Variation problem in algebra 2?
966420
well then\n1. y = 4/x \n by doubling the x value you will have to multiply by 1/2 the whole equation and thus you'll get \n2. y/2 = 4/2x => y/2 = 2/x\n\nanswer y is halved => C
965349
I believe you meant (2x^2 - 3x - 2 )/ (10 + x - 3x^2) right?\n\nWell we can rearrange the problem to look like (2x^2 - 3x - 2 )/ (- 3x^2+ x +10)\n\nWe now factor out the top and factor out the bottom.\n[(2x +1) (x-2)] / [(-3x-5)(x-2)]\n\nSince this is a fraction, we can cancel out the (x-2) because it is on top and the bottom.\n\nWe are left with:\n(2x +1) / (-3x-5)\n\nI think this is what you wanted, good luck and be careful of how you write your notation!
571026
Whenever you have the difference of two squares\n\nF^2-S^2\n\nit factors as\n\n(F-S)(F+S)\n\nMemorize that pattern and problems like this will be a breeze.
895900
This is essentially a maths question.\n\nP1 = P2V2 / V1\nV1 = P2V2 / P1\nP2 = P1V1 / V2\nV2 = P1V1 / P2
1127430
try this one http://www.algebrahelp.com/
372419
I fear you have a missing element here... do you want the differencial of 4P^2..etc because if so you need to know what 4P^2..etc equals, or is 4P^2..etc the differencial of another expression, because if so your question is a bit like saying what is the answer to x=2\n\nIf you want to differencial of 4P^2..etc then I suggest you solve the original equation for t which will make the algebra easier and then use the chain rule where dP/dt = [1/(dt/dP)]
381503
there is direct and indirect variation, direct is y=kx and indirect is xy=k or y=k/x.  the constant of varitaion is k, which determines what x and y are.  normally they will tell you x or y and k, and ask you to find the other variable.
520168
Try this:\n\nhttp://www.algebra.com/
1075715
Here are two sites that look into the problem:\n\nhttp://www.kw.igs.net/~jackord/bp/f6.html\nhttp://math.fullerton.edu/mathews/N310/projects/e6.htm
140559
Using the abc method… \n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nFe-->ax=2c------eq.1\nO-->ya+4b=12c+2d+e----eq.2\nH-->2b=2e----------eq.3\nS-->b=3c+d----------eq.4\n\nLet a=1\n\nFrom eq.1\nax=2c\nx/2=2/2c\nx/2=c-----eq.5\n\nForm eq.3\n2b=2e\n2/2b=2/2e\ne=b-----eq.6 \n\nFrom eq.2\nYa+4b=12c+2d+e\n1y+4b=12(x/2) +2d+b (Form eq.5&6)\nY+3b=6x+2d\n2d=y-6x+3b\nd=(y+3b-6x)/2------------eq.7\n\nFrom eq.4\nb=3c+d (but c=x/2)\nb=3(x/2) +d\nReplacing eq.4 by eq.7 we will get\nb=3x/2+(y+3b-6x) ½\nb= (3x+y+3b-6x) ½\n2[b=(y+3b-3x) ½]\n2b=y+3b-3x\n3x-y=b-------eq.8\n\nReplacing eq.6 by eq.8\ne=b\ne=3x-y\n\nReplacing eq.7 by eq.8\nd=(y+3b-6x)/2\nd=[y+3(3x-y)-6x]/2\nd=[y+9x-3y-6x]/2\nd= (3x-2y)/2\n\nFrom the above calculation we know that:\na=1, b=3x-y, c=x/2, d=(3x-2y)/2 & e=3x-y. \n\n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nWhen replacing the values we’ll get:\n[1]FexOy+[3x-y]H2SO4 -->[x/2]Fe2(SO4)3+[(3x-2y)/2]SO2 +[3x-y]H2O
723293
you have to use the couchy schuartz method\nim nit sure on the spelling \nbasicly you take one vector (V1) and use it then you take the second vecter (U1) minous vi times(V1 interproduct U1)/(magnatude V1^2) and use that total as V2 then you do a simular thing for the third and fourth vector and up if you have them by subtrating V1<Ua,V1>/(mag V1)^2 plus V2<Ua,V2>/(mag V2)^2 .... till you have done that for each vector \nin other words every vector will have a term for all of the new previous vectors subtracted from it then finaly you take all the new vectors and devide them by their magnatude to get unit vectors and they will then all be orthOGANOL UNIT VECTORS
798195
Don't be scared by all the words.  You just have to remove brackets and parentheses from\n2[x+(x+10)+(x+2(10))] \n\nand then collect the terms -- you should get 6x+60.\n\nThe last part just means do it again with 15 replacing 10.
380912
When dealing with inverse variation there are two different equations you can use, y= k/x or first y/new y= new x/ first x. So in this case you can either do I= k/d^2 then do I= k/2d^2 and compare the I's. Another way you can do it is I original/ I new= 2d^2/ d^2 then you compare the first I and the second I. Unfortunetly in this case you dont have numbers so you just look at the fact that in the firt equation your dividing by the d^2 or the 2d^2 and you look that in the first one your just dividing by d^2 or 12 inches, while in the second your dividing by 2d^2 or 4d or 24 inches, which is more. Then when you divide by more than the I is smaller or lest intense.
901005
Use the following equation and solve:\n\nP1V1/T1=P2V2/T2\n\nT2=P2V2T1/P1V1\n\nalso note the significant figures!!!!
195576
Y = U + VX^2\nVX^2 = Y - U\nX^2 = (Y-U)/V\nX = sqrt ( (Y-U)/V )
680155
i'd love to help but i'm having trouble with it myself\n\nuse ur math textbook it will have it unless ur math book say algebra and is red ( that my alg. book and it doesn't help) \n\nSRY good luck
370398
O.K.\n\nUse the substitution:\n\ny = uv , x = u/v\n==>\n(1+u^2)(u/v)(udv + vdu) + (1-u^2)uv(du/v - udv/v^2) = 0\n==>\n(1+u^2)(udv + vdu) + (1-u^2)(vdu-udv) = 0\n==>\n2vdu + u^2(2udv) = 0\n==>\ndu/u^3 + dv/v = 0\n\nIntegrating gives:\n\nv = (constant) e^(1/(2u^2))\n==>\nv^2 = A e^(1/u^2)\n\nReverse substituting x,y gives the answer as an implicit function:\n\ny/x = A exp(1/xy)