how to solve discrete mathematics problems?
826811
You can solve the discrete mathematics problem by following way.\n\n1. Define parameters\n2. Assign them their values. \nEg. A - He does his homework.\nB- If you do your homework ,you pass\n3. Make the use of these parameters and convert your problem to hypothesis.\n4. Make conclusion with given identities (eg. various tautologies) .\n\nI hope now you know.\n\nMy answer would have been precise if you would have stated type of problems.\n\nAnyways, I hope this helps
636772
k=1/4 or k=2 or many more possibilities
937047
Well, there may be more solutions here than usual, but the first solution I see is:\n(3 L,5 L,8 L)\n(0/3,0/5,8/8)\n(0/3,5/5,3/8)\n(3/3,2/5,3/8)\n(0/3,2/5,6/8)\n(2/3,0/5,6/8)\n(2/3,5/5,1/8)\n(3/3,4/5,1/8)\n(0/3,4/5,4/8)
533429
well..write them down ...try to derive them (if their derivation is simple enough or else skip it )......\n\ntry to connect each formula with other one with a hint ....a small hint will help u to recollect other one ....\n\ntry to do at least one problem based on each formula
653455
The best method to solve these type of problems is to sove them through VENN Diagrams. First draw a rectangle which is the sample space, then draw three circles intersecting each other wjich are E, F AND G. The common region between E & F is P[E n F] and so on. It is quite difficult to explain this in writing but I hope you have understood. Now by drawing Venn diagram I have tried to solve ur questions.\n\nPr[EuF'] = P[E] + P[F'] - P[E n F']\n         = 0.35 + (1 - 0.5) - (0.35 - 0.1)\n         = 0.6\n\nPr[F'nG']= 1 - P[F u G]\n         = 1 - ( P[F] + P[G] - P[F n G])\n         = 1 - ( 0.5 + 0.45 - 0.3)\n         = 0.35\n\nPr[E'nG] = P[G] - P[E n G] = 0.45 - 0.2 = 0.25\n\nI think above answers are correct and hope they help you!!
817920
Its crucial.\n\nBasically the field of algebra and discrete mathematics (combinatorics) lends naturally to the subject.\n\nAlgorithms or computer procedures are analysed in terms of their efficiency using maths.\n\nNumbers in computers are stored in binary.\n\nThe fundamental theorem of computer science is about the equivalence in dealing with certain types of problems. Problems whose solutions can be verified in polynomial time are called NP. Those that can be solved in polynomial time are called P. We would like to show P<NP but we cant. The best we can do is supply a bunch of equivalent problems in NP such that if any one of these key hard problems is in P, then P=NP. In practice that means these problems are impossible to solve precisely and we need an approximate solution.\n\nAnother area where math is used is correctness. Ideally people would like to prove a program is correct. In practice this is usually hell but to do so is mathemstical\n\nThere is an entire field of numerical analysis that involves the use of computers to approximate complex calculations.\n\nGraphic algorithms use lots of math and there is a whole field called computational geometry.\n\nCryptography which is the study of secret codes and encryption is all math.\n\nThe list goes on.\n\nIn addition mathematics requires one to think methodicaly and logically. This is fundamental to wroiting/designing s/w
28487
First scan the puzzle and see which numbers appear often.  You may be able to narrow down where that number goes into the 9x9 boxes that don't yet have it.  Then, look for lines that are almost complete and see if you can figure out which numbers fit where.  I pencil in tiny numbers that might fit in a box if there are only a few choices.  Then, look for 9x9 boxes that are almost full and fill in blanks if you can.  I find that I work them fastest by instinct or by using these steps in this order.  I limit the amount of time I can spend on these -  I have already spent too much time in the past year working these puzzles, but the time varies based on the difficulty.
723712
One way to solve this is to draw it on graph paper and count up the squares.  Howevr, I think I have a solution that doesn't resort to graphing:\n\nThe units 15 and 42 have a common divisor of 3.  So, if the diagonal starts at (0,0), it will also intersect (5,14) and (10,28).  Therefore, we can reduce this problem to testing a 5 by 14 rectangle and multplying by 3.\n\nFor the 5 by 14 rectangle, since this diagonal won't hit any vertices of the unit squares, just figure you will pass through 14 squares as you make your way across, and you will move up 4 squares.  So, 4 + 14 = 18 squares.  \n\nMultiple by 3 to get your answer for the 15 by 42 rectangle: 18 * 3 = 54.
963221
Discrete means individual and noncontinuous. There is "space" between consecutive elements. In discrete mathematics, you do a lot of counting. Some topics grouped under discrete mathematics include game theory and graph theory.
376854
I think practice is the only way. I use several different methods. But usually, I start with the obvious ones. Look within the boxes first to fit in the easy ones. Do that once again in case you missed any, then look at each row and column. Sometimes, if there looks like a few possibilities, you can work out if there is a number which can only go in one square which you might not see straight away. Sometimes, there will be squares where only 2 numbers are possible. If there are 2 squares in the same row/column/area which have the same 2 possiblities and then there is only one square remaining then you know the last square must be the other number. Some people don't think of that.
396946
right.  so, yeah, you're not going to solve the P & NP problem on yahoo answers.\n\nand to give you the proper answer to your question:\nfor all we know there could be, but we aren't even certain of this possibility.  it's conjectured that all NP-complete problems are not solvable in polynomial time, but whether this is a provable conjecture is unknown.
67171
Use the distributive property:\n4*5x + 4*(-9) =24\n20x + -36 = 24\n      +36   +36\n20x = 60\n--------\n20    20\nx=3\nI realize you already have the answer but sometimes the path is more important than the destination...
464622
Use any recursive algorithm.\n\nFor instance:\n\nf(x) = f(x) + 1
709282
Quick Math-Mind Games
966925
There are 3 solutions for ABCDEFGHJ -\n\n2 6 7 8 3 5 1 4 9\n2 6 7 9 5 3 1 4 8\n6 7 8 4 1 3 2 5 9
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.
575173
thats a geometric sequence with a common ratio of -1... you can get the common ratio by getting the first set of differences... thats 10-12... 12-11.. 11-15.. 15-12... then have the first set of difference: -2 1 -4 3.. then get the second set of difference... -2-1.. 1--4... -4-3... and u get -1 -1.. thats how u get to chek if its a geometric sequence by checking if they have the same second diff... chek the net and search for the formula for solving geometric sequences cuz i forgot bout those already.. goodluck!