how do i do i (as in square root of (-1)) to a negative power?
607947
i^-1= 1/i^1  you can get rid of the negative exponent by changing it to the reciprocal.
469613
two hints:\n\n1) anything raised to the 0 power gives 1\n\n2) anything raised to a negative power means 1/the same thing raised to the positive opposite power.\n\nEnjoy your homework.
298073
ok so when u divide  stuff like that, u just subtract the exponents.  \nr^-3 s^5 t^2\n---------------\nr^2 s t^-2\n\nso for the r, the new exponent would be r^(-3-2) = r^-5\nfor the s, s^(5-1) = s^4.  (s by itself is equal to s^1.)\nfor the t, t^(2-[-2]) = t^4\nso the new expression is  r^-5 s^4 t^4. \nand if you really wanted to, u could rewrite r^-5 as 1/r^5. cuz thatz how negative exponents work. (x^-2 = 1/x^2)\nso then the final answer would be s^4t^4/r^5.\nhope this helps. let me know if u need more help
559973
4i is the answer you seek,\nthe i is the square of -1 says any geek.\nThis is a abstract answer to solve your query,\nbut all negative roots include the 'i,' my deary.
860947
sqrt ( -1 * -1 ) doesn't equal sqrt(-1) * sqrt(-1).\n\nanother example:\n\nsqrt ( 5 * 5 ) = sqrt ( 25) = 5\n\nbut it doesn't equal sqrt(5) * sqrt(5) because this is an irrational answer..\n\nhope this helps
152568
2i. The character "i" denotes the square root of -1, which is not a real number. So, the sq. root of -4 is 2i.
307451
Try\n\nhttp://www.purplemath.com/modules/negative.htm
218778
The answer is -1\n\nIn Algebra, anytime you subtract a number, you change the number following the minus sign into its inverse (negative to a posotive and positive to a negative) and then add.  Therefore, your problem becomes -12+11, giving an answer of -1.
1122465
Just plug in 1 for x.\n√(1-1) = √0 = 0
51216
x = [-(-4) +/- root((-4)^2 - 4x1x(-5))]/2x1\n  = (4 +/- 6)/2\n  = 5 or -1
25136
1 = sqrt(1) = sqrt(-1*-1) = sqrt(-1)*sqrt(-1)=i*i = i^2 = -1\nSo 1 = -1.  Add 1 to both sides and 2 = 0.  Divide both sides by 2 and 1 = 0.
193948
The sq rt of x is x to the 1/2 power; 1 over sqrt x is x to the -1/2 power\n\nThen use your normal rule for exponents : d/dx (x^n) = n x^(n-1)\n\nd/dx x^-1/2 = -1/2 x^(-3/2)\n\n(This equals -1 over the square root of x cubed if you need to write it that way.)
15458
A negative number cannot have a square root.
1094108
Use a calculator and raise it to the 1/3 power.\nDon't try to do it by hand...
695417
So close:\n\ni^2 = (sqrt(-1))^2 = -1\n\nSo, to finish the first problem, \n\n14i + 4i^2 = 14i - 2\n\nSimilarly, to finish your second problem, replace i^2 = -1.
346355
The "principal" square root is the positive one, 4/5.\n\nThe "negative" square root of the fraction is -4/5, because\n\n-4/5 * -4/5 = 16/25 also.
445985
if by this you mean\n\n((3x^-2 * y)^-2)/((4xy^-2)^-1)\n\nthis becomes\n\n((3^-2) * x^(-2 * -2) * y^-2)/((4^-1) * x^-1 * y^(-2 * -1))\n\n((1/9)x^4 * y^-2)/((1/4) * x^-1 * y^2)\n\n((1/9)/(1/4)) * x^(4 - (-1)) * y^(-2 - 2)\n\n((1/9)*(4/1)) * x^(4 + 1) * y^(-2 + (-2))\n\n(4/9)x^5y^-4\n\nwithout using negative exponents, the answer would be\n\n(4x^5)/(9y^4)