does anyone understand ALGEBRA II?!?!?!? HELP!!!!!!?
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Look up "Algebra Solved 2005" or "algebrator", these two programs not only solve problems, but they give step by step solutions for understanding. they have tutorials and practice test. I went from knowing none to fairly advanced in a few weeks.
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It's bringing the √2 out from the brackets. \n\nMultiply 1+(3/2) by 2... what do you get? 2+(3)\nMultiply 1+(√3/2) by √2.. what do you get √2+(√3)
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I can help, I've taken College Algebra, Calc. I, Calc. II and I've also taken many other math courses, and I'm also studying to be a computer programmer, so if you need help you can contact me at assassin7262002@yahoo.com
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You're NOT dumb!  Everyone has different skills.  Thought you were going to send me a note when you got stuck??  I took alg II twice!\n\n1.  simply switch it to s=1-2t.  Now put that first one into the second , for s, which will give you the value of t when you solve it.  Plug in t's value into either and that will give you s.\n\n2.  do the same way\n\n3.  P=120\n     L=3w\n     P=L times W\n\n4/5:  take the number on the outside and multiply each number inside the (), then combine.\n\n7/8:  isolate the variable (letter) and get it just on one side
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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
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alright, you need to understand this much more than you need the correct answer. You won't have Yahoo! answers to help you on any tests.\n\nYou need help factoring. A real-life tutor who can talk to you and show you in person is best, but I'm going to try to help you understand. \n\na polynomial is in the form ax^2 + bx + c. This is from the formula of (Z*x + q) * (R*x + p). Don't let all these letters scare you. They are just place holders for numbers. \n\nThe only thing that can get the letter A in Ax^2 +bx +c is the first two parts of the "terms" you need to multiply together. With letters a= z*r. In order to get C, you only need to worry about the last parts in the terms. Using letters c= q*p. B is what makes sure you know what two terms were multiplied together. B = z*p + q*r. In order to "factor" out a term from a polynomial you have to find the set of z, p, q, and r that works. I'll show you now with numbers.\n\nLet's say you have the equation 2x^2 + 8x +8. See, it's in the form of ax^2 +bx + c. a=2, b=8, c=8. NOw you have to figure out the facotrs of 2 and 8 (which are a and c)\nfor a=2 the numbers that are multiplied together to get two(the factors) are 2 and 1. Since you know that only the first two parts of the things you multiply together have to get you a, you can set up something that looks like this: \n(2x + ?) *(x + ?) \n\nNow you need to find what two numbers fit into your equation to get you c=8. The factors of 8 are 1*8 or 2*4. Let's try putting 1*8 into the equation.\n(2x+1)* (x+8) = 2x^2 +17x + 8. That's not right, so lets try to reverse them.\n(2x+8)* (x+1) = 2x^2 +10x + 8. That's not right either, so we know only the factors of 2*4 will work in this equation.\nLet's try them\n\n(2x+2)*(x+4) = 2x^2 +10x +8. It may seem like we've tried each of the factors, but we can still reverse 2*4.\n\n(2x+4) * (x+2) = 2x^2 +8x+8. That's right. Once you get more practice and understand this better, you can "see" certain factors that have to be included just because "factoring" will be automatic. But, just to make sure you should always check you answer like I did when I showed you the example.\n\nNow that you hopefully understand the hard part, perfect squares are easy. All you do is take  the square root of 'a' and take the square root of 'c' and put them into the equation. When it is a PERFECT SQUARE this works because of: ax^2 + bx +c = (z*x+ q)^2. ax^2 = (zx)*(zx) so a = z^2, and the square root of a = z. c = (q)*(q) so c = q^2, and the square root of c= q. I'll use numbers now.\n\n100x^2 + 240x + 144. a = 100, so the quare root of 100 = 10. z=10. Then the square root of 144 is 12, so q=12. That means y=(10x+12)^2. Now, to find b, all you have to do is foil it out.
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ok first you have to distribute on each side, remember to watch the (-) (+) signs.\n\n- 4 (x + 2) - 11 = 2 (- 2x -3) - 13\n_________________________\n  -8x + (-8) - 11= - 4x -6 - 13\n\nnow that you are done distributing the 4 and 2, you must start isolating the X's, I like the x on the left, and so will work it that way. it makes no difference though....if you like it on the right, that is fine too. so I will start with the 4, and remember whatever you do to one side you must do to the other. \n\n    -8x + (- 8) - 11 = -4x - 6 - 13\n    +4x                     +4x\n__________________________\n     -4x + (-8) -11= - 6 -13\n\nremembering you cannot subtract with positive and negative numbers  so be careful of the signs and change/change the subtraction to plus signs. so let's do that next.\n\n  -4x + (-8)  + (-11)= -6 + (-13 )\n\nnow that you have changed the signs as above, do the operations on both sides.\n\n   -4x + (-19) = (-19)\n\nnow yo umust further isolate the X, so move the (-19) from left to right, again what you do to one side you must do to the other.\n\n    -4x + (-19) = (-19)\n             +19     + 19\n________________\n              -4x  = 0\n\nnow divide out the  -4, which into zero is zero, therefore x= 0, thus any number plugged in for the x would result in an "x=0" answer. \n\nhope this helped, for more help ask yahoo answers, or try www.purplemath.com, it is a great math site!
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If this is an algebra problem, they probably mean x/3 = (x+8)/7\n\nCross multiply to get 7x = 3x + 24 \n\nSubtract 3x from both sides to get 4x = 24\n\nDivide both sides by 4 to get x = 6\n\nCheck:  6/3 = 2; (6+8)/7 = 14/7 = 2
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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/
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ok. \n\nThe trick is this. SQUARE THE EQUATION.\nNote that : [{2 + (3^1/2)}^1/2]^x =  [{2 + (3^1/2)}^(x/2)] \n\nWe get\n\n[ {2 + (3^1/2)}^1/2]^x + [{2 - (3^1/2)}^1/2]^x = 4\n\n{2 + (3^1/2)}^x \n+ 2 [{2 + (3^1/2)}^(x/2)]*[{2 - (3^1/2)}^(x/2)]  \n+ {2 - (3^1/2)}^x = 16\n\nand [{2 + (3^1/2)}^(x/2)]*[{2 - (3^1/2)}^(x/2)]  equals to\n [ 4 - 3 ] ^(x/2) = 1.\n\nThus, we get,\n{2 + (3^1/2)}^x + 2 + {2 - (3^1/2)}^x = 16\n{2 + (3^1/2)}^x + {2 - (3^1/2)}^x = 14  ...(1)\n\nNow note that \n[{2 + (3^1/2)}^(x)]*[{2 - (3^1/2)}^(x)]  = [4 - 3]^x = 1\nmeans that,\n{2 - (3^1/2)}^x = 1 / [{2 + (3^1/2)}^(x)]\nSubstituiting this into (1), we get,\n{2 + (3^1/2)}^x + 1 / [{2 + (3^1/2)}^(x)]  = 14  \n\nLet y = [{2 + (3^1/2)}^(x)] \nThe equation above becomes,\n\n y + 1 / y = 14\n y^2 - 14y + 1 = 0 \n\nA quadratic equation in y.\nWhich gives y = [14 ± sqrt(192)]/2 \n              \nSince y = [{2 + (3^1/2)}^(x)], Thus,\n\n [{2 + (3^1/2)}^(x)] = [14 ± sqrt(192)]/2 \n                     = [14 ± 8sqrt(3)]/2\n                     =  7 ± 4*3(1/2)\nSo,\n    x = ln [ 7 ± 4*3^(1/2)] / ln [2 + (3^1/2)]\n      =   ± 2\n\nAlternately, \nOne can use the identity,\n{2 + (3^1/2)}^(2) = 7 + 4*3(1/2)\n\n{2 + (3^1/2)}^(-2) = 7 - 4*3(1/2)\n\nTo get x = ± 2 again.\n\nDone.\n\n \nThe below is ADDITIONAL information solving a slightly different problem though similar to the one you posted.\n\nSay we have a similar problem but x is a positive real.\nWe want to find solutions for \n [ 2 + {(3^1/2)}^1/2]^x + [2 - {(3^1/2)}^1/2]^x = 4\n\nBy observation, x = 1 is a solution since, \n\n[ {2 + (3^1/2)}^1/2] + [{2 - (3^1/2)}^1/2] = 4\n\nHow to show they arent any other solutions?\n\n( x > 0 )\n\nLet f(x) =[ {2 + (3^1/2)}^1/2]^x + [{2 - (3^1/2)}^1/2]^x - 4\nSo, our equation amounts for solutions for f(x) = 0.\nWe show that f(x) is increasing for x>0 and we are done.\nNote that when we show f(x) increasing, it means that \nf(x)=0 only at one value because for any x > y, increasing means f(x)> f(y).\nSince f(1) = 0, Thus,\nf(y) > f(1) = 0 for y > 1  and\nf(y) < f(1) = 0 for  0 < y < 1   \n\nAll that remains to be shown is that f(x) is increasing.\nSimplest way, is to use basic calculus. \nIts necessary and sufficent to show f'(x)>0 for x >0.\nUsing the fact for constants any constant m, if y = m^x,\nThen dy/dx = (m^x)(ln m) \n\nf'(x)=  [{2 + (3^1/2)}^1/2]^x *ln[{2 + (3^1/2)}^1/2]\n      + [{2 - (3^1/2)}^1/2]^x *ln[{2 - (3^1/2)}^1/2] \n\nNow, note that\n[{2 + (3^1/2)}^1/2] = 3.316074013\n[{2 - (3^1/2)}^1/2] = 0.683925987\nln[{2 + (3^1/2)}^1/2] = 1.198781557\nln[{2 - (3^1/2)}^1/2] = - 0.379905573\n\nSo,\nf'(x) = (3.316074013^x)*(1.198781557) \n        - (0.683925987^x)*(0.379905573)\n\nSo to show f'(x) >0, its equivalent to showing,\n\n(3.316074013^x)*(1.198781557) >(0.683925987^x)*(0.379905573)\n              (4.848586069)^x > (0.316909757)\n           x ln (4.848586069) > ln (0.316909757)\n                 1.57868713 x > -1.149138224\n                            x > - 0.727907513 \n\nwhich is true for all x > 0.\nThus, for x>0, the only solution to our modified\nproblem is x=1.\n\nAnd there is no "chance" in mathematics.\nIt must be reasoned out.\n\nRegards
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try this one http://www.algebrahelp.com/
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http://www.algebra.com
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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.
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http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut35_polyfun.htm\n\nhttp://www.sosmath.com/algebra/algebra.html\n\nI found these to be helpful.
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http://purplemath.com/modules/index.htm\n\nis an awesome resource for all kinds of algebra topics, even ones you may have forgotten and need to brush up on.\n\nTake a look at that one.\nUMRmathmajor
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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.....................
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Not just algebra...but anything in this world looks hard until we understand/learn it. So dont worry. \n\nNow, to give you some help regarding algebra. I I have answered a similar question before. Somebody asked what's the best way to teach algebra. I have answered them with a helpful link (see the link below), where they have video lessons of algebra for free. A professor teaches you algebra from scratch and gives real-world examples from time to time. He makes it look so easy that you not only would find algebra easy but interesting too. Here is the link.\n\nhttp://64.78.42.182/free-ed/Math/Algebra/Algebra01_VidLect.asp