How do you write #7/ x^-2 # with positive exponents?

2 Answers
Jan 7, 2016

Answer:

You should get #7x^2#

Explanation:

You need to change "floor" and send #x# from the ground to the first changing the sign of the exponent from #-# to #+#; so:
#7/x^(-2)=7x^(color(red)(+2))#

Jan 7, 2016

Answer:

#7*x^2#

Explanation:

The Exponent properties tell us that:

#a^-n=1/(a^n)#

then:

#7/(x^-2)=7/(1/x^2)=#

with #x!=0#

#=(x^2*7)/(cancel(x^2)*(1/cancel(x^2)))=7x^2#