PreCalc 11 Final Review Pack v1 Answer Section
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1 PreCalc Final Review Pack v Answer Section MULTIPLE CHOICE. ANS: A PTS: DIF: Difficult REF:. Arithmetic Sequences LOC:.RF9. ANS: C PTS: DIF: Difficult REF:. Arithmetic Sequences LOC:.RF9. ANS: C PTS: DIF: Difficult REF:. Arithmetic Series LOC:.RF9 4. ANS: A PTS: DIF: Easy REF:. Arithmetic Series LOC:.RF9 5. ANS: C PTS: DIF: Difficult REF:. Arithmetic Series LOC:.RF9 6. ANS: C PTS: DIF: Easy REF:. Geometric Sequences LOC:.RF0 7. ANS: D PTS: DIF: Difficult REF:. Geometric Sequences LOC:.RF0 8. ANS: A PTS: DIF: Moderate REF:. Geometric Sequences LOC:.RF0 9. ANS: D PTS: DIF: Moderate REF:.4 Geometric Series LOC:.RF0 0. ANS: D PTS: DIF: Moderate REF:.4 Geometric Series LOC:.RF0. ANS: D PTS: DIF: Moderate REF:.4 Geometric Series LOC:.RF0. ANS: C PTS: DIF: Easy REF:.5 Graphing Geometric Sequences and Series LOC:.RF9.RF0. ANS: A PTS: DIF: Moderate REF:.5 Graphing Geometric Sequences and Series LOC:.RF9.RF0 4. ANS: C PTS: DIF: Easy REF:.6 Infinite Geometric Series LOC:.RF0 5. ANS: A PTS: DIF: Easy REF:.6 Infinite Geometric Series LOC:.RF0 6. ANS: B PTS: DIF: Moderate REF:.6 Infinite Geometric Series LOC:.RF0 7. ANS: C PTS: 0 DIF: Moderate REF:. Absolute Value of a Real Number LOC:.AN 8. ANS: D PTS: 0 DIF: Moderate REF:. Adding and Subtracting Radical Expressions LOC:.AN
2 9. ANS: C PTS: 0 DIF: Difficult REF:. Simplifying Radical Expressions LOC:.AN 0. ANS: C PTS: 0 DIF: Easy REF:. Adding and Subtracting Radical Expressions LOC:.AN. ANS: A PTS: 0 DIF: Moderate REF:. Absolute Value of a Real Number LOC:.AN. ANS: D PTS: 0 DIF: Moderate REF:. Adding and Subtracting Radical Expressions LOC:.AN. ANS: C PTS: 0 DIF: Moderate REF:.4 Multiplying and Dividing Radical Expressions LOC:.AN 4. ANS: D PTS: 0 DIF: Moderate REF:. Factoring Polynomial Expressions LOC:.RF 5. ANS: D PTS: 0 DIF: Moderate REF:. Solving Quadratic Equations by Factoring LOC:.AN TOP: Algebra and Number 6. ANS: C PTS: 0 DIF: Difficult REF:.5 Solving Radical Equations LOC:.AN 7. ANS: A PTS: 0 DIF: Moderate REF:.4 Developing and Applying the Quadratic Formula LOC:.RF5 8. ANS: D PTS: 0 DIF: Easy REF:. Solving Quadratic Equations by Factoring LOC:.RF5 9. ANS: A PTS: 0 DIF: Moderate REF:.4 Developing and Applying the Quadratic Formula LOC:.RF5 0. ANS: C PTS: 0 DIF: Easy REF:.5 Interpreting the Discriminant LOC:.RF5. ANS: C PTS: 0 DIF: Moderate REF:.5 Interpreting the Discriminant LOC:.RF5. ANS: C PTS: 0 DIF: Moderate REF:.5 Interpreting the Discriminant LOC:.RF5. ANS: C PTS: 0 DIF: Moderate REF:.5 Interpreting the Discriminant LOC:.RF5 4. ANS: C PTS: 0 DIF: Moderate REF: 4. Properties of a Quadratic Function LOC:.RF4
3 5. ANS: D PTS: 0 DIF: Moderate REF: 4. Properties of a Quadratic Function LOC:.RF4 6. ANS: D PTS: 0 DIF: Easy REF: 4. Transforming the Graph of y = x^ LOC:.RF 7. ANS: B PTS: 0 DIF: Easy REF: 4. Transforming the Graph of y = x^ LOC:.RF 8. ANS: D PTS: 0 DIF: Easy REF: 4. Transforming the Graph of y = x^ LOC:.RF 9. ANS: D PTS: 0 DIF: Moderate REF: 4.4 Analyzing Quadratic Functions of the Form y = a(x - p)^ + q LOC:.RF 40. ANS: C PTS: 0 DIF: Moderate REF: 4.4 Analyzing Quadratic Functions of the Form y = a(x - p)^ + q LOC:.RF 4. ANS: B PTS: 0 DIF: Moderate REF: 4.5 Equivalent Forms of the Equation of a Quadratic Function LOC:.RF4 4. ANS: A PTS: 0 DIF: Moderate REF: 4.7 Modelling and Solving Problems with Quadratic Functions LOC:.RF4 4. ANS: C PTS: 0 DIF: Moderate REF: 4.6 Analyzing Quadratic Functions of the Form y = ax^ + bx + c LOC:.RF4 44. ANS: C PTS: 0 DIF: Easy REF: 5. Graphing Linear Inequalities in Two Variables LOC:.RF7 45. ANS: A PTS: 0 DIF: Easy REF: 5. Graphing Quadratic Inequalities in Two Variables LOC:.RF7 46. ANS: D PTS: 0 DIF: Easy REF: 5. Graphing Quadratic Inequalities in Two Variables LOC:.RF7 47. ANS: D PTS: 0 DIF: Easy REF: 5. Graphing Quadratic Inequalities in Two Variables LOC:.RF7 48. ANS: C PTS: 0 DIF: Moderate REF: 5.5 Solving Systems of Equations Algebraically LOC:.RF6 49. ANS: A PTS: 0 DIF: Moderate REF: 6. Angles in Standard Position in Quadrant LOC:.T TOP: Trigonometry Problem-Solving Skills
4 50. ANS: A PTS: 0 DIF: Moderate REF: 6. Constructing Triangles LOC:.T TOP: Trigonometry 5. ANS: C PTS: 0 DIF: Moderate REF: 7. Equivalent Rational Expressions LOC:.AN4 TOP: Algebra and Number 5. ANS: D PTS: 0 DIF: Moderate REF: 6. Angles in Standard Position in All Quadrants LOC:.T TOP: Trigonometry 5. ANS: D PTS: 0 DIF: Easy REF: 7. Equivalent Rational Expressions LOC:.AN4 TOP: Algebra and Number 54. ANS: B PTS: 0 DIF: Moderate REF: 7. Equivalent Rational Expressions LOC:.AN4 TOP: Algebra and Number 55. ANS: B PTS: 0 DIF: Moderate REF: 7. Equivalent Rational Expressions LOC:.AN4 TOP: Algebra and Number 56. ANS: A PTS: DIF: Moderate REF: 7. Multiplying and Dividing Rational Expressions LOC:.AN5 TOP: Algebra and Number 57. ANS: D PTS: DIF: Easy REF: 7. Multiplying and Dividing Rational Expressions LOC:.AN5 TOP: Algebra and Number 58. ANS: B PTS: 0 DIF: Moderate REF: 7.4 Adding and Subtracting Rational Expressions with Binomial and Trinomial Denominators LOC:.AN5 TOP: Algebra and Number 59. ANS: B PTS: 0 DIF: Easy REF: 7. Adding and Subtracting Rational Expressions with Monomial Denominators LOC:.AN5 TOP: Algebra and Number 60. ANS: D PTS: 0 DIF: Easy REF: 7. Adding and Subtracting Rational Expressions with Monomial Denominators LOC:.AN5 TOP: Algebra and Number 6. ANS: C PTS: 0 DIF: Easy REF: 7.4 Adding and Subtracting Rational Expressions with Binomial and Trinomial Denominators LOC:.AN5 TOP: Algebra and Number 6. ANS: B PTS: 0 DIF: Moderate REF: 8. Absolute Value Functions LOC:.RF Problem-Solving Skills 6. ANS: D PTS: 0 DIF: Moderate REF: 8. Absolute Value Functions LOC:.RF 64. ANS: C PTS: 0 DIF: Easy REF: 8. Absolute Value Functions LOC:.RF 65. ANS: C PTS: 0 DIF: Moderate REF: 8. Absolute Value Functions LOC:.RF 4
5 66. ANS: A PTS: 0 DIF: Difficult REF: 8. Absolute Value Functions LOC:.RF 67. ANS: C PTS: 0 DIF: Moderate REF: 8. Absolute Value Functions LOC:.RF 68. ANS: A PTS: 0 DIF: Moderate REF: 8. Absolute Value Functions LOC:.RF 69. ANS: D PTS: 0 DIF: Easy REF: 8. Solving Absolute Value Equations LOC:.RF 70. ANS: A PTS: 0 DIF: Easy REF: 8. Graphing Reciprocals of Linear Functions LOC:.RF 7. ANS: B PTS: 0 DIF: Moderate REF: 8. Absolute Value Functions LOC:.RF 7. ANS: D PTS: 0 DIF: Moderate REF: 8. Graphing Reciprocals of Linear Functions LOC:.RF 7. ANS: B PTS: 0 DIF: Easy REF: 8.4 Using Technology to Graph Reciprocals of Quadratic Functions LOC:.RF 74. ANS: A PTS: 0 DIF: Moderate REF: 8. Graphing Reciprocals of Linear Functions LOC:.RF 75. ANS: B PTS: 0 DIF: Easy REF: 8.4 Using Technology to Graph Reciprocals of Quadratic Functions LOC:.RF 76. ANS: B PTS: 0 DIF: Moderate REF: 8.4 Using Technology to Graph Reciprocals of Quadratic Functions LOC:.RF 77. ANS: C PTS: 0 DIF: Moderate REF: 8.5 Graphing Reciprocals of Quadratic Functions LOC:.RF 78. ANS: B PTS: 0 DIF: Moderate REF: 8.5 Graphing Reciprocals of Quadratic Functions LOC:.RF 79. ANS: B PTS: 0 DIF: Difficult REF: 8.5 Graphing Reciprocals of Quadratic Functions LOC:.RF 5
6 SHORT ANSWER 80. ANS: r =.5, so the sum is not finite. PTS: DIF: Easy REF:.6 Infinite Geometric Series LOC:.RF0 8. ANS: i) 4 ii) 6 iii) iv) 5 The values of the expressions from greatest to least are: 5, 6,, 4 PTS: 0 DIF: Moderate REF:. Absolute Value of a Real Number LOC:.AN 8. ANS: (7x 8y )(7x 8y ) PTS: 0 DIF: Moderate REF:. Factoring Polynomial Expressions LOC:.RF 8. ANS: 6 m PTS: 0 DIF: Moderate REF: 4.7 Modelling and Solving Problems with Quadratic Functions LOC:.RF4 KEY: Problem-Solving Skills Procedural Knowledge 84. ANS: A linear-quadratic system may have solutions, solution, or no solution. A quadratic-quadratic system may have solutions, solution, no solution, or infinite solutions. PTS: 0 DIF: Easy REF: 5.5 Solving Systems of Equations Algebraically LOC:.RF6 KEY: Communication Conceptual Understanding 85. ANS: a) coordinates of the vertex: (0, 50) b) x 0 c) x-intercepts: 45, 45 y-intercept: 50 PTS: 0 DIF: Easy REF: 4.4 Analyzing Quadratic Functions of the Form y = a(x - p)^ + q LOC:.RF 6
7 86. ANS: approximately 56. PTS: 0 DIF: Moderate REF: 6. Angles in Standard Position in Quadrant LOC:.T TOP: Trigonometry Problem-Solving Skills 87. ANS: a) The reference angle is 7. b) The other angles that have the same reference angle are: 87 5 PTS: 0 DIF: Moderate REF: 6. Angles in Standard Position in All Quadrants LOC:.T TOP: Trigonometry 88. ANS: 4m 6m mn n, m 0, n 0 4mn PTS: 0 DIF: Moderate REF: 7. Adding and Subtracting Rational Expressions with Monomial Denominators LOC:.AN5 TOP: Algebra and Number 89. ANS: 5x 4(x )(x ), x, x PTS: 0 DIF: Moderate REF: 7.4 Adding and Subtracting Rational Expressions with Binomial and Trinomial Denominators LOC:.AN5 TOP: Algebra and Number 90. ANS: Yes, the Sine Law can be used; KL =.0 cm. PTS: 0 DIF: Moderate REF: 6.4 The Sine Law LOC:.T TOP: Trigonometry 7
8 PROBLEM 9. ANS: The prices of the tickets form an arithmetic sequence with first terms $85, $77, $69,... The total cost of Joe s tickets is the sum of the first 8 terms of the arithmetic series: $85 $77 $69... Use:S n n[t d(n )] Substitute: n 8, t 85, d 8 8[(85) 8(8 )] S 8 S To buy ticket from each section, Joe will have to spend $456. PTS: DIF: Difficult REF:. Arithmetic Series LOC:.RF9 KEY: Communication Conceptual Understanding Problem-Solving Skills 9. ANS: a) S n n[t d(n )] Substitute: n 96, t 7, d 7 96[(7) 7(96 )] S 96 S b) t n t d(n ) Substitute: t 7, d 7, n 96 t (96 ) t 96 7 PTS: DIF: Moderate REF:. Arithmetic Series LOC:.RF9 Problem-Solving Skills 8
9 9. ANS: a) r is 5 5 Use:t n t r n Substitute: t n, r 5, n 4 4 Ê ˆ t Á 5 t 75 Ê t 75 ˆ Á 5 75 Ê t 75 ˆ Á 5 5 The first terms are: t 75, t 75, t 5 b) The sequence is convergent because the terms approach a constant value of 0. PTS: DIF: Moderate REF:. Geometric Sequences LOC:.RF0 KEY: Communication Conceptual Understanding Problem-Solving Skills 94. ANS: Write 5x kx 4 = (5x )(x h) Equate the constant terms. If 5x is a factor, then 4 h, so h Substitute this value for h in (5x )(x h), then expand. (5x )(x ) 5x 8x 4 The value of k is 8. PTS: 0 DIF: Difficult REF:. Factoring Polynomial Expressions LOC:.RF KEY: Communication Problem-Solving Skills 9
10 95. ANS: Write an inequality to represent the problem. Let x m represent the width of the garden. Then its length is (x 5) m. And its area, in square metres, is x(x 5). The area is at least 45 m. So, an inequality is: x(x 5) 45 Rearrange the inequality. x 5x 45 0 Use a graphing calculator to graph the corresponding quadratic function: y x 5x 45 Determine the x-intercepts: x and x From the graph, the inequality is greater than or equal to 0 for x or x Since the width of the garden is positive, the solution of the problem is: x The width of the garden is greater than or equal to approximately 4.7 m and its length is greater than or equal to approximately (4.7 5) m, or 9.7 m. Verify the solution. The area of the garden with dimensions 4.7 m and 9.7 m is: (4.7 m)(9.7 m) m This is greater than 45 m, so the possible dimensions are correct. PTS: 0 DIF: Difficult REF: 5. Solving Quadratic Inequalities in One Variable LOC:.RF8 KEY: Communication Problem-Solving Skills 96. ANS: Use the formula for the surface area, SA, of a cube: SA 6x, where x represents the edge length. Cube A has a surface area of M square units. For cube B: SA M 6x M x M For cube C: SA 9M 6x 9M x M x M x M The edge length of cube B is M units. The edge length of cube C is M units. PTS: 0 DIF: Moderate REF:. Simplifying Radical Expressions LOC:.AN Communication Problem-Solving Skills 0
11 97. ANS: a) Let x represent the first number and let y represent the second number. The statement that the sum of 5 times the first number and 6 times the second number is 0 can be modelled with the equation: 5x 6y 0 The statement that when twice the second number is subtracted from the square of the first number, the result is equal to 0 minus the first number can be modelled with the equation: x y 0 x Solve each equation for y to get a system: y 5 6 x y x x 0 b) Solve the system. Since the left sides of the equations are equal, the right sides must also be equal. 5 6 x x x 0 x x 5 6 x 0 0 x 6 x 5 6 x 0 0 x 8 x 0 0 Use the quadratic formula. x b b 4ac Substitute: a, b 8 a, c 0 8 Ê 8 ˆ Á 4()( 0) x () x So, x 6 or x 0 Substitute each value of x in equation. When x 6: y 5 6 ( 6) y 5 The solutions are: 6 and 5; 0 When x 0 : Ê y 5 ˆ Ê 0 ˆ Á 6 Á y 5 9 and 5 9 Verify the solutions using the statement of the problem.
12 For x 6 and y 5: The sum of 5 times 6 and 6 times 5 is 0. When times 5 is subtracted from ( 6), the result is equal to 0 minus 6. These numbers satisfy the problem statement. For x 0 and y 5 9 : The sum of 5 times 0 and 6 times 5 9 When times 5 9 is subtracted from Ê 0 Á is 0. ˆ These numbers satisfy the problem statement. So, the numbers are: 6 and 5; or 0 and 5 9, the result is equal to 0 minus 0. PTS: 0 DIF: Difficult REF: 5.5 Solving Systems of Equations Algebraically LOC:.RF6 KEY: Communication Problem-Solving Skills 98. ANS: Draw a labelled diagram to represent the problem. In right FLP, FP is the hypotenuse and FL is the side opposite P. So, use the sine ratio to determine the length of FL. sin P opposite hypotenuse sin P FL FP sin 57 FL 8. Solve the equation for FL. sin 57 FL 8. 8.sin 57 FL FL 6.79 The distance between the fishing boat and the lighthouse is approximately 6.8 km. PTS: 0 DIF: Moderate REF: 6. Angles in Standard Position in All Quadrants LOC:.T TOP: Trigonometry KEY: Communication Problem-Solving Skills
13 99. ANS: Let x hours represent the time it takes Ginelle to install a subfloor on her own. Then, the time it takes Tonya is (x ) hours. In h, Ginelle can install x of the subfloor and Tonya can install of the subfloor. x Working together, it takes 8 h to completely install the subfloor. So, an equation is: Ê 8 x ˆ Á x 8 x 8 x, x 0 Solve the equation. A common denominator is: x(x ) 8 x 8 x Ê 8 ˆ x(x ) Á x x(x ) Ê 8 ˆ Á x x(x ) 8(x ) 8(x) x(x ) 8x 96 8x x x x 4x 96 0 x or x 8 Since time cannot be negative, x. So, x 4 It would take Ginelle h to install a subfloor and it would take Tonya 4 h to install a subfloor. PTS: 0 DIF: Difficult REF: 7.6 Applications of Rational Equations LOC:.AN6 TOP: Algebra and Number Communication Problem-Solving Skills 00. ANS: a) I can equate the two reciprocals to determine if there is a value of x for which the y-values are the same. If there is, then the graphs intersect. b) The y-values are equal when: f(x) g(x) x x 5 x x 5 4x 8 x Substitute x in y x : y () y = The graphs of the reciprocal functions intersect at (, ). PTS: 0 DIF: Difficult REF: 8. Graphing Reciprocals of Linear Functions LOC:.RF Communication
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