Algebra  problem:Is the  set Q a group under the binary  multiplication operation?
824224
No, because 0 doesn't have an inverse.
133517
the solutions is 7,8,3,2 or 7,6,5,2. The problem works with both sets of numbers.
238182
p1    p2\n--- = ---\nq1    q2\n\n 5     2\n--- = ---\n 2    q2\n\nCross multiply,\n\n4/5
1083059
Boolean algebra is concerned with problems of the type of only two possible solutions. Yes, no, male, female, and can be represented by a 0 or a 1. A common use of this is a decision tree, where each answer/solution leads to another question with two possible answers. This continues under a desired answer is acheived. \n\nThis topic is also covered in symbolic logic. There are rules, and 0's and 1's are used exclusively. Here are some rules for addition and mutiplication.\n\nAddition:     + 0     1\n                   0 0     1\n                   1 1     1\nIn  this table the first row is added to the number in the first column. Note 1+1=1\n\nMutlipication    x   0  1\n                        0   0  0\n                        1   0  1\n\nExample: (0+1)x(1+1)=1x1=1\n\nHope this helps
749939
Nobody can help you because it's false.\n\nLet a1 = 1, a2 = 4, b1 = 2, b2 = 3, and n = 7\nThen a1 + a2 = 5, b1 + b2 = 5\nBut a1*a2 = 4, while b1*b2 = 6.\n\nYou are talking about modular arithmetic, right?
441742
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
771191
A possible proof:\n\nIt is easy to verify that AUB = A U (B-A). Since B is infinite and B is a subset of AUB, AUB is also infinite. However, A is finite, and if B-A is also finite then so is their union, which would lead to a contradiction (namely, AUB being finite and infinite at the same time), thus, B-A must be infinite.\n\nBy the way, after reading your previous question, I assumed that A and B were meant to be sets, not numbers.
38366
I think this is the answer\n\na^2-2ab-b^2\n-a^2(b-b^2)\n= -a^2
460347
When you have groups of terms like your problem you factor out the things that each group of terms share. \n\nFor your first part: (a^2c^2 + b^2c^2 ) the terms share a c^2 so you factor that out. \n\nyou have c^2 (a^2 + b^2)\n\nThe second terms (-4a^2d^2-4b^2d^2) share 4d^2 so you factor that out. \n\nSo you're left with c^2(a^2+b^2) - 4d^2 (a^2 + b^2). \n\nWhen you've done this part you can actually go farther because if you look at it both sections share (a^2 + b^2). \n\nso you can simplify the whole problem with \n\nc^2 - 4d^2 (a^2 + b^2).
359130
i) x is an element of B (or a member of B)\nii) A is a subset of B \niii) C is a subset of D\niv) The set A has exactly 2,4, and 6 as members.
1079282
If by a(i) and b(i) you mean "a times i" and "b times i", then:\n\nFor i = 0:\n\na*i / (b*i + 1) = 0 / 1 = 0\n\nSo you have:\n\nPo = 1 / (1 + 0 + (0 * (a / (b + 1))) + (0 * (a / (b + 1)) * (2a / (2b + 1))) + ...)\n\nPo = 1 / 1\n\nPo = 1\n\nIf by a(i) and b(i) you mean a and b are infinite sequences (then presumably a / b means "for all i, a(i) / b(1) < 1"), well then that is something else entirely.\n\nThis reminds me of the problem to expand and simplify the product of twenty-six binomial terms:\n\n(x - a) (x - b) (x - c) ... (x - z)
961456
assume b=boys ; g=gals\n\ng = (2/5)b\n \nb+(2/5)b =35\n(7/5)b=35\nb=25\n\nso , g=10 ; b=25
875476
(a + b - c)(a + b + c)\n((a+b)+c)((a+b)-c)\n(a+b)^2-c^2\na^2+2ab+b^2-c^2
786972
ABCDEFGHJ = 256973148\n\n\nFirstly...\nC - B = 1.......C = B + 1\nH - G = 3.......H = G + 3\n\nThe equation can be simplified as:\n\nA * [ (2B + 1)^(D - E) - F^(G(G+3)) ] / J = 10\n\n1. The difference between the numbers within the square brackets cannot be more than 100 since dividing by J, which is a single digit will not give 10 (assuming that A was 1). Therefore both numbers are supposed to be weighted similarly.\n\n2. Consider the value of 'A'. If the difference were even, then it must be multiplied by 5 before dividing in order to get 10. If it were 5, then 'A/J' has to scale it down accordingly. If it were odd, there are no possible values 'A' can take. Thus the difference can be either be even or a multiple of 5.\n\n3. I assume F is an odd number! The reasoning behind this is that the difference between the 2 numbers can be even [Exception is if the difference were an odd multiple of 5. If it doesn't work then consider the other alternative]. Since (2B + 1) raised to any power is odd, then F must be odd to obtain an even difference.\n\n4. Look at all powers that have values in the range of 1 - 256, since I assume that the values are small and close in value. I consider 1^2 to 15^2, 1^3 to 6^3, 1^4 to 4^4 and 1^5 to 3^5. Look at those with odd bases [also easier to start off with since trial and error must take place now].\n\n5. I obtain and test 11^2 - 3^4 = 121 - 81 = 40. From this, (2B+1) = 11-->B = 5, D-E=2, F = 3 and G^2 + 3G = 4 --> G=1\n\nusing 40...A = 2 and J = 8 in order to get a solution of 10\n\nIf G = 1 then H = 4 from original equation\nIf B = 5 then C = 6 from original equation\n\nThe remaining numbers are 7 and 9, which allows D = 9 and E = 7 to satisfy the final equation of unknowns [D-E=2]\n\nI hope you understand and it helps!
136789
That's an extremely good question. \n\nThe answer is simple. Before that, i will explain what your question means to those who don't get. Well, it means that we must get 24 using 1,3,4 and 6 any number of times however involving operations +,-,* and / only once. \n\n  Then, the answer is very simple :\n\n  {[(6*6)/(3-1)]+[6]}  =  24\n\n---------------------------------------------------------\nWOKRING :\n\n {[(6*6)/(3-1)]+[6]}   =  {[(36)/(2)]+[6]}\n                       =  18 + 6\n                       =  24\n\n----------------------------------------------------------\n\nI hope you post more questions like these. They evoke a lot of thought. I hope the solution serves th purpose.
402319
ya it is a^2 +b ^2 +2abcos(c)
729005
a.xa+xb\nb.21x^3+24x^2y\nc.3x(8x+5)+2(8x+5)\n=24x^2+15x+16x+10\n=24x^2+31x+10\nd.4x(2x+1)-6(2x+1)\n=8x^2+4x-12x-6\n=8x^2-8x-6\ne.8x(5x^2+7x+3)+2(5x^2+7x+3)\n=40x^3+56x^2+24x+10x^2+14x+6\n=40x^3+66x^2+38x+6\nf.x^2-1\ng.x(x^2+2x+4)-2(x^2+2x+4)\n=x^3+2x^2+4x-2x^2-4x-8\n=x^3-8