Anyone good at Algebra1 please help me!!!?
955861
area: l*w = 256\n-> 4w*w = 256\n->w(sqaure) = 64\n->w=8 (width)\n->l = 4w = 4*8 = 32(length). \n\nThere you go.
588927
easy.. -1 , -2 , 1,2,3.\n\n-1 x -2 =2\n2x2x3 = 12
1054627
assuming this is (3a+1)/(a^2-1) - 1/(a+1)\n\nfactorise the (a^2-1) to give (a+1)(a-1)\n\nmultiply the second fraction by (a-1)\n\n-->  (3a+1)/[(a+1)(a-1)] - 1(a-1)/[(a+1)(a-1)]\n\nfactorising this gives [(3a+1)-(a-1)]/[(a+1)(a-1)]\n\nwhich equals (2a+2)/[(a+1)(a-1)]\n\n= [2(a+1)]/[(a+1)(a-1)] \n\nthe (a+1)'s cancel out...\n\ntherefore = 2/(a-1)
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/
152888
I'm in Algebra 1 (I go to an online high school). I would be glad to help you if I can.
705864
Alright, don't listen to the above post...\n\nGrade 12 here, answering questions all day for mere fun.\n\n12)\ny - 3 = -1/3 (x + 4)\n\nEXPLAINED\n\nadd 3 to both sides\ny = -1/3 (x + 4) + 3\n\nexpand 1/3(x + 4) \n= x * 1/3 + 4 * -1/3\n= x/3 + 4/3\ny = -1/3x - 4/3 + 3\n\nTo add -4/3 and 3, use 3 = 9/3\ny = -1/3x - 4/3 + 9/3\ny = -1/3x + 5/3\n\nTrue, my bad. I originally added 4/3 instread of -4/3\n\n\n16)\n(-2, -1) (4, -4)\n\nm = Dy / Dx\nm = ((-1) - (-4)) / ((-2) - (4))\nm = 3 / -6\nm = -1/2\n\ny = mx + b\n(-1) = -1/2(-2) + b\n-1 = 1 + b\n-1 - 1 = b\n-2 = b\n\ny = -1/2x - 2\n\nCHECK\ny = -1/2(4) - 2\ny = -2 - 2\ny = -4\nTherefore it works.\n\n18)\n(0, 0) (2, 3)\n\nm = Dy / Dx\nm = ((3) - (0)) / ((2) - (0))\nm = 3/2\n\nb = 0\n\ny = mx + b\ny = (3/2)x + 0\ny = (3/2)x\n\nNote, the lower post copied my answers
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
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!
202280
http://www.aaamath.com/ Try this. Hope it works!!
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
594239
ok so you have p^2 + p, you can factor out p from both parts of the expression, it will end up like this: p(p + 1).  Hope that works.
913123
I think you can try this website \nhttp://www.algebra-test.com \nhttp://www.algebraavenue.com\nhttp://www.algebra-online.com/algebra-equation.html \nhttp://www.algebrahelp.com\nhttp://www.algebra-net.com\nhttp://www.algebra-online.com/solve-algebra-equation.html\nhttp://www.algebrahelp.com/lessons/equationbasics/pgw.htm\nhttp://www.algebrasolver.com\nhttp://www.edhelper.com/algebra.htm\nhttp://www.math.com/students/worksheet/algebra_sp.htm
53373
Here are a few sites you can check out:\nhttp://www.freemathhelp.com/\nhttp://www.mathgoodies.com/\n\nHope these sites can help you..
926616
2x+3<6x-1\n2x-6x<-1-3 \n-4x<-4\nx>1
4935
ur second equation is wrong. \n2a is not equal to a-squared (which u symbolized by ab, since a=b).
708644
Use the quadratic formula.\nx1 = 1\nx2 = -1/2
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)