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jsMath

Lösung 1.1:1a

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
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Zeile 1: Zeile 1:
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Because there no brackets nor multiplications/divisions, we calculate the expression from left to right,
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{{NAVCONTENT_START}}
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Because there are no brackets nor multiplications/divisions there is no subexpression that needs to be calculated first, so we can start calculating the expression from left to right. First we perform the subtraction of the two leftmost numbers
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:<math>\bbox[#FFEEAA;,1.5pt]{\,3-7\,}-4+6-5=\bbox[#FFEEAA;,1.5pt]{\,-4\,}-4+6-5</math>.
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{{NAVCONTENT_STEP}}
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The next step is to subtract the two new leftmost numbers
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:<math>\phantom{\bbox[#FFEEAA;,1.5pt]{\,3-7\,}-4+6-5}{}=\firstcbox{#FFEEAA;}{\,-4-4\,}{-8}+6-5</math>
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{{NAVCONTENT_STEP}}
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:<math>\phantom{\bbox[#FFEEAA;,1.5pt]{\,3-7\,}-4+6-5}{}=\secondcbox{#FFEEAA;}{\,-4-4\,}{-8}+6-5</math>
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{{NAVCONTENT_STEP}}
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and we continue in the same way to work with the two left-most terms in the expression that arises
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:<math>\phantom{\bbox[#FFEEAA;,1.5pt]{\,3-7\,}-4+6-5}{}=\firstcbox{#FFEEAA;}{\,-8+6\,}{-2}-5</math>
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{{NAVCONTENT_STEP}}
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:<math>\phantom{\bbox[#FFEEAA;,1.5pt]{\,3-7\,}-4+6-5}{}=\secondcbox{#FFEEAA;}{\,-8+6\,}{-2}-5</math>.
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{{NAVCONTENT_STEP}}
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Finally we have an expression which we can calculate in one step
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:<math>\phantom{\bbox[#FFEEAA;,1.5pt]{\,3-7\,}-4+6-5}{}=-2-5</math>
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{{NAVCONTENT_STEP}}
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:<math>\phantom{\bbox[#FFEEAA;,1.5pt]{\,3-7\,}-4+6-5}{}=-7</math>.
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{{NAVCONTENT_STOP}}

Version vom 06:39, 19. Sep. 2008

Because there are no brackets nor multiplications/divisions there is no subexpression that needs to be calculated first, so we can start calculating the expression from left to right. First we perform the subtraction of the two leftmost numbers

\displaystyle \bbox[#FFEEAA;,1.5pt]{\,3-7\,}-4+6-5=\bbox[#FFEEAA;,1.5pt]{\,-4\,}-4+6-5.

The next step is to subtract the two new leftmost numbers

\displaystyle \phantom{\bbox[#FFEEAA;,1.5pt]{\,3-7\,}-4+6-5}{}=\firstcbox{#FFEEAA;}{\,-4-4\,}{-8}+6-5
\displaystyle \phantom{\bbox[#FFEEAA;,1.5pt]{\,3-7\,}-4+6-5}{}=\secondcbox{#FFEEAA;}{\,-4-4\,}{-8}+6-5

and we continue in the same way to work with the two left-most terms in the expression that arises

\displaystyle \phantom{\bbox[#FFEEAA;,1.5pt]{\,3-7\,}-4+6-5}{}=\firstcbox{#FFEEAA;}{\,-8+6\,}{-2}-5
\displaystyle \phantom{\bbox[#FFEEAA;,1.5pt]{\,3-7\,}-4+6-5}{}=\secondcbox{#FFEEAA;}{\,-8+6\,}{-2}-5.

Finally we have an expression which we can calculate in one step

\displaystyle \phantom{\bbox[#FFEEAA;,1.5pt]{\,3-7\,}-4+6-5}{}=-2-5
\displaystyle \phantom{\bbox[#FFEEAA;,1.5pt]{\,3-7\,}-4+6-5}{}=-7.
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.1:1a“