Intermediate Algebra?
485664
Never buy a new textbook...it's a waste of money.  Goto half.com and see if they dont' have it.
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.
151739
algebra is the solution of equations like the above.  Let the "what" be expressed as w.  Then the question becomes\n\n  w * 3 = 4/3\n\nNow how do we solve this for w.  The trick is to operate on both sides of the equation the same function.  The caveat is that the function applied must have an inverse.  In this case multiply both sides by (1/3).\n\n(1/3)*3*w = (1/3)*(4/3)\n\nNow since 1/3 is multiplicative inverse of 3, the result is 1, and since 1 is the multiplicative identity, the result of the left hand side is w.  For the right hand side just do the multiplication\n\nw = 4/9\n\nThus 4/9 multiplied by 3 equals 4/3.
747794
Okay, it's "beginning algebra", not "begging algebra".  For a good start, check out "Elmentary Algebra" in Wikipedia, and check out its external links.
869966
http://www.math.psu.edu/tseng/class/Math140H/Sqrti.html\n\nAnd, BTW, i (or j) is not a 'complex' operator.  It is the 'imaginary unit' ☺\n\n\nDoug
605993
algebra is a broad topic. try to go back to ur notes and look for the formulas that your teacher has given to you. better to stick on what is in your notes and textbooks. try this also, ask someome who has taken this subject before...his information might help you...good luck.
516830
http://www.gateways2learning.com/Algebra/Algebra.htm\n\nAlso try here.......\n\nYou may want to reconsider having it solve the answer for you, always know how to do the breakdown, it will help you to know "how" to go about it when you take a test.\n\nGood luck.....................
878450
There are many for example, one even from Yahoo:\nhttp://education.yahoo.com/homework_help/math_help/algebra1\nhttp://www.coolmathalgebra.com/\nhttp://mathforum.org/library/drmath/sets/high_algebra.html
953801
Get a tutor, get your parents to help you, read through the chapter in your Algebra book that you're currently doing, if your school has extra help with teachers go to the help sessions, study with your friends, form a study group, and do the extra practice problems for the current chapter in the back of your book for learning how to do the equations and stuff. I am also in your same grade doing Algebra 1 Honors. Hehe, I should take my own advice!
843887
i'll try to help u as much as i can heres my email\nmz_princess39842003@hotmail.com\ndang i jus thought about it i am in the 8 takin algebra 2 so in  12th grade i take college courses but stil i'll help ya
890098
I enjoy algebra, and this is easily my favorite type of problem.  So I hope you, too, one day enjoy them for their inner beauty.\n\nFor now, let's tackle this "factoring problem."  We may wish to use the method of "synthetic division" which is arguably easier than "long division of polynomials."\n\nThink back to grade school where you learned that 6 = 2 * 3.  This simple statement enabled you to 'deconstruct' the number 6 as either "two times three" or "three times two".  Even if you forgot one of the 'factors' you could conjure it up again quickly by asking yourself, "What is six divided by two?" for example.\n\nIn problem (a), you are essentially asking, "What are the coefficients of P(x) if the only factor we know is (x-1)?"\n\nLet's set up synthetic division (see link below), by placing all the coefficients of P(x), and using 1 (the zero implied by (x-1)) as the divisor:\n\n1 ||  a  b  c  d\n___________\n.....a\n\n1 ||  a  b  c  d\n_____a_____\n.....a (a+b)\n\n1 ||  a  b        c       d\n_____a___(a+b)____\n.....a (a+b) (a+b+c)\n\n1 ||  a  b        c                d\n_____a___(a+b)___(a+b+c)_\n.....a (a+b) (a+b+c) (a+b+c+d)\n\nNote that this last term must equal 0 since it is assumed that (x-1) is a factor ... thus, the "remainder" term must equal zero!\n\nThere's the result you were after:  a+b+c+d = 0\n\nHope this helps you tackle the others!
721434
Yes.But just algebra 1 and then 2 other maths.English was also a requirement.
985294
you just keep on deriving.\n\nI find it easier to split things into exponents.\n\nI am assuming your question goes\n\n(1/(2x))  - 4 = g(x)\nor\n\n(1/2)*x^-1     - 4 = g(x)\n\ndifferentiate once\n\ng'(x) = -(1/2) * x^-2 \ng''(x) = -2 * (-1/2) * x^-3  = x^-3\ng'''(x) = -3 * x^-4  =  -3/(x^-4)  **ans**\n\nPlug 3 into this... bah\n\nFor reference please use parenthesis to make your problem clearer for us. I just solved the whole thing using the wrong g(x).\n\n\n===!using the correct g(x) which should be g(x) =1 / (2x -4)===\n\ng(x) = 1 / (2x -4) = (2x - 4)^-1  don't forget the chain rule\n\ng'(x) = -1*2*(2x - 4)^-2 = -2 * (2x -4)^-2    chain rule again\n\ng''(x) = -2*-2*2*(2x -4)^-3 = 8*(2x -4)^-3    chain rule again\n\ng'''(x) = -3*2*8(2x - 4)^-4 = -48 * (2x -4)^-4\n\ng'''(x) = -48/[(2x -4)^4]\ng'''(3) = -48 * (6 -4)^-4 = -48/ (2^4) = -48/16   = ***real ans***\n\ng'''(3) = -3
1169576
Try:\n\nhttp://www.sosmath.com/cyberexam/algebra/q001/1/1.html\n\nhttp://www.math.com/practice/Algebra.html\n\nhttp://www.glencoe.com/sec/math/studytools/prealglblr.shtml \n\nhttp://www.syvum.com/squizzes/algebra/
680155
i'd love to help but i'm having trouble with it myself\n\nuse ur math textbook it will have it unless ur math book say algebra and is red ( that my alg. book and it doesn't help) \n\nSRY good luck
817489
This is pretty basic algebra and if you are in an electrical certification program or any other program for that mater you may want to consider taking some math courses before you try to continue.  I'm not saying that you can't nuc your way through it, but it will be allot easier on you if you take a step back and strengthen your math skills before tackling an electrician course.
136386
Algebra 1\nGeometry\nAlgebra 2\nTrigonometry\nCalculus