Lösung 3.1:5a - Versionsgeschichte

Markus.bez um 09:54, 10. Jun. 2009

2009-06-10T09:54:17Z

← Nächstältere Version		Version vom 09:54, 10. Jun. 2009
Zeile 1:		Zeile 1:
-	Wir erweitern den Bruch mit <math>\sqrt{12}</math>, sodass wir den Nenner <math>\sqrt{12}\cdot\sqrt{12} = 12</math> erhalten	+	Wir erweitern den Bruch mit <math>\sqrt{12}</math>, sodass wir den Nenner <math>\sqrt{12}\cdot\sqrt{12} = 12</math> erhalten:

	{{Abgesetzte Formel\|\|<math>\frac{2}{\sqrt{12}} = \frac{2}{\sqrt{12}}\cdot \frac{\sqrt{12}}{\sqrt{12}} = \frac{2\sqrt{12}}{12} = \frac{2\sqrt{12}}{2\cdot 6} = \frac{\sqrt{12}}{6}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{2}{\sqrt{12}} = \frac{2}{\sqrt{12}}\cdot \frac{\sqrt{12}}{\sqrt{12}} = \frac{2\sqrt{12}}{12} = \frac{2\sqrt{12}}{2\cdot 6} = \frac{\sqrt{12}}{6}\,\textrm{.}</math>}}

-	Dieser Ausdruck kann weiter vereinfacht werden, indem wir 12 ~~wie~~ <math>12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3</math> schreiben	+	Dieser Ausdruck kann weiter vereinfacht werden, indem wir 12 als <math>12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3</math> schreiben

	{{Abgesetzte Formel\|\|<math>\frac{\sqrt{12}}{6} = \frac{\sqrt{2^2 3}}{6} = \frac{2\sqrt{3}}{6}= \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{\sqrt{12}}{6} = \frac{\sqrt{2^2 3}}{6} = \frac{2\sqrt{3}}{6}= \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.}</math>}}

Martin um 21:36, 25. Mär. 2009

2009-03-25T21:36:28Z

← Nächstältere Version		Version vom 21:36, 25. Mär. 2009
Zeile 5:		Zeile 5:
	Dieser Ausdruck kann weiter vereinfacht werden, indem wir 12 wie <math>12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3</math> schreiben		Dieser Ausdruck kann weiter vereinfacht werden, indem wir 12 wie <math>12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3</math> schreiben

-	{{Abgesetzte Formel\|\|<math>\frac{\sqrt{12}}{6} = \frac{2\sqrt{3}}{6}= \frac{\sqrt{~~2^2~~ 3}}{6} = \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{\sqrt{12}}{6} = \frac{\sqrt{2^2 3}}{6} = \frac{2\sqrt{3}}{6}= \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.}</math>}}

Martin um 21:36, 25. Mär. 2009

2009-03-25T21:36:03Z

← Nächstältere Version		Version vom 21:36, 25. Mär. 2009
Zeile 5:		Zeile 5:
	Dieser Ausdruck kann weiter vereinfacht werden, indem wir 12 wie <math>12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3</math> schreiben		Dieser Ausdruck kann weiter vereinfacht werden, indem wir 12 wie <math>12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3</math> schreiben

-	{{Abgesetzte Formel\|\|<math>\frac{\sqrt{12}}{6} = \frac{2\sqrt{3}}{6} = \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{\sqrt{12}}{6} = \frac{2\sqrt{3}}{6}= \frac{\sqrt{2^2 3}}{6} = \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.}</math>}}

Martin um 21:34, 25. Mär. 2009

2009-03-25T21:34:24Z

← Nächstältere Version		Version vom 21:34, 25. Mär. 2009
Zeile 1:		Zeile 1:
-	~~If we multiply the top and bottom of the fraction by~~ <math>\sqrt{12}</math>, ~~the new denominator will be~~ <math>\sqrt{12}\cdot\sqrt{12} = 12</math> ~~and we will get rid of the root sign in the denominator~~	+	Wir erweitern den Bruch mit <math>\sqrt{12}</math>, sodass wir den Nenner <math>\sqrt{12}\cdot\sqrt{12} = 12</math> erhalten

	{{Abgesetzte Formel\|\|<math>\frac{2}{\sqrt{12}} = \frac{2}{\sqrt{12}}\cdot \frac{\sqrt{12}}{\sqrt{12}} = \frac{2\sqrt{12}}{12} = \frac{2\sqrt{12}}{2\cdot 6} = \frac{\sqrt{12}}{6}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{2}{\sqrt{12}} = \frac{2}{\sqrt{12}}\cdot \frac{\sqrt{12}}{\sqrt{12}} = \frac{2\sqrt{12}}{12} = \frac{2\sqrt{12}}{2\cdot 6} = \frac{\sqrt{12}}{6}\,\textrm{.}</math>}}

-	~~This expression can be simplified even further if we write~~ <math>12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3</math> ~~and take <math>2^2</math> out from under the root, we get~~	+	Dieser Ausdruck kann weiter vereinfacht werden, indem wir 12 wie <math>12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3</math> schreiben

	{{Abgesetzte Formel\|\|<math>\frac{\sqrt{12}}{6} = \frac{2\sqrt{3}}{6} = \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{\sqrt{12}}{6} = \frac{2\sqrt{3}}{6} = \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.}</math>}}

Tekbot: hat „Solution 3.1:5a“ nach „Lösung 3.1:5a“ verschoben: Robot: moved page

2008-10-22T14:22:45Z

hat „Solution 3.1:5a“ nach „Lösung 3.1:5a“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:22, 22. Okt. 2008

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

2008-10-22T08:38:08Z

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

← Nächstältere Version		Version vom 08:38, 22. Okt. 2008
Zeile 1:		Zeile 1:
	If we multiply the top and bottom of the fraction by <math>\sqrt{12}</math>, the new denominator will be <math>\sqrt{12}\cdot\sqrt{12} = 12</math> and we will get rid of the root sign in the denominator		If we multiply the top and bottom of the fraction by <math>\sqrt{12}</math>, the new denominator will be <math>\sqrt{12}\cdot\sqrt{12} = 12</math> and we will get rid of the root sign in the denominator

-	{{~~Displayed math~~\|\|<math>\frac{2}{\sqrt{12}} = \frac{2}{\sqrt{12}}\cdot \frac{\sqrt{12}}{\sqrt{12}} = \frac{2\sqrt{12}}{12} = \frac{2\sqrt{12}}{2\cdot 6} = \frac{\sqrt{12}}{6}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{2}{\sqrt{12}} = \frac{2}{\sqrt{12}}\cdot \frac{\sqrt{12}}{\sqrt{12}} = \frac{2\sqrt{12}}{12} = \frac{2\sqrt{12}}{2\cdot 6} = \frac{\sqrt{12}}{6}\,\textrm{.}</math>}}

	This expression can be simplified even further if we write <math>12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3</math> and take <math>2^2</math> out from under the root, we get		This expression can be simplified even further if we write <math>12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3</math> and take <math>2^2</math> out from under the root, we get

-	{{~~Displayed math~~\|\|<math>\frac{\sqrt{12}}{6} = \frac{2\sqrt{3}}{6} = \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{\sqrt{12}}{6} = \frac{2\sqrt{3}}{6} = \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.}</math>}}

Tek um 11:18, 30. Sep. 2008

2008-09-30T11:18:47Z

← Nächstältere Version		Version vom 11:18, 30. Sep. 2008
Zeile 1:		Zeile 1:
-	If we multiply the top and bottom of the fraction by	+	If we multiply the top and bottom of the fraction by <math>\sqrt{12}</math>, the new denominator will be <math>\sqrt{12}\cdot\sqrt{12} = 12</math> and we will get rid of the root sign in the denominator
-	<math>\sqrt{12}</math>, the new denominator will be	+
-	<math>\sqrt{12}\~~centerdot~~ \sqrt{12}=12</math>	+
-	and we will get rid of the root sign in the denominator:	+

		+	{{Displayed math\|\|<math>\frac{2}{\sqrt{12}} = \frac{2}{\sqrt{12}}\cdot \frac{\sqrt{12}}{\sqrt{12}} = \frac{2\sqrt{12}}{12} = \frac{2\sqrt{12}}{2\cdot 6} = \frac{\sqrt{12}}{6}\,\textrm{.}</math>}}

-	<math>~~\frac{2}{\sqrt{~~12}}=~~\frac{~~2}{\~~sqrt{12}}\centerdot \frac{\sqrt{12}}{\sqrt{12}}~~=\~~frac{~~2\~~sqrt{12}}{12}~~=~~\frac{~~2\~~sqrt{12}}{~~2~~\centerdot 6}=\frac{\sqrt{12}}{6}~~</math>	+	This expression can be simplified even further if we write <math>12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3</math> and take <math>2^2</math> out from under the root, we get

-		+	{{Displayed math\|\|<math>\frac{\sqrt{12}}{6} = \frac{2\sqrt{3}}{6} = \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.}</math>}}
-	~~This expression can be simplified even further if we write~~	+
-	~~<math>\text~~{~~12}=\text~~{~~2}\centerdot \text{6}=\text{2}\centerdot \text{2}\centerdot \text{3}=\text{2}^{\text{2}}\centerdot \text{3 }</~~math>	+
-	~~and take~~	+
-	~~<math>\text{2}^{\text{2}}</math>~~	+
-	~~out from under the root, We get~~	+
-		+
-		+
-	<math>\frac{\sqrt{12}}{6}=\frac{2\sqrt{3}}{6}=\frac{2\sqrt{3}}{2\~~centerdot~~ 3}=\frac{\sqrt{3}}{3}.</math>	+

Ian um 14:28, 22. Sep. 2008

2008-09-22T14:28:49Z

← Nächstältere Version		Version vom 14:28, 22. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	If we multiply the top and bottom of the fraction by
-	<~~center~~> ~~[[Image~~:~~3_1_5a.gif]]~~ </~~center~~>	+	<math>\sqrt{12}</math>, the new denominator will be
-	{{~~NAVCONTENT_STOP~~}}	+	<math>\sqrt{12}\centerdot \sqrt{12}=12</math>
		+	and we will get rid of the root sign in the denominator:
		+
		+
		+	<math>\frac{2}{\sqrt{12}}=\frac{2}{\sqrt{12}}\centerdot \frac{\sqrt{12}}{\sqrt{12}}=\frac{2\sqrt{12}}{12}=\frac{2\sqrt{12}}{2\centerdot 6}=\frac{\sqrt{12}}{6}</math>
		+
		+
		+	This expression can be simplified even further if we write
		+	<math>\text{12}=\text{2}\centerdot \text{6}=\text{2}\centerdot \text{2}\centerdot \text{3}=\text{2}^{\text{2}}\centerdot \text{3 }</math>
		+	and take
		+	<math>\text{2}^{\text{2}}</math>
		+	out from under the root, We get
		+
		+
		+	<math>\frac{\sqrt{12}}{6}=\frac{2\sqrt{3}}{6}=\frac{2\sqrt{3}}{2\centerdot 3}=\frac{\sqrt{3}}{3}.</math>

Tekbot: Lösning 3.1:5a moved to Solution 3.1:5a: Robot: moved page

2008-09-09T09:12:23Z

Lösning 3.1:5a moved to Solution 3.1:5a: Robot: moved page

← Nächstältere Version

Version vom 09:12, 9. Sep. 2008

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

2008-08-21T06:53:19Z

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

← Nächstältere Version		Version vom 06:53, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_1_5a.gif]] </center>	+	<center> [[Image:3_1_5a.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}