Can somone help me with this algebra problem...I cant seem to get it?
147771
You have to write as a product of two terms.\nThis is always possible, as a consequence of an important theorem. \nIn general, if you have the expresion a*x^2 + b*x + c, where a,b,c are constants and x is a variable, here is the way you can put this expression in a product form.\nFirst calculate b^2-4*a*c (the so-called determinant). If this number is negative the expression can't be put in the product form (well at this stage).\nThen calculate x1=(2*a-sqrt(determinant))/2 and x2=(2*a+sqrt(determinant))/2. The result will be:\na*x^2 + b*x + c= (x-x1)*(x-x2).\nIn your case the determinant is 256 and sqrt(256)=16. then x1=-7 and x2=9 and that's it.
584912
http://www.algebrahelp.com/ I found this site very useful perhaps you should check it out http://www.algebrahelp.com/
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
795670
Your answer key is wrong, or you are not reading the right answer.\nFrom the data you presented above y = x^3 is OK.
617282
are u sure its x2? shouldn't it be 2x+10x=(a number) usually there is a number at the end of the equation and u have 2 solve for "x". add details to ur question and then i may be able 2 help u.
807552
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!
80315
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
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)
936579
your question seems to be incomplete.\nso far....\nwidth=x\nlength= 2x-1
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
752879
ok re arranged the second question so that the 6 is factored out so its x= such and such then sub the x equals such and such into the first equation and solve then do the same thing to Y in the second equation
849885
I understand your frustration in receiving no Answer yet to this question, but since I do not have any direct knowledge of how to perform the calculation(s) you need ... I hope the link below serves you well.  It appears to explain the mathematics behind all the concepts you mention.
1089643
http://www.algebra.com
326220
You actually need to solve this a different way.\nYou can't multiply both sides by x+3 because this could be positive or it could be negative...you don't know.\nInstead, bring everything to one side by subtracting 2, then get a common denominator.\n(x - 2)/(x + 3) <= 2\n(x - 2)/(x + 3) - 2 <= 0\n(x - 2)/(x + 3) -2(x+3)/(x+3) <= 0\n[(x - 2) - 2(x+3)]/(x + 3) <= 0\n[(x - 2) - 2x - 6)]/(x + 3) <= 0\n(-x - 8)/(x + 3) <= 0\nYour important values are -8 and -3\nTest values less than -8, between -8 and -3, and greater than -3.\nYou will see that you get negative values for\nx <= -8 and x >= -3
745882
C is correct but not complete... only the 3 is factored out... the real answer would be:\n\n3(x-1)(x+4)
851212
hmm im not quite in college yet, but i hope to be help.\ndo you know the values for f, h, v, and a?\nif you do, id substitute it in.\nthen square both sides. \nand then solve.
154380
I posted these sources earlier to someone else who was looking for math help.\n\nGood Luck!!