We know that:
[tex]$\sqrt[a]{x^b}=x^{\frac{b}{a}}[/tex]
so:
1)
[tex]$(\sqrt[3]{t^4})\cdot(t^{\frac{-1}{2}})=(t^{\frac{4}{3}})\cdot(t^{\frac{-1}{2}})=
t^{\frac{4}{3}+\frac{-1}{2}}=t^{\frac{8}{6}+\frac{-3}{6}}=t^{\frac{8-3}{6}}=\boxed{t^{\frac{5}{6}}=\sqrt[6]{t^5}}[/tex]
2)
[tex]x^{-\frac{2}{5}}=\dfrac{1}{x^{\frac{2}{5}}}=\boxed{\dfrac{1}{\sqrt[5]{x^2}}}[/tex]
Answer 4)
