 1 + 3π₯ = 3.25π₯ 1 = 0.25π₯ π₯ = 4 hours 16. JUN β18 [37] 10π + 25π = 1755 π + π = 90 or π = 90 β π 10(90 β π) + 25π = 1755 900 β 10π + 25π = 1755 15π = 855 π = 57 No, because $20.98 Γ 1.08 > 90 Γ 0.25. 17. AUG β18 [37] π = 4π + 6 π β 3 = 7(π β 3) 4π + 6 β 3 = 7π β 21 3π = 24 π = 8 π = 4(8) + 6 = 38 38 + π₯ = 3(8 + π₯) 38 + π₯ = 24 + 3π₯ 2π₯ = 14 π₯ = 7
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18. JUN β19 [37] 4π + 3π = 16.53 5π + 4π = 21.11 No, because 5(2.49) + 4(2.87) β 21.11 Eq. 1 Γ 4 16π + 12π = 66.12 Eq. 2 Γ β3 β15π β 12π = β63.33 π = 2.79 4(2.79) + 3π = 16.53 3π = 5.37 π = 1.79 19. AUG β19 [37] 3.75π΄ + 2.5π· = 35 π΄ + π· = 12 3.75(12 β π·) + 2.5π· = 35 45 β 3.75π· + 2.5π· = 35 45 β 1.25π· = 35 β1.25π· = β10 π· = 8 π΄ + 8 = 12, so π΄ = 4 7(2π΄ + π·) = 7(16) = 112 eggs 112 Γ· 12 = 9 dozen $2.50 Γ 9 = $22.50 4.6 Word Problems β Systems of Inequalities 1. JUN β17 [11] Ans: 1 2. AUG β18 [10] Ans: 2 3. MAY β13 [6] π₯ + π¦ β€ 800 and 6π₯ + 9π¦ β₯ 5000; yes, 6(440) + 9π¦ β₯ 5000 2640 + 9π¦ β₯ 5000 9π¦ β₯ 2360 π¦ β₯ 262 440 + 263 β€ 800 β
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4. JUN '15 [35]
50π₯ + 500π¦ β₯ 2500 and π₯ + π¦ β€ 15; any point in the solution set, such as (4,7) for 4 printers, 7 computers 5. AUG '14 [37] π₯ + π¦ β€ 15 and 4π₯ + 8π¦ β₯ 80
Solution stated such as (3,10) = 3 hrs. of babysitting and 10 hrs. at the library 6. JAN '16 [37] π₯ + π¦ β€ 200 and 12.5π₯ + 6.25π¦ β₯ 1500
No, because the point (30,80) is not in the shaded area labelled S
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7. AUG β16 [35] π₯ + π¦ β€ 200 12π₯ + 8.50π¦ β₯ 1000 12π₯ + 8.50(50) β₯ 1000 12π₯ + 425 β₯ 1000 12π₯ β₯ 575 π₯ β₯ β 47.9 Minimum of 48 tickets at the door 8. JAN β18 [35] 2π + 1.5π β₯ 500 π + π β€ 360 2(144) + 1.5π β₯ 500 288 + 1.5π β₯ 500 1.5π β₯ 212 π β₯ 141 At least 142 bottles must be sold
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Chapter 5. Polynomials
5.1 Polynomial Expressions 1. JUN β16 [2] Ans: 4 2. JUN β18 [19] Ans: 3 3. JUN β19 [5] Ans: 1 4. AUG β19 [3] Ans: 4 5. JAN β20 [24] Ans: 4 6. AUG β16 [28] No, the leading coefficient is the coefficient of the term with the highest power, β2. 7. AUG β17 [31] 5π₯ + 4π₯ (2π₯ + 7) β 6π₯ β 9π₯ = 5π₯ + 8π₯ + 28π₯ β 6π₯ β 9π₯ = 8π₯ + 22π₯ β 4π₯ 5.2 Add and Subtract Polynomials 1. JUN '14 [3] Ans: 2 2. JAN '15 [28] β2π₯ + 6π₯ + 4 3. JUN β17 [25] 5π₯ β 10 5.3 Multiply Polynomials 1. JAN '15 [10] Ans: 2 2. AUG '15 [9] Ans: 3 3. AUG '15 [24] Ans: 4 4. JAN '16 [10] Ans: 3 5. JUN β16 [10] Ans: 2 6. AUG β16 [12] Ans: 3 7. JAN β17 [7] Ans: 4 8. JAN β18 [13] Ans: 3 9. JUN β18 [3] Ans: 2 10. AUG β18 [13] Ans: 1
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11. AUG β18 [24] Ans: 3 12. JAN β19 [11] Ans: 3 13. AUG β19 [12] Ans: 2 14. AUG '14 [28] x 5 2x2 2x3 10x2 7x 7x2 35x β10 β10x β50 2π₯ + 17π₯ + 25π₯ β 50 15. JUN '15 [28] (2π₯ β 5π₯ + 7) π₯ = π₯ β π₯ + π₯ 16. JUN β19 [26] πΆ = 3π₯ + 4 β 3(2π₯ + 6π₯ β 5) = 3π₯ + 4 β 6π₯ β 18π₯ + 15 = β3π₯ β 18π₯ + 19 17. JAN β20 [28] 3π₯ + 21π₯ β 4π₯ β 28 β π₯ 2.75π₯ + 17π₯ β 28 5.4 Divide a Polynomial by a Monomial There are no Regents exam questions on this topic.
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Chapter 6. Irrational Numbers
6.1 Simplifying Radicals There are no Regents exam questions on this topic. 6.2 Operations with Radicals [NG] There are no Regents exam questions on this topic. 6.3 Rationalizing Denominators [NG] There are no Regents exam questions on this topic. 6.4 Closure 1. JUN '14 [13] Ans: 3 2. AUG '14 [1] Ans: 1 3. JUN '15 [8] Ans: 2 4. AUG '15 [22] Ans: 2 5. JAN '16 [4] Ans: 1 6. JAN β18 [8] Ans: 3 7. JAN β19 [3] Ans: 3 8. JUN β19 [7] Ans: 1 9. JAN '15 [25] correct; 4.2 is rational and β2 is irrational, and the sum of a rational and irrational is always irrational 10. JUN β16 [26] 3β2 β 8β18 = 24β36 = 24(6) = 144 144 is an integer and all integers are rational. 11. AUG β16 [29] The sum is 7β2, which is irrational. 7 is rational and β2 is irrational, and the product of a rational and irrational is always irrational.
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12. JAN β17 [28] No. The second fraction is irrational because it is the quotient of an irrational and a rational. The sum is irrational because it is the sum of a rational and an irrational. 13. JUN β17 [27] Irrational. 7 is rational and β2 is irrational, and the difference of a rational and irrational is always irrational. 14. AUG β17 [25] a is irrational. b and c are rational (c = 15). a + b is irrational because the sum of a rational and irrational is always irrational, and b + c is rational because the sum of two rationals is always rational. 15. JUN β18 [31] Rational; β16 = 4, so it is rational, and is a ratio of two integers, so it is also rational. The product of two rationals is always rational. (The product is .) 16. AUG β19 [30] No. For example, the product of irrational numbers β2 and β8 is β16, which is the rational number 4. 17. JAN β20 [30] Product is irrational. β3 is irrational and β9 = 3 is rational, and the product of an irrational number and a rational number is always irrational.
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Chapter 7. Univariate Data
7.1 Types of Data There are no Regents exam questions on this topic. 7.2 Frequency Tables and Histograms There are no Regents exam questions on this topic. 7.3 Central Tendency 1. AUG '14 [4] Ans: 3 2. JUN '15 [20] Ans: 3 3. JAN β18 [16] Ans: 1 7.4 Distribution 1. JAN β17 [20] Ans: 4 7.5 Standard Deviation 1. AUG '15 [19] Ans: 1 2. JUN β19 [22] Ans: 1 3. JAN β19 [31] Los Angeles because the standard deviation for LA (β3.64) is less than the standard deviation for Miami (β7.23) 7.6 Percentiles and Quartiles 1. JUN '14 [19] Ans: 3 2. JUN β16 [20] Ans: 3 3. JUN β17 [15] Ans: 4
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4. AUG β17 [34]
For 29 players, the upper quartile would be taller than 21 heights (29 Γ 0.75 = 21.75). The 22nd value is in the 6.4 β 6.5 interval. 5. AUG β18 [31] 4th Period because the IQR and π are greater for 4th Period. 7.7 Box Plots 1. JAN '15 [14] Ans: 4 2. AUG β16 [3] Ans: 4 3. JUN β18 [5] Ans: 2 4. AUG β19 [15] Ans: 1 5. JAN β20 [22] Ans: 4 6. JUN '14 [32]
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Chapter 8. Bivariate Data
8.1 Two-Way Frequency Tables 1. JUN β16 [15] Ans: 4 2. JAN β17 [5] Ans: 2 3. JUN β18 [9] Ans: 1 4. AUG β18 [14] Ans: 2 5. JUN β19 [12] Ans: 2 6. AUG β19 [6] Ans: 2 7. JAN β20 [10] Ans: 3 8. JAN '15 [26] = 25% 9. JAN '16 [30] = ; Γ 351 = 234 234 males 10. JUN β17 [29]
8.2 Scatter Plots There are no Regents exam questions on this topic. 8.3 Correlation and Causality 1. JAN β17 [13] Ans: 2 2. AUG β17 [8] Ans: 2 3. AUG β18 [21] Ans: 3 8.4 Identify Correlation in Scatter Plots 1. JUN β16 [4] Ans: 2 8.5 Lines of Fit 1. AUG '14 [21] Ans: 4
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2. JAN β19 [1] Ans: 2 3. SEP '13 [7] π¦ = 0.05π₯ β 0.92 4. AUG β16 [33] π¦ = 17.159π₯ β 2.476 π¦ = 17.159(0.65) β 2.476 β 8.7 5. JAN β18 [34] π¦ = β8.5π₯ + 99.2; the y-intercept represents the original length of the rope; the slope represents how much shorter the rope gets (8.5 cm) after each knot. 8.6 Correlation Coefficients 1. JUN '14 [11] Ans: 3 2. JUN '15 [16] Ans: 2 3. AUG β16 [6] Ans: 2 4. JAN β17 [3] Ans: 4 5. JUN β17 [14] Ans: 1 6. AUG β17 [22] Ans: 1 7. JAN '15 [35] 0.94; it shows a strong positive relationship between the calories and mg of sodium 8. AUG '15 [36] π¦ = 0.16π₯ + 8.27; 0.97, a strong association 9. JAN '16 [35] π(π‘) = β58π‘ + 6182; β0.94; yes, because it is close to β1 10. JAN β18 [31] π¦ = 0.81π₯ + 15.19; 0.92; there is a high positive correlation between mathematics and physics scores. 11. JUN β18 [36] π¦ = 0.96π₯ + 23.95; 0.92; there is a high positive correlation between scoring 85 or better on the math and English exams. 12. JAN β19 [34] π¦ = 1.9π₯ + 29.8; π = 0.3, which represents a weak correlation between a dogβs mass and height.
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13. JUN β19 [35] π¦ = 7.79π₯ + 34.27; π = 0.98, which represents a high positive correlation between hours spent studying and test scores. 14. AUG β19 [35] π¦ = β7.76π₯ + 246.34; π = β0.88. There is a negative correlation; as the distance from Times Square increases, the cost of a room decreases. 15. JAN β20 [35] π(π) = β0.79π + 249.86; π = β0.95. There is a strong negative correlation; as the sales price increases, the number of new homes available decreases. 8.7 Residuals [CC] 1. SEP '13 [3] Ans: 2 2. JAN '16 [24] Ans: 3 3. SEP '13 [14] π¦ = 6.32π₯ + 22.43
The equation is a good fit for the data because the points on the residual plot are scattered without a pattern.
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4. AUG '14 [31]
Poor fit because there is a pattern in the residuals 5. JUN '15 [31] Graph A; no pattern shows
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Chapter 9. Introduction to Functions
9.1 Recognize Functions 1. JUN '15 [4] Ans: 3 2. AUG '15 [11] Ans: 2 3. JUN β17 [9] Ans: 3 4. JAN β18 [4] Ans: 2 5. JUN β18 [11] Ans: 4 6. JAN β19 [7] Ans: 4 7. JUN β19 [3] Ans: 4 8. AUG β19 [2] Ans: 4 9. JAN β20 [4] Ans: 2 10. JAN '15 [27] (β4,1) because the input β4 would lead to two different outputs, which a function cannot have 11. JAN '16 [26] no, it is not a function because for π₯ = 2, there are two different values of y. 12. JAN β17 [32] Neither is correct. Nora is wrong; a circle is not a function because it fails the vertical line test. Miaβs reason is wrong; a circle is not a function because the same x-value maps to multiple values of y. 13. AUG β18 [26] III and IV are functions. I has two y-values for x = 6, and II has two y-values for x = 1 and x = 2.
9.2 Function Graphs 1. JUN '14 [20] Ans: 1 2. AUG β18 [5] Ans: 1 3. AUG β18 [18] Ans: 1 9.3 Evaluate Functions 1. JUN '15 [15] Ans: 1 2. AUG '15 [12] Ans: 3
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3. AUG β16 [5] Ans: 2 4. AUG β16 [11] Ans: 4 5. JAN β17 [10] Ans: 1 6. JUN β17 [5] Ans: 1 7. AUG β17 [4] Ans: 3 8. JUN β18 [2] Ans: 4 9. JUN β18 [8] Ans: 4 10. AUG β18 [11] Ans: 4 11. JAN β19 [2] Ans: 1 12. JAN β19 [21] Ans: 4 13. JUN β19 [2] Ans: 2 14. JAN β20 [1] Ans: 2 15. JAN '16 [32] π(5) = 8 β 2 = 256 π(5) = 2 = 2 = 256 The functions are equal since they produce the same values for all inputs, t. This is shown by the fact that 8 β 2 = 2 β 2 = 2 16. AUG β19 [25] π(β2) = β4(β2) β 3(β2) + 2 = β8 9.4 Features of Function Graphs 1. JUN '14 [9] Ans: 3 2. JAN β17 [21] Ans: 1 3. JUN β17 [1] Ans: 3 4. JUN β18 [20] Ans: 3 9.5 Domain and Range 1. JUN '14 [2] Ans: 4 2. JUN '14 [17] Ans: 4 3. AUG '14 [11] Ans: 1 4. AUG '14 [23] Ans: 2 5. JAN '15 [6] Ans: 2 6. JUN '15 [9] Ans: 4 7. JAN '16 [15] Ans: 1 8. JAN '16 [19] Ans: 2
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9. JUN β16 [23] Ans: 4 10. AUG β16 [20] Ans: 2 11. JAN β17 [19] Ans: 4 12. AUG β17 [10] Ans: 1 13. JAN β18 [12] Ans: 3 14. JUN β18 [16] Ans: 3 15. JUN β18 [21] Ans: 2 16. AUG β18 [6] Ans: 2 17. JAN β19 [14] Ans: 2 18. JAN β19 [17] Ans: 4 19. JUN β19 [20] Ans: 4 20. AUG β19 [21] Ans: 4 21. JAN β20 [18] Ans: 1 22. JAN β20 [21] Ans: 4 23. JUN '14 [30] yes, because each number in the domain leads to a unique number in the range 24. AUG β17 [29] Since fractions of cookies may be eaten, the domain is continuous. 9.6 Absolute Value Functions 1. JUN β16 [22] Ans: 2 2. JAN β17 [12] Ans: 1 3. AUG β17 [2] Ans: 2 4. AUG β17 [18] Ans: 2 5. SEP '13 [10]
range π¦ β₯ 0; increasing for π₯ > β1
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6. JAN β17 [33]
Yes, because the graph of π(π₯) intersects the graph of π(π₯) at π₯ = β2. 7. JAN β18 [25]
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Chapter 10. Functions as Models
10.1 Write a Function from a Table 1. AUG β16 [4] Ans: 4 2. AUG '15 [25] β(π) = 1.5(π β 1) + 3 or β(π) = 1.5π + 1.5 3. AUG '15 [32] d 1 2 3 4 5 π(π) 30 32 34 36 38 π(π) = 2(π β 1) + 30 or π(π) = 2π + 28 π(6) = 2(6) + 28 = 40 4. JAN β17 [35] π = . = 0.75 π(π₯) = 0.75(π₯ β 4) + 7.50 π(π₯) = 0.75π₯ + 4.50 Each card costs 75Β’ and start-up costs were $4.50.
10.2 Graph Linear Functions 1. SEP '13 [8]
πΆ(π₯) = π₯ π₯ = 180 10π₯ = 540 π₯ = 54
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2. AUG β17 [37] 7 sodas cost $3.50, so they have $25 left to spend on food. 1.25π₯ + 2.5π¦ = 25
There are 11 combinations, represented by the dots in the graph. 10.3 Rate of Change for Linear Functions 1. AUG '15 [2] Ans: 3 2. JAN '16 [2] Ans: 2 3. AUG β16 [15] Ans: 4 4. JUN β17 [4] Ans: 2 5. JAN '16 [29] The slope is the amount paid per month and the y-intercept is the initial cost. 6. JUN β16 [30] There are 2 inches of snow every 4 hours. 7. AUG β17 [33] a) 10 hrs. 55(4) + 65(π‘ β 4) = 610 220 + 65π‘ β 260 = 610 65π‘ = 650 π‘ = 10 b) 0.3 hrs. 55(2) + 65(π‘ β 2) = 610 110 + 65π‘ β 130 = 610 65π‘ = 630 π‘ β 9.7 10 β 9.7 = 0.3
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10.4 Average Rate of Change 1. MAY β13 [1] Ans: 4 2. JUN '14 [18] Ans: 1 3. AUG '14 [14] Ans: 4 4. JAN '15 [21] Ans: 1 5. JUN '15 [11] Ans: 3 6. AUG '15 [15] Ans: 1 7. JAN '16 [13] Ans: 4 8. JUN β16 [3] Ans: 1 9. AUG β16 [1] Ans: 1 10. AUG β17 [5] Ans: 2 11. JAN β18 [24] Ans: 2 12. JAN '16 [28] from 1960β1965, because the decrease of 0.15 degrees is the largest change among the intervals (it has the steepest slope) 13. JAN β17 [31] = 68 mph 14. JAN β18 [36] The domain may include fractions of hours, and the number of hours cannot be negative; 0 < π‘ < 6; = β = β15; the business sold 15 less pairs of shoes per hour 15. AUG β18 [27] . . = β0.475 16. JUN β19 [29] β 2.9 and β 2.8, so the interval from 1 a.m. to 12 noon has the greater rate. 10.5 Functions of Time 1. JUN '15 [2] Ans: 4 2. AUG β19 [18] Ans: 1
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3. MAY β13 [7]
3.5 4. AUG '15 [28]
5. JUN β16 [35] 762 β 192 = 570 miles 92 β 32 = 60 minutes = 9.5 miles per minute π¦ = 9.5π₯ 192 + 9.5(120 β 32) = 1028 miles 6. JUN β17 [34] D to E because his speed was slower; Craig may have stopped to eat; β 32.9 mph 7. JAN β19 [28] 2 < π‘ < 6 and 14 < π‘ < 15 because horizontal lines have a slope of zero.
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8. JAN β19 [36]
6 am to 4 pm; = β3 9. AUG β19 [36]
.. β 0.59 10.6 Systems of Functions 1. JAN '16 [17] Ans: 1 2. AUG β18 [19] Ans: 3 3. AUG β19 [14] Ans: 3 4. SEP '13 [15] π(π₯) = 120π₯ and π(π₯) = 70π₯ + 1600 120π₯ = 70π₯ + 1600 50π₯ = 1600 π₯ = 32 π(35) = 4200, π(35) = 4050, so Green Thumb is less expensive
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5. MAY β13 [8] π΄(π₯) = 1.50π₯ + 6 and π΅(π₯) = 2π₯ + 2.50 1.50π₯ + 6 = 2π₯ + 2.50 3.50 = 0.50π₯ 7 = π₯, so 7 rides π΄(5) = 1.50(5) + 6 = 13.50 and π΅(5) = 2(5) + 2.50 = 12.50, so B has a lower cost. 6. AUG '14 [27] 185 + 0.03π₯ = 275 + 0.025π₯ 0.005π₯ = 90 π₯ = 18,000 7. JAN '15 [31] 36 + 15π₯ = 48 + 10π₯ 5π₯ = 12 π₯ = 2.4 8. AUG β16 [30] β3 and 1, because the two functions intersect at (β 3,4) and (1,3). 10.7 Combine Functions 1. AUG '14 [6] Ans: 2 2. JUN β16 [25] π(π₯) = 2(2π₯ + 1) β 1 = 2(4π₯ + 4π₯ + 1) β 1 = 8π₯ + 8π₯ + 2 β 1 = 8π₯ + 8π₯ + 1
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Chapter 11. Exponential Functions
11.1 Exponential Growth and Decay 1. JAN '15 [4] Ans: 1 2. JAN '15 [8] Ans: 1 3. JUN '15 [17] Ans: 2 4. AUG '15 [7] Ans: 3 5. JAN '16 [3] Ans: 3 6. JAN '16 [8] Ans: 4 7. JUN β16 [17] Ans: 2 8. AUG β16 [17] Ans: 1 9. AUG β16 [24] Ans: 2 10. JAN β17 [24] Ans: 3 11. JUN β17 [12] Ans: 2 12. AUG β17 [14] Ans: 2 13. AUG β17 [16] Ans: 2 14. AUG β17 [21] Ans: 3 15. JAN β18 [2] Ans: 3 16. JAN β19 [12] Ans: 4 17. JUN β19 [23] Ans: 2 18. JAN β20 [2] Ans: 1 19. JAN β20 [14] Ans: 2 20. JUN '14 [26] rate of decay; number of milligrams of the substance at the start 21. AUG '14 [26] π΅ = 3000(1.042) 22. JUN '15 [29] 600(1.016) β 619.35 23. AUG '15 [30] 5%; in the decay function π¦ = π(1 β π) , r represents the percent of change, 1 β π = 0.95, so π = 0.05 = 5% 24. JUN β17 [28] 15%; in the decay function π¦ = π(1 β π) , r represents the percent of change, 1 β π = 0.85, so π = 0.15 = 15%
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25. JUN β18 [33] There are 20 rabbits at the start, and their population is growing at 1.4% per day. ( ) ( ) β 0.8 26. AUG β18 [34] π(π‘) = 25000(0.815) π(3) β π(4) β 2503.71 27. JAN β19 [33] π = 450(1.025) No, because (1.025) < 2. 28. AUG β19 [34] π΄(π‘) = 5000(1.012) π΄(32) β π΄(17) β 1200 11.2 Graphs of Exponential Functions 1. AUG '14 [10] Ans: 3 2. JAN '15 [15] Ans: 3 3. JUN β19 [16] Ans: 2 4. JAN '15 [32] π¦ = 0.25(2) , by entering (x,y) coordinates, for integer values of x from 2 to 5, into calculator, then STAT / ExpReg 5. JUN '15 [36] π¦ = 80(1.5) ; 3,030,140; No, because the number would grow so large it would be more than the number of potential customers 6. JAN β19 [29]
Yes, π(4) > π(4) because 2 β 7 > 1.5(4) β 3.
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11.3 Rewrite Exponential Expressions 1. JAN '15 [19] Ans: 4 2. JUN '15 [13] Ans: 2 3. JUN β16 [14] Ans: 3 4. JAN β17 [14] Ans: 2 5. JAN β18 [21] Ans: 4 6. AUG β18 [1] Ans: 2 7. JAN β19 [23] Ans: 4 8. AUG β19 [13] Ans: 2 9. JAN β20 [19] Ans: 3 11.4 Compare Linear and Exponential Functions 1. JUN '14 [6] Ans: 4 2. JUN '14 [15] Ans: 3 3. AUG '14 [12] Ans: 3 4. JAN '15 [5] Ans: 3 5. AUG '15 [18] Ans: 3 6. JAN '16 [16] Ans: 4 7. JAN '16 [23] Ans: 1 8. JUN β16 [6] Ans: 1 9. JUN β16 [21] Ans: 3 10. AUG β16 [18] Ans: 1 11. JAN β17 [11] Ans: 3 12. JUN β17 [7] Ans: 1 13. JUN β17 [21] Ans: 3 14. AUG β17 [17] Ans: 1 15. JAN β18 [5] Ans: 1 16. JUN β18 [14] Ans: 4 17. AUG β18 [2] Ans: 1 18. AUG β18 [23] Ans: 4 19. JAN β19 [16] Ans: 3 20. JUN β19 [6] Ans: 1 21. JUN β19 [11] Ans: 3 22. AUG β19 [7] Ans: 2 23. JAN β20 [17] Ans: 3
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24. SEP '13 [13] π¦ = 836.47(2.05) ; The data appear to grow at an exponential rate; π¦ = 836.47(2.05) β 3515 25. JUN '14 [35] π΄(π) = 175 β 2.75π 175 β 2.75π = 0 π = 63. 63 63 weeks; she won't have enough money for 64 rentals 26. AUG '15 [27] exponential; the function does not grow at a constant rate, it is close to a function with a common ratio of 1.25 27. JAN '16 [25] linear; there is a constant rate of change (a slope of β1.25). 28. AUG β16 [27] exponential, because the function does not grow at a constant rate. 29. JUN β17 [36] π(π₯) = 100π₯ + 10 and π(π₯) = 10(2) Both, since π(7) = 100(7) + 10 = 710 and π(7) = 10(2) = 1280. 30. JUN β18 [26] Yes, because π(π₯) does not have a constant rate of change. 31. JAN β19 [26] Linear, because the function grows at a constant rate (π = 87).
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Chapter 12. Sequences
12.1 Arithmetic Sequences 1. JUN '14 [24] Ans: 2 2. AUG '14 [16] Ans: 2 3. JUN β16 [13] Ans: 3 4. AUG β16 [10] Ans: 1 5. JUN β18 [7] Ans: 1 6. AUG β18 [20] Ans: 4 7. JUN β19 [19] Ans: 2 8. JAN β20 [8] Ans: 2 12.2 Geometric Sequences 1. AUG β19 [24] Ans: 1 2. AUG β17 [26] Yes, because the sequence has a common ratio, 3. 12.3 Recursively Defined Sequences [CC] 1. JUN '14 [21] Ans: 4 2. AUG '14 [24] Ans: 4 3. JAN '15 [20] Ans: 3 4. JUN '15 [22] Ans: 3 5. AUG '15 [14] Ans: 1 6. JAN '16 [18] Ans: 3 7. JAN β17 [8] Ans: 1 8. JUN β17 [18] Ans: 2 9. AUG β17 [15] Ans: 3 10. JAN β18 [18] Ans: 3 11. JUN β18 [24] Ans: 3 12. JAN β19 [19] Ans: 2 13. AUG β19 [19] Ans: 1 14. JAN β20 [23] Ans: 1 15. AUG β18 [32] 0, β1, 1, 1, 1
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Chapter 13. Factoring
13.1 Factor Out the Greatest Common Factor There are no Regents exam questions on this topic. 13.2 Factor a Trinomial 1. AUG '14 [15] Ans: 1 2. JUN β18 [10] Ans: 1 3. AUG β18 [3] Ans: 3 4. JAN β20 [12] Ans: 4 5. AUG '14 [25] (π₯ + 6)(π₯ + 4) or (π₯ + 4)(π₯ + 6), so 4 and 6 13.3 Factor the Difference of Perfect Squares 1. JUN '15 [3] Ans: 2 2. JUN β16 [1] Ans: 3 3. JUN β17 [6] Ans: 3 4. AUG β17 [3] Ans: 3 5. JAN β18 [9] Ans: 3 6. AUG β18 [7] Ans: 3 7. JUN β19 [1] Ans: 4 8. AUG β19 [8] Ans: 3 13.4 Factor a Trinomial by Grouping [CC] 1. JAN β19 [6] Ans: 1 2. JUN β19 [17] Ans: 3 13.5 Factor Completely 1. JAN '15 [22] Ans: 3 2. JAN '16 [12] Ans: 3 3. AUG β16 [8] Ans: 2 4. JAN β17 [1] Ans: 2 5. AUG β19 [16] Ans: 3
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6. JAN β20 [6] Ans: 3 7. JAN β20 [16] Ans: 2 8. JUN '14 [31] (π₯ + 7)(π₯ β 1) = (π₯ + 7)(π₯ + 1)(π₯ β 1)
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Chapter 14. Quadratic Functions
14.1 Solve Simple Quadratic Equations 1. JUN '14 [23] Ans: 1 2. AUG '14 [3] Ans: 3 3. JAN '15 [16] Ans: 1 4. JUN '15 [19] Ans: 2 5. JAN β17 [15] Ans: 4 6. SEP '13 [6] π₯ β 4 = 0 π₯ β 8 = 0 π₯ = 8 π₯ = Β±2β2 7. AUG '15 [35] π = π = 2π = 2 = 2 . β 5 8. JUN β16 [33] a) π»(1) = β16(1) + 144 = 128 π»(2) = β16(2) + 144 = 80 128 β 80 = 48 ft. b) 0 = β16π‘ + 144 16π‘ = 144 π‘ = 9 π‘ = Β±3 3 secs (reject β3)
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9. AUG β17 [27] π = ππ β 3π = ππ β = π π = (reject negative radius) 10. JAN β18 [30] π = π = 11. JAN β19 [32] 4π₯ = 80 π₯ = 20 π₯ = Β±β20 12. JUN β19 [28] 5π₯ = 180 π₯ = 36 π₯ = Β±6 13. AUG β19 [31] 6π₯ = 42 π₯ = 7 π₯ = Β±β7
14.2 Solve Quadratic Equations by Factoring 1. JUN '15 [10] Ans: 4 2. AUG '15 [13] Ans: 4 3. JAN '16 [9] Ans: 1 4. JUN β16 [12] Ans: 1 5. JAN β17 [2] Ans: 3 6. JUN β18 [4] Ans: 3
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7. JUN '14 [33] π(π₯) = 9π₯ β 3π₯ β 3 + π₯ + 4π₯ + 19 π(π₯) = π₯ + 10π₯ + 16 π₯ + 10π₯ + 16 = 0 (π₯ + 8)(π₯ + 2) = 0 {β8,β2} 8. JAN '16 [27] π¦ β 6π¦ + 9 = 4π¦ β 12 π¦ β 10π¦ + 21 = 0 (π¦ β 3)(π¦ β 7) = 0 {3,7} 9. AUG β16 [36] 0 = (π΅ + 3)(π΅ β 1) 0 = (8π₯ + 3)(8π₯ β 1) β , Janice substituted B for 8x, resulting in a simpler quadratic to factor. Once factored, she could replace the 8x for B. 10. JAN β18 [26] π₯ β 4π₯ + 3 = 0 (π₯ β 1)(π₯ β 3) = 0 {1,3} 11. JAN β19 [27] π₯ β 8π₯ β 9 = 0 (π₯ β 9)(π₯ + 1) = 0 {9,β1} I factored the trinomial. 14.3 Find Quadratic Equations from Given Roots 1. MAY β13 [3] Ans: 2 2. JUN '14 [12] Ans: 3 14.4 Equations with the Square of a Binomial 1. AUG '14 [18] Ans: 4 2. JUN '15 [21] Ans: 1 3. AUG '15 [23] Ans: 3
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4. JUN β16 [19] Ans: 3 5. JAN β18 [14] Ans: 1 6. JAN β20 [15] Ans: 3 7. AUG β16 [31] (π₯ β 3) β 49 = 0 (π₯ β 3) = 49 π₯ β 3 = Β±7 π₯ = 3 Β± 7 {β4,10} 14.5 Complete the Square 1. JUN '14 [8] Ans: 2 2. JUN '14 [10] Ans: 2 3. JAN '15 [17] Ans: 4 4. JUN '15 [23] Ans: 1 5. AUG '15 [20] Ans: 1 6. JAN '16 [14] Ans: 2 7. JAN β17 [22] Ans: 1 8. JUN β17 [22] Ans: 2 9. JUN β18 [12] Ans: 3 10. JAN β19 [15] Ans: 1 11. AUG '14 [32] 9, π = = = 9 12. AUG β17 [32] π₯ β 6π₯ + 9 = 15 + 9 (π₯ β 3) = 24 π₯ β 3 = Β±β24 π₯ = 3 Β± 2β6 3 β 2β6, 3 + 2β6 13. AUG β18 [30] π₯ + 4π₯ + 4 = 2 + 4 (π₯ + 2) = 6 π₯ + 2 = Β±β6 π₯ = β2 Β± β6
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14. JAN β20 [31] π₯ β 8π₯ = β6 π₯ β 8π₯ + 16 = β6 + 16 (π₯ β 4) = 10 π₯ β 4 = Β±β10 π₯ = 4 Β± β10 14.6 Quadratic Formula and the Discriminant 1. SEP '13 [2] Ans: 1 2. JUN β18 [22] Ans: 3 3. JUN β19 [21] Ans: 2 4. AUG '15 [29] no real solutions; the discriminant π β 4ππ = (β2) β 4(1)(5) = β16 is negative 5. JAN '16 [34] completing the square or quadratic formula; π₯ = Β±β = Β± ( )( )( ) β β2 Β± 1.3 β {β3.3,β0.7} 6. JUN β17 [35] 2π₯ + 3π₯ + 10 = 4π₯ + 32 2π₯ β π₯ β 22 = 0 π₯ = Β±β = Β± ( ) ( )( )( ) β 0.25 Β± 3.33 β {β3.1,3.6} Chose the quadratic formula because the equation could not be solved by factoring, and completing the square would require fractions. 7. JUN β18 [27] π₯ = Β±β = Β± ( )( )( ) β 0.5 Β± 2.29 β {β2.8,1.8} 8. AUG β18 [28] Irrational, since the discriminant π β 4ππ = 3 β 4(2)(β10) = 89 is not a perfect square.
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14.7 Solve Quadratics with π β π [CC] 1. JAN '15 [3] Ans: 4 2. AUG β16 [19] Ans: 1 3. AUG β18 [16] Ans: 2 4. SEP '13 [5] 8π + 20πβ 12 = 0 Divide by 4: 2π + 5πβ 3 = 0 2π βπ + 6πβ 3 = 0 π(2πβ 1) + 3(2πβ 1) = 0 (π + 3)(2πβ 1) = 0 β3, 5. JAN '15 [29] 4π₯ β 12π₯ β 7 = 0 π β 12π β 28 = 0 (π + 2)(π β 14) = 0 π = β2 or π = 14 π₯ = β or π₯ = β , 6. JUN β16 [28] 2π₯ + 5π₯ β 42 = 0 π + 5π β 84 = 0 (π + 12)(π β 7) = 0 π = β12 or π = 7 π₯ = β or π₯ = β6, Yes, as shown by solving the equation by the transforming method. 14.8 Word Problems β Quadratic Equations 1. AUG '14 [9] Ans: 3 2. JUN β16 [24] Ans: 2 3. AUG β17 [23] Ans: 4
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4. JUN '14 [34] (2π₯ + 16)(2π₯ + 12) = 396, the length of the garden plus walkway times the width of the garden plus walkway, 4π₯ + 56π₯ β 204 = 0 4(π₯ + 14π₯ β 51) = 0 4(π₯ + 17)(π₯ β 3) = 0 width = 3 m 5. AUG '14 [36] π₯(π₯ + 40) = 6000 π₯ + 40π₯ β 6000 = 0 (π₯ + 100)(π₯ β 60) = 0 60 and 100 6. JAN '15 [37] (2π₯)(π₯ β 3) = 1.25π₯ ; the product of the new length and width give an area that is 1.25 larger; 2π₯ β 6π₯ = 1.25π₯ 0.75π₯ β 6π₯ = 0 π₯(0.75π₯ β 6) = 0 π₯ = 0 or π₯ = 8 1.25(8) = 80 80 square meters 7. JUN '15 [27] 60 + 5π₯ = π₯ + 46 0 = π₯ β 5π₯ β 14 0 = (π₯ + 2)(π₯ β 7) π₯ = β2 or π₯ = 7 Set the equations equal and solve for a positive x. 8. JUN '15 [32] π€(2π€) = 34 2π€ = 34 π€ = 17 π€ = β17 β 4.1 9. AUG '15 [31] 0 β€ π‘ β€ 4, the roots 0 and 4 represent the start time and when it hits the ground in 4 seconds
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10. AUG '15 [37] (2π₯ + 8)(2π₯ + 6) = 100 4π₯ + 28π₯ β 52 = 0 π₯ + 7π₯ β 13 = 0 If x is the width of the frame, add 2x to each dimension to account for both sides, then multiply to find the area π₯ = Β± ( )( )( ) = β8.5 or 1.5 11. JAN '16 [36] width = = 24 β π₯ π₯(24 β π₯) = 108 24π₯ β π₯ = 108 βπ₯ + 24π₯ β 108 = 0 π₯ β 24π₯ + 108 = 0 (π₯ β 6)(π₯ β 18) = 0 {6,18} dimensions are 6 and 18 meters 12. JUN β18 [29] β16π‘ + 256 = 0 16π‘ = 256 π‘ = 16 π‘ = 4 (reject β4) 13. JAN β20 [36] length = + 6 π€ + 6 = 432 + 6π€ = 432 2 + 2[6π€] = 2[432] π€ + 12π€ = 864 π€ + 12π€ + 36 = 864 + 36 (π€ + 6) = 900 π€ + 6 = Β±30 π€ = β6 Β± 30 width π€ = 24 (reject π€ = β36) length = + 6 = 18
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Chapter 15. Parabolas
15.1 Find Roots Given a Parabolic Graph 1. MAY β13 [2] Ans: 3 2. AUG '14 [5] Ans: 4 3. JAN β17 [6] Ans: 4 4. JAN β19 [9] Ans: 3 5. JUN β19 [18] Ans: 3 6. JUN β17 [33] π₯ + 3π₯ β 18 = 0 (π₯ + 6)(π₯ β 3) = 0 {β6,3} The zeros are the x-intercepts on the graph of π(π₯). 7. JAN β18 [32] yes, because the x-intercepts (roots) are β2 and 3. 15.2 Find Vertex and Axis Graphically There are no Regents exam questions on this topic. 15.3 Finding Vertex and Axis Algebraically 1. JUN '15 [14] Ans: 3 2. AUG '15 [21] Ans: 4 3. JAN '16 [22] Ans: 3 4. JUN β16 [11] Ans: 2 5. JAN β18 [23] Ans: 2 6. JUN β18 [13] Ans: 2 7. AUG β19 [17] Ans: 3 8. AUG β19 [23] Ans: 4
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9. AUG '14 [29] the vertex for f is (1,6) so the maximum is 6; for the vertex of g, π₯ = = 4 π¦ = β (4) + 4(4) + 3 = 11 the vertex is (4,11) so the maximum is 11; therefore, π(π₯) has the larger maximum 10. JAN '16 [33] π‘ = ( ) = 2 maximum at the vertex where π‘ = 2 seconds; β16π‘ + 64π‘ + 80 = 0 β16(π‘ β 4π‘ β 5) = 0 β16(π‘ + 1)(π‘ β 5) = 0 π‘ = 5 (reject β1) decreases from the vertex until it hits the ground at β(π‘) = 0, or when 2 < π‘ < 5 11. AUG β17 [36] π₯ = ( ) = 4 β(4) = β16(4) + 128(4) + 9000 = 9256 (4, 9256) The y-coordinate is the pilotβs maximum height above the ground after being ejected. 9256 β 9000 = 256, so she was 256 feet above the aircraft. 12. JAN β18 [29] (π₯ + 3)(π₯ β 5) = 0 π₯ β 2π₯ β 15 = 0 π₯ = ( )( ) = 1 (Alternate solution: π₯ = = 1) 15.4 Graph Parabolas 1. AUG β16 [13] Ans: 3 2. JUN β17 [24] Ans: 1 3. JAN β18 [11] Ans: 2 4. JAN β19 [13] Ans: 1
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5. JUN β19 [14] Ans: 3 6. AUG β19 [10] Ans: 1 7. JAN β20 [13] Ans: 2 8. JUN '15 [37]
(75,25), maximum height of 25 feet at a distance of 75 feet; no, 45 yds = 135 ft and β(135) = 9. 9. AUG '15 [33]
π(20) = 400 and π(20) = 1,048,576, so π(π₯) is greater
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10. JUN β16 [27] π₯ = ( )( ) = 2 π¦ = (2) β 4(2) β 1 = β5
π₯ = 2 11. JAN β17 [36]
(2.5,55) The ball reaches a maximum height of 55 units at 2.5 seconds. 12. JUN β17 [26]
(β3,9)
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13. AUG β18 [36]
If the width is 9 ft, its area is 162 ft2. 14. JAN β20 [33]
= 32 ft/sec 15.5 Vertex Form 1. JAN '16 [1] Ans: 2 2. JAN '16 [7] Ans: 4 3. JUN β16 [16] Ans: 3 4. JUN β17 [17] Ans: 3 5. AUG β19 [11] Ans: 1
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6. JUN β19 [32] π¦ = π₯ β 2π₯ β 8 π¦ + 8 = π₯ β 2π₯ π¦ + 8 + 1 = π₯ β 2π₯ + 1 π¦ + 9 = (π₯ β 1) π¦ = (π₯ β 1) β 9 Vertex is (1,β9) 15.6 Vertex Form with aβ 1 [CC] 1. AUG β16 [21] Ans: 3 2. JUN β17 [16] Ans: 4 3. JAN '15 [36] maximum because of the negative coefficient of π₯ ; π¦ β 9 = βπ₯ + 8π₯ π = β = β16 π¦ β 9 β 16 = βπ₯ + 8π₯ β 16 π¦ β 25 = β(π₯ β 8π₯ + 16) π¦ = β(π₯ β 4) + 25
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Chapter 16. Quadratic-Linear Systems
16.1 Solve Quadratic-Linear Systems Algebraically 1. AUG '15 [17] Ans: 2 2. JAN β19 [18] Ans: 3 3. JUN β17 [31] π₯ = π₯ π₯ β π₯ = 0 π₯(π₯ β 1) = 0 {0,1} 16.2 Solve Quadratic-Linear Systems Graphically 1. JAN β17 [23] Ans: 2 2. JAN β18 [10] Ans: 3 3. JUN '14 [37]
3 because the graphs intersect where π₯ = 3; site A because the costs π΄(π₯) < π΅(π₯) at π₯ = 2
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Chapter 17. Cubic and Radical Functions
17.1 Cubic Functions 1. JAN '15 [24] Ans: 1 2. JUN '15 [12] Ans: 2 3. AUG '15 [4] Ans: 1 4. AUG β16 [23] Ans: 1 5. JUN β17 [10] Ans: 3 6. AUG β17 [7] Ans: 1 7. AUG β17 [19] Ans: 3 8. JAN β18 [6] Ans: 4 9. JUN β18 [18] Ans: 2 10. JUN β19 [8] Ans: 3 11. JAN β20 [9] Ans: 2 12. AUG β18 [25] Set each factor equal to zero and solve for x, or graph the function and find the x-intercepts. β3, 1, 8. 13. JAN β19 [30] 3π₯ + 21π₯ + 36π₯ = 0 3π₯(π₯ + 7π₯ + 12) = 0 3π₯(π₯ + 4)(π₯ + 3) = 0 {0,β4,β3} 17.2 Square Root Functions 1. JUN β17 [3] Ans: 4 2. JUN '14 [25]
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Chapter 18. Transformations of Functions
18.1 Translations 1. AUG '15 [1] Ans: 2 2. JAN '16 [20] Ans: 1 3. JAN β18 [19] Ans: 2 4. AUG β18 [8] Ans: 3 5. JAN β19 [10] Ans: 3 6. JUN β19 [4] Ans: 2 7. AUG β19 [5] Ans: 1 8. JAN β20 [7] Ans: 4 9. JUN '14 [28] (4,β1), function is shifted 2 to the right 10. AUG '14 [33]
shifted 2 units down; shifted 4 units to the right 11. JUN '15 [25]
shifted 3 units to the right
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12. JUN β16 [32] π(π₯) = π₯ + 2π₯ β 4 because π(π₯) is a translation 4 units down. 13. AUG β16 [26] Translated 2 units right and 3 units down 14. JUN β17 [32] π(π₯) is a translation a units right and β(π₯) is a translation a units down. 15. JUN β18 [28]
18.2 Reflections There are no Regents exam questions on this topic. 18.3 Stretches 1. AUG '14 [17] Ans: 1 2. JAN '15 [12] Ans: 2 3. JAN β17 [17] Ans: 2 4. AUG β17 [6] Ans: 1
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Chapter 19. Discontinuous Functions
19.1 Piecewise Functions 1. AUG '14 [22] Ans: 2 2. AUG '15 [16] Ans: 2 3. AUG β18 [15] Ans: 4 4. SEP '13 [12]
Since according to the graph, 8 pencils cost $14 and 10 pencils cost $12.50, the cashier is correct. 5. JAN '15 [30]
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6. JUN '15 [34] 15(52 β 40) + 400 = 580 10(38) = 380 580 β 380 = $200 15(π₯ β 40) + 400 = 445 15π₯ β 600 + 400 = 445 15π₯ = 645 π₯ = 43 Solve 15(π₯ β 40) + 400 = 445 for x 7. JUN β16 [36]
1, because there is one point of intersection 8. JUN β18 [32]