What is ,(2/3)^2*(3/2)^{-1}*2^{-1}*(1/4)^0 ?

The solution to ,(2/3)^2*(3/2)^{-1}*2^{-1}*(1/4)^0 is 4/27

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simplify ,(2/3)^2(3/2)^{-1}2^{-1}*(1/4)^0

simplify

(\frac{2}{3})^{2} \cdot (\frac{3}{2})^{- 1} \cdot 2^{- 1} \cdot (\frac{1}{4})^{0}

\frac{4}{27}

Decimal

0.14814 \dots

Solution steps

(\frac{2}{3})^{2} (\frac{3}{2})^{- 1} \cdot 2^{- 1} (\frac{1}{4})^{0}

(\frac{2}{3})^{2} = \frac{4}{9}

= \frac{4}{9} (\frac{3}{2})^{- 1} \cdot 2^{- 1} (\frac{1}{4})^{0}

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

= \frac{4}{9} \cdot \frac{2}{3} \cdot 2^{- 1} (\frac{1}{4})^{0}

Apply exponent rule:

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

2^{- 1} = \frac{1}{2}

= \frac{4}{9} \cdot \frac{2}{3} \cdot \frac{1}{2} (\frac{1}{4})^{0}

Apply exponent rule:

a^{0} = 1, a \neq = 0

(\frac{1}{4})^{0} = 1

= \frac{4}{9} \cdot \frac{2}{3} \cdot \frac{1}{2} \cdot 1

Multiply the numbers:

\frac{4}{9} \cdot \frac{2}{3} \cdot \frac{1}{2} \cdot 1 = \frac{4}{9} \cdot \frac{2}{3} \cdot \frac{1}{2}

= \frac{4}{9} \cdot \frac{2}{3} \cdot \frac{1}{2}

Factor the number:

\frac{4}{9} = (\frac{2}{3})^{2}

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

(\frac{2}{3})^{2} \frac{2}{3} = (\frac{2}{3})^{2 + 1}

= (\frac{2}{3})^{2 + 1} \frac{1}{2}

Add the numbers:

2 + 1 = 3

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

(\frac{2}{3})^{3} = \frac{2 ^{3}}{3 ^{3}}

= \frac{2 ^{3}}{3 ^{3}} \cdot \frac{1}{2}

2^{3} = 8

= \frac{8}{3 ^{3}} \cdot \frac{1}{2}

3^{3} = 27

= \frac{8}{27} \cdot \frac{1}{2}

\frac{8}{27} \cdot \frac{1}{2} = \frac{4}{27}

= \frac{4}{27}

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What is ,(2/3)^2*(3/2)^{-1}*2^{-1}*(1/4)^0 ?
The solution to ,(2/3)^2*(3/2)^{-1}*2^{-1}*(1/4)^0 is 4/27