What is the algebraic expression?
921331
First, find your final value.  This is what your expression equals.  \n\nSo, you can see that the problem states that there are 17 ducks left.  Good start.  We can begin with:\n??? = 17\n\nNow, we tackle each part of the problem.  We see that we started with 28 ducks.  After that, 16 ducks fly in, so we need to incorporate 28 + 16.  So now we have:\n28 + 16 ??? = 17\n\nNow, we need to see what is being asked.  It is being asked how many ducks left.  Since ducks flying in would be addition, then ducks flying out would be subtraction.  We are being asked how many ducks flew out, so this is our unknown (we'll call it x).  Now we can write it as:\n28 + 16 - x = 17\n\nTo find the answer, you solve for x by getting it by itself on the equal sign.  Add 28 and 16 to get 44:\n44 - x = 17\n\nAdd x to each side and subtract 17 from each side:\n44 - 17 - x + x = 17 - 17 + x\n27 = x\n\nSo, 27 ducks flew out.\n\nNote that I added x to both sides above.  This is normally unnecessary, but I didn't know how far along you were.  For future reference, you may want to consider this step:\n44 - x = 17\n\nSubtract 44 from each side:\n44 - 44 - x = 17 - 44\n-x = -27\n\nMultiply each side by -1:\n(-x)(-1) = (-27)(-1)\nx = 27
87985
The expression was not set equal to anything.\n\nIf you're doing simple factoring in algebra class, the \nsimplest factors would be (b+4) (b+3).\n\nIf the expression was set to equal 0, solve for b, then your answer set would include -4 and -3.\n\nHope this helps!
106150
I am not sure what you mean. I assume you mean what is tan(x+a)=sin(x+a)/cos(x+a)\n=[sin(a)cos(x)+sin(x)cos(a)] / [cos(x)cos(a)-sin(x)sin(a)]\n=[tan(a)+tan(x)] / [1-tan(x)tan(a)]\n\nSo tan(x), if the argument x is translated by a, becomes this expression.
1089643
http://www.algebra.com
1053338
I'm not certain what the expression is.\nI'm getting 2 + [y²/(x^4 + x^3*y)]\n\nI assumed that the total denominator is the x²+xy.  So I'm ignoring that for now.\n\nThen the total numerator has a least common denominator of x².  So I got (2x^4 + 2x^3*y+y^2) / x².  Now go back and multiply all that by 1/(x²+xy)\nthe denominator is x^4+x^3*y.  You see the first two terms in the numerator?  they are this new denominator *2.  So we can move them out and have 2+ the remainder [y² / (x^4+x^3*y)]\n\n-edit-\n\nThen put answer D.  You don't understand how to phrase math, so none of us can understand the question.  However, (x+y) is definitely a factor in the denominator and D is the only answer that has it.
972642
(A^2+2ab+b^2)/(a+b) + (a^2-b^2)/(a-b) + (a-b)/(a+b)/(a^2-b^2)/a+b)\n(a+b)^2/(a+b)+(a+b)(a-b)/(a-b)+ (a-b)/(a+b)*(a+b)/(a^2-b^2)\na+b+a+b+(a-b)/((a+b)(a-b))\n2a+2b+1/(a-b)\n\nif I got all the ( ) in the right places.
533029
Let's call :\n(a) : x + 2y < 3\n(b) : 2x - 3y < 6\n\nWith the 1st one:\nx + 2y < 3\n<=> 2x + 4y < 6 \n\nThe 2nd one is 2x - 3y < 6\nThat means : \n(a) - (b)\n<=> 4y - (-3y) < 0 \nor 7y < 0 \nor y < 0\n\nWe have also: 2x - 3y < 6 <=> (2/3).2x - (2/3).3y < (2/3).6\n<=> (4/3)x -  2y < 4\nThat means:\n(a) + (b)\n<=> x + (4/3)x < 7\n<=> (7/3)x < 7\n<=> x < 3\n\nSo : x < 3 and y < 0
973940
A previous answer was incorrect. It is\n\n-4x(x+3y)(x-2y)
51216
x = [-(-4) +/- root((-4)^2 - 4x1x(-5))]/2x1\n  = (4 +/- 6)/2\n  = 5 or -1
971943
A variable expression is simply an expression -- a colllection of values and operators -- that contains at least one "variable" -- which is a quantity whose value has not yet been determined.\n\nFor example:\n\n1 + 9/3 is an expression.  It can be simplified to a single value, 4.\n\n1 + x is a variable expression.  It cannot be simplified to a single value, because 'x' represents an unknown quantity, a quantity that can vary.
883411
lets just simplyfy al of them :)\nThe expression is : -3m -2 + 5m -8 =  2m-10\n\nThe expression is : 4(3a - 5) + 2(-3a + 7) = 12a-20 -6a +14\n                                                                       =6a-6\nThe expression is : 2/3(1/2x + 9) + 3 = x/3 + 6 +3 = x/3 +9\n\nThe expression is : x/4 - 1/8 - x/6 + 1/6 = x/12 + 1/24\n\nThe expression is : 2ab + a - ab - 7 = ab + a -7.\n\nso i would say,, The expression is : 2ab + a - ab - 7 is closest to being in its simplest form than the rest
123892
3(x – 2)^2 – 3(x – 2) – 6\n3((x-2)^2 - (x-2) -2)\n3(x^2-4x+4-x+2-2)\n3(x^2-5x+4)\n\n3(x-4)(x-1)\n\nYay!!
634747
What you want is express \n\nv = a*v1 + b*v2 + c*v3 \n\nWhere a, b and c are, of course, scalars. The coordinate vector will then be (a, b, c).\n\nNow how do we do this: \n\n(2, -1, 3) = a(1, 0, 0) + b(2, 2, 0) + c(3, 3, 3) \n\nEither you just look at it and try to guess what a b and c can be, or you write this up as a set of 3 equations in a,b and c: \n\n2 = 1a + 2b + 3c\n-1 = 0a + 2b + 3c\n3 = -0a + 0b + 3c \n\n...and then solve it: \n\nFrom the last equation we see that c = 1, then using the second equation, -1 = 2b + 3  <=>    -4 = 2b  <=>  -2 = b\n\nAnd substituting into the first, \n\n2 = 1a + 2(-2) + 3(1) = a - 4 + 3 = a - 1 \n\nthus 3 = a. \n\nSo the coordinate vector is (3, -2, 1) and with these numbers the criteria are (supposedly) satisfied. :)
856605
Your ultimate goal of solving algebraic equations is to isolate variables on one side (which are the letters, symbols, greek letters, etc.), and the numbers on the other side.\n\nFor instance:\n2n+12=5n\nYou want to get the 'n's on one side, so you would substract 2n from each side\n\n2n + 12 = 5\n-2n             -2n\n          12 = 3n\n\nThen you want to find what the variable is, since the variable is connected to a number (this is called the coefficient) by multiplication, you would isolate this number, by the opposite of multiplying, which is dividing.\n\nRemember to divide BOTH sides.\n\n12  =  3n\n----      ----\n 3         3\n\n4 = n\n\nAnd your answer is n=4.\n____________________\n\nSome problems will involve additional steps to inorder to isolate the variable.\n4(x-5)+2=x+3 \n\nUsing the distributive propery, distribute 4 to each term in the parentheses (x and -5). So it should look like this:\n4x - 4(5) + 2 = x + 3\n\nMultiply 4(5)\n\n4x - 20 + 2 = x + 3\n\nCombine -20 + 2\n\n4x - 18 = x + 3\n\n**You have just simplified the left side (multiply and add what ever you are able to.)\n\nIsolate the variable by subtracting an x on both sides.\n\n4x - 18 = x + 3\n-x           -x\n3x - 18 = 3\n\nIsolate the number by adding 18 to both sides.\n\n3x - 18 = 3\n     +18     +18\n3x = 21\n\nDivide\n\n3x =21\n---   ----\n3      3\n\nx = 7\n\nDo the same for this one:\n4(y-2/4)=9(y+1/3) \n\nDistribute    4y - 4(2/4) = 9(y) + 9(1/3)\nSimplify       4y - 2 = 9y + 3\nIsolate Variables:\n4y - 2 = 9y + 3\n-4y        -4y\n     -2 = 5y + 3\nIsolate Numbers:\n-2 = 5y +3\n-3           -3\n-5  = 5y\nDivide:\n-5 = 5y\n---    ---\n5      5\n-1 =y\n\nGood luck, try http://www.regentsprep.org/ and click on Math A if you need more help!\n\nhttp://www.mathematicshelpcentral.com/ may also help you!
1079811
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!
488305
Just do a search for "algebra." You'll find a lot of web sites, including:\nhttp://mathworld.wolfram.com/topics/Algebra.html\nhttp://www.purplemath.com/\nhttp://www.algebrahelp.com/
529729
(x2)-3 (x-2) over (x2) -4?\nIs this what you mean?\n(x^2)^(-3)*x^(-2)/(x^2)^(-4)\n= x^(-6)*x^(-2)/x^(-8)\n= x^(-6-2-(-8))\n= x^(-8+8)\n= x^0 \n= 1