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If x = cos θ and y = sin3θ, show that `y(d^2y)/(dx^2) + (dy/dx)^2` = 3sin2θ(5cos2θ – 1)
Concept: undefined >> undefined
The negation of p ∧ (q → r) is ______________.
Concept: undefined >> undefined
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Without using truth tabic show that ~(p v q)v(~p ∧ q) = ~p
Concept: undefined >> undefined
Without using the truth table show that P ↔ q ≡ (p ∧ q) ∨ (~ p ∧ ~ q)
Concept: undefined >> undefined
If A = {2, 3, 4, 5, 6}, then which of the following is not true?
(A) ∃ x ∈ A such that x + 3 = 8
(B) ∃ x ∈ A such that x + 2 < 5
(C) ∃ x ∈ A such that x + 2 < 9
(D) ∀ x ∈ A such that x + 6 ≥ 9
Concept: undefined >> undefined
Rewrite the following statement without using if ...... then.
If a man is a judge then he is honest.
Concept: undefined >> undefined
Rewrite the following statement without using if ...... then.
It 2 is a rational number then `sqrt2` is irrational number.
Concept: undefined >> undefined
Rewrite the following statement without using if ...... then.
It f(2) = 0 then f(x) is divisible by (x – 2).
Concept: undefined >> undefined
Without using truth table prove that:
(p ∨ q) ∧ (p ∨ ∼ q) ≡ p
Concept: undefined >> undefined
Without using truth table prove that:
(p ∧ q) ∨ (∼ p ∧ q) ∨ (p ∧ ∼ q) ≡ p ∨ q
Concept: undefined >> undefined
Without using truth table prove that:
∼ [(p ∨ ∼ q) → (p ∧ ∼ q)] ≡ (p ∨ ∼ q) ∧ (∼ p ∨ q)
Concept: undefined >> undefined
Using rules in logic, prove the following:
p ↔ q ≡ ∼(p ∧ ∼q) ∧ ∼(q ∧ ∼p)
Concept: undefined >> undefined
Using rules in logic, prove the following:
∼p ∧ q ≡ (p ∨ q) ∧ ∼p
Concept: undefined >> undefined
Using rules in logic, prove the following:
∼ (p ∨ q) ∨ (∼p ∧ q) ≡ ∼p
Concept: undefined >> undefined
Using the rules in logic, write the negation of the following:
(p ∨ q) ∧ (q ∨ ∼r)
Concept: undefined >> undefined
Using the rules in logic, write the negation of the following:
p ∧ (q ∨ r)
Concept: undefined >> undefined
Using the rules in logic, write the negation of the following:
(p → q) ∧ r
Concept: undefined >> undefined
Using the rules in logic, write the negation of the following:
(∼p ∧ q) ∨ (p ∧ ∼q)
Concept: undefined >> undefined
Without using truth table prove that (p ∧ q) ∨ (∼ p ∧ q) v (p∧ ∼ q) ≡ p ∨ q
Concept: undefined >> undefined
Without using truth table, prove that : [(p ∨ q) ∧ ∼p] →q is a tautology.
Concept: undefined >> undefined
