Please explain an elementary algebra question for me...?
966094
Consecutive numbers differ by 1.\nIf the first one is n, then the second one is n+1\nThat means the third one is the second number plus one (n+1)+1 = n+2\nThe fourth one is (n+2)+1 = n+3\nand the fifth one is (n+3)+1 = n+4\nAdd them altogether\nn + n+1 + n+2 + n+3 + n+4 = 5n + 10
622612
Joel, you need some help.  This is one of the easiest questions I have ever heard!  You should know this, man!  You're gonna fail your test if you keep cheating on here and don't actually figure out how it works.  Good luck, brotha'\n\nL = -2 or L(x) = -2
491455
(x^2+4x+1)/(x-2) = x+2+3/(x-2)\n\nx  : x^2+4x+1-x^2-2x =2x+1\n2  : 2x+1 -2x-4 =3
58450
use the distributive property (poor u, u seem to be having a lot of trouble with your math homework tonight!! hope this helps:)\n\nthe property states that a(b+c)=ab+ac, basically you take the "a" and multiply it by the other 2, or distribute it.  say a=2, b=3, and c=1.  your equation would be 2(3+1) and, using the distributive property, make it become 6+2 which equals 8.  so its the same when your using x.  3(-4x+7)= -12x+21!!! yay!! hope this helped!
145326
i'm good at math, but what exactly is the quustion??? there isn't any... what is it, a equation?\nany way you can post it to sciences and mathematics > mathematics, im sure someone there can help you...\n\n>>>ok one minute i'm working on it...\n\n>>>there you go:\nlets do the first: (x-3)(x-1)(x+2)\nfirst we open the last two parentheses as in usual opening, and the (x-3) remains as it is. we get (x-3)[x^2+2x-x-2]\nequals (x-3)(x^2+x-2) then we open the two parentheses left. first we multiply by the first member (x) then the second (-3). then we get:\nx*x^2+x*x+x*(-2)-3*x^2-3*x-3*(-2)\nx^3 + x^2 - 2x -3x^2 -3x + 6\nand finally:\nx^3 -2x^2 -5x + 6\n\nthe second is the same way, i'm doing it, good luck :)\n\n>>>the second: (x-5)(x+1)(x-2)\nfirst we open the last two parentheses as in usual opening, and the (x-5) remains as it is. we get (x-5)[x^2-2x+x-2]\nequals (x-5)(x^2-x-2) then we open the two parentheses left. first we multiply by the first member (x) then the second (-5). then we get:\nx*x^2+x*(-x)+x*(-2)-5*x^2-5*(-x)-5*(-2)\nx^3 -x^2 -2x -5x^2 +5x +10\nand finally:\nx^3 -6x^2 +3x + 10\n\ni hope i helped...\n\n>>>oh i leave the first one to the last because i have to leave one of the three, and it really doesn't matter which one i leave. for example on the first i left the first parenthese (x-3)[(x-1)(x+2)] but i could leave the last [(x-3)(x-1)](x+2) or the middle (x-1)[(x-3)(x+2)] and i'd still get the same answer. i have to leave one of the three bcuz i can't open them all together, it's complicated and hard, and by opening two ane leave one it becomes much easier.
953940
Hi again \n\nY = √2x – 3 \n2x – 3 = y ²\nX =( y ² + 3 )/ 2\nf –¹ (x)  =( y ² + 3 )/ 2  =>   f –¹ (x)  = (x ² + 3 )/ 2   &  x ≥ 0 \n\nGood Luck \n try to solve them yourslef again if you got wrong , am sure you can solve them correctly , if you just try your best
134065
Your statement translates to the following algebraic equation.\na+b=a-b=a*b\nSolving a+b=a-b, we get b = 0; \nSubstituting b=0 and solving a+b=a*b, we get a = 0;\nSo the two numbers are zeroes.
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!
832515
(1) any of the following may be the root of the expression,+1,-1,+2,-2,+3,-3,+6 or -6.\nNow putting the value of x=-1,see ifthe value of the expression becomes zero or not?If it becomes zero,then x=1 is one factor of the expression,if it does not become zero try with =1,then -2.+2 and so on.\nLuckily for us,with x=-1,the value of the expression becomes zero\nTherfore,x+1 is one factor\nx^4++x^3-7x^2-13x-6\n=x^3(x+1)-7x(x+1)-6(x+1)\n=(x+1)(x^3-7x-6)\n=(x+1){x^2(x+1)-x(x+1)-6(x+1)}\n=(x+1)(x+1)(x^2-x-6)\n=(x+1)^2(x-3)(x+2)\n(2)Will you please check up the second one once more and if there is any correction send it to me through Yahoo answers.
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!
888151
Basic math/algebra. You are taking two known numbers and finding the unknown number. \nE=P/I  solve for I    Think like simple Algebra I is X. So you want to work the problem and get a final answer x=?  or   I=?\nHope this helps.
120236
6x\n\n3x^2 --> 3*2 x^2-1 --> 6x\n\nUpdate: You are deriving with respect to x, right?
924405
Is it possible to have 2 equal signs in a problem?\nI am only in algebra but I haven't seen one like that before.\nSorry I couldn't help.
1137023
(2x-9)(2x+9)\n\na guess... its been a long time since I did that stuff
574042
we need to multiply something by (x+1) so that the result would be x^2-1. but we know x^2-1 = (x-1)(x+1).\nso we have to multiply (x-1) with (x+1) to get x^2-1.
973940
A previous answer was incorrect. It is\n\n-4x(x+3y)(x-2y)
4935
ur second equation is wrong. \n2a is not equal to a-squared (which u symbolized by ab, since a=b).