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Identify the hypothesis and conclusion of each conditional; write the converse of the conditional; decide whether the conditional and converse are true.

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Presentation on theme: "Identify the hypothesis and conclusion of each conditional; write the converse of the conditional; decide whether the conditional and converse are true."— Presentation transcript:

1 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. 1. If you attend CB West, then you are in high school. 2. If 4x = 20, then x = 5.

2 1.If you attend CB West, then you are in high school. Hypothesis: You attend CB West. Converse: If you are in high school, then you attend CB West. Conclusion: You are in high school. False Counterexample: You could attend CB South.

3 2. If 4x = 20, then x = 5. Hypothesis: 4x = 20. Converse: If x = 5, then 4x = 20. True Conclusion: x = 5

4 1) Hypothesis: 2x – 1 = 5 Conclusion: x = 3 2) Hypothesis: Shes smart Conclusion: Im a genius 3) Hypothesis: 8y = 40 Conclusion: y = 5 4) Hypothesis: S is the midpoint of RT Conclusion: RS = ½RT 5) Hypothesis: m 1 = m 2 Conclusion: 1 2 6) Hypothesis: 1 2 Conclusion: m 1 = m 2

5 1) Hypothesis: 3x – 7 = 32 Conclusion: x = 13 2) Hypothesis: Im not tired Conclusion: I cant sleep 3) Hypothesis: You will Conclusion: Ill try 4) Hypothesis: m 1 = 90 Conclusion: 1 is a right angle 5) Hypothesis: a + b = a Conclusion: b = 0 6) Hypothesis: x = -5 Conclusion: x² = 25 7)B is between A and C if and only if AB + BC = AC 8) m AOC = 180 if and only if AOC is a straight angle.

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7 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 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 DE ; D D Symmetric Property: --If DE FG, then FG DE Transitive Property: --If D B and B C, then D C.

10 Introduction to Proofs StatementsReasons All reasons must be: 1. Given 2. Postulates 3. Theorems 4. Definitions 5. Properties Given: Prove: What you know What youre trying to prove All of your math steps go here.

11 Example 1: Given: 3x – 10 = 20 Prove: x = 10. StatementsReasons 1. 3x – 10 = x – = x = Given 2. Addition Property 3. Division Property

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

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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 AQC, then m AQB + m BQC = m AQC. If AQC is a straight angle and B is a point not on AC, then m AQB + m BQC = 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.

16 StatementsReasons Given: FL = AT Prove: FA = LT F LA T 1. Given 2. Reflexive Property 3. Addition Property 5. Substitution Example 3: 4. Segment Addition Postulate 1. FL = AT 2. LA = LA 3.FL + LA = AT + LA 4. FL + LA = FA LA + AT = LT 5. FA = LT Sometimes, this step is not necessary. You could just go right to the next step.

17 Example 4: Given: m AOC = m BOD Prove: m 1 = m 3 Statements Reasons 1. m AOC = m BOD 2.m AOC = m 1 + m 2 m BOD = m 2 + m 3 3.m 1 + m 2 = m 2 + m 3 4. m 2 = m 2 5. m 1 = m 3 1. Given 2. Angle Addition Postulate 3. Substitution 5. Subtraction Property 4. Reflexive A O B C D Sometimes, this step is not necessary. You could just go right to the next step.

18 StatementsReasons Example 5: 1. RS = PS1. Given 2. RS + ST = PS + ST2. Addition Property 5. RS + ST = RT PS + SQ = PQ 5. Segment Addition Postulate 6. RT = PQ 4. Substitution R P Q T S 4. RS + ST = PS + SQ 3. ST = SQ 3. Given 6. Substitution Given: RS = PS and ST = SQ. Prove: RT = PQ


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