How do you map a relation in algebra?
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r u 4 r3al ?!! \n\nwhat mapping ? what equasion?  deckard maybe ? \n\nmore info please.
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It *is* awfully hard to write algebra expressions on Yahoo Answers, isn't it? Oh well, we do what we can.\n\nAddressing your first question, let's write it out like a long division problem.  You'll have to imagine the horizontal bar above the larger polynomial, because I can't format it properly here:\n\nx^2 + 3x - 2  )  x^4 + x^3 - 5x^2 + 13x - 6\n\nOur first step is to look at the first term in each polynomial, and divide the smaller into the larger.  x^4 / x^2 = x^2, so the first term we write in our quotient (above the bar) is x^2.  You can write it directly above the x^4, if you wish, or above the 5x^2 term, your preference.\n\nNow, we're going to multiply that "partial quotient" by the entire divisor (x^2 + 3x - 2), which gives us x^4 + 3x^3 - 2x^2.  Write that "partial product" below the first three terms of the original dividend (x^4 + x^3 - 5x^2).  Our next step is to subtract the partial product: (x^4 + x^3 - 5x^2) - (x^4 + 3x^3 - 2x^2) = -2x^3 - 3x^2.  We then drop down the next term from the dividend, the 13x term, and repeat:\n\n-2x^3 / x^2 = -2x (add that term to our quotient)\n\n-2x * (x^2 + 3x - 2) = -2x^3 - 6x^2 + 4x\n\n(-2x^3 - 3x^2 + 13x) - (-2x^3 - 6x^2 + 4x) = 3x^2 + 9x\n\nFinally, we drop down the last term from the dividend, -6, and repeat once more.  I'm sorry that I can't write this out the way your math book can... I really hope Yahoo Answers creates some sort of way of inputting formulas, or at least monospacing text entry!\n\n3x^2 / x^2 = 3 (so we add a +3 term to our quotient)\n\n3 * (x^2 + 3x - 2) = 3x^2 + 9x - 6\n\n(3x^2 + 9x - 6) - (3x^2 + 9x - 6) = 0... no remainder!\n\nSo our final quotient is x^2 - 2x + 3, with no remainder.\n===\nNow, when doing the 2nd problem, you'll want to write the larger polynomial, the dividend, as x^4 - 3x^3 + 0x^2 + 6x - 5.  We incorporate that extra 0x^2 term so that we never find ourselves trying to subtract unlike terms.\n\nSorry there's no better way to write this out on Yahoo Answers.  Hope that helped anyway!
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"The Complete Idiot's Guide to Algebra" \n\nRotten title - great resource
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Here you go!\n\nhttp://www.math.com/homeworkhelp/Algebra.html
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http://www.anagramgenius.com/archive/algebr.html\n\nedit: sorry didn't answer your real question. I don't know of a program that does what you want, but it shouldn't be too hard to write it. \n\nEdit: Contact me at algebra@envisagecapital.com if you would like the solution. I wrote a quick program to do what you wanted.
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Without knowing the relationship between successive elements, I think the best you can do is this:\n[k=0, n-1]∑(x_(k+1) - x_k)²\n[k=0, n-1]∑(x_(k+1)² - 2x_k*x_(k+1) + x_k²)\n[k=0, n-1]∑x_(k+1)² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n[k=1, n]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n\nx_0 + x_n + 2[k=1, n-1]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1))
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You need to use the distributive property where:\n8(4w+x-6y)=\n8(4w)+8(x)-8(6y)=       \n32w+8x-48y
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Let us say you have a certain coordinate system C with the basis vectors:\nex= (1,0) and ey=(0,1)\n. ey↑\n. . .C →\n. . . . ex\nand you want to express points according to another coordinate system C' with the basis vectors\ne'x = (-1,0), and e'y =( 0,-1), the coordinates of the new basis vectors is according to C\n. ex'0(-1,0)\n←C'\n. . .↓ey'=(0,-1)\n\nThe relationship between the (x,y) in C and (x',y') in C' can be expressed by a Matrix mapping.\n. . . . . . .(-1 o) ..(x)\n(x',y') =  (0, -1) *(y) \n\nremember it is matrix multiplication. This was of cause a very simple example\n\nif we take a point (5,-1) in C its coordinates will be:\n\n. . . . . . . . . . . . . . (5)\n. . . (x',y') = (-1,0)*(-1) = (-5,1)\n. . . . . . . . .(0 ,-1)\n\naccording to the new basis ex',ey' in C'\nI hope its might have helped you \nthe matrix I have use is of cause \n\nE= (-1, 0)\n. . .(0 ,-1)\n\nI this simple situation I could also have used this very simple matrix = (-1,-1)\nand then the mapping would have looked like this:\n\n. . . . . .(-1)\n(x',y' )=(-1)*(x.y)\n\nAgain remember its matrix multiplication.
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Transform each term:\ny" => s^2*Y(s) - s*x(0) - x'(0)\ny' => s*Y(s) - x(0)\ny  => Y(s)\ne(2x) => 5/(s-2)  \nThen substitute back into the equation and that is the transform.  To solve the equation, you collect terms and get an expression in terms of Y(s).  This is the transform of y(x).\n\nThe algebra will undoubtedly involve a partial fractions expansion.\n\nIn any event the solution will look something like\n\ny(x) = A*e(-3x) + B*e(-2x) + C*e(2x) with A, B, C depending on any Initial conditions.
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You can do it by hand or with a calculator. Either way, it would look like this:\n\n2x + y = 3 \n(move the 2x to the right side of the equation, so only y is on the left side)\ny = 3 - 2x\n\nNow put 1, 2, 3, etc. in for x, and see what y equals:\n\nwhen x=1, y=1\nwhen x=2, y=(-1)\nwhen x=3, y=(-3)\nand so on.....\n\nNow put these values into a table:\n\nx --- y\n1 --- 1\n2 --- (-1)\n3 --- (-3)
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I wish you the best of luck, I know how hard algebra is. Here are a couple of websites that will hopefully help:\n\nhttp://www.algebrahelp.com/\nhttp://www.math.com/homeworkhelp/Algebra.html\nhttp://www.purplemath.com/\nhttp://www.helpalgebra.com/\nhttp://www.algebra.com/\nhttp://www.mathleague.com/help/algebra/algebra.htm\nhttp://www.freemathhelp.com/algebra-help.html\nhttp://www.gomath.com/algebra.html\nhttp://archives.math.utk.edu/topics/algebra.html
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Domain ~ x values\nRange ~ y values\n\nYou have to work the function and figure them out.
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a (x+y) + b (x*y) = z
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Google the word algebra help.  Purple math is great, but there are other sites also.  I am sorry I do not have the names at the moment.
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To find the inverse of this relation, reverse the coordinates of the ordered pairs.\n\nThe answer to this would be\n {(18, -14), (-18, 3), (-1, -2)}
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i have  a book about algebra..... and some website...see if they could help u
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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