How Do You Balence out this equation?
832005
mn : what a great resource!\n\nhmmm... is it a redox question?
51150
UGH!!!  Thx for the effort (I was the one to post that question, as a part of a solution to another problem I found here on Yahoo! Answers.)  Incidentally, I made a mistake when simplifying the other problem and it should be xln(2)-2ln(x)-ln(86400) = 0\n\nI don't know any way to get rid of the ln(x) unless maybe it's using calculus techniques?  I'm just glad this problem is making other ppl trouble about it as much as I did :-).
488850
Depress the cubic to eliminate the second order term.\n\nsee\nhttp://www.karlscalculus.org/cubic.html
712145
f(x) = ba^x + c\nSince it passes through the point P(1,4)\nf(1) = ba + c = 4\nSince the y-intercept is 2 (ie it passes through (0, 2)\nf(0) = b + a = 2\nThe statement "horizontal asymptote y=2" renders a solution impossible. I think you meant horizontal asymptote y=1\nThen lim x→-∞  f(x) = c → c = 1\nSo b + 1 = 2 → b = 1\nba + c = 4 → a = 3\n\nThus f(x) = 3*x + 1
859744
(x + 4y)(x - 4y)\nI answered 2 of your questions. Do the rest of the questions yourself.
167128
There's an infinite number of answers.\n\nIf you do it all out, they both equal the same line...\nI multiplied the second equation by a -4 and got 0=0.\n\nIf you just multiply it by a 4 you'd see that it's the same exact line.
60599
This is not an equation.  Its garbage.  Both side of the equation have to equal each other.\n-35=.000075 this does not compute
986332
y=ln[sec5x+tan5x]\ny'={1/[sec5x+tan5x]}[5sec5xtan5x+5sec^25x]\n=5sec5x[sec5x+tan5x]/[sec5x+tan5x]\n=5sec5x
1051508
Your equation would look like x^2-ax+b.  Which means when you solve the equation you would two answers one of which will contain a plus sign other will contain a minus sign since the middle term is negative.  Thus the answer is C.
658148
given 4a^2-12ab+9b^2\n    =(2a)^2-2(2a)(3b)+(3a)^2\n    =(2a-3b)^2
288520
cmon dont do this again, give the other equation
518727
Multiply the first equation by 2.\n\nSubtract the second equation from the new first equation.\n\nSolve for y.\n\nTake your value for y and substitute it into the original first equation.\n\nSolve for x.
22859
Subract the two terms on the left giving:\n-2t < 16\n\nAdd 2t to both sides giving\n0 < 16 + 2t\n\nSubtract 16 from both sides\n\n-16 < 2t\n\nDivide both sides by 2\n\n-8 < t
464039
Your equation would be:\n\n= ((1/p) - (1/q))/(1/pq)\n\n= ((q - p)/pq) * pq\n\n= q - p\n\nHope I am clear.
872047
Remember that you need to isolate your variable and that anything you do to one side of the equation needs to be done to the other side, as well.  \n\nFirst problem:\n-7r - 8 = -5r + 3\n\nAdd 5r to both sides to nullify the -5r on the right:\n-7r - 8 + 5r = -5r + 3 + 5r\n\nAlso, add 8 to both sides to nullify the -8 on the left.  This leaves only the r terms on the left:\n-7r - 8 + 5r + 8 = -5r + 3 + 5r + 8\n-2r = 11\n\nDivide both sides by -2 to get r all by itself:\n-2r / (-2) = 11 / (-2)\nr = -11/2\n\nSecond problem:\nDistribute through.  Multiply everything out so you can isolate p.\n0.3(p + 4) = 2.7p\n0.3p + 4(0.3) = 2.7p\n0.3p + 1.2 - 0.3p = 2.7p - 0.3p (I combined two steps here by subtracting 0.3p from both sides).\n1.2 = 2.4p\n1.2 / 2.4 = 2.4p / 2.4\np = 0.5\n\nThird problem:\nI'm assuming you mean (5/6)q, which means the q is in the numerator.  The way you have it written, it could mean 5/(6q).  If you mean that, let us know.  For now, I'll assume (5/6)q.\n\nI'll do something a little different here.  Fractions can be a pain to work with, so I'll get rid of them.  I'll multiply the entire equation by 6.\n\n4q + 19 = 5/6q\n6(4q + 19) = 6(5/6q)\n24q + 114 = 5q\n24q - 5q + 114 - 114 = 5q - 5q - 114\n19q = -114\n19q / 19 = -114 / 19\nq = -6\n\nHope that helps you with the rest of your assignment.
617758
Actually, all you need is:\n\n2 H2O2 ======> 2 H2O + O2\n\nif you have water on both sides of the equation, they "cancel" out.  Just like in algebra:\n\na + b = a + c  is the same as b = c
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.