i definitely have a tough time in analyzing math equations especially involving ungrouped real nos. i don't know which of the nos. to solve first.
126391
Probably easier to show by example.\n4x2+8-:-4-1 (-:- = divide)\nmultiply any numbers first, then divide, then add and subtract.\n\nYou could add brackets in if that helps you.\n(4x2)+(8-:-4)-1 = 9
184831
can do. (I will put sh and ch instead of sinh and cosh)\nTake sh of the first equation.\nx=sh(arch(y^2))\nsquare it\nx^2=sh^2(arch(y^2))\napply the hint sh^2=ch^2-1\nx^2=ch^2(arch(y^2))-1\ngiven ch(arch(x))=x\nx^2=y^4-1=(y^2-1)(y^2+1) (*)\nby symmetry, for the second equation\ny^2=x^4-1 (**)\nPutting (**) into (*)\nx^2=(x^4-2)*x^4\nwe see for x=0 it is true.\nfor x!=0\nx^(-2)=(x^4-2).\nWe see there is a positive and a negative solution (knowing what the graphs of both functions look like.\nThey are symmetrical with respect to x=0.\n\nYou can also back down a few steps to when we had\n1=(x^4-2)x^2\nTransform it into\nx^6-2x^2-1=0\nlet t=x^2\nt^3-2t-1=0\nAnd this is solvable using Cardano's method.\nIt's just numbers from here. I don't like numbers.\nFor illumination, check the source
575918
vic4ovo = \nYou can submit it to Mathworld, which you can find at\nmathworld.wolfram.com. \nAny competent graduate student ought to be able to evaluate your equations.
396924
The same equation, in this case, cannot have two different answers. So there is no solution to this system (to understand it, try graphing it)
675863
this has happened to me before on my 89. it's usually when the expression is too complicated--that is, it would take the calculator more memory than it has to solve it. so i'd recommend trying to first simplify the expression and then enter it. For instance, don't leave things in terms like sin(2) when you can just get an approximate number for that. maybe that's it?
130686
Well. Do you know how to work rational expressions in general? The first link below is a useful general resource and has a section on rational expressions and polynomial long division; the second is specifically on rational expressions at college intro-level.\n\nI used to tutor math and I found that most people's difficulty with word problems was not the actual math, it was extracting the appropriate equations from the words. Once they had the equations, they could solve them (assuming they'd learned the basics). \n\nI'd find it pretty difficult to give you general guidelines on "how to solve word problems with rational expressions" but the main trick with word problems is to figure out a) what variables are involved, and assign each of them a letter (e.g., speed of first train is x, time it has been travelling is y, speed of second train is z...), b) sort out the relationship between them (e.g. x = 2z/5), and c) make sure you know what *exactly* is being asked. It is not at all uncommon for word problems to include extraneous information (colour of train, age of driver) to muddle the issue.\n\nYour best bet is almost certainly to go ask your teacher, professor, TA, tutor, study group, or classmate-who-understands-it for help. They will be better able to figure out where your problem is if they can watch you work through an actual example.
56977
Pay more attention in class! And assuming it is an ideal gas, use PV=nRT. Since there is not change in volume you can basically drop it from the equation and make P=nRT. If you like algebra you can leave it in but it will cancel out eventually. You could solve for 'n' first using initial P and T, and then plug and chug.
457497
I know this is a little late (since your test is already over), but here is one site:\nhttp://www.edhelper.com/algebra.htm\nI found this by searching for algebra 2 problems on google.\nIf you do that, you can find more.
551385
Gopal has the correct answer! \n\nSo... becareful with some other results... and... never use:\n0.833 instead of 5/6... it's a mistake!\n\nFinished Algebra with 19!
464660
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
429503
We can use faux-algebra to solve these. You probably haven't taken algebra yet. With algebra these would be extremely easy to solve. It's why algebra was invented.\n\na. 16 x ___ has to be -42, because if we add 51 to it we get 9. Thus, ____ has to be -42/16, or -21/8.\n\nb. 2 x ____ has to be 22 - 9 x _____, because if we subtract 8 from it we get 14 - 9 x ____. And 22 has to be 11 x ____, because if we subtract 9 x ____ from it we get 2 x ____. Thus, ____ has to be 22/11, or 2.\n\nc. 9 x ___ has to be 3 x ____ + 3, because if we add 2 to it we get 3 x ___ + 5.  And 3 has to be 6 x ____, becuase if we add 3 x ____ to it we get 9 x ____. Thus, ____ has to be 3/6, or 1/2.\n\nWith algebra, no such thinking would be required. You just plug in the formulas and grind away. You'll see when you study algebra.
425095
Answer:\nThe problem to solve is: \nRemember, with complex variables, keep like terms involving i together...\n \nAdding Multiplication Sign: y*x2-4x-5\nAdding Multiplication Sign: y*x*2-4x-5\nAdding Multiplication Sign: y*x*2-4*x-5\nPoly Parsing: y*x*2-4*x-5\nSplitting y*x*2-4*x-5 \ny*x*2-4*x - 5 \n[Left Side]Poly Parsing: y*x*2-4*x\nSplitting y*x*2-4*x \ny*x*2 - 4*x \n[Left Side]Poly Parsing: y*x*2\nMultiply Term y with Term x equals Term xy\nMultiply Term xy with Term 2 equals Term 2xy\nAddTerm: Term (2xy)\nPoly =  2xy\nRepace with Reduced Left Side \ny*x*2 - 4*x \n2xy - 4*x \n[Right Side]Poly Parsing: 4*x\nMultiply Term 4 with Term x equals Term 4x\nAddTerm: Term (4x)\nPoly =  4x\nRepace with Reduced Right Side \ny*x*2 - 4*x \n2xy - 4x \nBecause of the minus sign 4x becomes -4x\n\nPoly Adding 2xy with -4x...\nAddTerm: Term (2xy)\nAddTerm: Term (-4x)\n... 2xy-4x\nPoly =  2xy-4x\nRepace with Reduced Left Side \ny*x*2-4*x - 5 \n2xy-4x - 5 \n[Right Side]Poly Parsing: 5\nAddTerm: Term (5)\nPoly =  5\nRepace with Reduced Right Side \ny*x*2-4*x - 5 \n2xy-4x - 5 \nBecause of the minus sign 5 becomes -5\n\nPoly Adding 2xy-4x with -5...\nAddTerm: Term (2xy)\nAddTerm: Term (-4x)\nAddTerm: Term (-5)\n... 2xy-4x-5\nPoly =  2xy-4x-5\nThe problem to solve is: yx2-4x-5 \n2xy-4x-5
83086
You see, it would be easier to post *exactly* what you have to do because at the moment you are not providing enough information on the problem.  For example, it seems to me that you maybe solving for integer solutions (x,y), with integers a,b,c.
80588
In the future, us ^ for exponents. It's pretty much standard for typed equations. But I digress....\n\n6. 10[(4 + 5).2. - 12)]\n10[9^2 - 12)]\n10[81 - 12]\n10 * 69\n690\n\n-------------\n\n9. 7.0.\n\nAny number raised to the zero will equal 1.\n\n-------------\n\n27. 3(2a + 4) + 2(4a + 2)\n6a + 12 + 8a + 4\n14a + 16\n\n-------------\n\n43. 32 - (-15) \n\nSubtracting a negative is the same as adding so...\n32 - (-15) = 32 + 15 = 47\n\n-------------\n\nX. Evaluate the expessions. b = 6, c = -3.\n2b - c\n\n2(6) - (-3) = 12 + 3 = 15\n\n-------------\n\nXIV. Sequences and tables. look at the table below and awser the questions.\n_____\nn ! t!\n1 ! 4!\n2 ! 8!\n3 !12!\n4 !16!\n5 !20!\n------\n\n67. Write and equation in the for of t = ____ that describes the table above.\n\nt is always 4 times n, therefore t = 4n\n\n-------------\n\nXIV. Sequences and tables continued. Look at the table below.\n_____\nn ! t!\n1 ! 4!\n2 ! 7!\n3 !10!\n4 !13!\n5 !16!\n-----\n69. Write and equation in the form of t = ____\nthat describes the table above.\n\nThis on is trickier but notice that the value of t increases at 3 times the rate of n. At that point, we know the 'slope' is 3. Then, the 'intersection' would have to be 1 to make the equations true. So....\n\nt = 3n+1
121130
Oh yes i have cracked it at last....\n\nsolution...\n\nLet x=tan-1 (y) (i mean tan inverse)\nthen tan x = y,and cot x = 1/y (coz tan-1 (x) = cot-1 (1/x))\n\nthen ur given equation's Left hand side part (LHS) will be\n\n(1/y)/(1-y) + (y)/(1-(1/y))\n\n= 1/(y-y2)+ y/(1-(1/y))\n\n= 1/(y-y2) + y2/(y-1)\n\n= y2(y-1) - 1/y(y-1)\n\n= y3/y(y-1)\n\nI hope u understood till this step.\n\nNow we have (a3-b3) = (a-b) (a2+ab+b2)\n\nthen the LHS becomes...\n\n(y-1)(y2+y+1)/y(y-1)\n\n= (y2+y+1)/y\n\n= y+1+(1/y)\n\nsubstitute back y=tanx, 1/y = cot x and leave 1 as it is...\n\nU have LHS = 1 + tan x + cot x...Thats ur RHS !!!
220857
Algebra can get difficult but i get used to it and i grasped it pretty quickly.  But when i first learn some of the concepts i need to do the homework and look at it for a while.  Then, it just sticks to me.  I can understand how people can find it really confusing and difficult.
173450
Write the equation for K.  You know K, [NO2}, so solve for the only variable you don't know [N2O4].  By the way, these are gasses so it is not a solubility product.  That involves a solid.