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3 years ago

H=8 is in the solution set for h-2>6 True False

Mathematics

2 answers:

lianna [129]3 years ago

7 0

Answer:

False

Step-by-step explanation:

H=8

8-2>6

Simplify

8-2=6

6>6

It should be 6=6

KonstantinChe [14]3 years ago

5 0

Step-by-step explanation:

h - 2 > 6 <em>add 2 to both sides</em>

h - 2 + 2 > 6 + 2

h > 8

h = 8 is not greater than 8. Therefore your answer is: False.

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Answer:

The statement $(A\rightarrow \lnot B)\land (B\rightarrow A)$ is a contingency.

The statement $(P\rightarrow \lnot P)\land P$ is a contradiction.

Step-by-step explanation:

A tautology is a proposition that is always true.

A contradiction is a proposition that is always false.

A contingency is a proposition that is neither a tautology nor a contradiction.

a) To classify the statement $(A\rightarrow \lnot B)\land (B\rightarrow A)$ , you need to use the logic laws as follows:

$(A\rightarrow \lnot B)\land (B\rightarrow A) \equiv$

$\equiv (\lnot A \lor\lnot B)\land(\lnot B \lor A)$ by the logical equivalence involving conditional statement.

$\equiv (\lnot B\lor \lnot A )\land(\lnot B \lor A)$ by the Commutative law.

$\equiv \lnot B \lor (\lnot A \land A)$ by Distributive law.

$\equiv \lnot B \lor (A \land \lnot A)$ by the Commutative law.

$\equiv \lnot B \lor F$ by the Negation law.

Therefore the statement $(A\rightarrow \lnot B)\land (B\rightarrow A)$ is a contingency.

b) To classify the statement $(P\rightarrow \lnot P)\land P$ , you need to use the logic laws as follows:

$(P\rightarrow \lnot P)\land P \equiv$

$\equiv (\lnot P \lor \lnot P)\land P$ by the logical equivalence involving conditional statement.

$\equiv P \land (\lnot P \lor \lnot P)$ by the Commutative law.

$\equiv (P \land \lnot P) \lor (P \land \lnot P)$ by Distributive law.

$\equiv F \lor F \equiv F$ by the Negation law.

Therefore the statement $(P\rightarrow \lnot P)\land P$ is a contradiction.

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