What is (4/5)^3*(3/10)^{-2}*(4/3)^{-4} ?

The solution to (4/5)^3*(3/10)^{-2}*(4/3)^{-4} is 9/5

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simplify (4/5)^3(3/10)^{-2}(4/3)^{-4}

simplify

(\frac{4}{5})^{3} \cdot (\frac{3}{10})^{- 2} \cdot (\frac{4}{3})^{- 4}

\frac{9}{5}

Decimal Notation

1.8

Solution steps

(\frac{4}{5})^{3} (\frac{3}{10})^{- 2} (\frac{4}{3})^{- 4}

(\frac{4}{5})^{3} = \frac{64}{125}

= \frac{64}{125} (\frac{3}{10})^{- 2} (\frac{4}{3})^{- 4}

(\frac{3}{10})^{- 2} = \frac{100}{9}

= \frac{64}{125} \cdot \frac{100}{9} (\frac{4}{3})^{- 4}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

(\frac{4}{3})^{- 4} = \frac{1}{( \frac{4}{3} ) ^{4}}

= \frac{64}{125} \cdot \frac{100}{9} \cdot \frac{1}{( \frac{4}{3} ) ^{4}}

(\frac{4}{3})^{4} = \frac{256}{81}

= \frac{64}{125} \cdot \frac{100}{9} \cdot \frac{1}{\frac{256}{81}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{64}{125} \cdot \frac{100}{9} \cdot \frac{81}{256}

\frac{64}{125} \cdot \frac{100}{9} \cdot \frac{81}{256} = \frac{9}{5}

= \frac{9}{5}

5 + - 9 - 7 + - 121

\frac{6}{11 + 3}

(x + 9) 2

\frac{1}{8} x^{2}

lo g_{3} (5) + lo g_{3} (0.6)

What is (4/5)^3*(3/10)^{-2}*(4/3)^{-4} ?
The solution to (4/5)^3*(3/10)^{-2}*(4/3)^{-4} is 9/5