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| author | Florian Fischer <florian.fischer@muhq.space> | 2024-02-07 17:04:43 +0100 |
|---|---|---|
| committer | Florian Fischer <florian.fischer@muhq.space> | 2024-02-07 17:04:43 +0100 |
| commit | e0578e9de40b8729031efe28c85f71281b315427 (patch) | |
| tree | d0445a8b107292d6a5f3a89e0a00ff39713c3518 /Mathe/7 | |
| parent | c23a230ce1c6257886c945e1167d5f157a7c039d (diff) | |
| download | unterricht-e0578e9de40b8729031efe28c85f71281b315427.tar.gz unterricht-e0578e9de40b8729031efe28c85f71281b315427.zip | |
fix mistake in solution
Diffstat (limited to 'Mathe/7')
| -rw-r--r-- | Mathe/7/1/2/Aufgaben/binom_luecken.tex | 2 |
1 files changed, 1 insertions, 1 deletions
diff --git a/Mathe/7/1/2/Aufgaben/binom_luecken.tex b/Mathe/7/1/2/Aufgaben/binom_luecken.tex index 755eee9c..7f94ce62 100644 --- a/Mathe/7/1/2/Aufgaben/binom_luecken.tex +++ b/Mathe/7/1/2/Aufgaben/binom_luecken.tex @@ -14,6 +14,6 @@ Ergänze die Lücken so, dass die Terme jeweils äquivalent sind. \begin{enumerate}[a), itemsep=0pt] \item $(a-4b)^2 = a^2 - 8ab + 16b^2$ \item $(9+\frac{1}{4}x)(9-\frac{1}{4}x) = 81 - \frac{1}{16}x^2$ -\item $-\frac{1}{3}x(9+\frac{1}{2}y^2) = 3x - \frac{1}{6}xy^2$ +\item $\frac{1}{3}x(9-\frac{1}{2}y^2) = 3x - \frac{1}{6}xy^2$ \end{enumerate} \end{Lsg} |
