PLEASE HELP!!...i need help with these algebra problems...?
953232
Problem 1:\n\nif t=3 then 0.18t = 0.54, so\n\nAoe^-0.54 = 40, amount in 3 days.  Solve for Ao:\n\nAo = 40e^0.54 = 68.64.\n\nNote e^-x = 1/e^x (change the sign of the exponent when inverting), I divided both sides by e^-0.54 to isolate Ao.\n\nProblem 2:\n\nlog 5^(1/5) = (1/5)log 5 = (1/5) x 0.7 = 0.14.\n\nI just guessed what you meant by the second one, assuming 1/5 was an exponent.
571127
Try going to this website. The will explain it step by step and you can ask all the math questions you want. www.algebra.com
930340
PROBLEM 1:\n\nsqrt(x^3) = 8\n\nSquare both sides:\nx^3 = 64\n\nNow take the cuberoot of both sides:\nx = cuberoot(64)\n\nSolve:\nx = 4\n\nPROBLEM 2:\n\ncuberoot(x^2) =4 \n\nCube both sides:\nx^2 = 4^3\n\nSimplify:\nx^2 = 64\n\nTake the square root of both sides:\nx = ±8
128596
im great with algebra, ive been doing it for the past 4 years now non stop, email me and ill be more than happy to show you how to do it step by step and answer you questions\nzinger_38@yahoo.com
972476
........._______________\nx - 5 ) x³ - 3x² - 13x + 15 ( x²\n......... x³ - 5x²\n......... -------- ....↓\n............... 2x² - 13x ....... ( 2x\n............... 2x² - 10x\n............... ----------- ....↓ \n............... ....... -3x + 15. (-3\n............... ....... -3x + 15\n............... ....... -----------\n............... ....... -----------\n\nSo P(x) = x³ - 3x² - 13x + 15\n= (x - 5)(x² + 2x - 3)
821200
(x²+4x+4) / (2x-3) / (x²-4) / (2x²-7x+6)\n\nYou have 4 terms and three divisions, so it's a little unclear if you mean (a/b) / (c/d) or what.  I will try to help to the extent I can.\n\nLet's assume you mean:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ]\n\nFactor your x² equations to see if you can cancel terms:\n\nx²+4x+4 = (x+2)(x+2)\nx²-4 = (x+2)(x-2)\n2x²-7x+6 = (2x-3)(x-2)\n\nWoohoo - things are looking good!  Now we can rewrite our original problem using these factors:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ] = \n\n[ (x+2)(x+2) / (2x-3) ] / [ (x+2)(x-2) / (2x-3)(x-2) ] =\n \n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\nnotice we can cancel out three terms: (2x-3), (x-2) and (x+2)\n\n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\n(x+2)\n\nHope that helped!\n\n~ ♥ ~
945604
1.adding-2f-7\n-6f-2f+7-7<2f-2f-9-7\n-8f<-16\ndividing by -8\nx>2\n\n2.5/8g-4>2/3g+6\nadding 4\n5/8g>2/3g+10\nmultiplying by 24\n15g>16g+240\nadding -16g\n-g>240\ndividing by -1\ng<-240\n\n3.sum incomplete\n\n4.-2 1/9d+3/5<-3 2/9d-4 2/5\n-19/9d+3/5<-29/9d-22/5\nmultiplying by45\n-95d+27<-145d-198\nadding 145d\n50d+27<-198\nadding -27\n50d<-225\ndividng by 50\nd<-5 1/2
1082751
1)   f(x) = x+1/x =(x^2+1)/x  \n      g(x) = x-2/x =(x^2-2)/x\n(f+g)(x) = (x^2+x^2+1-2)/x = (2*x^2-1)/x = 2*x -1/x\n\n2)  f(x) = -2/x\n     g(x) = (x+1) / (x-3)\n(f+g)(x) = -2/x + (x+1)/(x-3) = {-2*(x-3)+x*(x+1)}/x*(x-3)\n              = (-2x+6+x^2-x)/(x^2-3x) =(x^2-x+6)/x^2-3x
905618
(A) 3+2(x-1)\n\n3+2x+2\n2x+5\n\n(B) 5-2(x+3)\n\n5-2x-6\n-2x-1\n\n(C) 1+2(x-1)\n\n1+2x-2\n2x-1\n\n(D) 6-3(w-2)\n\n6-3w+6\n-3w+12
1039099
This is the answer i got when i worked out the problem\n1. You have to distribute the (x + 2) & (x + 1)....you do the by using the FOIL method....Firsts Outsides Insides Last and you get: x + 2 = x^2 + 3x + 2\n\n2. Get rid of the 2 on the left side by subtracting it on the right side and you end up with : x = x^2 + 3x\n\n3. Now you can move the x on the left to the right side too so you have: 0 = x^2 + 2x\n\n4. Pull out and x leaving you with 0= x (x+2)\n\n5. x = 0 and x = -2\n\nMy algebra is rusty but I think this is how it's done.
851212
hmm im not quite in college yet, but i hope to be help.\ndo you know the values for f, h, v, and a?\nif you do, id substitute it in.\nthen square both sides. \nand then solve.
703982
1.\n3x + 6/((x + 1)(x + 2)) Factor 3 out of the top.\n[3(x + 2)]/[(x + 1)(x + 2)] x+2's cancel\n3/(x + 1) <-- This is simplified.\n\nI don't have the energy to do the other ones right now.  :/
827502
1.6a^2b^2(1+3ab)\n\n2.(y-4)(y-1)\n\n3.(2b-1)(b+7)\n\n4.3(x^2-9y^2)\n=3(x+3y)(x-3y)
102673
If you can't get an answer here, try asking your Algebra teacher for extra help after school tomorrow.  It's sure to add to your grade that you asked for the help.
826369
I have found sites like the following helpful:\n\nhttp://www.algebra.com/\nhttp://www.saab.org/mathdrills/factor.cgi\nhttp://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=algebra&s2=factor&s3=basic\n\nThe last one is a solution program, but it can be helpful to check your answers.
873410
1.  (x + 9)(x - 9)\n2.  (t + 2)(t - 2)\n3.  (b + 12)(b - 12)
706334
To solve #1, you  have to use partial fractions.  \n\nYou have to convert \n\n** x / [(3x-1)(x+2)] = A/(3x-1) + B/(x+2) for some constants A,B.\n\nTaking the integral of x / [(3x-1)(x+2)] would be the same as taking the integral of A/(3x-1) + B/(x+2), so let's do that.\n\nIntegral (A/(3x-1) + B/(x+2))dx = \nIntegral (A/(3x-1))dx + Integral (B/(x+2))dx\n\nAlways pull out constants.\n\nA*Integral(1/(3x-1))dx + B*Integral(1/(x+2))dx\nA * 1/3 * ln |3x - 1| + B * ln|x+2| + C\n\nNow, solve for A and B using the equation marked as **.\n\nx / [(3x-1)(x+2)] = A/(3x-1) + B/(x+2)\n\nFirst, multiply both sides by (3x-1)(x+2) to get\n\nx = A(x+2) + B(3x -1)\n\nRemember that the above is a statement that is __true for all x___, so in order to solve for A and B, we can let x be ANY value.\n\nFor simplicity purposes, we're going to let x=-2 and then x=1/3.\n\nx = -2:  \n\n-2 = A(-2 + 2) + B(3(-2) - 1)\n-2 = B(-7)\nB = 2/7\n\nx = 1/3:\n\n1/3 = A(1/3 + 2) + B(3(1/3) -1)\n1/3 = A(7/3)\nA = 1/7\n\nTherefore, substitute A=1/7 and B=2/7 back into here:\n\nA * 1/3 * ln |3x - 1| + B * ln|x+2| + C\n\nto get\n\n1/7 * 1/3 * ln |3x - 1| + 2/7 * ln|x+2| + C\n7/3 * ln |3x - 1| + 2/7 * ln|x+2| + C\n\nTo solve #2:\n\nUse integration by parts.\n\nlet u = arcsin(3x)                  dv = dx\ndu = 1/sqrt(1-(3x)^2) * 3         v = x\n\nSo our new integral becomes\n\nx*arcsin(3x) - Integral [ (3x)/sqrt(1 - (3x)^2) ] dx\n\nAfter that it becomes substitution.