Name: Class: Date: B. The twentieth term is A. D. There is not enough information.
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1 Class: Date: Chapter 2 Review 1. Based on the pattern, what are the next two terms of the sequence? 9, 15, 21, 27,... A. 33, 972 B. 39, 45 C. 162, 972 D. 33, What conjecture can you make about the twentieth term in the pattern A, B, A, C, A, B, A, C? A. The twentieth term is C. C. The twentieth term is B. B. The twentieth term is A. D. There is not enough information. 3. What conjecture can you make about the sum of the first 10 odd numbers? A. The sum is C. The sum is B. The sum is D. The sum is What conjecture can you make about the sum of the first 40 positive even numbers? 2 = 2 = = 6 = = 12 = = 20 = = 30 = 5 6 A. The sum is C. The sum is B. The sum is D. The sum is What conjecture can you make about the product of 13 and 8,888,888? = = 11, = 115, ,888 = 1,155,544 A. 115,555,544 C. 1,155,555,544 B. 1,115,555,444 D. 11,155,555, Alfred is practicing typing. The first time he tested himself, he could type 45 words per minute. After practicing for a week, he could type 48 words per minute. After two weeks he could type 51 words per minute. Based on this pattern, predict how fast Alfred will be able to type after 4 weeks of practice. A. 61 words per minute C. 54 words per minute B. 57 words per minute D. 51 words per minute 1
2 7. What is a counterexample for the conjecture? Conjecture: Any number that is divisible by 4 is also divisible by 8. A. 24 B. 40 C. 12 D What is the conclusion of the following conditional? A number is divisible by 2 if the number is even. A. The sum of the digits of the number is divisible by 2. B. If a number is even, then the number is divisible by 2. C. The number is even. D. The number is divisible by Identify the hypothesis and conclusion of this conditional statement: If two lines intersect at right angles, then the two lines are perpendicular. A. Hypothesis: The two lines are perpendicular. Conclusion: Two lines intersect at right angles. B. Hypothesis: Two lines intersect at right angles. Conclusion: The two lines are perpendicular. C. Hypothesis: The two lines are not perpendicular. Conclusion: Two lines intersect at right angles. D. Hypothesis: Two lines intersect at right angles. Conclusion: The two lines are not perpendicular. 10. Write this statement as a conditional in if-then form: All triangles have three sides. A. If a triangle has three sides, then all triangles have three sides. B. If a figure has three sides, then it is not a triangle. C. If a figure is a triangle, then all triangles have three sides. D. If a figure is a triangle, then it has three sides. 11. What is the converse of the following conditional? If a point is in the fourth quadrant, then its coordinates are negative. A. If a point is in the fourth quadrant, then its coordinates are negative. B. If a point is not in the fourth quadrant, then the coordinates of the point are not negative. C. If the coordinates of a point are not negative, then the point is not in the fourth quadrant. D. If the coordinates of a point are negative, then the point is in the fourth quadrant. 2
3 12. Which conditional has the same truth value as its converse? A. If x = 7, then x 7. B. If a figure is a square, then it has four sides. C. If x 17 = 4, then x = 21. D. If an angle has a measure of 80, then it is acute. 13. For the following true conditional statement, write the converse. If the converse is also true, combine the statements as a biconditional. If x = 7, then x 2 = 49. A. If x 2 = 49, then x = 7. True; x 2 = 49 if and only if x = 7. B. If x 2 = 49, then x = 7. True; x = 7 if and only if x 2 = 49. C. If x 2 = 49, then x = 7. False D. If x 2 = 7, then x = 49. False 14. What is the converse of the following true conditional? If the converse is true, rewrite the statements as a biconditional. If either is false, give a counterexample. If two lines are parallel, they do not intersect. A. If two lines do not intersect, they are parallel. One statement is false. If two lines do not intersect, they could be skew. B. If two lines do not intersect, they are parallel. One statement is false. If two lines are parallel, they may intersect twice. C. If two lines do not intersect, they are parallel. Both statements are true. Two lines are parallel if (and only if) they do not intersect. D. If two lines do not intersect, they are not parallel. Both statements are true. Two lines are not parallel if (and only if) they do not intersect. 3
4 15. What is the value of x? Identify the missing justifications. m PQR x 5, m SQR x 7, and m PQS 100. m PQR m SQR m PQS x 5 + x 7 = 100 2x 12 = 100 2x = 112 x = 56 a. b. Substitution Property c. Simplify d. e. Division Property of Equality A. Angle Addition Postulate; Subtraction Property of Equality B. Angle Addition Postulate; Addition Property of Equality C. Protractor Postulate; Addition Property of Equality D. Protractor Postulate; Subtraction Property of Equality Use the given property to complete the statement. 16. Transitive Property of Congruence If CD EF and EF GH, then. A. EF EF C. EF GH B. CD EF D. CD GH 4
5 17. Complete the two-column proof. Given: 12x 6y 5; x 5 Prove: 65 6 y 12x 6y 5; x y 5 6y 65 y 65 6 a. b. c. d. 65 y e. 6 A. a. Given b. Substitution Property c. Addition Property of Equality d. Division Property of Equality e. Symmetric Property of Equality B. a. Given b. Substitution Property c. Addition Property of Equality d. Addition Property of Equality e. Symmetric Property of Equality C. a. Given b. Symmetric Property of Equality c. Addition Property of Equality d. Division Property of Equality e. Reflexive Property of Equality D. a. Given b. Substitution Property c. Addition Property of Equality d. Division Property of Equality e. Reflexive Property of Equality 18. Complete the two-column proof. Given: x Prove: x 10 x 9 11 a. 5 x 5 2 b. x 10 c. A. a. Given b. Subtraction Property of Equality c. Division Property of Equality B. a. Given b. Subtraction Property of Equality c. Multiplication Property of Equality C. a. Given b. Addition Property of Equality c. Multiplication Property of Equality D. a. Given b. Addition Property of Equality c. Division Property of Equality 5
6 19. What is the value of x? A. 16 B. 120 C. 60 D What is the value of x? A. 84 B. 36 C. 120 D m Find m 4. A. 150 B. 30 C. 160 D. 20 6
7 22. Find the values of x and y. A. x = 15, y = 17 C. x = 68, y = 112 B. x = 112, y = 68 D. x = 17, y = Write the converse of the statement. If the converse is true, write true; if not true, provide a counterexample. If x = 4, then x 2 = What is the value of x? Identify the missing justifications. m AOC 150 m AOB m BOC m AOC 2x 6(x 3) 150 2x 6x x x 168 x 21 a. b. c. d. e. f. 7
8 25. Solve for x. Justify each step. 4x
9 Chapter 2 Review Answer Section 1. ANS: D PTS: 1 DIF: L3 TOP: 2-1 Problem 1 Finding and Using a Pattern KEY: pattern inductive reasoning 2. ANS: A PTS: 1 DIF: L3 TOP: 2-1 Problem 2 Using Inductive Reasoning KEY: inductive reasoning pattern conjecture 3. ANS: B PTS: 1 DIF: L4 TOP: 2-1 Problem 3 Collecting Information to Make a Conjecture KEY: inductive reasoning conjecture pattern 4. ANS: C PTS: 1 DIF: L3 TOP: 2-1 Problem 3 Collecting Information to Make a Conjecture KEY: inductive reasoning pattern conjecture 5. ANS: A PTS: 1 DIF: L3 TOP: 2-1 Problem 3 Collecting Information to Make a Conjecture KEY: pattern inductive reasoning conjecture 6. ANS: B PTS: 1 DIF: L3 TOP: 2-1 Problem 4 Making a Prediction KEY: conjecture inductive reasoning word problem problem solving 7. ANS: C PTS: 1 DIF: L2 TOP: 2-1 Problem 5 Finding a Counterexample KEY: conjecture counterexample 8. ANS: D PTS: 1 DIF: L3 REF: 2-2 Conditional Statements OBJ: To recognize conditional statements and their parts TOP: 2-2 Problem 1 Identifying the Hypothesis and the Conclusion KEY: conditional statement conclusion 1
10 9. ANS: B PTS: 1 DIF: L3 REF: 2-2 Conditional Statements OBJ: To recognize conditional statements and their parts TOP: 2-2 Problem 1 Identifying the Hypothesis and the Conclusion KEY: conditional statement hypothesis conclusion 10. ANS: D PTS: 1 DIF: L2 REF: 2-2 Conditional Statements OBJ: To recognize conditional statements and their parts TOP: 2-2 Problem 2 Writing a Conditional KEY: hypothesis conclusion conditional statement 11. ANS: D PTS: 1 DIF: L2 REF: 2-2 Conditional Statements OBJ: To write converses, inverses, and contrapositives of conditionals NAT: CC G.CO.9 CC G.CO.10 CC G.CO.11 TOP: 2-2 Problem 4 Writing and Finding Truth Values of Statements KEY: conditional statement converse of a conditional 12. ANS: C PTS: 1 DIF: L3 REF: 2-2 Conditional Statements OBJ: To write converses, inverses, and contrapositives of conditionals NAT: CC G.CO.9 CC G.CO.10 CC G.CO.11 TOP: 2-2 Problem 4 Writing and Finding Truth Values of Statements KEY: conditional statement converse of a conditional truth value 13. ANS: C PTS: 1 DIF: L3 REF: 2-3 Biconditionals and Definitions OBJ: To write biconditionals and recognize good definitions NAT: CC G.CO.9 CC G.CO.10 CC G.CO.11 G.1.c TOP: 2-3 Problem 1 Writing a Biconditional KEY: conditional statement converse of a conditional biconditional statement 14. ANS: A PTS: 1 DIF: L3 REF: 2-3 Biconditionals and Definitions OBJ: To write biconditionals and recognize good definitions NAT: CC G.CO.9 CC G.CO.10 CC G.CO.11 G.1.c TOP: 2-3 Problem 1 Writing a Biconditional KEY: conditional statement converse of a conditional biconditional statement counterexample 15. ANS: B PTS: 1 DIF: L3 REF: 2-5 Reasoning in Algebra and Geometry OBJ: To connect reasoning in algebra and geometry NAT: CC G.CO.9 CC G.CO.10 CC G.CO.11 G.5.b TOP: 2-5 Problem 1 Justifying Steps When Solving an Equation KEY: Properties of Equality Angle Addition Postulate deductive reasoning 16. ANS: D PTS: 1 DIF: L3 REF: 2-5 Reasoning in Algebra and Geometry OBJ: To connect reasoning in algebra and geometry NAT: CC G.CO.9 CC G.CO.10 CC G.CO.11 G.5.b TOP: 2-5 Problem 2 Using Properties of Equality and Congruence KEY: Properties of Congruence Transitive Property 2
11 17. ANS: A PTS: 1 DIF: L3 REF: 2-5 Reasoning in Algebra and Geometry OBJ: To connect reasoning in algebra and geometry NAT: CC G.CO.9 CC G.CO.10 CC G.CO.11 G.5.b TOP: 2-5 Problem 3 Writing a Two-Column Proof KEY: Properties of Equality Proof Two-column Proof 18. ANS: B PTS: 1 DIF: L2 REF: 2-5 Reasoning in Algebra and Geometry OBJ: To connect reasoning in algebra and geometry NAT: CC G.CO.9 CC G.CO.10 CC G.CO.11 G.5.b TOP: 2-5 Problem 3 Writing a Two-Column Proof KEY: Properties of Equality proof Two-column Proof 19. ANS: D PTS: 1 DIF: L3 REF: 2-6 Proving Angles Congruent OBJ: To prove and apply theorems about angles NAT: CC G.CO.9 G.5.b TOP: 2-6 Problem 1 Using the Vertical Angles Theorem KEY: vertical angles Vertical Angles Theorem 20. ANS: D PTS: 1 DIF: L2 REF: 2-6 Proving Angles Congruent OBJ: To prove and apply theorems about angles NAT: CC G.CO.9 G.5.b TOP: 2-6 Problem 1 Using the Vertical Angles Theorem KEY: vertical angles Vertical Angles Theorem 21. ANS: B PTS: 1 DIF: L2 REF: 2-6 Proving Angles Congruent OBJ: To prove and apply theorems about angles NAT: CC G.CO.9 G.5.b TOP: 2-6 Problem 1 Using the Vertical Angles Theorem KEY: Vertical Angles Theorem vertical angles 22. ANS: A PTS: 1 DIF: L4 REF: 2-6 Proving Angles Congruent OBJ: To prove and apply theorems about angles NAT: CC G.CO.9 G.5.b TOP: 2-6 Problem 1 Using the Vertical Angles Theorem KEY: Vertical Angles Theorem vertical angles supplementary angles multi-part question 23. ANS: If x 2 = 16, then x = 4. False; if x 2 = 16, then x can be equal to 4. PTS: 1 DIF: L3 REF: 2-2 Conditional Statements OBJ: To write converses, inverses, and contrapositives of conditionals NAT: CC G.CO.9 CC G.CO.10 CC G.CO.11 TOP: 2-2 Problem 4 Writing and Finding Truth Values of Statements KEY: conditional statement converse of a conditional counterexample truth value 3
12 24. ANS: a. Angle Addition Postulate b. Substitution Property c. Distributive Property d. Simplify e. Addition Property of Equality f. Division Property of Equality PTS: 1 DIF: L3 REF: 2-5 Reasoning in Algebra and Geometry OBJ: To connect reasoning in algebra and geometry NAT: CC G.CO.9 CC G.CO.10 CC G.CO.11 G.5.b TOP: 2-5 Problem 1 Justifying Steps When Solving an Equation KEY: proof deductive reasoning Properties of Equality multi-part question 25. ANS: 4x 9 99 Given 4x Addition Property of Equality 4x 108 Simplify 4x Division Property of Equality x = 27 Simplify PTS: 1 DIF: L4 REF: 2-5 Reasoning in Algebra and Geometry OBJ: To connect reasoning in algebra and geometry NAT: CC G.CO.9 CC G.CO.10 CC G.CO.11 G.5.b TOP: 2-5 Problem 1 Justifying Steps When Solving an Equation KEY: Properties of Equality proof deductive reasoning 4
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