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ehidna [41]
2 years ago
6

Solve these One Step Equations

Mathematics
2 answers:
sukhopar [10]2 years ago
5 0

\huge \bf༆ Answer ༄

Let's solve ~

<h3>Example 1 : </h3>

\qquad \looparrowright \sf \: d + 1 = 5

\qquad \looparrowright \sf \: d = 5 - 1

\qquad \looparrowright \sf \: d = 4

<h3>Example 2 : </h3>

\qquad \looparrowright \sf \: x + 11 = 3

\qquad \looparrowright \sf \: x = 3 - 11

\qquad \looparrowright \sf \: x =  - 8

<h3>Example 3 : </h3>

\qquad \looparrowright \sf \: c + 4 = 5

\qquad \looparrowright \sf \: c = 5 - 4

\qquad \looparrowright \sf \: c = 1

<h3>Example 4 : </h3>

\qquad \looparrowright \sf \: b - 12 = 49

\qquad \looparrowright \sf \: b = 49 + 12

\qquad \looparrowright \sf \: b = 61

<h3>Example 5 : </h3>

\qquad \looparrowright \sf \: z - 5 = 12

\qquad \looparrowright \sf \: z = 12 + 5

\qquad \looparrowright  \: z = \sf17

<h3>Example 6 : </h3>

\qquad \looparrowright \sf \: p - 30 = 42

\qquad \looparrowright \sf \: p = 42 + 30

\qquad \looparrowright \sf \: p = 72

Ad libitum [116K]2 years ago
4 0
<h3><u>Question</u><u>:</u><u>-</u></h3>

Solve these One Step Equations

Ex) d + 1 = 5 Ex) x + 11 = 3 Ex) c + 4= 5

Ex) b – 12 = 49 Ex) z – 5 = 12 Ex) p – 30 = 42

<h3><u>Answer</u><u>:</u><u>-</u></h3>

  • d + 1 = 5

=> d = 5-1

=> d = 4

_________________

  • x + 11 = 3

=> x = 3-11

=> x = -8

_________________

  • c + 4 = 5

=> c = 5-4

=> c = 1

__________________

  • b - 12 = 49

=> b = 49+12

=> b = 61

__________________

  • z - 5 = 12

=> z = 12+5

=> z = 17

__________________

  • p-30 = 42

=> p = 42+30

=> p = 72

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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    

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(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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