Lösung 1.3:1a

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We calculate
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We calculate <math>2^{3}</math> first and then <math>3^{2}</math>,
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<math>2^{3}</math>
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first and then
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<math>3^{2}</math>
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<math>\begin{align}
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& 2^{3}=2\centerdot 2\centerdot 2=4\centerdot 2=8, \\
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& \\
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& 3^{2}=3\centerdot 3=9 \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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2^{3} &= 2\cdot 2\cdot 2 = 4\cdot 2 = 8\,,\\[5pt]
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3^{2} &= 3\cdot 3=9\,\textrm{.}
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\end{align}</math>}}
Hence,
Hence,
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{{Displayed math||<math>2^{3}\cdot 3^{2}=8\cdot 9=72\,</math>.}}
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<math>2^{3}\centerdot 3^{2}=8\centerdot 9=72</math>
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Version vom 12:39, 22. Sep. 2008

We calculate \displaystyle 2^{3} first and then \displaystyle 3^{2},

Vorlage:Displayed math

Hence,

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.3:1a“