// Numbas version: finer_feedback_settings {"name": "\u00dcbungen (Lektion1)", "metadata": {"description": "

Übungsbeispiele zur 1. Lektion

", "licence": "Creative Commons Attribution 4.0 International"}, "duration": 0, "percentPass": "0", "showQuestionGroupNames": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", ""], "questions": [{"name": "Keilschriftzahlen", "extensions": [], "custom_part_types": [], "resources": [["question-resources/num22074.png", "/srv/numbas/media/question-resources/num22074.png"], ["question-resources/num22641.png", "/srv/numbas/media/question-resources/num22641.png"], ["question-resources/num25532.png", "/srv/numbas/media/question-resources/num25532.png"], ["question-resources/num35112.png", "/srv/numbas/media/question-resources/num35112.png"], ["question-resources/num40169.png", "/srv/numbas/media/question-resources/num40169.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Andreas Vohns", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3622/"}], "tags": [], "metadata": {"description": "

Umwandlung von keilschriftzahlen in Dezimalzahl und Dezimalbruch, Zufallswerte.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Die Baylonischen Keilschriftzahlen wurden sowohl für natürliche Zahlen als auch für Bruchzahlen verwendet, dabei ergeben sich Mehrdeutigkeiten in der Zahldarstellung.

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{image('resources/question-resources/'+chosenimage2)}

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Zur Vereinfachung gehen wir davon aus, dass zwischen den hier dargestellten drei Zifferngruppen keine leeren Stellenwerte vorkommen.

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\n

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", "advice": "

a) Dargestellt sind die Ziffern $\\var{chosenfact[0]},\\var{chosenfact[1]},\\var{chosenfact[2]}$.

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b) Es ergibt sich $\\var{chosenfact[0]}\\cdot 60^2+\\var{chosenfact[1]}\\cdot 60^1+\\var{chosenfact[2]}\\cdot 60^0=\\var{whole}$.

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c) Es ergibt sich $\\var{chosenfact[0]}\\cdot 60^0+\\var{chosenfact[1]}\\cdot 60^{-1}+\\var{chosenfact[2]}\\cdot 60^{-2}\\approx\\var{fraction}$

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Ermitteln Sie die in den drei dargestellten Zifferngruppen dargestellten Ziffern ($0<z_n<60$).

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1. Zifferngruppe: $z_1=$ [[0]]
2. Zifferngruppe: $z_2=$[[1]]
3. Zifferngruppe: $z_3=$[[2]]

\n

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Welche ganze Zahl (im Dezimalsystem) ergibt sich, wenn die letzte Zifferngruppe Einer darstellt? 

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", "minValue": "whole", "maxValue": "whole", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": "3", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Welcher Dezimalbruch ergibt sich, wenn die erste Zifferngruppe Einer darstellt? 

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Bitte Beistrich (,) als Dezimaltrennzeichen verwenden!

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\n

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Wirzelziehen nach der baylonischen Methode.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Ermitteln Sie für die Zahl $N=\\var{N}$ nach der babylonischen Methode mit Startwert $a_0=\\var{a0}$ näherungsweise die Wurzel durch zweifache Iteration.

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Bitte  jeweils Beistrich (,) als Dezimaltrennzeichen verwenden!

", "advice": "

Mit Startwert $a_0=\\var{a0}$ ergibt sich für $N=\\var{N}$

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    \n
  1. $B_1=a_0^2-N=\\var{a0}^2-\\var{N}=\\var{B1}$
    und $a_1=a_0-\\frac{1}{2}\\cdot\\frac{B_1}{a_0}=\\var{a0}-\\frac{1}{2}\\cdot\\frac{\\var{B1}}{\\var{a0}}\\approx\\var{precround(a1,5)}$.
  2. \n
  3. $B_2=a_1^2-N\\approx\\var{precround(a1,5)}^2-\\var{N}\\approx\\var{precround(B2,5)}$
    und $a_2=a_1-\\frac{1}{2}\\cdot\\frac{B_2}{a_1}\\approx\\var{precround(a1,5)}-\\frac{1}{2}\\cdot\\frac{\\var{precround(B2,5)}}{\\var{precround(a1,5)}}\\approx\\var{precround(a2,5)}$.
  4. \n
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Anders als oben dargestellt sollte mit ungerundeten Werten oder Bruchzahlen als Zwischenergebnissen weitergerechnet werden.

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Erste Iteration: $a_1=$[[0]]

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Zweite Iteration: $a_2=$[[1]]

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Rechnen mit der Näherungsformel und exakt.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Gegeben ist das folgende Viereck (bitte etwas Geduld beim Laden des Applets haben):

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{geogebra_applet('https://www.geogebra.org/m/b7qqv88a',defs)}

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", "advice": "

a) Die Näherungsformel für allgemeine Vierecke lautet: $\\frac{(a+c)}{2}\\cdot\\frac{(b+d)}{2}$, eingesetzt:

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$\\frac{(\\var{a}+\\var{c})}{2}\\cdot\\frac{(\\var{b}+\\var{d})}{2}\\approx\\var{A_approx}$

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b) Die Flächenformel für ein rechtwinkliges Trapez lautet: $c\\cdot\\frac{(b+d)}{2}$, eingesetzt:

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$\\var{c}\\cdot\\frac{(\\var{b}+\\var{d})}{2}\\approx\\var{A_precise}$

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Parameter 1

", "templateType": "anything"}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "random(8..16)/4", "description": "

Parameter 2

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Ermitteln Sie den Flächeninhalt des Vierecks mit der babylonischen Näherungsformel für allgemeine Vierecke!

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A=[[0]]

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Bitte Beistrich (,) als Dezimaltrennzeichen verwenden!

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Ermitteln Sie den Flächeninhalt mit der für dieses Viereck exakten Formel.

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A=[[0]]

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Bitte Beistrich (,) als Dezimaltrennzeichen verwenden!

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