true or false if p, q, r, and s, are all true?
123509
not p implies q is TRUE because q is true is enough for  p implies q to be true ( p is false is also enough and in your case not p is always false).\nr impliies negation of s means true implies false and that is impossible- the statement is FALSE\nP.S. true and false is false
90519
1. p'q\n\n2. p'\n\n3. pq' + p'\n\nI think i got the correct ones !
137705
Well, the basic problem you've run into is that the conjecture isn't true.  If a = 6, b = 3, and c = 9, then a*b = 18, and 18/9 = 2, but neither 6 nor 3 is divisible by 9.\n\nTo prove that something is true you must prove it is true for all possible values of the variables, in this case a, b, and c.  However, to prove something is false, you need only supply a single counter-example.  Thus the above example disproves the conjecture that for all integers a, b, and c if a*b is divisible by c then a or b is also divisible by c.\n\nNote that a ≠ cm and b ≠ cn does not imply that ab ≠ c(cmn).  For example, if a = 6, b = 8, c = 2, m = 4, and n = 3, then cm = 2*3 = 8 ≠6 so cm ≠ a, and cn = 2 * 3 = 6 ≠ 8 so cn ≠ b.  However, (cm)(cn) = 2 * 4 * 2 * 3 = 48, and a * b = 6 * 8 = 48, so ab = (cm)(cn).  This proves that you cannot multiply not-equalities (i.e. statements with "doesn't equal).\n\nAlso, concluding that if c does not divide a or b implies that c does not divide ab is not the same as concluding that if c does divide ab it must also divide a or b.  This logical fallacy is equivalent to the statement that for all plants, not a tree implies not a rock, so being a rock implies being a tree.  The first statement is true: all plants that are not trees are also not rocks.  However, the second statement is clearly false, as no rocks are also trees.
559127
1. yes.\n\n2. one\n\n3. 2 equal segments (this one's a theorem)
417388
Just expand the right hand side:\n2x^2 - 12x + p = q(x^2 - 2xr + r^2) + 10\n2x^2 - 12x + p = qx^2 - 2qrx + qr^2 + 10.\nNow, for that to be always true, the coefficients have to match up.\nSo q = 2, 2qr = 12, qr^2 + 10 = p.\nSo q = 2, r = 3, p = 28.
397113
a) true\nb) false\nc) true\nd) true\ne) false
450855
How about setting it up first in a truth table? \n\nIf you do, you'll see there's a counterexample. So there's no way to logically derive A from assumptions 1 and 2.
444172
1.  C  Correct   \n  2.  B  Correct   \n  3.  A  Correct   \n  4.  D  Correct   \n  5.  C  Correct   \n  6.  B  A  PAGE 17-23  \n  7.  C  Correct   \n  8.  A  Correct   \n  9.  B  Correct   \n  10.  C  Correct   \n  11.  D  Correct   \n  12.  A  Correct   \n  13.  D  Correct   \n  14.  A  Correct   \n  15.  C  Correct   \n  16.  D  Correct   \n  17.  B  C  PAGE 1-23  \n  18.  A  B  PAGE 17-23  \n  19.  B  Correct   \n  20.  A  Correct   \n  21.  C  Correct   \n  22.  C  Correct   \n  23.  B  Correct   \n  24.  C  Correct   \n  25.  B  Correct
133898
1 is answer A, 2 is not answer B, C or D
263929
A permutation. P(n,r)=(n!)/((n-r)!). C(n,r)=(n!)/(r!(n-r)!)\nor (P(n,r))/(r!). x/something is smaller than x itself,(unless there's negatives or decimals, but we don't have to worry about that because r is positive) so the permutation is biggger than the combination.
45624
B.\n\nAll of the rest are false.
291631
b)\nbecause when you plug each value in for x you get a true equation
646227
B, C, and E are all true.
455413
P - - - - Q - - - - R - S - - - - - - - - T\n\nPR = RT (midpoint)\nPQ = QR (midpoint)\n\nQS = 10, PS = 19\n\nPQ = PS - QS\nSo PQ = 19 - 10 = 9\nQR = 9\nPR = PQ + QR = 18\nRT = PR = 18\n\nSo the complete line PT = PR + RT = 18 + 18 = 36\n\nST = 36 - PS\nST = 36 - 19\nST = 17
37830
i believe c is the answer - not 100%
210680
64) b\n\n65)b\n\n66)c\n\n90 percent sure
41456
There is many answers\na=1 b=2 c=3 is true\nalso a=b=c=sqrt(3) is true but not integers\n\nHow we can get them? \na,b,c must satisfy abc=a+b+c\n\nWhy?\natan(a) + atan(b) + atan (c) =pi\nlet atan(b)+atan(c) = y     (eq1)\n\natan(a) + y =pi\ntaking tan to the 2 sides\n\ntan [atan(a) + y] =tan(pi) = 0\n[tan(atan(a)) +tan (y) ]/[1- tan(atan(a))*tan (y) ]=0\n\nso a + tan(y)=0                       (eqn2)\nbut tan(y) = tan (atan (b) +atan(c))    from eqn1\n=(b+c)/(1-bc)\n\nsubstitue in eqn2\na+(b+c)/(1-bc)=0\nmultiply by (1-bc)\na(1-bc)+b+c=0\na+b+c-abc=0\n***********\n*a+b+c=abc*\n***********\n\na=1 b=2 c=3 is true since 1+2+3 =1*2*3 =6\na=b=c=sqrt(3) is true since 3sqrt(3)=srqt(3)^3