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8 answers

x^-2

2006-11-27 14:26:07 · answer #1 · answered by 6 · 1 0

Rule of thumb: you could flow bases with exponents around the branch image, yet once you do it, the sign of the exponent transformations. permit's do an occasion right here. 5^(-2) think of of this as 5^(-2)/a million What you're able to do is flow the 5^(-2) around the dividing line, and once you gain this, it turns into useful, like so: a million/(5^2) right here is yet another occasion: (2^8)*(3^(-5)) / (8^(-2)) subsequently, what you're able to do is flow the three^(-5) DOWN for the duration of the dividing line, and eight^(-2) UP for the duration of the dividing line, and subsequently, the ability will change signs and indicators. [(2^8)*(8^2)]/(3^5) and finally, one extra occasion: m^(-3)/[n^(-a million)x^(-5)] we are in a position to flow the n^(-a million) and x^(-5) UP for the duration of the dividing line the place they are going to become useful exponents, whilst transferring m^(-3) down the place it's going to become useful. n*(x^5)/m^3 it incredibly is the right comparable theory right here. a million/x^3 = x^(-3)

2016-12-14 07:51:34 · answer #2 · answered by ? 4 · 0 0

I assume you mean 1/(x^2) in which case the answer would be x^-2

2006-11-27 14:26:32 · answer #3 · answered by Lucan 3 · 0 0

1/(x^2) = x^(-2)

if the exponent is on the bottom, just put a negative sign in front of it.

ex. (x+5)/x^3 = (x+5)*(x^-3)

hope this helps

2006-11-27 14:28:12 · answer #4 · answered by kdesky3 2 · 0 0

(x)^-2

2006-11-27 14:29:19 · answer #5 · answered by zgraf 4 · 0 0

x^(-2) reads x to the negative second.

2006-11-27 14:26:31 · answer #6 · answered by Anonymous · 0 0

one over x square

2006-11-27 14:27:05 · answer #7 · answered by KO 3 · 0 0

WHAT WAS THAT AGAIN???? LOL

2006-11-27 14:27:02 · answer #8 · answered by MOMOF3 2 · 0 0

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