Question

I WILL MARK YOU BRAINLYEST if you answer all of my problems !!!!!!!!!!!

Answer 1

$a^{-n}=\dfrac{1}{a^n}\\\\a^{-1}=\dfrac{1}{a}\\\\(ab)^n=a^nb^n\\\\(a^n)^m=a^{nm}\\----------------------\\\\10\#\ (6x)^{-2}=\dfrac{1}{(6x)^2}=\dfrac{1}{6^2x^2}=\boxed{\dfrac{1}{36x^2}}\\\\11\#\ \left(\dfrac{x^2}{y^3}\right)^{-1}=\dfrac{1}{\frac{x^2}{y^3}}=\boxed{\dfrac{y^3}{x^2}}\\----------------------$

$\sqrt[n]{a}=b\iff b^n=a\\-----------------------\\\\17\#\ \left(\sqrt[3]{27}\right)^4=3^4=\boxed{81}\\\\\sqrt[3]{27}=3\ because\ 3^3=27\\\\19\#\ \left(\sqrt[4]{625}\right)^3=5^3=\boxed{125}\\\\\sqrt[4]{625}=5\ because\ 5^4=625\\-----------------------$

$a^{\frac{1}{n}}=\sqrt[n]{a}\\\\a^n\cdot a^m=a^{n+m}\\-----------------------\\\\22\#\ 25^{\frac{3}{2}}=25^{1+\frac{1}{2}}=25^1\cdot25^{\frac{1}{2}}=25\cdot\sqrt{25}=25\cdot5=\boxed{125}\\\\\sqrt{25}=5\ because\ 5^2=25\\\\23\#\ (-27)^{\frac{4}{3}}=(-27)^{1+\frac{1}{3}}=(-27)^1\cdot(-27)^{\frac{1}{3}}=-27\cdot\sqrt[3]{-27}\\=-27\cdot(-3)=\boxed{81}\\\\\sqrt[3]{-27}=-3\ because\ (-3)^3=-27\\------------------------$

$a^{\frac{m}{n}}=\sqrt[n]{a^m}\\----------------------\\\\24\#\ \left(\sqrt[5]{32}\right)^7=\boxed{32^{\frac{7}{5}}}\\\\or\ \left(\sqrt[5]{32}\right)^7=\boxed{2^7}\\\\\sqrt[5]{32}=2\ because\ 2^5=32\\\\25\#\ \left(\sqrt[3]{125}\right)^2=\boxed{125^{\frac{2}{3}}}\\\\or\ \left(\sqrt[3]{125}\right)^2=\boxed{5^2}\\\\\sqrt[3]{125}=5\ because\ 5^3=125$