The answer is \(\frac{32 r^{2} }{25 t^{9} }\)
To calculate this, we will use several rules:
\((x^{a} )^{b} =x^{a*b} \\
x^{-a}= \frac{1}{ x^{a} } \\
(x*y) ^{a} = x^{a} * y^{a} \\
( \frac{x}{y} ) ^{a} = \frac{ x^{a} }{y^{a} }\)
Now, let's implement these rules in our expression:
\((2t^{-3})^{3} (0.4r)^{2} = 2^{3} * (t^{-3})^{3} *(0.4) ^{2} * r^{2} =8* t^{-3*3} *( \frac{4}{10} ) ^{2} * r^{2} = \\
=8* t^{-9} *( \frac{2}{5} ) ^{2} * r^{2}=8* \frac{1}{ t^{9} } * \frac{2^{2} }{5^{2} }* r^{2} = 8* \frac{1}{ t^{9} } * \frac{4 }{25}* r^{2}= \frac{8*4* r^{2} }{ t^{9}*25 } = \\
= \frac{32 r^{2} }{25 t^{9} }\)
