Wiki-Quellcode von Lösung Akkuentladung
Zuletzt geändert von akukin am 2025/05/31 13:13
Verstecke letzte Bearbeiter
| author | version | line-number | content |
|---|---|---|---|
 |
1.1 | 1 | (%class=abc%) |
| |
3.1 | 2 | 1. (((Gerade durch {{formula}}(0|200){{/formula}} und {{formula}}(14|0){{/formula}} → {{formula}}y=-\frac{200}{14}x+200{{/formula}} |
| 3 | [[image:Akkuentladung.png||width="300" style="float: left"]] | ||
| 4 | |||
| 5 | |||
| 6 | |||
| 7 | |||
| 8 | |||
| 9 | |||
| 10 | |||
| 11 | |||
| 12 | |||
| 13 | |||
| 14 | |||
| 15 | |||
| 16 | |||
| 17 | |||
| 18 | |||
| 19 | |||
| 20 | |||
| 21 | |||
| 22 | |||
| 23 | |||
| 24 | |||
| 25 | ))) | ||
| |
1.1 | 26 | 1. {{formula}}-\frac{200}{14}\cdot 9+200=71,4{{/formula}} |
| |
3.1 | 27 | Der Akku enthält also 71,4 mA nach 9 Tagen. |
| |
1.1 | 28 | 1. 80% weg heißt 40 mA drin |
| |
3.1 | 29 | {{formula}}-\frac{200}{14}\cdot x+200=40 \implies x=11,2{{/formula}} |
| |
1.1 | 30 | Nach 11,2 Tagen sind 80% weg. |