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Warm Up Identify 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.

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**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. False Counterexample: You could attend CB South.

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**2. If 4x = 20, then x = 5. Hypothesis: 4x = 20. Conclusion: x = 5**

Converse: If x = 5, then 4x = 20. True

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**HOMEWORK ANSWERS – PG.34 Hypothesis: 2x – 1 = 5 Conclusion: x = 3**

Hypothesis: She’s smart Conclusion: I’m a genius Hypothesis: 8y = 40 Conclusion: y = 5 Hypothesis: S is the midpoint of RT Conclusion: RS = ½RT Hypothesis: m1 = m2 Conclusion: 1 2 Hypothesis: 1 2 Conclusion: m1 = m2

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**HOMEWORK ANSWERS – PG.35 Hypothesis: 3x – 7 = 32 Conclusion: x = 13**

Hypothesis: I’m not tired Conclusion: I can’t sleep Hypothesis: You will Conclusion: I’ll try Hypothesis: m1 = 90 Conclusion: 1 is a right angle Hypothesis: a + b = a Conclusion: b = 0 Hypothesis: x = -5 Conclusion: x² = 25 B is between A and C if and only if AB + BC = AC mAOC = 180 if and only if AOC is a straight angle.

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Section 2-2 Introduction to Proofs

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**Properties from Equality**

Addition Property: If a = b, then a + c = b + c. Subtraction Property: If a = b, then a - c = b - c. Multiplication Property: If a = b, then ca = cb. Division Property: If a = b and c ¹ 0, then

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**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.

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**Properties of Congruence**

Reflexive Property: DE ; ÐD Symmetric Property: --If FG, then DE Transitive Property: --If ÐB and ÐC, then ÐC.

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**Introduction to Proofs**

Given: Prove: What you know What you’re trying to prove Statements Reasons All reasons must be: 1. “Given” 2. Postulates 3. Theorems 4. Definitions 5. Properties All of your math steps go here.

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**Statements Reasons 1. Given 2. Addition Property 3. Division Property**

Example 1: Given: 3x – 10 = 20 Prove: x = 10. Statements Reasons Given 1. 3x – 10 = 20 2. 3x – = 3x = 30 2. Addition Property 3. Division Property

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**Statements Reasons 1. Given 2. Substitution 4. Division Property**

Example 2: Given: 3x + y = 22; y = 4 Prove: x = 6 Statements Reasons 1. 3x + y = 22; y = 4 Given 2. Substitution 2. 3x + 4 = 22 3. 3x = 18 3. Subtraction Property 4. x = 6 4. Division Property

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NOTE: You will NEVER use the reason “PROVE” as a justification in your proofs.

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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. A B C

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**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. A Q C B A Q C B If you used this statement above, it would also be acceptable to use “definition of a linear pair” as your reason.

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**F T L A Statements Reasons FL = AT LA = LA FL + LA = AT + LA**

Example 3: F T Given: FL = AT Prove: FA = LT L A Statements Reasons FL = AT LA = LA FL + LA = AT + LA FL + LA = FA LA + AT = LT 5. FA = LT Given Sometimes, this step is not necessary. You could just go right to the next step. 2. Reflexive Property 3. Addition Property 4. Segment Addition Postulate 5. Substitution

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**Statements Reasons mÐAOC = mÐBOD mÐAOC = mÐ1 + mÐ2 mÐBOD = mÐ2 + mÐ3**

Example 4: Given: mÐAOC = mÐBOD Prove: mÐ1 = mÐ3 B A 1 2 3 D O Statements Reasons mÐAOC = mÐBOD mÐAOC = mÐ1 + mÐ2 mÐBOD = mÐ2 + mÐ3 mÐ1 + mÐ2 = mÐ2 + mÐ3 mÐ2 = mÐ2 5. mÐ = mÐ3 Given 2. Angle Addition Postulate 3. Substitution Sometimes, this step is not necessary. You could just go right to the next step. Reflexive 5. Subtraction Property

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**P R S Q T Statements Reasons 1. RS = PS 1. Given 2. RS + ST = PS + ST**

Example 5: P Given: RS = PS and ST = SQ. Prove: RT = PQ R S Q T Statements Reasons RS = PS Given 2. RS + ST = PS + ST 2. Addition Property ST = SQ Given 4. RS + ST = PS + SQ 4. Substitution 5. RS + ST = RT PS + SQ = PQ 5. Segment Addition Postulate RT = PQ 6. Substitution

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Chapter 2 Review Reasoning and Proof.

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