Use the commutative property to rewrite each of the following equations.  Check your work by finding the answe
916199
Commutative Property means that you can change the order of the numbers and you will get the same answer.\n\n1. 13 x 87 = 1131  and  87 x 13 = 1131\n\nAssociative Property means that you can group the numbers in any way and you will get the same answer.\n\n5.  (22 + 76) + 91 = 189   and  22 + (76 + 91) = 189\n\nThis should help you solve the rest of your questions.
245176
u*(u-2)+v*(u-2)\n(u+v)*(u-2)
508029
The equation is:\n(x+a+b+c)^2 = 2x^2+(x+a)^2-(x+b)^2-(x+c)^2\nA simple rearrangement yields\n(x+a+b+c)^2 - x^2 = (x+a)^2-(x+b)^2 + x^2-(x+c)^2\nFactorize using the identity a^2-b^2=(a+b)(a-b)\n(x+a+b+c+x)(x+a+b+c-x) = (x+a+x+b)(x+a-x-b) + (x+x+c)(x-x-c)\nwhich can be written as\n(2x+a+b+c)(a+b+c) = (2x+a+b)(a-b)-c(2x+c)\n==> 2x(a+b+c)+(a+b+c)^2 = 2x(a-b)+(a-b)(a+b)-2xc-c^2\n==> 2x(a+b+c-a+b+c) = (a+b+c)^2 + a^2-b^2-c^2\n==> 4x(b+c) = a^2+b^2+c^2+2ab+2bc+2ca+a^2-b^2-c^2\n            = 2(a^2+ab+bc+ca) = 2(a+b)(c+a)\n==> x = (a+b)(c+a)/[2*(b+c)]
457175
2 + (a + b)\ngiven\n\n2 + (b + a)\ncommutative prop. of addition\n\n(2 + b) + a\nassociative prop. of addition\n\n(b + 2) + a\ncommutative prop. of addition\n\nb + (2 + a)\nassociative prop. of addition\n\n^_^
850884
using chain rule and product rule\n\ny'= 3x^(2)*tan(4x)\n         +4x^(3)*sec^(2)(4x)\n  = x^2(3tan(4x)+4x*sec^(2)(4x) \n\ni hope that this helps
848345
I assume that this is s = (a-rl) / (1-r).\n\nFirst, multiply both sides of the equation by 1-r.\ns(1-r) = a-rl\n\nUse the distributive property on s(1-r)\ns-sr = a-rl\n\nGet the r terms on one side of the equation and the others opposite it.\nrl-sr = a-s\n\nUse the distributive property to factor out the r.\nr(l-s) = a-s\n\nDivide both sides by l-s\n\nr = (a-s) / (l-s)
195201
i think u typed the question wrong, can u please correct it and if u have 2 unknowns, please give 2 equations, else u cant get a definite solution
1150392
No problem, i just finished a calc3 course in college.\n\nYour problem is the same as 125 * (6x^2 + 5x + 7)^(-1/3)\nYou have to use the chain rule so you get\n125 * (-1/3) * (12x + 5) * (6x^2 +5x + 7)^(-4/3)\n\nthen you can simplify how you like from there
531852
you have to use instinct, there is no clear rules that states what the expressions must look like (except if the problem insist you on giving the answer in a specific form). \n\nPerhaps one of the guidelines is make variables that have same indices grouped into one. So x + 3x becomes 4x, group them out don't leave them straying.
928663
Are you asking for help or an answer?\n\nTake any 2 equations and multiply all of the terms or one or both by numbers that will make one of the terms in each equation have the same coefficient. Now subtract one from the other and the term will disappear. Now take the resulting equation and do the same with it against the third. Keep doing this until you are down to one term and solve for the variable. Now substitute the variable into the equation and solve for the other variables.
919864
1.  ax + xb\n2.  21x^3 + 24x*2y\n3.  24x^2 + 31x + 10\n4.  24x^2 - 8x - 6\n5.  40x^3 + 66x^2 + 38x + 6\n6.  x^2 - 1\n7.  x^3 - 8
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.
864870
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
869887
If you have 2 equations, just solve for one variable for one of the equations. Take that equation and plug it into the same variable for the other equation. You will solve for that variable, and plug in that number to any of the 2 equations to solve for the other variable.
714493
This *is* the way to write it so that x is the input and w is the output. If you want to solve for x in terms of w (i.e. have w the input and x the output), you are better off doing numerical approximations than using the cubic formula (it is really, really nasty).
913056
The trick here is to find three equations where three of the variables are expressed functions of another variable. \nSo for example we could express x, y and w functions of z\nIf you add the first equation to the third and substract the second, you'll find: y= -3-z. We'll call this new equation A\nIf you add the second and the fourth equation and replace y by its value found in (A), you'll have: w= 7/4 + 1/4 z, which we'll call (B)\nNow take the first equation and replace w and y by their value from (A) and (B): \nx= -17/2 -7/2 z, this is the equation (C).\nNow that you have your three equations A,B and C you can substitute them in one of the four initial equations, for example the third one. You'll find:\nz=1\ny= -4\nw=2\nx= -12\nThis is not the entire solution (you have to detail how to get to A, B and C) but you should be able to do it (I don't want to be the only one working here!)
778763
1) (a - b)(x - y)\n2) (x^9 + y^3)(x^9 - y^3)\n3) (6x + 5)(x + 8)\n4) (x - 9)(x - 7)