Question

-4x^-3y^2z^-4/2x^2y^-1z^-6

Answer 1

I'm getting
y^2/-2x^5z^10
Is this correct?

Answer 2

\[(-4x{}^{\wedge}-3y{}^{\wedge}2z{}^{\wedge}-4)/(2x{}^{\wedge}2y{}^{\wedge}-1z{}^{\wedge}-6) \]\[\left(-\frac{4 y^2}{x^3 z^4}\right)/\left(\frac{2 x^2}{y z^6}\right)=-\frac{2 y^3 z^2}{x^5} \]

Answer 3

Could you please explain step by step how you solved?

Answer 4

\[((-4x{}^{\wedge}-3)(y{}^{\wedge}2)(z{}^{\wedge}-4))/((2x{}^{\wedge}2)(y{}^{\wedge}-1)(z{}^{\wedge}-6)) \]\[\left(\left(-\frac{4}{x^3}\right)\left(y^2\right)\left(\frac{1}{z^4}\right)\right)/\left(\left(2 x^2\right)\left(\frac{1}{y}\right)\left(\frac{1}{z^6}\right)\right) \]\[-\frac{4 y^2}{x^3 z^4}/\frac{2 x^2}{y z^6} \]\[-\frac{4 y^2}{x^3 z^4}*\frac{y z^6}{2 x^2} \]\[-\frac{2 y^3 z^2}{x^5} \]

Answer 5

I used the negative exponent rule. If "a" is a real number other than "0" and "n" is an integer, then
a^-n=1/a^n and 1/a^-n=a^n
I solved with the same answer but seems I used a different method.

Answer 6

Thanks