Prove by venn diagram that : An(BuC)=(AnB)u(AnC)?
403308
It is correct, but there's no way to paste diagrams here. Just draw the intersecting circles (i.e. 7 regions) and then shade the appropriate areas.
202912
let o: X->Y be the bijection.  if alpha is bijection from u to v, then o(u) -> o(v) should also be a bijection from Y to Y.
819996
T1=m1*a, no matter are there m2 & m3 or not.\nT2=(m1+m2)*a, no matter whether m1 & m2 are connected with a string or are in one lump, the presence of m3 being unimportant too.
1128579
abc=V\n2(ab+ac+bc)=SA
909313
if u meant make x the subjct then tis is how u wd go abt it.\n\na^2 - ax = b^2 - bx\nax-bx=a^2 - b^2\nx(a-b)= a^2-b^2\nx        =(a^2-b^2)/(a-b)\nx        =[(a-b)(a+b)]/(a-b)    (using difference of two squares)\nx        =a+b
304526
let a-x=u\nso, -dx=du\n\nthen ∫1/(a-x)^2dx= \n-∫1/[(u)^2]du=\n-∫u^(-2)du=\n\n-[u^(-2+1)]/(-2+1)=\n\n-u^(-1)/-1=\nu^(-1)=\n\n1/u + C=\n\n1/(a-x) + C
300657
let a = b  \nMultiply both sides by a \na² = ab  \n Add (a² - 2ab) to both sides \na² + a² - 2ab = ab + a² - 2ab  \n Factor the left, and collect like terms on the right \n2(a² - ab) = a² - ab  \n Divide both sides by (a² - ab)\n2 = 1\n(The place where the proof goes wrong is the very last step.  If a² - ab = 0, then you would be dividing by zero, which is undefined.)
1178662
a (x+y) + b (x*y) = z
469247
A = A\n\nThis is one.
336800
let a = b\na² = ab (Multiply both sides by a)\na² + a² - 2ab = ab + a² - 2ab (Add {a² - 2ab} to both sides)\n2(a² - ab) = a² - ab (Factor the left, and collect like terms on the right)\n2 = 1 (Divide both sides by {a² - ab})\n\nOnce you've got that settled, you can start on 2 + 2 = 5:\n\n2 = 1 + 1, so therefore\n\nx = 2 + 2 goes to\n\nx = 2 + (1 + 1)\n\nBecause you now have proof that 1 = 2\n\nx = 2 + (1 + 2)\n\nx = 2 + 3\n\nx = 5\n\nSubstituting the original equation for x:\n\n2 + 2 = 5\n\nBut also, since 1=2 and 2=1 then 1+1=4 and 2+2=2
441742
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
972809
Consider the left-hand side(LHS) [cot(x) + 1] / [cot(x) – 1]. We want to prove that this is equal to the right-hand side (RHS), that is, [1 + tan(x)] / [1 – tan(x)]. Thus, we have the following:\n\nLHS = [cot(x) + 1] / [cot(x) + 1]\n        =  { [cot(x) + 1] / [cot(x) – 1] }* { tan(x) / tan(x) }.\n\nNOTE: We multiply tan(x) to cot(x) + 1 and the other tan(x) is multiplied to cot(x) – 1.\n\n        =  { tan(x) [cot(x) + 1] } / { tan(x) [cot(x) – 1] }\n\n        =  { tan(x) cot(x) + tan(x)] } / { tan(x) cot(x) – tan(x)] }, by distributive property of multiplication over addition (subtraction)\n\n        =  { 1 + tan(x)] } / { 1 – tan(x)] }, since tan(x) cot(x) = 1 by reciprocal identity.\n\n        =  RHS.\n\nIndeed, [cot(x) + 1] / [cot(x) – 1] = [1 + tan(x)] / [1 – tan(x)]
245176
u*(u-2)+v*(u-2)\n(u+v)*(u-2)
722973
K=positive then U=negative K=U=A and B=equal.
885883
for example\na/b\n---- = ad/bc\nc/d\n \n\na/b + c/d = (ad +cb)/ad\n\na/b -c/d =( ad- bc) /ad\n\n(a/b)(c/d)= ac/bd\n\nx
725621
6Rb(s)+N2(g)--->2Rb3N(s)
966444
Sorry I'm not english spoken\n\nIl suffit de choisir deux vecteurs u et v non nuls et non liés.\nu ∧ v ≠ 0\nOn constate alors que :\n(u ∧ u) ∧ v = 0 ∧ v = 0\nSupposons que :\nu ∧ (u ∧ v) = 0\nAlors\nu et u ∧ v sont liés\nCe qui est absurde car :\nu ∧ v ⊥ u\nDonc :\nu ∧ (u ∧ v) ≠ 0\nD'où :\n(u ∧ u) ∧ v ≠ u ∧ (u ∧ v)