limit theorem proofs

Since a velocity gradient exists across the tube, the mean velocity of the liquid is to be considered. The viscous drag which comes into play when the liquid is in motion, is not taken into account. In above conservation principle, part of K.E. is converted into heat..

Suppose P is different from O, then in triangles OPM and OPN, OP=OP (common side) (given) ( given) (AAS congruency theorem) Hence, PM=PN ( CPCT) If P coincides with O, then it is equidistant from AB and CD, since they intersect at O P is equidistant from AB and ..

x.(x+y) = (x + 0)(x + y) 2(a) (Refer to Axion in previous topic) = x + (0.y) Axiom 4(a) = x + (y.0) Axiom 3(b) = x + 0 (Theorem 2) = x Axiom 2..

When p(x) is divided by x-a, R = p(a) (by remainder theorem) p(x) = (x-a).q(x)+p(a) (Dividend = Divisor x quotient + Remainder Division Algorithm) But p(a) = 0 is given. Hence p(x) = (x-a).q(x) Conversely if x-a is a factor of p(x) then p(a)=0. p(x) = (x-a).q(x) + R If (x-a) is a facto..

The proof follows from theorem 2, P( f ) C = 1 - P( f ) = 1 - 1 = ..

Since the diagonals AC and BD are equal, ABCD is a rectangle - - -(i) (Diagonal property of rectangle) Since the diagonals are perpendicular to each other. ABCD is a rhombus. AB=AD - - -(ii) ABCD is a rectangle. (from i) With consecutive sides equal. (from ii) ABCD is a square. (by definition of a ..

By definition of complement of an element it is sufficient to prove xy + (x' + y') = 1 and xy(x' + y') = 0 xy + (x' + y') = (xy + x') + y' (Associativity of +) = (x + x')(y + x') + y' (Axiom 4b) = 1(y + x') + y' (Axiom 5) = y + x' + y' (2b) or 1 . x = x = y + y' + x' (Axiom 3a) = 1 + x' (Axiom 5) ..

(i) ( Theorem) From (i) and (ii), ..

Let AC and BD intersect at right angles at O. In triangles AOD and COD, AO=OC (diagonals bisect each other) OD=OD (common side) AD = DC i.e., the adjacent sides are equal. By definition, ABCD is a rhombus. Hence the theorem is prov..

i) AD is the median i.e., BD = DC In D ABK, F is the mid-point of AB . (given) G is the mid-point of AK. (by construction) FG || BK or GC || BK ...(1) By theorem on line joining the mid-points of any two sides of a triangle. Similarly in D ACK, GE || CK or GB || CK From (1) and (2)..