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⤋