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Marat540 [252]
3 years ago
9

Simplify 5m-2(6m-5)+2a. -7m-8b. -7m+12c. 17m+12d. 17m-8

Mathematics
1 answer:
fgiga [73]3 years ago
7 0

The answer is B. -7m+12.


First distribute the -2 to 6m and -5.

5m-2(6m-5)+2

5m+(-2*6m)+(-2*-5)+2

5m+-12m+10+2


After that, combine the like terms.

5m+-12m+10+2

-5m+-12m=-7m

10+2=12


The simplified expression is -7m+12.

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Step-by-step explanation:

Given: [∀x(L(x) → A(x))] →

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To prove, we shall follow a proof by contradiction. We shall include the negation of the conclusion for

arguments. Since with just premise, deriving the conclusion is not possible, we have chosen this proof

technique.

Consider ∀x(L(x) → A(x)) ∧ ¬[∀x(L(x) ∧ ∃y(L(y) ∧ H(x, y)) → ∃y(A(y) ∧ H(x, y)))]

We need to show that the above expression is unsatisfiable (False).

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E.I with respect to y,

(L(a) ∧ (L(b) ∧ H(a, b))) ∧ (∀y(¬A(y) ∧ ¬H(a, y))), for some b

U.I with respect to y,

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Since P ∧ Q is P, drop L(a) from the above expression.

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Apply distribution

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Note: P ∧ ¬P is false. P ∧ f alse is P. Therefore, the above expression is simplified to

(L(b) ∧ H(a, b) ∧ ¬A(b))

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==================================================

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Now add straight down

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