Subtract ` p - 2q + r " from the sum of " 10 p - r and 5p + 2q `

Subtract : 5p - 3q - 2r " from " 7p + 2q - 5r

Subtract 3pq(p-q) from 2pq(p+q)

Subtract 3p q\ (p-q) from 2p q\ (p+q)

if p: q :: r , prove that p : r = p^(2): q^(2) .

Consider the equation p = 5-2q - 3 If p = |r| + 5 , then number of possible ordered pair (r,q) is, are

Given that q^2-p r ,0,\ p >0, then the value of |p q p x+q y q r q x+r y p x+q y q x+r y0| is- a. zero b. positive c. negative \ d. q^2+p r

The incorrect statement is (A) p →q is logically equivalent to ~p ∨ q. (B) If the truth-values of p, q r are T, F, T respectively, then the truth value of (p ∨ q) ∧ (q ∨ r) is T. (C) ~(p ∨ q ∨ r) = ~p ∧ ~q ∧ ~r (D) The truth-value of p ~ (p ∧ q) is always T

Sum of the first p, q and r terms of an A.P are a, b and c, respectively.Prove that a/p(q-r)+b/q(r-p)+c/r(p-q)=0

The sum of the first p , q , r terms of an A.P. are a , b , c respectively. Show that a/p(q-r)+b/q(r-p)+c/r(p-q)=0

The sum of the first p , q , r terms of an A.P. are a , b , c respectively. Show that a/p(q-r)+b/q(r-p)+c/r(p-q)=0