Question

Please help on b (left of c) and c !!!

Answer 1

$(x^5y^4)^{ \frac{1}{2} }= \sqrt{x^5y^4} =x^2y^2 \sqrt{x} \\ \\ \\ (x^2y^{-1})(x^{-3}y)^0= (x^2y^{-1})*1= \frac{x^2}{y}$

Answer 2

Alrighty


remember
$(ab)^c=(a^c)(b^c)$
and
$x^\frac{m}{n}=\sqrt[n]{x^m}$
and
$(x^m)^n=x^{mn} and [tex]x^0=1$ for all real numbers x
and
$x^{-m}=\frac{1}{x^m}$


b.
$(x^5y^4)^\frac{1}{2}=((x^5)^\frac{1}{2})((y^4)^\frac{1}{2})$=
$(x^\frac{5}{2})(y^\frac{4}{2})=(\sqrt{x^5})(\sqrt{y^4})=x^2y^2\sqrt{x}$

c.
x^0=1
so
that (x^-3y)^0=1
because exponents first in pemdas
so we are left with
x^2y^-1
$x^2y^{-1}=(x^2)(y^{-1})=(x^2)(\frac{1}{y^1})=\frac{x^2}{y}$