For numbers 6 - 7. Given: If p, q, r, and s denote the following propositions. p: 1+1+3 r: Alexander exists x2 – 2x + 1 = 0, where x = 1 s: Mouse is an output device q: " If Alexander exists or x2 – 2x + 1 = 0, where x = 1 then it is not true that 1+1 = 3 and mouse is not an output device, " can be symbolized as (rVq) → - (~p^~s) В. 6. (rVq) (rVq) → (~p^~s) А. C. (rVq) → ~ (~ pas) D. (s v d ) - 7. "Either mouse is an output device and x² – 2x + 1 +0, where x = 1, or 1+ 1 = 3 and Alexander does not exists" can be symbolized as (s^q) V(~p ^ ~r) (s^ q) V (p ^~r) (S^~q) V (p ^~r) (S^~q) V (~p ^~r) A. C. В. D. 8. It combines two or more propositions into a single proposition connectives С. logic truth table А. В. D. operations
For numbers 6 - 7. Given: If p, q, r, and s denote the following propositions. p: 1+1+3 r: Alexander exists x2 – 2x + 1 = 0, where x = 1 s: Mouse is an output device q: " If Alexander exists or x2 – 2x + 1 = 0, where x = 1 then it is not true that 1+1 = 3 and mouse is not an output device, " can be symbolized as (rVq) → - (~p^~s) В. 6. (rVq) (rVq) → (~p^~s) А. C. (rVq) → ~ (~ pas) D. (s v d ) - 7. "Either mouse is an output device and x² – 2x + 1 +0, where x = 1, or 1+ 1 = 3 and Alexander does not exists" can be symbolized as (s^q) V(~p ^ ~r) (s^ q) V (p ^~r) (S^~q) V (p ^~r) (S^~q) V (~p ^~r) A. C. В. D. 8. It combines two or more propositions into a single proposition connectives С. logic truth table А. В. D. operations
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Question
Write the letter of correct answers
Transcribed Image Text:For numbers 6 – 7. Given: If p, q, r, and s denote the following propositions.
1 + 1 # 3
x2 – 2x + 1 = 0, where x = 1
p:
r:
Alexander exists
Mouse is an output device
1 then it is not true that 1 + 1 = 3 and
q:
s:
6.
If Alexander exists or x2 – 2x + 1 = 0, where x
%3D
mouse is not an output device, " can be symbolized as
(r Vq) → -(pas)
(rVq) → - (~p^ ~s)
(rVq) → - (~pas)
A.
С.
В.
D.
(rVq) → (- p^~s)
>
7. " Either mouse is an output device and x2 – 2x + 1 +0, where x = 1, or 1+ 1 = 3 and
%3D
Alexander does not exists'" can be symbolized as
(saq) V(~p ^~r)
(s a q) V (p A~r)
(SA-q) V (p A ~r)
(SA ~q) V (~ p ^~r)
A.
С.
В.
D.
8. It combines two or more propositions into a single proposition
logic
truth table
А.
С.
connectives
В.
D.
operations
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