Associative Law
Trending Questions
Q. The logical statement [∼(∼p∨q)∨(p∧r)]∧(∼q∧r) is equivalent to :
- ∼p∨r
- (∼p∧∼q)∧r
- (p∧r)∧∼q
- (p∧∼q)∨r
Q. The Boolean expression (p⇒q)∧(q⇒∼p) is equivalent to
- ∼p
- ∼q
- p
- q
Q. The Boolean expression (p ∧∼q)⇒(q ∨∼p) is equivalent to
- q⇒p
- p⇒q
- p⇒∼q
- ∼q⇒p
Q. Let ∗, □∈{∧, ∨} be such that Boolean expression (p∗∼q)⇒(p□q) is a tautology. Then:
- ∗=∧, □=∧
- ∗=∨, □=∧
- ∗=∨, □=∨
- ∗=∧, □=∨
Q. The statement (p∧∼q)∨q∨(∼p∧q) is equivalent to
- p∧q
- p∨q
- p∨∼q
- ∼p∧q
Q. The statement (p∧∼q)∨q∨(∼p∧q) is equivalent to
- p∧q
- p∨q
- p∨∼q
- ∼p∧q
Q.
State True=1 and False=0
If z1, z2, z3 are three distinct complex numbers and p, q, r are three positive real numbers such that p|z2−z3|=q|z3−z1|=r|z1−z2| then p2z2−z3+q2z3−z1+r2z1−z2=0. Q. The expression (p ∧∼q)∨q∨(∼p∧q) is equivalent to
- p∧q
- p∨q
- p ∨∼q
- ∼p∧q
Q. Find (3A−2B). If A=(2i+3j−k) and B=(2i−5j+2k).
Q. ∼(p∨q)∨(∼p∧q) is logically equivalent to
- ∼p
- qp
- q
- ∼q
Q. Write the value of ∫dxax+1.
Q. The expression (p ∧∼q)∨q∨(∼p∧q) is equivalent to
- p∨q
- p ∨∼q
- ∼p∧q
- p∧q
