can anybody calculate (15)^12 mod 17 for me? ..please((( is the answer 13))?
606056
Great question!  I learned something answering this one.  Click on start- all programs- accessories - calculator.  Then get the scientific calculator. (View-Scientific) This will allow you to evaluate 15^12 (without having the answer rounded), hit mod, and enter 17.   The answer is 16.
981364
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!
528657
x^4 - 8x^2 + 15 = 0\n(x^2 - 5)(x^2 - 3) = 0\nSo x^2 = 5 or x^2 = 3.\nSo x = sqrt(5) or x = -sqrt(5) or x = sqrt(3) or x = -sqrt(3).\n\nTo purdue_engineer:\nIf it makes you feel any better, I've been screwed a bit lately.  One person copied my answer and he ended up getting best answer.  Another person was wrong but bashed my answer and so she got best answer instead of me!\n\nBut anyway, sorry!  I'm a teacher, and I just can't help myself.
167162
I do agree with Anniemc46.\nIf the denominator is 1+(tanhx)^2 then it is easy to do.\n\nI have done and uploaded to my web site.\nPlease check the url I provided in the sourse.\n\nThanking U,\nThuvalpakshi.
528929
i dont have a calculator, but if you do the equation is sin12 then multiply by 15 then divide by 12 for feet
271312
Let the two numbers be x,y \n\nTwo numbers multiplied together is - 15\n==> xy = -15 --> (1)\n\nTwo numbers added together is 2\n==> x + y = 2 --> (2)\n\nUsing (2)\nx + y = 2 ==> x = 2 - y --> (2A)\n\nSubstituting the value of "x" from (2A) in (1) we get\n\n(2 - y)y = -15\n==> 2y - y^2 = - 15\n==> y^2 - 2y - 15 = 0\n==> y^2 - 5y + 3y - 15 = 0  \n==> y(y - 5) + 3(y - 5) = 0  \n==> (y + 3)(y - 5)= 0 \n==> (y + 3) = 0 (Or) (y - 5)= 0 \n==> y = - 3 (Or) y = 5 \n\n-------------------------------------------------------\n[y^2 - 2y - 15 = 0\n(Coefficient of y^2)*(Constant) = (1)(-15) = -15\n\nFactorize -15 in such a way that the sum of the factors is equal to the coefficient of y\n\n-15 = - 1 * + 15 [(-1) + (+ 15) = +14] Na\n-15 = + 1 * - 15 [(+1) + (- 15) = -14] Na\n-15 = + 3 * - 5 [(+3) + (- 5) = -2] Ha ... ha ... ha\n-15 = - 3 * + 5 [(-3) + (+ 5) = +2] Na\n--------------------------------------------------\n(y + 3)(y - 5)= 0 \nProduct of two factors is zero\n==> \nOne of the factors should be equal to zero for the product to be equal to zero.\n==> (y + 3) = 0 (Or) (y - 5)= 0 \n-------------------------------------------------------\n\nNote: \n(2) and (2A) are two forms of the same equation.  Therefore, when the value of "x" as in (2A) should not be substituted in (2) as it amounts to using the same equation for solving for values of two variables which would not give you a result.... try to confirm this..\n\n\nhttp://www.schoolingkids.com/\n\nhttp://www.futureaccountant.com/
559085
Let me check... I need to see the full equation by hovering... give me a sec.\n\nOkay you have:\n= (5a-12)/(a^2-8a+15) - (3a-2)/(a^2-8a+15)\n\nFortunately the denominators are the same, so you just have to subtract the numerators and keep the same denominator:\n= [ (5a-12) - (3a - 2) ] / (a^2-8a+15)\n\nTurn -(3a-2) into -3a + 2:\n= ( 5a - 12 - 3a + 2 ) / (a^2-8a+15)\n\nGroup the like terms:\n= ( 5a - 3a - 12 + 2 ) / (a^2-8a+15)\n\nSimplify:\n= ( 2a - 10 ) / (a^2 -8a + 15)\n\nFactor out a 2 in the front\n= 2 ( a - 5 ) / (a^2 -8a + 15)\n\nFactor the denominator:\n= 2 ( a - 5 ) / [ (a - 3)(a - 5) ]\n\nCancel the ( a -5 ) from numerator and denominator:\n= 2 / (a - 3)\n\nThis is different from your answer.  I'm not sure where you went off, but hopefully this explanation will help.  If you need more details, add a comment to your question and I'll check back.\n\nThe answer is: 2 / (a - 3)
822221
Using mods...\n\nn^3 + n - 4 = 0 mod 7\n\nn^3 + n = 4 mod 7\n\nn(n^2+1) = 4 mod 7\n\nIf n=0 mod 7 then n(n^2+1) = 0 mod 7, doesn't work.\nIf n=1 mod 7 then n(n^2+1) = 1(1^2+1) mod 7 = 2 mod 7, doesn't work.\nIf n=2 mod 7 then n(n^2+1) = 2(4+1) mod 7 = 3 mod 7, doesn't work.\nIf n=3 mod 7 then n(n^2+1) = 3(9+1) mod 7 = 2 mod 7, doesn't work.\nIf n=4 mod 7 then n(n^2+1) = 4(16+1) mod 7 = 5 mod 7, doesn't work.\nIf n=5 mod 7 then n(n^2+1) = 5(25+1) mod 7 = 130 mod 7 = 4 mod 7 - YESSSS!\nIf n=6 mod 7 then n(n^2+1) = 6(36+1) mod 7 = 222 mod 7 = 5 mod 7, doesn't work.\n\nSo, n must be equal to 7m+5 for some integer m.  This gives us:\nn^3 + n - 4 =\n(7m+5)^3 + 7m+5 - 4 =\n7^3m^3 + 3*7^2m^2*5 + 3*7m*5^2 + 5^3 + 7m + 1 = \na whole bunch of multiples of 7 + 5^3 + 1 = \nmultiples of 7, plus 126 =\nmultiples of 7, plus 18*7.
920734
2xy(-15x^2y^4)+8x^2)-5/-4xy^3\n(-1/2)[-15x^2y^4+8x^2-5)/y^2\n(-1/2y^2)[-15x^2y^4+8x^2-5]
55351
rt11guru is correct up to the point where he says:\n\nX = ((2x2)^15)x((3x3)^6 x 3)x((5x5)^2 x 5)x((7x7)x7)\n\nThe number of possibilities is then given by:\n\nN = (16 * 7 * 3 * 2) - 1 = 671\n\nYou need to subtract 1 to account for the case where the divisor has no prime factors (and so equals 1).  If 1 counts as a solution (perfect square and divisor) the number of solutions is 672.
712544
ok so you know 15 can only be broken down to 5*3, -5*-3, 15*1 and -15*-1\ntherefore when you break it up\n(x+3)(x+5) using FOIL (I assume you know how to do that) k would equal 8. For (x-3)(x-5) k would be -8. Do the same for the set of (x+15)(x+1) and (x-15)(x-1)
1147832
The previous poster has the right method, but use 11 as your prime. You then need a divisor of 2*7*13 which is 1 modulo 11. This shows that there is only one 11-Sylow subgroup, so it is normal.
814352
(12x^2 +16x) - (12x-7)- (4x-5)\n12x^2 +16x -12x +7 -4x +5\n\n12x^2+12\n\nthat is what I got....\nI think you might be able to put 12 on the other side of the = sign and turn it into 12x^2 = -12 and then since both sides have 12 in common you can divide both sides by 12 and you will have x^2 = -1 then you can add 1 to both sides and you will have x^2 +1 but I think I'm getting off track by that point....it's been a while since I had to do this stuff.
631080
You can noy factor it out... it makes it easier to figure out when you put it into lowest terms... so it would be \n\n   x^2+7x-25..... so you cant factor it.
484601
The previous poster didn't see the whole problem. This is a question about modular arithmetic, not standard algebra. The equations are to be done modulo n and the solutions are to be integers.\n\nBy adding 3 times the first to the second, we need 5x=8. By subtracting twice the first from the second, we get -5y=-2, so 5y=2. So all that is required is that 5 be invertible modulo n, i.e., that 5 does not divide n. For example, if n=4, x=0 and y=2 works.
203508
Two.\n\nYou can rewrite it as x^2 = 15 or x = sqrt(15)\n\nx solves to (rounded) 3.87 or -3.87
746284
Well, it is 96750 ^ 2 = (430 * 15^2)^2.  Not sure what else you might be looking for.