Is it possible to proof that the following formulas are correct?
543816
Yes it is !!\n\nFor proof of volume of cone, pls go to :\n\nhttp://mathforum.org/library/drmath/view/53646.html\n\n\nThe following site has got the proof for area of circle :-\n\n\nhttp://64.233.161.104/search?q=cache:FbBn10NweQkJ:www.artofproblemsolving.com/LaTeX/Examples/AreaOfACircle.pdf+area%2Bcircle%2Bproof&hl=en&gl=in&ct=clnk&cd=1
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.
603804
apply formula for it\nits easy
946513
Subject: Re: Proving trig identiies...small error..thisis the right one\nFrom: \nDate: Monday June 14, 2004 at 08:35:54\n\nRHS= 2csc^2x=2(cot^2x+1) \n\nnote :-(csc^2x=cot^2x+1) \n\n=2cot^2x+2 \n\n=2cos^2x/sin^2x +2 \n\nnote :-((cot^2x=cos^2x/sin^2x) \n\n=2cos^2x/(1-cos^2x) +2 \n\nnote :-((sin^2x=1-cos^2x) \n\n=2cos^2x +2(1-cos^2x) /(1-cos^2x) \n\n=2cos^2x +2 -2cos^2x) /(1-cos^2x) \n\n=2/(1-cos^2x) \n\n=2/(1-cosx)(1+cosx) \n\nnote :-((1-cos^2x=(1-cosx)(1+cosx)) \n\n\nTHEREFORE lHS =2/(1-cosx)(1+cosx) (1-cos^2x=(1-cosx)(1+cosx)) \nnow for the RHS we get \n\nRHS= 1/1+COSX + 1/1-COSX \n\n=(1-cosx) + ( 1 + cosx)/(1-cosx)(1-cosx) \n\nnote :-(( through expansion) \n\n=2/(1-cosx)(1+cosx) \n\nnote :-(cosx on the numerator cancel out) \n\ntherefore RHS= LHS \n\nTHEREFORE 2csc^2x= 1/1+cosx + 1/1-cosx \n\nas required\n\n\n\nCurrent Thread\n   Proving trig identiies   a girl - Monday June 14, 2004 at 08:06:52 \n   Re: Proving trig identiies   - Monday June 14, 2004 at 08:31:28 \n--> Re: Proving trig identiies...small error..thisis the right one   - Monday June 14, 2004 at 08:35:54 <--
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
909722
The only case that I could think of that A is not equal to B where AC=BC is when C=0.  Other than that A must equal B for the expression to be valid
1006089
your math is 100% correct!
849885
I understand your frustration in receiving no Answer yet to this question, but since I do not have any direct knowledge of how to perform the calculation(s) you need ... I hope the link below serves you well.  It appears to explain the mathematics behind all the concepts you mention.
467967
this is probably not a perfect answer but this site may give you some clues.\n\nhttp://www.gomath.com/Questions/question.php?question=4751\n\naltrnatively type 'prove v*v = 2gr*r/s+v0*v0-2gr' into a search engine and see what comes up.
737009
a is true\nb is true\nc is not true\nd is true\ne is true\n\nBalanced equation is:\nCa(OH)2 + 2HCl ---> CaCl2 + 2H2O\n\nIonic equation is:\nCa(OH)2(s) + 2H3O(+)(aq) ---> Ca(+2)(aq) + 4H2O(l)
629091
Sorry, but the formulas do not present well here.  Cannot discern operations.
620893
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
775434
1) x= 2\n2) correct\n3)correct\n4)correct, u got the rest right!
103563
Curiouscat is way kewl.....good answer on her part.\n\nE = mc^2 has been proven valid over and over again. Altho, in science, everything is theory, because someone may come along and find a flaw or correction.\n\nHowever up the present, this equation is rock solid.
775741
I cannot prove it formally but I can mention the steps\n\na | u(n)\n\na | u(n+1)\n\nso a | u(n+1) - u(n) if a devided b and a devided c then it devides b-c\n\nso a | u(n-1)\n\nthere by we reach a | u(0) and a|u(1)\n\nu(0) does not give any infomation as u(0) =0\n\nso a | u(1) = 1\nso a= +/-1 \nu(0) = 0\nu(1)  = 1\n\nnext terms are u(2) = 1,u(3) = 2 ,u(4) = 3, then 5,8,13,
650155
simple\nfirst\nwhen we apply ∑ we can write the series as\n(a_2 - a_1) + (a_3 - a_2) + (a_4 - a_3).... (a_n - a_n-1) (a_n+1 - a_n)\n\nwhen u add \nthe terms starting frm a_2(except a_1 & a_n+1) get cancelled with the correseponding terms and u get\na_n+1 + a_1
541116
3=2 did get a correct answer...maybe you need to review your rules of square roots.