Answer to this equation: 9633 over 1- (1 over 1.073) power of 7?
828957
why can't you shove this direcly in a calculator?\n\nif "9633/[1-(1/1.073)^7]" is the question then the answer is 24742.09183\n\nif "[9633/(1-(1/1.073))]^7" is the question then the answer is 0.1140956153e37\n\nI think the first one is what ur looking for
647999
21, 4^0 = 1, not 0\n\n0.32, 0.32, 1/3
386770
the 5th roots of the cubes of the square root of the seventh power of (3 - i) is equal to:\nR = {{sqrt[(3 - i)^7]}^3}^(1/5)\nR = {(3 - i)^7}^(1/10)\n\nexpanding (3 - i)^7:\nby Euler formula: e^(ix) = cos(x) + i sin(x)\n3 = 3[cos(2*pi) + i sin(2*pi)] = 3e^(2*pi*i)\ni = [cos(pi/2) + i sin(pi/2)] = e^(pi/2)\n\n(3 - i)^7 = [3e^(2*pi*i) - e^(pi/2)]^7\n(3 - i)^7 = [e^(pi/2)]^7 * [3e^(4*pi) - 1]^7\n(3 - i)^7 = [3e^(4*pi) - 1]^7 * [cos(7*pi/2) + i sin(7*pi/2)]\n(3 - i)^7 = -i * [3e^(4*pi) - 1]^7\n\nthen\nR = {-i * [3e^(4*pi) - 1]^7}^(1/10)\nR = { [3e^(4*pi) - 1]^7 * [cos(pi/2) - i sin(pi/2)] }^(1/10)\n\nby De Moivre's formula if z = A[cos(x) + i sin(x)] then\nz^(1/n) = A^(1/n) * { cos[(x + 2*k*pi)/n] + i sin [(x + 2*k*pi)/n] };   where k varies from 0 to n-1 to give n-th roots\n\nR = [3e^(4*pi) - 1]^7]^(1/10) * { cos[(pi/2 + 2*k*pi)/10] + i sin [(pi/2 + 2*k*pi)/10] }; where k varies from o to 9
26076
=7(1-x)+2x(1-x)\n=7-7x+2x-2x^2\n=7-5x-2x^2
194129
1-.42^7=0.99769460666752\n\nboris had the right idea but made an arithmetic error.
16160
its 1, 2, sqrt (3). i dont know what u mean by 6,6 over 3,12
22440
Well, the power is of course 99, but if you are looking for the coefficient, it is -5050.  Since terms that are constant multiples of X^99 must be multiples of all X's except for one, one simply adds the second terms in each binomial {-1,-2,...,-100}.  The sum is -5050
858645
Find the square root of 7 squared minus 3 squared.\nI got 6.3 for your #16 question
685997
u should write it like\n7.52*10^-4
209966
Set up an equation:\n\n72 = .6 * X\n\n72/.6 = X\n\nX = 120
710821
I'm sure it isn't correct, though I'm not totally sure what you've done.  It looks to me as if you've raised both sides of \naⁿ + bⁿ= cⁿ\nto the power 1/n.  The trouble is, when you do this the left side is not equal to\n(aⁿ)^(1/n) + (bⁿ)^(1/n)\n\nYou can see this if you consider\n(5^3 + 2^3)^(1/3)= 133^(1/3), which is just slightly over 5, definitely not equal to\n(5^3)^(1/3) + (2^3)^(1/3)\n= 5 + 2\n= 7
555506
1.54 x 10^7\n\njust ignore the brackets since everything is multiplication.\nmultiply 77 x .02 = 1.54\nand add exponents becuase its multiplication 4+3 = 7\n\ntherefore 1.54x10^7
1133641
Divide 204 by 7 and get the answer... use a calculator. It's greater than zero.
164974
The non-brute-force solution is to notice that:\n\n   a b c a b (c+1) = (abc) * 1001 + 1\n\nSo the number needs to be a square which is 1 more than a multiple of 1001.\n\nIn number theory, we write this as:\n\n    z^2 == 1 (mod 1001)\n\nNow, we factor 1001 into primes as 1001 = 13 * 11 * 7\nand we can say that the above equation is the same as the three equations:\n\n    z^2 == 1 (mod 7)\n    z^2 == 1 (mod 11)\n    z^2 == 1 (mod 13)\n\nNow, for prime numbers, z^2 == 1 (mod p) can be solved only for z = +/- 1 (mod p)\n\nSo we get eight possible triples of equations:\n\n    z == (1,1,1)    (mod 7,11,13)\n    z == (1,1,-1)   (mod 7,11,13)\n    z == (1,-1,1)   (mod 7,11,13)\n    z == (-1,1,1)   (mod 7,11,13)\n    z == (1,-1,-1)  (mod 7,11,13)\n    z == (-1,1,-1)  (mod 7,11,13)\n    z == (-1,-1,1)  (mod 7,11,13)\n    z == (-1,-1,-1) (mod 7,11,13)\n\nNow, the equation:\n\n    z == (p,q,r)   (mod 7,11,13)\n\nis solved via the Chinese remainder theorem, and you get the following values for z:\n\n     z = 1,     z^2 = 000001\n     z = 155,   z^2 = 024025\n     z = 274,   z^2 = 075076\n     z = 428,   z^2 = 183184\n     z = 573,   z^2 = 328329\n     z = 727,   z^2 = 528529\n     z = 846,   z^2 = 715716\n     z = 1000,  z^2 = 1000000 (obviously, not a solution)\n\n[ You can think of the 1,000,000 case as when a=b=c=9. ]
178588
Rearrange as (-7)(-11) m^(-6) m^(-4) u^(-4). Then combine the two coefficients into 77, and add exponents for the same base m, to get m^(-10). The final result is \n77 m^(-10) u^(-4). By the way, you should ask your teacher or consult the book to see if exponents should be made positive. In that case write the answer as 77 / [m^10 u^4].
942305
Okay. You have a few things that you have to consider.\n\nNegative exponents: moves it to the denominator\nFractional exponents: the denominator tells you what the root is, the numerator tells you what the root is rasied to\n\nSo: (x+1)^-2/3 rewritten would be:\n\n1/[cuberoot(x+1)^2]\n\nwhen x is seven you have\n1/[cuberoot(8)^2]\n1/[2^2]\n1/4
108399
use binomial theorem, it can be applied to any power.if you don't know this theorem, i'll send it to you.