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3 years ago

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What are the missing statement and reason in step 2 of the proof? Statement: ∠ABC ≅ ∠BCD, ∠BAD ≅ ∠ADC Reason: For parallel lines

cut by a transversal, corresponding angles are congruent. Statement: ΔABC ≅ ΔCDA Reason: SSS criterion for congruence Statement: AB ≅ CD, BC ≅ AD Reason: given Statement: AB ∥ CD, BC ∥ AD Reason: definition of a parallelogram Statement: ∠BAC ≅ ∠ACD Reason: Alternate Interior Angles Theorem

1 answer:

ICE Princess25 [194]3 years ago

7 0

Answer:

Statement: AB=CD, BC=AD

Reason: By definition of parallelogram .

Statement: AC=AC

Reason: Reflexive property of equality.

Step-by-step explanation:

1.Statement: $\angle ABC\cong \angle BCD,\angle BAD\cong \angle ADC$ .

Reason: For parallel lines cut by a transversal , corresponding angles are congruent.

2.Statement:AB=CD,BC=AD

Reason: By definition of parallelogram.

3.Statement: AC=AC

Reason: Reflexive property of equality.

4.Statement: $\triangle ABC\cong \triangle CDA$ .

Reason: SSS criterion for congruence.

5.Statement: $AB\cong CD, BC\cong AD$ .

Reason: Given .

6.Statement: $AB\parallel CD, BC\parallel AD$ .

Reason: By definition of parallelogram.

7.Statement: $\angle BAC\cong \angle ACD$ .

Reason: Alternate Interior Angles Theorem.

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3 years ago

Let C(x) be the statement "x has a cat," let D(x) be the statement "x has a dog," and let F(x) be the statement "x has a ferret.

jek_recluse [69]

Answer:

$\mathbf{a)} \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\\\mathbf{b)} \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee \; F(x)\\\mathbf{c)} \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)\\\mathbf{d)} \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)\\\mathbf{e)} \left((\exists x\in X)C(x) \right) \wedge \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)$

Step-by-step explanation:

Let $X$ be a set of all students in your class. The set $X$ is the domain. Denote

$C(x) - ' \text{$x $ has a cat}'\\D(x) - ' \text{$x$ has a dog}'\\F(x) - ' \text{$x$ has a ferret}'$

$\mathbf{a)}$

Consider the statement '<em>A student in your class has a cat, a dog, and a ferret</em>'. This means that $\exists x \in X$ so that all three statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)$

$\mathbf{b)}$

Consider the statement '<em>All students in your class have a cat, a dog, or a ferret.' </em>This means that $\forall x \in X$ at least one of the statements C(x), D(x) and F(x) is true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee F(x)$

$\mathbf{c)}$

Consider the statement '<em>Some student in your class has a cat and a ferret, but not a dog.' </em>This means that $\exists x \in X$ so that the statements C(x), F(x) are true and the negation of the statement D(x) . We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)$

$\mathbf{d)}$

Consider the statement '<em>No student in your class has a cat, a dog, and a ferret..' </em>This means that $\forall x \in X$ none of the statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as a negation of the statement in the part a), as follows

$\neg \left( \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\right) \iff \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)$

$\mathbf{e)}$

Consider the statement '<em> For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.' </em>

This means that for each of the statements C, F and D there is an element from the domain $X$ so that each statement holds true.

We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left((\exists x\in X)C(x) \right) \wedge \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)$

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4 years ago

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frutty [35]

Answer:

Step-by-step explanation:

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For instance,

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3 years ago