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

Answer 1

Eva Abramson

#7*x^2#

The properties of the exponent inform 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#

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Answer 2

Julia Adams

You should get #7x^2#

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))#

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Answer 3

Ethan Adkins

To write ( \frac{7}{x^{-2}} ) with positive exponents, we can move the term with the negative exponent to the denominator and change the exponent to positive by using the reciprocal:

[ \frac{7}{x^{-2}} = 7 \cdot x^{2} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.