Lösung 3.1:4b

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{{NAVCONTENT_START}}
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By writing
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<center> [[Image:3_1_4b.gif]] </center>
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<math>0.0\text{27 }</math>
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{{NAVCONTENT_STOP}}
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as
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<math>\text{27}\cdot \text{1}0^{-\text{3}}</math>, where
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<math>\text{27}=\text{3}\cdot \text{3}\cdot \text{3}=\text{3}^{\text{3}}</math>
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and
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<math>10^{-3}=\left( 10^{-1} \right)^{3}=0.1^{3}</math>
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we see that
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<math>\begin{align}
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& \sqrt[3]{0.027}=\sqrt[3]{27\centerdot 10^{-3}}=\sqrt[3]{27}\centerdot \sqrt[3]{10^{-3}}=\sqrt[3]{3^{3}}\centerdot \sqrt[3]{0.1^{3}} \\
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& =3\centerdot 0.1=0.3 \\
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\end{align}</math>
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where we have used
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<math>\sqrt[3]{a^{3}}=\left( a^{3} \right)^{\frac{1}{3}}=a^{3\centerdot \frac{1}{3}}=a^{1}=a</math>

Version vom 13:58, 22. Sep. 2008

By writing \displaystyle 0.0\text{27 } as \displaystyle \text{27}\cdot \text{1}0^{-\text{3}}, where \displaystyle \text{27}=\text{3}\cdot \text{3}\cdot \text{3}=\text{3}^{\text{3}} and \displaystyle 10^{-3}=\left( 10^{-1} \right)^{3}=0.1^{3} we see that


\displaystyle \begin{align} & \sqrt[3]{0.027}=\sqrt[3]{27\centerdot 10^{-3}}=\sqrt[3]{27}\centerdot \sqrt[3]{10^{-3}}=\sqrt[3]{3^{3}}\centerdot \sqrt[3]{0.1^{3}} \\ & =3\centerdot 0.1=0.3 \\ \end{align}


where we have used \displaystyle \sqrt[3]{a^{3}}=\left( a^{3} \right)^{\frac{1}{3}}=a^{3\centerdot \frac{1}{3}}=a^{1}=a

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:4b“