Lösung 1.2:1d - Versionsgeschichte

Moni um 13:08, 8. Aug. 2009

Moni — Sat, 08 Aug 2009 13:08:36 GMT

← Nächstältere Version		Version vom 13:08, 8. Aug. 2009
Zeile 1:		Zeile 1:
-	Mit drei Brüchen, müssen wir jeweiligen Bruch mit dem Produkt des anderen Nenners erweitern, sodass alle Brüche denselben Nenner haben,	+	Mit drei Brüchen, müssen wir jeweiligen Bruch mit dem Produkt des anderen Nenners erweitern, sodass alle Brüche denselben Nenner haben

	{{Abgesetzte Formel\|\|<math>\frac{1\cdot 4\cdot 5}{3\cdot 4\cdot 5}+\frac{1\cdot 3\cdot 5}{4\cdot 3\cdot 5}+\frac{1\cdot 3\cdot 4}{5\cdot 3\cdot 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{1\cdot 4\cdot 5}{3\cdot 4\cdot 5}+\frac{1\cdot 3\cdot 5}{4\cdot 3\cdot 5}+\frac{1\cdot 3\cdot 4}{5\cdot 3\cdot 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}\,</math>.}}

Martin um 11:33, 26. Okt. 2008

Martin — Sun, 26 Oct 2008 11:33:16 GMT

← Nächstältere Version		Version vom 11:33, 26. Okt. 2008
Zeile 1:		Zeile 1:
-	~~When we have three fractions involved in an addition~~, ~~we need to multiply the top and bottom of each fraction by the product of the other fraction's denominators so that all fractions have a common denominator~~,	+	Mit drei Brüchen, müssen wir jeweiligen Bruch mit dem Produkt des anderen Nenners erweitern, sodass alle Brüche denselben Nenner haben,

	{{Abgesetzte Formel\|\|<math>\frac{1\cdot 4\cdot 5}{3\cdot 4\cdot 5}+\frac{1\cdot 3\cdot 5}{4\cdot 3\cdot 5}+\frac{1\cdot 3\cdot 4}{5\cdot 3\cdot 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{1\cdot 4\cdot 5}{3\cdot 4\cdot 5}+\frac{1\cdot 3\cdot 5}{4\cdot 3\cdot 5}+\frac{1\cdot 3\cdot 4}{5\cdot 3\cdot 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}\,</math>.}}

-	~~When all three fractions have a common denominator, they can easily be added up~~	+	Mit gemeinsamen Nenner können wir die Brüche einfach addieren

	{{Abgesetzte Formel\|\|<math>\frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{20+15+12}{60}=\frac{47}{60}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{20+15+12}{60}=\frac{47}{60}\,</math>.}}

Tekbot: hat „Solution 1.2:1d“ nach „Lösung 1.2:1d“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 13:26:04 GMT

hat „Solution 1.2:1d“ nach „Lösung 1.2:1d“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 13:26, 22. Okt. 2008

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

Tekbot — Wed, 22 Oct 2008 08:13:48 GMT

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

← Nächstältere Version		Version vom 08:13, 22. Okt. 2008
Zeile 1:		Zeile 1:
	When we have three fractions involved in an addition, we need to multiply the top and bottom of each fraction by the product of the other fraction's denominators so that all fractions have a common denominator,		When we have three fractions involved in an addition, we need to multiply the top and bottom of each fraction by the product of the other fraction's denominators so that all fractions have a common denominator,

-	{{~~Displayed math~~\|\|<math>\frac{1\cdot 4\cdot 5}{3\cdot 4\cdot 5}+\frac{1\cdot 3\cdot 5}{4\cdot 3\cdot 5}+\frac{1\cdot 3\cdot 4}{5\cdot 3\cdot 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}\,</math>.}}	+	{{Abgesetzte Formel\|\|<math>\frac{1\cdot 4\cdot 5}{3\cdot 4\cdot 5}+\frac{1\cdot 3\cdot 5}{4\cdot 3\cdot 5}+\frac{1\cdot 3\cdot 4}{5\cdot 3\cdot 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}\,</math>.}}

	When all three fractions have a common denominator, they can easily be added up		When all three fractions have a common denominator, they can easily be added up

-	{{~~Displayed math~~\|\|<math>\frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{20+15+12}{60}=\frac{47}{60}\,</math>.}}	+	{{Abgesetzte Formel\|\|<math>\frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{20+15+12}{60}=\frac{47}{60}\,</math>.}}

Tek um 07:36, 19. Sep. 2008

Tek — Fri, 19 Sep 2008 07:36:52 GMT

← Nächstältere Version		Version vom 07:36, 19. Sep. 2008
Zeile 1:		Zeile 1:
	When we have three fractions involved in an addition, we need to multiply the top and bottom of each fraction by the product of the other fraction's denominators so that all fractions have a common denominator,		When we have three fractions involved in an addition, we need to multiply the top and bottom of each fraction by the product of the other fraction's denominators so that all fractions have a common denominator,

-	~~<center>~~<math>\frac{1\~~centerdot~~ 4\~~centerdot~~ 5}{3\~~centerdot~~ 4\~~centerdot~~ 5}+\frac{1\~~centerdot~~ 3\~~centerdot~~ 5}{4\~~centerdot~~ 3\~~centerdot~~ 5}+\frac{1\~~centerdot~~ 3\~~centerdot~~ 4}{5\~~centerdot~~ 3\~~centerdot~~ 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}</math>~~</center>~~	+	{{Displayed math\|\|<math>\frac{1\cdot 4\cdot 5}{3\cdot 4\cdot 5}+\frac{1\cdot 3\cdot 5}{4\cdot 3\cdot 5}+\frac{1\cdot 3\cdot 4}{5\cdot 3\cdot 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}\,</math>.}}

	When all three fractions have a common denominator, they can easily be added up		When all three fractions have a common denominator, they can easily be added up

-	~~<center>~~<math>\frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{20+15+12}{60}=\frac{47}{60}</math>~~</center>~~	+	{{Displayed math\|\|<math>\frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{20+15+12}{60}=\frac{47}{60}\,</math>.}}

Ian um 12:08, 11. Sep. 2008

Ian — Thu, 11 Sep 2008 12:08:58 GMT

← Nächstältere Version		Version vom 12:08, 11. Sep. 2008
Zeile 1:		Zeile 1:
	When we have three fractions involved in an addition, we need to multiply the top and bottom of each fraction by the product of the other fraction's denominators so that all fractions have a common denominator,		When we have three fractions involved in an addition, we need to multiply the top and bottom of each fraction by the product of the other fraction's denominators so that all fractions have a common denominator,

-		+	<center><math>\frac{1\centerdot 4\centerdot 5}{3\centerdot 4\centerdot 5}+\frac{1\centerdot 3\centerdot 5}{4\centerdot 3\centerdot 5}+\frac{1\centerdot 3\centerdot 4}{5\centerdot 3\centerdot 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}</math></center>
-	<math>\frac{1\centerdot 4\centerdot 5}{3\centerdot 4\centerdot 5}+\frac{1\centerdot 3\centerdot 5}{4\centerdot 3\centerdot 5}+\frac{1\centerdot 3\centerdot 4}{5\centerdot 3\centerdot 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}</math>	+

	When all three fractions have a common denominator, they can easily be added up		When all three fractions have a common denominator, they can easily be added up

-		+	<center><math>\frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{20+15+12}{60}=\frac{47}{60}</math></center>
-	<math>\frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{20+15+12}{60}=\frac{47}{60}</math>	+

Ian um 12:05, 11. Sep. 2008

Ian — Thu, 11 Sep 2008 12:05:49 GMT

← Nächstältere Version		Version vom 12:05, 11. Sep. 2008
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-	{{~~NAVCONTENT_START~~}}	+	When we have three fractions involved in an addition, we need to multiply the top and bottom of each fraction by the product of the other fraction's denominators so that all fractions have a common denominator,
-	~~<center> [[Image:1_2_1d.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\frac{1\centerdot 4\centerdot 5}{3\centerdot 4\centerdot 5}+\frac{1\centerdot 3\centerdot 5}{4\centerdot 3\centerdot 5}+\frac{1\centerdot 3\centerdot 4}{5\centerdot 3\centerdot 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}</math>
		+
		+	When all three fractions have a common denominator, they can easily be added up
		+
		+
		+	<math>\frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{20+15+12}{60}=\frac{47}{60}</math>

Tekbot: Lösning 1.2:1d moved to Solution 1.2:1d: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:15:43 GMT

Lösning 1.2:1d moved to Solution 1.2:1d: Robot: moved page

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Version vom 08:15, 9. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 06:24:29 GMT

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

← Nächstältere Version		Version vom 06:24, 21. Aug. 2008
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Lina — Wed, 26 Mar 2008 09:49:29 GMT

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