Question

How do you solve this? pre calculus; (-0.008) ^2/3

Answer 1

`a^(m)/(n)=\sqrt[n]{a^m}\\---------------\\\\(-0.008)^(2)/(3)=\sqrt[3]{(-0.008)^2}=\left(\sqrt[3]{-0.008}\right)^2=\left(-0.2\right)^2=0.04`

Answer 2

0.008 is actually 8/1000, therefore -0.008 is -8/1000.


`( (-8)/(1000) )^{ (2)/(3) } = \sqrt[3]{( (-8)/(1000))^(2) } = \sqrt[3]{ (64)/(1,000,000) }= (4)/(100)=0.04= (2)/(50)= (1)/(25)`