it is a problem or should i say equation that i find a difficult one!\nwould you please help me?
198876
(tanA + tanB + tanC - tanAtanBtanC) / (1 - tabAtabB - tabBtanC - tanCtanA)\ngets nasty fast, so remember these rules\ntanX = sinX / cosX\nsin(X+Y) = sinXcosY + cosXsinY\ncos(X+Y) = cosXcosY - sinXsinY\n\n(tanA + tanB + tanC - tanAtanBtanC) / (1 - tabAtabB - tabBtanC - tanCtanA)\n=[cosAcosBcosC * (tanA + tanB + tanC - tanAtanBtanC)] / [(1 - tabAtabB - tabBtanC - tanCtanA) * cosAcosBcosC]\n=[sinAcosBcosC + sinBcosAcosC + sinCcosAcosB - sinAsinBsinC] / [cosAcosBcosC - sinAsinBcosC - sinBsinCcosA - sinCsinAcosB]\n=[(sinAcosC + cosAsinC)cosB + sinB(cosAcosC - sinAsinC)] / \n[cosC(cosAcosB - sinAsinB) - sinC(sinAcosB + cosAsinB)]\n=[sin(A+C)cosB + sinBcos(A+C)] / [cosCcos(A+B) - sinCsin(A+B)]\n=sin(A+B+C) / cos(A+B+C)\n=tan(A+B+C)
386429
First of all try to give a correct version of the problem before you want to get it solved
484842
Well, you got me stumped!\n\n-------------------------------------\n\nI started working it out and got pretty far.\n\nThe above given information might be incorrect.  Maybe?  I don't know.\n\nI tried it several times.  I even double and triple checked my work!\n\n---------------------------------------------\n\nThis is as far as I got and then I got stumped:\n(D-E) (B+C) = (10J + AFGH) / A\n\n------------------------------------------------------\n\nGreat question!\n\nI look forward to the continuing discussion.\n\nI love the way you worded your question.  Right on!!!\n\nKeep those challenge questions coming!  ;o}
445079
Well hey, you need help, ask the teacher! I know how to do this, with help from my nerdy Calculus BC friend, but it's really not that hard once you think about it...
584488
say her age is X now then..4 years hence her age= X+4\nthen the equation which you need to solve going by the problem statement would be :\n\n2(X+4)-3(X-10) = X\nor solving this \n2X+8-3X+30=X\nor\n38= 2X\nhence X = 19\nthe choice c is right
603947
well, ther answer would oBViously be.....I don't know......or maybe it's just too oBVious and I don't quite see it.\n\n\nnice question, took me a minute. Would "Obvious" work in your HW?
391172
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
513271
wellz why not just help me with that matrix problem? :oP
10918
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.
488296
are you solving for y or n?
514653
Ummm...\nThe equation of algebra, eh?\nWhat on earth are teachers teaching kids these days?!
382050
Looks as though this is some kind of assignment. No? Maybe you ought to think it all through and provide your own answers. You can do it!!\n\nChow!!
263621
u jammed the equation together.\ndo u mean\n6x^3 + ( 6/x - 2 * x^2 - x/3x2 - 3x + 3)\nits really hard to see ur question. can u write it out with spaces...\nhurry up i am waiting... lol
517891
You can't. Any equation like that where the answer isn't reasonably obvious is actually impossible to solve exactly. For example, if you had 2^a + a = 4, its impossible to get an exact answer (using square roots, logs, or whatever you want). Your only hope is to get an approximate answer via graphing or other approximation algorithms.\nOf course, you were lucky in that case :)
347228
i dont see anything wrong the way u solved it(mathematically).if there was any condition or state is mentioned about that thing in time or wrt time then u have to reconsider the problem.
90027
Add them together and divide by 5.\n\nWow. I could seriously have you people do my homework for me. Every night. I'll type out the problems and see how you all do. :P JP
83086
You see, it would be easier to post *exactly* what you have to do because at the moment you are not providing enough information on the problem.  For example, it seems to me that you maybe solving for integer solutions (x,y), with integers a,b,c.