Presentation on theme: "Warm Up Identify the hypothesis and conclusion of each conditional; write the converse of the conditional; decide whether the conditional and converse."— Presentation transcript:
1 Warm UpIdentify the hypothesis and conclusion of each conditional; write the converse of the conditional; decide whether the conditional and converse are true or false. If false, provide a counterexample.If you attend CB West, then you are in high school.If 4x = 20, then x = 5.
2 False If you attend CB West, then you are in high school. Hypothesis: You attend CB West.Conclusion: You are in high school.Converse: If you are in high school, then you attend CB West.FalseCounterexample: You could attend CB South.
3 2. If 4x = 20, then x = 5. Hypothesis: 4x = 20. Conclusion: x = 5 Converse: If x = 5, then 4x = 20.True
4 HOMEWORK ANSWERS – PG.34 Hypothesis: 2x – 1 = 5 Conclusion: x = 3 Hypothesis: She’s smartConclusion: I’m a geniusHypothesis: 8y = 40Conclusion: y = 5Hypothesis: S is the midpoint of RTConclusion: RS = ½RTHypothesis: m1 = m2Conclusion: 1 2Hypothesis: 1 2Conclusion: m1 = m2
5 HOMEWORK ANSWERS – PG.35 Hypothesis: 3x – 7 = 32 Conclusion: x = 13 Hypothesis: I’m not tiredConclusion: I can’t sleepHypothesis: You willConclusion: I’ll tryHypothesis: m1 = 90Conclusion: 1 is a right angleHypothesis: a + b = aConclusion: b = 0Hypothesis: x = -5Conclusion: x² = 25B is between A and C if and only if AB + BC = ACmAOC = 180 if and only if AOC is a straight angle.
7 Properties from Equality Addition Property: If a = b, thena + c = b + c.Subtraction Property: If a = b, thena - c = b - c.Multiplication Property: If a = b, thenca = cb.Division Property:If a = b and c ¹ 0, then
8 Properties of Equality Substitution Property: if a = b, then either a or b may be substituted for the other in any equation or inequality.Reflexive Property: a = a.Symmetric Property: if a = b, then b = a.Transitive Property: if a = b and b = c, then a = c.
9 Properties of Congruence Reflexive Property: DE ; ÐDSymmetric Property:--If FG, then DETransitive Property:--If ÐB and ÐC, then ÐC.
10 Introduction to Proofs Given:Prove:What you knowWhat you’re trying to proveStatementsReasonsAll reasons must be:1. “Given”2. Postulates3. Theorems4. Definitions5. PropertiesAll of your math steps go here.
12 Statements Reasons 1. Given 2. Substitution 4. Division Property Example 2: Given: 3x + y = 22; y = 4Prove: x = 6StatementsReasons1. 3x + y = 22; y = 4Given2. Substitution2. 3x + 4 = 223. 3x = 183. Subtraction Property4. x = 64. Division Property
13 NOTE:You will NEVER usethe reason “PROVE”as a justification inyour proofs.
14 Because you use postulates, properties, definitions, and theorems for your reasons in proofs, it is a good idea to review the vocabulary and postulates that we learned in Unit 1.Segment Addition Postulate –If point B is between points A and C, then AB + BC = AC.ABC
15 Angle Addition Postulate – If point B lies in the interior of ÐAQC, then mÐAQB + mÐBQC = mÐAQC.If ÐAQC is a straight angle and B is a point not on AC, then mÐAQB + mÐBQC = 180.AQCBAQCBIf you used this statement above, it would also be acceptable to use “definition of a linear pair” as your reason.
16 F T L A Statements Reasons FL = AT LA = LA FL + LA = AT + LA Example 3:FTGiven: FL = ATProve: FA = LTLAStatementsReasonsFL = ATLA = LAFL + LA = AT + LAFL + LA = FALA + AT = LT5. FA = LTGivenSometimes, this step is not necessary. You could just go right to the next step.2. Reflexive Property3. Addition Property4. Segment Addition Postulate5. Substitution
17 Statements Reasons mÐAOC = mÐBOD mÐAOC = mÐ1 + mÐ2 mÐBOD = mÐ2 + mÐ3 Example 4:Given: mÐAOC = mÐBOD Prove: mÐ1 = mÐ3BA123DOStatementsReasonsmÐAOC = mÐBODmÐAOC = mÐ1 + mÐ2 mÐBOD = mÐ2 + mÐ3mÐ1 + mÐ2 = mÐ2 + mÐ3mÐ2 = mÐ25. mÐ = mÐ3Given2. Angle Addition Postulate3. SubstitutionSometimes, this step is not necessary. You could just go right to the next step.Reflexive5. Subtraction Property
18 P R S Q T Statements Reasons 1. RS = PS 1. Given 2. RS + ST = PS + ST Example 5:PGiven: RS = PS and ST = SQ.Prove: RT = PQRSQTStatementsReasonsRS = PSGiven2. RS + ST = PS + ST2. Addition PropertyST = SQGiven4. RS + ST = PS + SQ4. Substitution5. RS + ST = RTPS + SQ = PQ5. Segment Addition PostulateRT = PQ6. Substitution