Question

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Answer 1

Answer:

$\mathrm{Simplify}\:\left(\left(64\right)^{\dfrac{2}{3}}\times \dfrac{2^{-2}}{7^0}\right)^{-\dfrac{1}{2}}:\quad \dfrac{1}{2}\quad \left(\mathrm{Decimal:\quad }\:0.5\right)$

Step-by-step explanation:

$\left(\left(64\right)^{\dfrac{2}{3}}\dfrac{2^{-2}}{7^0}\right)^{-\dfrac{1}{2}}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\dfrac{1}{a^b}$

$=\dfrac{1}{\left(64^{\dfrac{2}{3}}\times \dfrac{2^{-2}}{7^0}\right)^{\dfrac{1}{2}}}$

$\mathrm{Apply\:rule}\:a^0=1,\:a\ne \:0$

$7^0=1$

$=\dfrac{1}{\left(64^{\dfrac{2}{3}}\times \dfrac{2^{-2}}{1}\right)^{\dfrac{1}{2}}}$

$\mathrm{Apply\:rule}\:\dfrac{a}{1}=a$

$\dfrac{2^{-2}}{1}=2^{-2}$

$=\dfrac{1}{\left(2^{-2}\times \:64^{\dfrac{2}{3}}\right)^{\dfrac{1}{2}}}$

$\left(64^{\dfrac{2}{3}}\times \:2^{-2}\right)^{\dfrac{1}{2}}$

$\mathrm{Simplify}\:64^{\dfrac{2}{3}}\times \:2^{-2}$

$\mathrm{Factor\:integer\:}64=2^6$

$=\left(2^6\right)^{\dfrac{2}{3}}\times \:2^{-2}$

$\mathrm{Apply\:exponent\:rule}:\quad \left(a^b\right)^c=a^{bc}$

$\left(2^6\right)^{\dfrac{2}{3}}=2^{6\times \dfrac{2}{3}}$

$=2^{6\times \dfrac{2}{3}}\times \:2^{-2}$

$\mathrm{Refine}$

$=2^4\times \:2^{-2}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^b\times \:a^c=a^{b+c}$

$2^4\times \:2^{-2}=\:2^{4-2}$

$=2^{4-2}$

$\mathrm{Subtract\:the\:numbers:}\:4-2=2$

$=2^2$

$=\left(2^2\right)^{\dfrac{1}{2}}$

$\mathrm{Apply\:exponent\:rule}:\quad \left(a^b\right)^c=a^{bc}$

$=2^{2\times \dfrac{1}{2}}$

$=2$

$=\dfrac{1}{2}$

Answer 2

Answer:-

Refer the attachment