Finite Mathmatics Linear Problem?
403252
let x be the number from A\nlet y be the number from B\nlet z be the number from C\nthey have 500 pounds\nso\n2x+2y+3z=500\nand we have 1000 hours to produce \nso \n4x+5y+5z=1000\nsolve this system with respect to x u will get the following\n-2(2x+2y+3z=500)\n   4x+5y+5z=1000\nadd them u get \ny-z=0\nthen y=z\n5(2x+2y+3z=500)\n-2(4x+5y+5z=1000)\nu add and get\n2x+5z=500\nthen z=(500-2x)/5=100-2x/5\nz=y=100-2x/5\nlet f(x)=500x+600y+1000z=500x+y(1000+600) since y=z\n        =500x+1600y\nsubstitute y and z\nu will get\nf(x)=500x+1600(100-2x/5)=500x+160000-640x\n    =160000-140x\ndifferentiate f(x)\nu get\nf'(x)=-140\nso it is maximal when x=0\nthen y=z=100\nand profit =f(0)=160000$\nf(x)=
1066144
Without the knowledge of where point e is, this problem is insolvable.\n\nI suspect, however, from a brief diagramming session, that you will need to solve for the length of ab using the Pythagorean Theorem, and set up similar triangles.
708619
Please state the requirements of the mathematic's work project.\n\n- - - - - - - -s-
958467
Replace x with 2:\nf(2)=2^3-4(2)^2+14(2)-20\nf(2)=8-16+28-20\nf(2)=-8+8\nf(2)=0
867649
That is an equation, but y is a linear function of x.
390304
THE POSSIBILITIES:\n\n1.1:1→1*1=1\n2.1:2→1*2=2\n3.1:3→1*3=3\n4.1:4→1*4=4\n5.1:5→1*5=5 √\n6.1:6→1*6=6\n\n7.2:1→2*1=2\n8.2:2→2*2=2\n9.2:3→2*3=5\n10.2:4→2*4=8\n11.2:5→2*5=10 √\n12.2:6→2*6=12\n\n13.3:1→3*1=3\n14.3:2→3*2=6\n15.3:3→3*3=9\n16.3:4→3*4=12\n17.3:5→3*5=15 √\n18.3:6→3*6=18\n\n19.4:1→4*1=4\n20.4:2→4*2=8\n21.4:3→4*3=12\n22.4:4→4*4=16\n23.4:5→4*5=20 √\n24.4:6→4*6=24\n\n25.5:1→5*1=5    √\n26.5:2→5*2=10   √ \n27.5:3→5*3=15   √\n28.5:4→5*4=20   √\n29.5:5→5*5=25   √\n30.5:6→5*6=30   √\n\n31.6:1→6*1=6\n32.6:2→6*2=12\n33.6:3→6*3=18\n34.6:4→6*4=24\n35.6:5→6*5=30   √\n36.6:6→6*6=36\n\nProbability = No. of required possibilities/No. of total possible events =11/36\nSo the right answer is D
464622
Use any recursive algorithm.\n\nFor instance:\n\nf(x) = f(x) + 1
511948
There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.
510279
Jimbo, you gave up too soon......\n\nV = L*W*H = L*W*x in this problem\n\nV = (6-2x)*(4-2x)*x\n\nV  oila!
1137447
2 linear equations\n\n X  + 3Y - 1 = 0\n4X + 7Y + 5 = 0\n\nFROM (1)   4X + 12Y - 4 = 0\nFROM (2)   4X +  7Y + 5 = 0\n\nSUBTRACT\n                          5Y - 9 = 0\n\n                          5Y = 9\n\n                            Y = 1.8\n \n  FROM (1)      X + 5.4 - 1 = 0\n\n                        X + 4.4 =   0\n\n                        X =  -4.4
930026
wellz why not just help me with that matrix problem? :oP
1178298
Read this: http://en.wikipedia.org/wiki/Linear_equations
1135008
This question! Isn't it under mathmatics? LOL!
916972
When all the variables cancel out, then you have an infinite number of solutions, or no solutions.  Here, you have none.\n\nThis problem shows you that you need to check for linear independence.  That is, none of your equations should be linear combinations of any others.  What is a linear combination?  It's when you add equations which may or may not be multiplied by constants.  If you get something like this, then you can't solve your system, as you'll have fewer equations than unknowns.\n\nIn your case, 3[equation1]-[equation3] yields 0=-1.  This is false, so your system has no solution.\n\nIf there was a 3 instead of a 4, you would have an infinite number of solutions, where they satisfied the first and second equations.
159422
Is this a trick question? For a given system of linear equations, there is one solution that will yield a particular answer. There are infinite possible systems that will yield that answer (x=-2). There are also multiple ways to arrive at the same  correct answer for a given system. I think we need more info to be of any real help...
642018
Wonderful.\nTHis is some sort of simultaneous equation.\nAccording to the principle of this equation, we try to rationalize one term of the equation so as to get one unknown term. We can either try finding the x term or the y term. For your question as to find the y term.\n\nFirstly, we rationalize the x terms by multipling them by the inverts. then it will result to \n                                    12x + 20y = 32\n                                    12x + 18y = 39\nThe we apply the subtraction rule. ie. we multiply all the terms beneath by the negative (-) sign. This will result to \n                                    12x + 20y = 32\n                                   -12x - 18 = -39\nWe later subtract so as to attain\n                                2y = -7\nWe apply the law of linear equation by dividing both sides by the coefficient which in this case is 2.\n                              2y/2 = -7/2\nThis will eventually cancel the 2 in the first term therby leavin y alone. \n The answer will be in either of the following cases through simplification.  -7/2, -3 1/2 or -3.5
40850
-3 <4x\n x < -3/4\n solution set : {- infinity ,.............,-3/4}