I need information about the calculations of Fermat Number 2^(2^m)+ 1 please help me, urgent?
145253
Take a look at this link. You should find all you need about Fermat Numbers there.
920470
The answer is:  4.47, and that's all there is to it!\n\nOk Vanessa, all you have to do is pull up your calculator (Start>Accessories).\nI am not suggesting you don't work it out in your head first: it's just basic. Check the answer on the calc, or, if you get stuck....go back.... to POWERS OF HIGH HOLY ROOTISM   SQUARE YOU KNOW, 1x1+(2x2)+2+(10-5)/1=       Simple:\n1+4+2+5/1=12
858963
The answer is actually much more complex than one would think! The correct answer though is ...\n\n((2^n)(n-.5)!)/(2(.5!))\n\nThat looks kind of messy, so in English that is "2 to the n times n minus 1 half factorial over 2 times 1 half factorial."\n\nPlugging in 1,3,5,7 for n on a graphing calculator, we get 1,3,15,105, which are the correct first four terms in the sequence.
134270
m2-m1 = -2.50 log(B2/B1)\nthis is the formula - \n\n?????
21504
Its 2^0+2^1+2^2+2^3+....+2^63.\nThe sum of powers of 2 till 64. \n\nthe speed of a 64bit processor?
141990
(assuming the second 2 is an exponent) 2*2*2*7*5*g*h*h \nyou really should learn to do this if you are ever going to get through school though
830444
d.) 2m^2(6m^6+3m^4+7)
386376
My computer gives me this 4 real results, there are two more Imaginary results.\n\n1) -1\n2) 1.8698\n3) -0.260\n4) -2.688\n\nImaginary roots are\n\n5) 1.039 + j1.40\n6) 1.039 - j1.40
640891
I'm guessing you meant 17x^3 - 10x^2 - 6x +4 and 2x^2- 6x +2\nIf that's right, you just add like powers of x and their coefficients to get 17x^3 - 8x^2 -12x +6.
1094108
Use a calculator and raise it to the 1/3 power.\nDon't try to do it by hand...
899137
let us evaluate this upto n terms\n\nwe get 2^(n) -1\n\n\nnow this to be divisible by 9\n\n2^n mod 9 = 1\n\n2^1 mod 9 = 2\n2^2 mod 9 = 4\n2^ 3 mod 9 = 8 or -1 \n\nso 2^6 mod 9 = 1\n\nso (2^6n)\nso 2^6-1 mod 9 = 0\n\nso 6k  terms sum is dvisible by 0\n\nso n = 6, 12 , 18 so on of the form 6k
386814
They are correct.  Observe the order of operations or PEMDAS\n\nParenthesis\nExponents\nMultiplication-Division\nAddition-Subtraction\n\nso first multiply   -89 * 658 =   -58562\nThen add 2   =                    -58560\nThen add 6   =                    -58554\nThen subtract 89 (or add -89) =   -58643
1053403
Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
962594
you sure you have this problem right ???????\n\nhere's what i came up with\n\nm + ( m + 3 ) = ( m - 2 ) - ( m - 1 ) \nthis is wrong \n\nm + ( m + 3 ) = 2m + 3\n\n( m - 2 ) - ( m - 1 ) \n\n= ( m - 2 ) ( m + 12 ) \n\n= m^2  + 1m - 2m  + 2 \n\n= m^2 - 1m + 2
834003
You have:\n\n2(-3^1 - 1) + 2(-3^1 - 1) + 2(-3^1 - 1) + 2(-3^1 - 1) + 2(-3^1 - 1) + 2(-3^1 - 1)\n\n3^1 = 3\n3^2 = 9\n3^3 = 27\n3^4 = 81\n3^5 = 243\n3^6 = 729\n\n2(-3 - 1) + 2(-9 - 1) + 2(-27 - 1) + 2(-81 - 1) + 2(-243 - 1) + 2(-729 - 1)\n\nOk....I assume you can add and subtract and all that stuff. You can take it from here.
729558
I worked this problem. I did get both sides to zero. I don't even know how to begin explaining it.\n\nFor the sake of typing this with some readability, let (k+1) = b and (k+2) = c.\n\nFirst you subtract (2^c)/(k+3) from both sides to get 0 on one side. Then to get a common denominator, you multiply top and bottom of the first term by (k+3) and top and bottom of the third term by (k+2). You get:\n\n[ (k+3)(2^b) + (k+1)(2^b) - (k+2)(2^c) ] / [ (k+2)(k+3) ] = 0\n\nThen you distribute out the terms:\n\n[ k(2^b) + 3(2^b) + k(2^b) + (2^b) - k(2^c) - 2(2^c) ] / [ (k+2)(k+3) ]\n\nThen you factor out 2^b from the first four terms:\n\n[ (2^b)(k + 3 + k + 1) - k(2^c) - 2(2^c) ] / [ (k+2)(k+3) ]\n\nThen you combine like terms on the left four. On the right two, you reduce the exponent by one and multiply the term by two. Since c is one more than b, this turns (2^c) into (2^b). You can do this since it's an exponent on two:\n\n[ (2k + 4)(2^b) - 2k(2^b) - 4(2^b) ] / [ (k+2)(k+3) ]\n\nYou can factor 2^b out of the right two terms to get:\n\n[ (2k + 4)(2^b) - (2k + 4)(2^b) ] / [ (k+2)(k+3) ] = 0\n\nThe two terms in the numerator cancel out:\n\n0 / [ (k+2)(k+3) ] = 0\n\nAnd:\n\n0 = 0\n\nI must admit that I liked this problem. I don't do math for fun, but every once in a while I like a challenge.\n\nI hope this helped.
904822
Hello Dear Blue \n\n2x^3+8x^2 - 5x + 5 ÷ x-2\n\n▪ Step1\n( 2x^2 ) ( x-2 ) = 2x^3 - 4x^2 ; { let Q1 = 2x^2 }\n2x^3+8x^2 - 5x + 5 - ( 2x^3 - 4x^2 ) = \n2x^3+8x^2 - 5x + 5 - 2x^3 + 4x^2 = +12x^2 - 5x + 5 {remainder} \n\n▪ Step 2\n+12x^2 - 5x + 5 ÷ x - 2 \n(x - 2) (12x) = 12x^2 - 24x ; { let Q2 = +12x }\n+12x^2 - 5x + 5 - ( 12x^2 - 24x  ) =\n+12x^2 - 5x + 5 - 12x^2 + 24x = +19x +5 {remainder}\n\n▪ Step 3 \n19x +5 ÷ x -2\n(x -2) (+19) = +19x - 38; { let Q3 = +19}\n19x +5  - (+19x - 38) =\n19x +5  -19x + 38 = +43 {remainder}\n\n▪ Step 4\nSum Q1, Q2, Q3\nQ1 + Q2 + Q3\n2x^2 +12x +19\n\nSo;\n(2x^3+8x^2-5x+5) ÷ (x - 2)    = >\n\nQ(x) OR quotient = 2x^2 +12x +19 \nR(x) OR Remainder is + 43 \n\nGood Luck dear.