Can any one prove 2+2=5?
927994
* Start with the identity\n          − 20 = − 20\n\n    * Express both sides in slightly different, yet equivalent ways\n          25 − 45 = 16 − 36\n\n    * Factor both sides\n          5^2 - 5 &#92;times 9 = 4^2 - 4 &#92;times 9\n\n    * Add the same thing to both sides\n          5^2 - 5 &#92;times 9 + &#92;frac{81}{4} = 4^2 - 4 &#92;times 9 + &#92;frac{81}{4}\n\n    * Now factor both sides again\n          &#92;left(5 - &#92;frac{9}{2}&#92;right)^2 = &#92;left(4 - &#92;frac{9}{2}&#92;right)^2\n\n    * Take the square root of both sides\n          5 - &#92;frac{9}{2} = 4 - &#92;frac{9}{2}\n\n    * Cancel the common term\n          5 = 4\n\n    *Proof \n           2+2=5
389436
5x4+4+4+5+4/2\n=20+4+4+5+2\n=35
537227
you do what is in the parenthesis first.   \n5(5+2)\ntake the 5+2 first which equals 7.  Then multiply that by 5 which is 35.\nNext take each side of the plus sign 5(5) and 5(2) and do them seperatly.  5(5)=25 and 5(2)=10  then put the plus in and 25+10= 35   Your done  35=35
590233
Hint:\nDo a simple substitution u = x-5\nWhich means x = u + 5\n\nYou should get\n\n(2/5)(x-5)^(5/2) + 2(x-5)^(3/2) + C
599876
no can do it equals 2
703605
n^5 - n\n= n(n^4-1)\n= n(n^2-1)(n^2+1)\n= n(n-1)(n+1)(n^2+1)\n\nWe must prove that one of these four factors is even; that one is a multiple of 3; and one is a multiple of 5.  The product will be multiple of 2*3*5 = 60.\n\nThree of the factors are consecutive numbers (n-1, n, n+1) as factors.  One of these must be a multiple of 3 and one of them must be even.  Therefore the result is a multiple of 6.\n\nIf one of n-1, n or n+1 is a multiple of 5, the proof is over.  Assume then that none of them is.  Then either n=5i+2 or n=5i+3 for some integer i.  The factor (n^2 + 1) then becomes\n(5i+2)^2 + 1\n= 25i^2 + 20i + 4 +1\n= 25i^2 +20i + 5\n= 5 * (5i^2 + 4i + 1)\n\nor\n\n(5i+3)^2 + 1\n= 25i^2 + 30i + 9 +1\n= 25i^2 +30i + 10\n= 5 * (5i^2 + 6i + 2)\n\nBoth of these are multiples of 5 - therefore, if none of (n-1), n or (n+1) is a multiple of 5, then n^2+1 is.\n\nQED.\n\nJust saw the hint to go for induction, let's try it!\n\nfor n=2 we have n^5 - n = 32-2 = 30 OK.\n\nAssume it works for n, i.e. n^5-n is a multiple of 30, then\n(n+1)^5 - (n+1)\n= n^5 + 5n^4 + 10 n^3 + 10 n^2 + 5n + 1 - n + 1\n= (n^5 - n) + 5n(n^3 + 2n^2 + 2n + 1)\n= (multiple of 30) + 5n(n+1)(n^2 + n + 1).\nn(n+1) is certainly even.  Is it a multiple of 3? if not this means n=3i+1 for some i.  Therefore the third factor equals (3i+1)^2 + 3i + 1 + 1 = 9i^2 + 9i + 3 is a multiple of 3.  So that big ugly thing is a multiple of 5, 2 and 3 - and thus of 30.\n\nWorks too for negatives too, since (-n)^5 - (-n) = - (n^5-n).\n\nEDIT: Want a third proof???\n\nIf n= r mod a (meaning r is the remainder of n/a) then n^5-n = (r^5 -r) mod a\n\nSetting a=2:\nIf r=0 then n^5-n = 0^5-0 mod 2 = 0 mod 2\nIf r=1 then n^5-n = 1^5-1 mod 2 = 0 mod 2\nHence n^5-n is a multiple of 2 in all cases\n\nFor a=3\nIf r=0 then n^5-n = 0^5-0 mod 3 = 0 mod 3\nIf r=1 then n^5-n = 1^5-1 mod 3 = 0 mod 3\nIf r=2 then n^5-n = 2^5-2 mod 3 = 30 mod 3 = 0 mod 3\nHence n^5-n is a multiple of 3 in all cases\n\nFor a=5\nIf r=0 then n^5-n = 0^5-0 mod 5 = 0 mod 5\nIf r=1 then n^5-n = 1^5-1 mod 5 = 0 mod 5\nIf r=2 then n^5-n = 2^5-2 mod 5 = 30 mod 5 = 0 mod 5\nIf r=3 then n^5-n = 3^5-3 mod 5 = 240 mod 5 = 0 mod 5\nIf r=4 then n^5-n = 4^5-4 mod 5 = 1020 mod 5 = 0 mod 5\nHence n^5-n is a multiple of 5 in all cases.\n\nBeing a multiple of 2, 3 and 5, n^5-n is therefore a multiple of 30.
958973
y=5x+2\ny=5(2)+2\ny=10+2\ny=12
19885
(5) x (2x^2-3x+5) I think
1062229
Technically there are only 4 answers that will work, and they all rely on parentheses to modify the natural precedence rules:\n\n(5+5/5)*5\n(5/5+5)*5\n5*(5+5/5)\n5*(5/5+5)\n\nAs you can see they are all variations on (5 + 1) * 5, where you use 5/5 = 1.  So there is really just one answer.\n\nYou can easily solve problems like this yourself using the link below...\n\nFor example, can you make 24 using 3, 3, 8, 8?  Try it before you cheat... also, note that the solver allows % (mod) as an operator, so you may have to filter out those results.
838051
5(x – 2)^2 = 3\n(x – 2)^2 =3/5\nx -2 =+- sqrt(3/5)\nx=2+- sqrt(3/5)
114816
-3+5=5-3y+4y       \n   2=5+y\n  2-5=y \n   -3=y\n because...          2=5-3(-3)+4(-3)\n                     2=5+9-12 \n                     2= 14-12         \n                     2=2
771836
Simple substitution.\n\n5(3) - 2(2) + 4(4)\n15 - 4 + 16\n27
18182
2x^2 + 5x + 20x + 50\n2x^2 + 25x + 50
988599
2^1999 x 5^2000 = \n2^1999 x 5^1999 x 5 =\n(2 x 5)^1999 x 5 =\n5 x 10^1999\nThis is 5 followed by 1999 zeroes.\nThis an exact aresult.
811163
(2 * 20) - 35 + 5 = 10
168603
easy if that is the question\n2x+3x=5\nsince they have the same variable\nadd them \n5x=5\nthen divide by 5(no. to reduce the variable with 1 as its coefficient) \nx=1\n\nlonger method\n\n2(x)+3(x)=5\nby distributive property we know that (2+3)x=2(x)+3(x)\nthen: \n(2+3)x=5\n(5)x=5 divide by 5 \nx=1
347304
I'll show you a slightly different way.\nI am assuming you know basic algebra.\n\nSee,       5b < -10\n      5b + 10 < -10 + 10    ( by adding 10 both sides )\n     5b + 10  < 0\n    5*( b + 2 ) < 0         ( Factorize out 5 )   \n\nNow, just think for a second. The two numbers "5" and \n"(b +2) " when multiplied, we require a negative ( i.e < 0 )\nresult. Since 5 is positive, we must have that ( b + 2 ) must be negative. Why? because by convention only a positive number multiplied by a negative number yields a negative number and vice versa.\n\nThus, ( b + 2 ) negative means that \n      b + 2 < 0  \n   b + 2 -2 < 0 - 2   ( by substracting 2 both sides )\n          b < -2