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Nady [450]
3 years ago
6

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

Mathematics
2 answers:
Murrr4er [49]3 years ago
4 0
(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}
vesna_86 [32]3 years ago
3 0
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}
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