Lösung 4.3:9 - Versionsgeschichte

Tek: Replaced figure with metapost figure

Tek — Tue, 25 Aug 2009 10:45:28 GMT

Replaced figure with metapost figure

← Nächstältere Version		Version vom 10:45, 25. Aug. 2009
Zeile 19:		Zeile 19:
	{{Abgesetzte Formel\|\|<math>\cos 80^{\circ}\cdot\cos 40^{\circ}\cdot \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\cos 80^{\circ}\cdot\cos 40^{\circ}\cdot \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}}\,\textrm{.}</math>}}

-	~~[[Image~~:~~4_3_9~~.~~gif\|\|right]]~~	+	<center>{{:4.3.9 - Solution - The unit circle with angles 20° and 160°}}</center>

	Zeichnen wir den Winkel <math>160^{\circ}</math> im Einheitskreis, sehen wir, dass der Winkel dieselbe ''y''-Koordinate wie der Winkel <math>20^{\circ}</math> hat und daher denselben Sinus. Also erhalten wir		Zeichnen wir den Winkel <math>160^{\circ}</math> im Einheitskreis, sehen wir, dass der Winkel dieselbe ''y''-Koordinate wie der Winkel <math>20^{\circ}</math> hat und daher denselben Sinus. Also erhalten wir

Markus.bez um 13:10, 19. Jun. 2009

Markus.bez — Fri, 19 Jun 2009 13:10:46 GMT

← Nächstältere Version		Version vom 13:10, 19. Jun. 2009
Zeile 3:		Zeile 3:
	{{Abgesetzte Formel\|\|<math>\sin 160^{\circ} = 2\cos 80^{\circ}\sin 80^{\circ}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\sin 160^{\circ} = 2\cos 80^{\circ}\sin 80^{\circ}\,\textrm{.}</math>}}

-	Wir verwenden jetzt die Doppelwinkelfunktion für den Faktor <math>\sin 80^{\circ}</math>, nachdem wir den Faktor math>\cos 80^{\circ}</math> behalten möchten	+	Wir verwenden jetzt die Doppelwinkelfunktion für den Faktor <math>\sin 80^{\circ}</math>, nachdem wir den Faktor <math>\cos 80^{\circ}</math> behalten möchten:

-	{{Abgesetzte Formel\|\|<math>2\cos 80^{\circ}\sin 80^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\sin 40^{\circ}\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>2\cos 80^{\circ}\sin 80^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\sin 40^{\circ}\,.</math>}}

	Wir verwenden noch einmal die Doppelwinkelfunktion, dieses Mal für den Faktor <math>\sin 40^{\circ}</math>		Wir verwenden noch einmal die Doppelwinkelfunktion, dieses Mal für den Faktor <math>\sin 40^{\circ}</math>
Zeile 11:		Zeile 11:
	{{Abgesetzte Formel\|\|<math>2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot\sin 40^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot 2\cos 20^{\circ}\sin 20^{\circ}\,\textrm{·}</math>}}		{{Abgesetzte Formel\|\|<math>2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot\sin 40^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot 2\cos 20^{\circ}\sin 20^{\circ}\,\textrm{·}</math>}}

-	Also haben wir gezeigt dass	+	Also haben wir gezeigt, dass

	{{Abgesetzte Formel\|\|<math>\sin 160^{\circ} = 8\cos 80^{\circ}\cdot \cos 40^{\circ}\cdot \cos 20^{\circ}\cdot\sin 20^{\circ}</math>}}		{{Abgesetzte Formel\|\|<math>\sin 160^{\circ} = 8\cos 80^{\circ}\cdot \cos 40^{\circ}\cdot \cos 20^{\circ}\cdot\sin 20^{\circ}</math>}}

-	~~oder anders~~ geschrieben,	+	Anders geschrieben:

	{{Abgesetzte Formel\|\|<math>\cos 80^{\circ}\cdot\cos 40^{\circ}\cdot \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\cos 80^{\circ}\cdot\cos 40^{\circ}\cdot \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}}\,\textrm{.}</math>}}
Zeile 21:		Zeile 21:
	[[Image:4_3_9.gif\|\|right]]		[[Image:4_3_9.gif\|\|right]]

-	Zeichnen wir den Winkel <math>160^{\circ}</math> im Einheitskreis, sehen wir dass der Winkel ~~denselben~~ ''y''-~~Koordinaten~~ wie der Winkel <math>20^{\circ}</math> hat, und daher denselben Sinus. Also erhalten wir	+	Zeichnen wir den Winkel <math>160^{\circ}</math> im Einheitskreis, sehen wir, dass der Winkel dieselbe ''y''-Koordinate wie der Winkel <math>20^{\circ}</math> hat und daher denselben Sinus. Also erhalten wir

	<center><math>\sin 20^{\circ} = \sin 160^{\circ}\,\textrm{.}</math></center>		<center><math>\sin 20^{\circ} = \sin 160^{\circ}\,\textrm{.}</math></center>

-	~~und also~~ haben wir die Gleichung	+	Damit haben wir die Gleichung

	<center><math>\cos 80^{\circ} \cos 40^{\circ} \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}} = \frac{1}{8}\,\textrm{.}</math></center>		<center><math>\cos 80^{\circ} \cos 40^{\circ} \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}} = \frac{1}{8}\,\textrm{.}</math></center>
-
-	gezeigt.

Martin um 13:29, 5. Apr. 2009

Martin — Sun, 05 Apr 2009 13:29:50 GMT

← Nächstältere Version		Version vom 13:29, 5. Apr. 2009
Zeile 1:		Zeile 1:
-	~~Using the formula for double angles on~~ <math>\sin 160^{\circ}</math> ~~gives~~	+	Wir verwenden die Doppelwinkelfunktion für <math>\sin 160^{\circ}</math>

	{{Abgesetzte Formel\|\|<math>\sin 160^{\circ} = 2\cos 80^{\circ}\sin 80^{\circ}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\sin 160^{\circ} = 2\cos 80^{\circ}\sin 80^{\circ}\,\textrm{.}</math>}}

-	~~On the right-hand side, we see that the factor~~ <math>\~~cos~~ 80^{\circ}</math> ~~has appeared~~, ~~and if we use the formula for double angles on the second factor (<~~math>\~~sin~~ 80^{\circ}</math>),	+	Wir verwenden jetzt die Doppelwinkelfunktion für den Faktor <math>\sin 80^{\circ}</math>, nachdem wir den Faktor math>\cos 80^{\circ}</math> behalten möchten

	{{Abgesetzte Formel\|\|<math>2\cos 80^{\circ}\sin 80^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\sin 40^{\circ}\,,</math>}}		{{Abgesetzte Formel\|\|<math>2\cos 80^{\circ}\sin 80^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\sin 40^{\circ}\,,</math>}}

-	~~we obtain a further factor <math>\cos 40^{\circ}</math>. A final application of the formula for double angles on~~ <math>\sin 40^{\circ }</math> ~~gives us all three cosine factors,~~	+	Wir verwenden noch einmal die Doppelwinkelfunktion, dieses Mal für den Faktor <math>\sin 40^{\circ}</math>

	{{Abgesetzte Formel\|\|<math>2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot\sin 40^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot 2\cos 20^{\circ}\sin 20^{\circ}\,\textrm{·}</math>}}		{{Abgesetzte Formel\|\|<math>2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot\sin 40^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot 2\cos 20^{\circ}\sin 20^{\circ}\,\textrm{·}</math>}}

-	~~We have thus succeeded in showing that~~	+	Also haben wir gezeigt dass

	{{Abgesetzte Formel\|\|<math>\sin 160^{\circ} = 8\cos 80^{\circ}\cdot \cos 40^{\circ}\cdot \cos 20^{\circ}\cdot\sin 20^{\circ}</math>}}		{{Abgesetzte Formel\|\|<math>\sin 160^{\circ} = 8\cos 80^{\circ}\cdot \cos 40^{\circ}\cdot \cos 20^{\circ}\cdot\sin 20^{\circ}</math>}}

-	~~which can also be written as~~	+	oder anders geschrieben,

	{{Abgesetzte Formel\|\|<math>\cos 80^{\circ}\cdot\cos 40^{\circ}\cdot \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\cos 80^{\circ}\cdot\cos 40^{\circ}\cdot \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}}\,\textrm{.}</math>}}

	[[Image:4_3_9.gif\|\|right]]		[[Image:4_3_9.gif\|\|right]]
-	~~If we draw the unit circle, we see that~~ <math>160^{\circ}</math> ~~makes an angle of~~	+
-	~~<math>20^{\circ}</math> with the negative~~ ''x''-~~axis, and therefore the angles~~	+	Zeichnen wir den Winkel <math>160^{\circ}</math> im Einheitskreis, sehen wir dass der Winkel denselben ''y''-Koordinaten wie der Winkel <math>20^{\circ}</math> hat, und daher denselben Sinus. Also erhalten wir
-	<math>20^{\circ}</math> ~~and <math>160^{\circ}</math> have the same ''y''-coordinate in the unit circle~~, ~~i.e~~.	+

	<center><math>\sin 20^{\circ} = \sin 160^{\circ}\,\textrm{.}</math></center>		<center><math>\sin 20^{\circ} = \sin 160^{\circ}\,\textrm{.}</math></center>

-	~~This shows that~~	+	und also haben wir die Gleichung

	<center><math>\cos 80^{\circ} \cos 40^{\circ} \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}} = \frac{1}{8}\,\textrm{.}</math></center>		<center><math>\cos 80^{\circ} \cos 40^{\circ} \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}} = \frac{1}{8}\,\textrm{.}</math></center>
		+
		+	gezeigt.

Tekbot: hat „Solution 4.3:9“ nach „Lösung 4.3:9“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 15:09:44 GMT

hat „Solution 4.3:9“ nach „Lösung 4.3:9“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 15:09, 22. Okt. 2008

Tekbot: Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

Tekbot — Wed, 22 Oct 2008 08:57:38 GMT

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 08:57, 22. Okt. 2008
Zeile 1:		Zeile 1:
	Using the formula for double angles on <math>\sin 160^{\circ}</math> gives		Using the formula for double angles on <math>\sin 160^{\circ}</math> gives

-	{{~~Displayed math~~\|\|<math>\sin 160^{\circ} = 2\cos 80^{\circ}\sin 80^{\circ}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\sin 160^{\circ} = 2\cos 80^{\circ}\sin 80^{\circ}\,\textrm{.}</math>}}

	On the right-hand side, we see that the factor <math>\cos 80^{\circ}</math> has appeared, and if we use the formula for double angles on the second factor (<math>\sin 80^{\circ}</math>),		On the right-hand side, we see that the factor <math>\cos 80^{\circ}</math> has appeared, and if we use the formula for double angles on the second factor (<math>\sin 80^{\circ}</math>),

-	{{~~Displayed math~~\|\|<math>2\cos 80^{\circ}\sin 80^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\sin 40^{\circ}\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>2\cos 80^{\circ}\sin 80^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\sin 40^{\circ}\,,</math>}}

	we obtain a further factor <math>\cos 40^{\circ}</math>. A final application of the formula for double angles on <math>\sin 40^{\circ }</math> gives us all three cosine factors,		we obtain a further factor <math>\cos 40^{\circ}</math>. A final application of the formula for double angles on <math>\sin 40^{\circ }</math> gives us all three cosine factors,

-	{{~~Displayed math~~\|\|<math>2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot\sin 40^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot 2\cos 20^{\circ}\sin 20^{\circ}\,\textrm{·}</math>}}	+	{{Abgesetzte Formel\|\|<math>2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot\sin 40^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot 2\cos 20^{\circ}\sin 20^{\circ}\,\textrm{·}</math>}}

	We have thus succeeded in showing that		We have thus succeeded in showing that

-	{{~~Displayed math~~\|\|<math>\sin 160^{\circ} = 8\cos 80^{\circ}\cdot \cos 40^{\circ}\cdot \cos 20^{\circ}\cdot\sin 20^{\circ}</math>}}	+	{{Abgesetzte Formel\|\|<math>\sin 160^{\circ} = 8\cos 80^{\circ}\cdot \cos 40^{\circ}\cdot \cos 20^{\circ}\cdot\sin 20^{\circ}</math>}}

	which can also be written as		which can also be written as

-	{{~~Displayed math~~\|\|<math>\cos 80^{\circ}\cdot\cos 40^{\circ}\cdot \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\cos 80^{\circ}\cdot\cos 40^{\circ}\cdot \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}}\,\textrm{.}</math>}}

	[[Image:4_3_9.gif\|\|right]]		[[Image:4_3_9.gif\|\|right]]

Tek um 11:31, 10. Okt. 2008

Tek — Fri, 10 Oct 2008 11:31:48 GMT

← Nächstältere Version		Version vom 11:31, 10. Okt. 2008
Zeile 1:		Zeile 1:
-	Using the formula for double angles on ~~sin~~	+	Using the formula for double angles on <math>\sin 160^{\circ}</math> gives
-	<math>160^{\circ }</math>	+
-	gives	+

		+	{{Displayed math\|\|<math>\sin 160^{\circ} = 2\cos 80^{\circ}\sin 80^{\circ}\,\textrm{.}</math>}}

-	<math>~~\sin 160^{\circ }=2~~\cos 80^{\circ }\sin 80^{\circ }</math>	+	On the right-hand side, we see that the factor <math>\cos 80^{\circ}</math> has appeared, and if we use the formula for double angles on the second factor (<math>\sin 80^{\circ}</math>),

		+	{{Displayed math\|\|<math>2\cos 80^{\circ}\sin 80^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\sin 40^{\circ}\,,</math>}}

-	~~On the right-hand side, we see that the factor~~	+	we obtain a further factor <math>\cos 40^{\circ}</math>. A final application of the formula for double angles on <math>\sin 40^{\circ }</math> gives us all three cosine factors,
-	~~<math>\cos 80^{\circ }</math>~~	+
-	~~has appeared, and if we use the formula for double angles on the second factor (~~	+
-	~~<math>\sin 80^{\circ }</math>~~	+
-	),	+
-		+
-		+
-	~~<math>2\cos 80^{\circ }\sin 80^{\circ }=2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\sin 40^{\circ }</math>~~	+
-		+
-		+
-	we obtain a further factor	+
-	<math>\cos 40^{\circ }</math>. A final application of the formula for double angles on	+
-	<math>\sin 40^{\circ }</math>	+
-	gives us all three cosine factors:	+
-		+
-		+
-	~~<math>2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\centerdot \sin 40^{\circ }=2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\centerdot 2\cos 20^{\circ }\sin 20^{\circ }</math>~~	+

		+	{{Displayed math\|\|<math>2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot\sin 40^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot 2\cos 20^{\circ}\sin 20^{\circ}\,\textrm{·}</math>}}

	We have thus succeeded in showing that		We have thus succeeded in showing that

-		+	{{Displayed math\|\|<math>\sin 160^{\circ} = 8\cos 80^{\circ}\cdot \cos 40^{\circ}\cdot \cos 20^{\circ}\cdot\sin 20^{\circ}</math>}}
-	<math>\sin 160^{\circ }=8\cos 80^{\circ }\~~centerdot~~ \cos 40^{\circ }\~~centerdot~~ \cos 20^{\circ }\sin 20^{\circ }</math>	+
-		+

	which can also be written as		which can also be written as

		+	{{Displayed math\|\|<math>\cos 80^{\circ}\cdot\cos 40^{\circ}\cdot \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}}\,\textrm{.}</math>}}

-	<math>~~\cos 80~~^{\circ }~~\centerdot \cos 40~~^{\circ }~~\centerdot \cos~~ 20^{\circ }~~=\frac{\sin~~ 160^{\circ ~~}}{8\sin 20^{\circ }~~}</math>	+	[[Image:4_3_9.gif\|\|right]]
		+	If we draw the unit circle, we see that <math>160^{\circ}</math> makes an angle of
		+	<math>20^{\circ}</math> with the negative ''x''-axis, and therefore the angles
		+	<math>20^{\circ}</math> and <math>160^{\circ}</math> have the same ''y''-coordinate in the unit circle, i.e.

-		+	<center><math>\sin 20^{\circ} = \sin 160^{\circ}\,\textrm{.}</math></center>
-	~~If we draw the unit circle, we see that~~	+
-	<~~math~~>~~160^{\circ }</math>~~	+
-	~~makes an angle of~~	+
-	<math>20^{\circ }~~</math>~~	+
-	~~with the negative~~	+
-	~~<math>x</math>~~	+
-	~~-axis, and therefore the angles~~	+
-	~~<math>20^{~~\~~circ }</math>~~	+
-	~~and~~	+
-	~~<math>~~160^{\circ }~~</math>~~	+
-	~~have the same~~	+
-	~~<math>y</math>~~	+
-	~~-coordinate in the unit circle~~, ~~i.e.~~	+
-		+
-	~~<math>~~\~~sin 20^~~{~~\circ }=\sin 160^{\circ~~ }</math>.	+
-		+
-		+
-	~~[[Image:4_3_9.gif\|~~center]]	+

	This shows that		This shows that

-		+	<center><math>\cos 80^{\circ} \cos 40^{\circ} \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}} = \frac{1}{8}\,\textrm{.}</math></center>
-	<math>\cos 80^{\circ }~~\centerdot~~ \cos 40^{\circ }~~\centerdot~~ \cos 20^{\circ }=\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}=\frac{1}{8}</math>	+

Ian um 11:31, 30. Sep. 2008

Ian — Tue, 30 Sep 2008 11:31:02 GMT

← Nächstältere Version		Version vom 11:31, 30. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	Using the formula for double angles on sin
-	<~~center~~> ~~[[Image:4_3_9~~-1(2).~~gif]]~~ </~~center~~>	+	<math>160^{\circ }</math>
-	{{~~NAVCONTENT_STOP~~}}	+	gives
-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image:4_3_9~~-~~2(2).gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	<math>\sin 160^{\circ }=2\cos 80^{\circ }\sin 80^{\circ }</math>
		+
		+
		+	On the right-hand side, we see that the factor
		+	<math>\cos 80^{\circ }</math>
		+	has appeared, and if we use the formula for double angles on the second factor (
		+	<math>\sin 80^{\circ }</math>
		+	),
		+
		+
		+	<math>2\cos 80^{\circ }\sin 80^{\circ }=2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\sin 40^{\circ }</math>
		+
		+
		+	we obtain a further factor
		+	<math>\cos 40^{\circ }</math>. A final application of the formula for double angles on
		+	<math>\sin 40^{\circ }</math>
		+	gives us all three cosine factors:
		+
		+
		+	<math>2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\centerdot \sin 40^{\circ }=2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\centerdot 2\cos 20^{\circ }\sin 20^{\circ }</math>
		+
		+
		+	We have thus succeeded in showing that
		+
		+
		+	<math>\sin 160^{\circ }=8\cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }\sin 20^{\circ }</math>
		+
		+
		+	which can also be written as
		+
		+
		+	<math>\cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }=\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}</math>
		+
		+
		+	If we draw the unit circle, we see that
		+	<math>160^{\circ }</math>
		+	makes an angle of
		+	<math>20^{\circ }</math>
		+	with the negative
		+	<math>x</math>
		+	-axis, and therefore the angles
		+	<math>20^{\circ }</math>
		+	and
		+	<math>160^{\circ }</math>
		+	have the same
		+	<math>y</math>
		+	-coordinate in the unit circle, i.e.
		+
		+	<math>\sin 20^{\circ }=\sin 160^{\circ }</math>.
		+

	[[Image:4_3_9.gif\|center]]		[[Image:4_3_9.gif\|center]]
		+
		+	This shows that
		+
		+
		+	<math>\cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }=\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}=\frac{1}{8}</math>

Tekbot: Lösning 4.3:9 moved to Solution 4.3:9: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:59:23 GMT

Lösning 4.3:9 moved to Solution 4.3:9: Robot: moved page

← Nächstältere Version

Version vom 09:59, 9. Sep. 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

Tekbot — Thu, 21 Aug 2008 07:16:49 GMT

Robot: Automated text replacement (-[[Bild: +[[Image:)

← Nächstältere Version		Version vom 07:16, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_3_9-1(2).gif]] </center>	+	<center> [[Image:4_3_9-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_3_9-2(2).gif]] </center>	+	<center> [[Image:4_3_9-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

-	[[~~Bild~~:4_3_9.gif\|center]]	+	[[Image:4_3_9.gif\|center]]

Madde um 10:16, 19. Aug. 2008

Madde — Tue, 19 Aug 2008 10:16:24 GMT

← Nächstältere Version		Version vom 10:16, 19. Aug. 2008
Zeile 5:		Zeile 5:
	<center> [[Bild:4_3_9-2(2).gif]] </center>		<center> [[Bild:4_3_9-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+
		+	[[Bild:4_3_9.gif\|center]]