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#1**0 **

I do not know if the variable m stands for meters or not. If it does, how can something be both kilograms and meters?

\(a = \dfrac{1 (kg) m^{-1}}{1 (kg) m^{-2}} \Rightarrow a = \dfrac{1 (kg) m^{-1}}{\dfrac{1}{1 (kg) m^{2}}}\). if we multiply the numerator and denominator \(1 (kg) m^{2}\)by , we have \(a = {1 (kg^2) m^{1}} \). You can adjust my answer according to your variables.

- PM

PartialMathematician Dec 17, 2018

#2**0 **

Note to PM

a = b/e

b = 1 kg m-1

e = 1 kg m-2

what is a? including units

PM, your current answer is \(a = {1 (kg^2) m^{1}}\)

It is NOT correct. Can you work out why?

Melody Dec 17, 2018

#3**0 **

I guess you could also do this:

Take away kg from the numerator and denominator of \(a = \dfrac{(kg)m^{-1}}{(kg)m^{-2}}\) to become \(a = \dfrac{m^{-1}}{m^{-2}}\).

This equals \(a = {m^1}\) because \(m^{-1} / m^{-2} = m^{-1} \cdot m^{2} \Rightarrow m^1\).

- PM

PartialMathematician
Dec 17, 2018

#5**0 **

Ok, I wasn't 100% sure if I was supposed to cancel out the Kg since the answer was supposed to include units.

PartialMathematician
Dec 17, 2018