Lösung 2.2:5e

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The line should go through the points
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<math>\left( 5 \right.,\left. 0 \right)</math>
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and
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<math>\left( 0 \right.,\left. -8 \right)</math>
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which must therefore satisfy the equation of the line
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<math>y=kx+m</math>, i.e.
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<math>0=k\centerdot 5+m</math>
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From the other equation, we get
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<math>m=-8</math>
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and substituting this into the first equation gives
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<math>0=5k-8\ \Leftrightarrow \ k={8}/{5}\;</math>
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The gradient of the line is
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<math>{8}/{5}\;</math>.
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[[Image:S1_2_2_5_e.jpg|400px]]
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Version vom 10:05, 18. Sep. 2008

The line should go through the points \displaystyle \left( 5 \right.,\left. 0 \right) and \displaystyle \left( 0 \right.,\left. -8 \right) which must therefore satisfy the equation of the line \displaystyle y=kx+m, i.e.


\displaystyle 0=k\centerdot 5+m


From the other equation, we get \displaystyle m=-8 and substituting this into the first equation gives


\displaystyle 0=5k-8\ \Leftrightarrow \ k={8}/{5}\;


The gradient of the line is \displaystyle {8}/{5}\;.