Lösung 1.2:2a - Versionsgeschichte

Moni: Rechtschreibung

Moni — Sat, 08 Aug 2009 13:09:29 GMT

Rechtschreibung

← Nächstältere Version		Version vom 13:09, 8. Aug. 2009
Zeile 9:		Zeile 9:
	{{Abgesetzte Formel\|\|<math>\frac{1}{2\cdot 3}+\frac{1}{2\cdot 5}\,</math>,}}		{{Abgesetzte Formel\|\|<math>\frac{1}{2\cdot 3}+\frac{1}{2\cdot 5}\,</math>,}}

-	Nachdem beide Nenner den Faktor 2 enthalten, müssen wir die beiden Brüche nicht mit ~~diesen~~ Faktor erweitern.	+	Nachdem beide Nenner den Faktor 2 enthalten, müssen wir die beiden Brüche nicht mit diesem Faktor erweitern.

	{{Abgesetzte Formel\|\|<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\cdot \frac{5}{5}+\frac{1}{10}\cdot \frac{3}{3}=\frac{5}{30}+\frac{3}{30}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\cdot \frac{5}{5}+\frac{1}{10}\cdot \frac{3}{3}=\frac{5}{30}+\frac{3}{30}\,</math>.}}

	Also ist der kleinster gemeinsamer Nenner 30.		Also ist der kleinster gemeinsamer Nenner 30.

Martin um 09:03, 2. Jun. 2009

Martin — Tue, 02 Jun 2009 09:03:37 GMT

← Nächstältere Version		Version vom 09:03, 2. Jun. 2009
Zeile 9:		Zeile 9:
	{{Abgesetzte Formel\|\|<math>\frac{1}{2\cdot 3}+\frac{1}{2\cdot 5}\,</math>,}}		{{Abgesetzte Formel\|\|<math>\frac{1}{2\cdot 3}+\frac{1}{2\cdot 5}\,</math>,}}

-	~~Nachde~~ beide Nenner den Faktor 2 enthalten, müssen wir die beiden Brüche nicht mit diesen Faktor erweitern.	+	Nachdem beide Nenner den Faktor 2 enthalten, müssen wir die beiden Brüche nicht mit diesen Faktor erweitern.

	{{Abgesetzte Formel\|\|<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\cdot \frac{5}{5}+\frac{1}{10}\cdot \frac{3}{3}=\frac{5}{30}+\frac{3}{30}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\cdot \frac{5}{5}+\frac{1}{10}\cdot \frac{3}{3}=\frac{5}{30}+\frac{3}{30}\,</math>.}}

	Also ist der kleinster gemeinsamer Nenner 30.		Also ist der kleinster gemeinsamer Nenner 30.

Martin um 11:45, 26. Okt. 2008

Martin — Sun, 26 Oct 2008 11:45:43 GMT

← Nächstältere Version		Version vom 11:45, 26. Okt. 2008
Zeile 1:		Zeile 1:
-	~~A common way to calculate the expression in the exercise is to multiply top and bottom of each fraction by the other fraction's denominator, so as to obtain a common denominator~~,	+	Um einen gemeinsamen Nenner zu finden, reicht es jeweiligen Bruch mit den Nenner des anderen Bruches zu erweitern.

	{{Abgesetzte Formel\|\|<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\cdot \frac{10}{10}+\frac{1}{10}\cdot \frac{6}{6}=\frac{10}{60}+\frac{6}{60}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\cdot \frac{10}{10}+\frac{1}{10}\cdot \frac{6}{6}=\frac{10}{60}+\frac{6}{60}\,</math>.}}

-	~~However, this gives a common denominator~~, 60, ~~which is larger than it really needs to be~~.	+	Dieser Gemeinsame Nenner, 60, ist aber nicht der kleinster gemeinsamer Nenner.

-	~~If we instead divide up the fractions' denominators into their smallest possible integral factors~~,	+	Indem wir die Nenner in ihre Primfaktoren zerlegen, können wir den kleinsten gemeinsamen Nenner bestimmen.

	{{Abgesetzte Formel\|\|<math>\frac{1}{2\cdot 3}+\frac{1}{2\cdot 5}\,</math>,}}		{{Abgesetzte Formel\|\|<math>\frac{1}{2\cdot 3}+\frac{1}{2\cdot 5}\,</math>,}}

-	~~we see that both denominators contain the factor~~ 2 ~~and it is then unnecessary to include that factor when we multiply the top and bottom of each fraction~~.	+	Nachde beide Nenner den Faktor 2 enthalten, müssen wir die beiden Brüche nicht mit diesen Faktor erweitern.

	{{Abgesetzte Formel\|\|<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\cdot \frac{5}{5}+\frac{1}{10}\cdot \frac{3}{3}=\frac{5}{30}+\frac{3}{30}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\cdot \frac{5}{5}+\frac{1}{10}\cdot \frac{3}{3}=\frac{5}{30}+\frac{3}{30}\,</math>.}}

-	~~This gives the lowest common denominator (LCD),~~ 30.	+	Also ist der kleinster gemeinsamer Nenner 30.

Tekbot: hat „Solution 1.2:2a“ nach „Lösung 1.2:2a“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 13:26:54 GMT

hat „Solution 1.2:2a“ nach „Lösung 1.2:2a“ 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 08:14:08 GMT

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

← Nächstältere Version		Version vom 08:14, 22. Okt. 2008
Zeile 1:		Zeile 1:
	A common way to calculate the expression in the exercise is to multiply top and bottom of each fraction by the other fraction's denominator, so as to obtain a common denominator,		A common way to calculate the expression in the exercise is to multiply top and bottom of each fraction by the other fraction's denominator, so as to obtain a common denominator,

-	{{~~Displayed math~~\|\|<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\cdot \frac{10}{10}+\frac{1}{10}\cdot \frac{6}{6}=\frac{10}{60}+\frac{6}{60}\,</math>.}}	+	{{Abgesetzte Formel\|\|<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\cdot \frac{10}{10}+\frac{1}{10}\cdot \frac{6}{6}=\frac{10}{60}+\frac{6}{60}\,</math>.}}

	However, this gives a common denominator, 60, which is larger than it really needs to be.		However, this gives a common denominator, 60, which is larger than it really needs to be.
Zeile 7:		Zeile 7:
	If we instead divide up the fractions' denominators into their smallest possible integral factors,		If we instead divide up the fractions' denominators into their smallest possible integral factors,

-	{{~~Displayed math~~\|\|<math>\frac{1}{2\cdot 3}+\frac{1}{2\cdot 5}\,</math>,}}	+	{{Abgesetzte Formel\|\|<math>\frac{1}{2\cdot 3}+\frac{1}{2\cdot 5}\,</math>,}}

	we see that both denominators contain the factor 2 and it is then unnecessary to include that factor when we multiply the top and bottom of each fraction.		we see that both denominators contain the factor 2 and it is then unnecessary to include that factor when we multiply the top and bottom of each fraction.

-	{{~~Displayed math~~\|\|<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\cdot \frac{5}{5}+\frac{1}{10}\cdot \frac{3}{3}=\frac{5}{30}+\frac{3}{30}\,</math>.}}	+	{{Abgesetzte Formel\|\|<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\cdot \frac{5}{5}+\frac{1}{10}\cdot \frac{3}{3}=\frac{5}{30}+\frac{3}{30}\,</math>.}}

	This gives the lowest common denominator (LCD), 30.		This gives the lowest common denominator (LCD), 30.

Tek um 07:42, 19. Sep. 2008

Tek — Fri, 19 Sep 2008 07:42:21 GMT

← Nächstältere Version		Version vom 07:42, 19. Sep. 2008
Zeile 1:		Zeile 1:
	A common way to calculate the expression in the exercise is to multiply top and bottom of each fraction by the other fraction's denominator, so as to obtain a common denominator,		A common way to calculate the expression in the exercise is to multiply top and bottom of each fraction by the other fraction's denominator, so as to obtain a common denominator,

-		+	{{Displayed math\|\|<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\cdot \frac{10}{10}+\frac{1}{10}\cdot \frac{6}{6}=\frac{10}{60}+\frac{6}{60}\,</math>.}}
-	<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\~~centerdot~~ \frac{10}{10}+\frac{1}{10}\~~centerdot~~ \frac{6}{6}=\frac{10}{60}+\frac{6}{60}</math>	+

	However, this gives a common denominator, 60, which is larger than it really needs to be.		However, this gives a common denominator, 60, which is larger than it really needs to be.
Zeile 8:		Zeile 7:
	If we instead divide up the fractions' denominators into their smallest possible integral factors,		If we instead divide up the fractions' denominators into their smallest possible integral factors,

-		+	{{Displayed math\|\|<math>\frac{1}{2\cdot 3}+\frac{1}{2\cdot 5}\,</math>,}}
-	<math>\frac{1}{2\~~centerdot~~ 3}+\frac{1}{2\~~centerdot~~ 5}</math>	+

	we see that both denominators contain the factor 2 and it is then unnecessary to include that factor when we multiply the top and bottom of each fraction.		we see that both denominators contain the factor 2 and it is then unnecessary to include that factor when we multiply the top and bottom of each fraction.

-		+	{{Displayed math\|\|<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\cdot \frac{5}{5}+\frac{1}{10}\cdot \frac{3}{3}=\frac{5}{30}+\frac{3}{30}\,</math>.}}
-	<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\~~centerdot~~ \frac{5}{5}+\frac{1}{10}\~~centerdot~~ \frac{3}{3}=\frac{5}{30}+\frac{3}{30}</math>	+
-		+

	This gives the lowest common denominator (LCD), 30.		This gives the lowest common denominator (LCD), 30.

Ian um 12:22, 11. Sep. 2008

Ian — Thu, 11 Sep 2008 12:22:14 GMT

← Nächstältere Version		Version vom 12:22, 11. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	A common way to calculate the expression in the exercise is to multiply top and bottom of each fraction by the other fraction's denominator, so as to obtain a common denominator,
-	<~~center~~> ~~[[Image:1_2_2a~~.~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\centerdot \frac{10}{10}+\frac{1}{10}\centerdot \frac{6}{6}=\frac{10}{60}+\frac{6}{60}</math>
		+
		+	However, this gives a common denominator, 60, which is larger than it really needs to be.
		+
		+	If we instead divide up the fractions' denominators into their smallest possible integral factors,
		+
		+
		+	<math>\frac{1}{2\centerdot 3}+\frac{1}{2\centerdot 5}</math>
		+
		+	we see that both denominators contain the factor 2 and it is then unnecessary to include that factor when we multiply the top and bottom of each fraction.
		+
		+
		+	<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\centerdot \frac{5}{5}+\frac{1}{10}\centerdot \frac{3}{3}=\frac{5}{30}+\frac{3}{30}</math>
		+
		+
		+	This gives the lowest common denominator (LCD), 30.

Tekbot: Lösning 1.2:2a moved to Solution 1.2:2a: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:16:23 GMT

Lösning 1.2:2a moved to Solution 1.2:2a: Robot: moved page

← Nächstältere Version

Version vom 08:16, 9. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 06:25:09 GMT

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

← Nächstältere Version		Version vom 06:25, 21. Aug. 2008
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-	<center> [[~~Bild~~:1_2_2a.gif]] </center>	+	<center> [[Image:1_2_2a.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Lina: Ny sida: {{NAVCONTENT_START}} Bild:1_2_2a.gif {{NAVCONTENT_STOP}}

Lina — Wed, 26 Mar 2008 09:50:48 GMT

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