What is (1/2*(2/7)^{-3})^2 ?

The solution to (1/2*(2/7)^{-3})^2 is 117649/256

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simplify (1/2*(2/7)^{-3})^2

simplify

(\frac{1}{2} \cdot (\frac{2}{7})^{- 3})^{2}

\frac{117649}{256}

Decimal Notation

459.56640 \dots

Solution steps

(\frac{1}{2} (\frac{2}{7})^{- 3})^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

(\frac{1}{2} (\frac{2}{7})^{- 3})^{2} = (\frac{1}{2})^{2} ((\frac{2}{7})^{- 3})^{2}

= (\frac{1}{2})^{2} ((\frac{2}{7})^{- 3})^{2}

(\frac{1}{2})^{2} = \frac{1}{4}

= \frac{1}{4} ((\frac{2}{7})^{- 3})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}, a \geq 0

((\frac{2}{7})^{- 3})^{2} = (\frac{2}{7})^{- 3 \cdot 2}

= \frac{1}{4} (\frac{2}{7})^{- 3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{1}{4} (\frac{2}{7})^{- 6}

Apply exponent rule:

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

(\frac{2}{7})^{- 6} = \frac{1}{( \frac{2}{7} ) ^{6}}

= \frac{1}{4} \cdot \frac{1}{( \frac{2}{7} ) ^{6}}

(\frac{2}{7})^{6} = \frac{64}{117649}

= \frac{1}{4} \cdot \frac{1}{\frac{64}{117649}}

Apply the fraction rule:

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

= \frac{1}{4} \cdot \frac{117649}{64}

\frac{1}{4} \cdot \frac{117649}{64} = \frac{117649}{256}

= \frac{117649}{256}

0 = - 16 t^{2} + 80 t

5 x^{2} - 20 x

14 x^{2} - 17 x - 6

- 6 + 9 (8 - 2 b)

\frac{a}{3} \geq - 9