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suter [353]
2 years ago
7

Can someone help me. I’ve tried doing this on paper. I don’t know what I’m doing wrong

Mathematics
2 answers:
IrinaK [193]2 years ago
6 0
ANSWER: x=12
EXPLANATION:
9+x/6=11
(54+x)/6=11 (finding common denominator)
x/6=11-54/6 (rearranging terms)
x/6=(66-54)/6 (finding common denominator so 6 gets cancelled from both the sides)
x=12
Nadya [2.5K]2 years ago
3 0

{ \huge \mathfrak \color {pink}{ \tt{answer}}}

\large \color{cyan}{ \tt{x = 12}}

Step-by-step explanation:

\color {cyan}{ \tt{9 +  \frac{x}{6 = 11}}}

  • multiply both sides of the equation by 6

{ \tt{54 + x = 66}}

  • move the constant to the right hand side and change its sign

{ \tt{x = 66 - 54}}

  • substract the numbers

\huge \color{magenta}{ \tt{ = \:  \:  x = 12}}

\large \color{brown}{ \tt{hope \:  \: it \:  \: helps}}

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Recall, we have five connectives in propositional logic ¬, ∧, ∨, →, ↔ (negation, conjunction, disjunction, conditional and bicon
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(a) ¬(p→¬q)

(b) ¬p→q

(c) ¬((p→q)→¬(q→p))

Step-by-step explanation

taking into account the truth table for the conditional connective:

<u>p | q | p→q </u>

T | T |   T    

T | F |   F    

F | T |   T    

F | F |   T    

(a) and (b) can be seen from truth tables:

for (a) <u>p∧q</u>:

<u>p | q | ¬q | p→¬q | ¬(p→¬q) | p∧q</u>

T | T |  F  |   F     |    T       |  T

T | F |  T  |  T      |    F       |  F

F | T |  F  |  T      |    F       |  F

F | F |  T  |  T      |    F       |  F

As they have the same truth table, they are equivalent.

In a similar manner, for (b) p∨q:

<u>p | q | ¬p | ¬p→q | p∨q</u>

T | T |  F  |   T     |    T    

T | F |  F  |   T     |    T    

F | T |  T  |   T     |    T    

F | F |  T  |   F     |    F    

again, the truth tables are the same.

For (c)p↔q, we have to remember that p ↔ q can be written as (p→q)∧(q→p). By replacing p with (p→q) and q with (q→p) in the answer for part (a) we can change the ∧ connector to an equivalent using ¬ and →. Doing this we get ¬((p→q)→¬(q→p))

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