Negation of \( p \land (q \land \neg (p \land q)) \) is:}
To find the negation of the statement (¬p∧q)∨(p∧¬q), we will follow the steps of logical negation and apply De Morgan's laws.
1. Write down the original statement:
S=(¬p∧q)∨(p∧¬q)
2. Apply negation to the entire statement:
¬S=¬((¬p∧q)∨(p∧¬q))
3. Use De Morgan's Law: According to De Morgan's laws, the negation of a disjunction is the conjunction of the negations:
¬S=¬(¬p∧q)∧¬(p∧¬q)
4. Apply De Morgan's Law to each part:
¬(¬p∧q)=¬(¬p)∨¬(q)=p∨¬q
¬(p∧¬q)=¬(p)∨¬(¬q)=¬p∨q
5. Combine the results:
¬S=(p∨¬q)∧(¬p∨q)
6. Final expression:
The negation of the original statement is: ¬S=(p∨¬q)∧(¬p∨q)
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to:
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is: