Lösung 1.2:3b - Versionsgeschichte

Moni um 13:12, 8. Aug. 2009

Moni — Sat, 08 Aug 2009 13:12:53 GMT

← Nächstältere Version		Version vom 13:12, 8. Aug. 2009
Zeile 7:		Zeile 7:
	\end{align}</math>}}		\end{align}</math>}}

-	und alle Nenner haben den Faktor <math>2\cdot 2\cdot 2=8</math> gemeinsam,	+	und alle Nenner haben den Faktor <math>2\cdot 2\cdot 2=8</math> gemeinsam

	{{Abgesetzte Formel\|\|<math>\frac{1}{3\cdot 8}+\frac{1}{5\cdot 8}-\frac{1}{2\cdot 8}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{1}{3\cdot 8}+\frac{1}{5\cdot 8}-\frac{1}{2\cdot 8}\,</math>.}}

-	Also erweitern wir die Brüche nur mit den nicht-gemeinsamen Faktoren, 2, 3 und 5.	+	Also erweitern wir die Brüche nur mit den nicht-gemeinsamen Faktoren 2, 3 und 5.

	{{Abgesetzte Formel\|\|<math>\frac{1\cdot 2\cdot 5}{3\cdot 8\cdot 2\cdot 5}+\frac{1\cdot 2\cdot 3}{5\cdot 8\cdot 2\cdot 3}-\frac{1\cdot 3\cdot 5}{2\cdot 8\cdot 3\cdot 5}=\frac{10}{240}+\frac{6}{240}-\frac{15}{240}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{1\cdot 2\cdot 5}{3\cdot 8\cdot 2\cdot 5}+\frac{1\cdot 2\cdot 3}{5\cdot 8\cdot 2\cdot 3}-\frac{1\cdot 3\cdot 5}{2\cdot 8\cdot 3\cdot 5}=\frac{10}{240}+\frac{6}{240}-\frac{15}{240}\,</math>.}}

-	Der kleinster gemeinsamer Nenner ist also 240, und der Ausdruck ist	+	Der kleinster gemeinsamer Nenner ist also 240 und der Ausdruck ist

	{{Abgesetzte Formel\|\|<math>\frac{10}{240}+\frac{6}{240}-\frac{15}{240}=\frac{10+6-15}{240}=\frac{1}{240}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{10}{240}+\frac{6}{240}-\frac{15}{240}=\frac{10+6-15}{240}=\frac{1}{240}\,</math>.}}

Martin um 12:14, 26. Okt. 2008

Martin — Sun, 26 Oct 2008 12:14:18 GMT

← Nächstältere Version		Version vom 12:14, 26. Okt. 2008
Zeile 1:		Zeile 1:
-	~~If we divide the denominators in succession by 2, we see that~~	+	Die Primfaktoren des Nenners sind

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 7:		Zeile 7:
	\end{align}</math>}}		\end{align}</math>}}

-	~~i.e. they all have a factor~~ <math>2\cdot 2\cdot 2=8</math> ~~in common~~,	+	und alle Nenner haben den Faktor <math>2\cdot 2\cdot 2=8</math> gemeinsam,

	{{Abgesetzte Formel\|\|<math>\frac{1}{3\cdot 8}+\frac{1}{5\cdot 8}-\frac{1}{2\cdot 8}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{1}{3\cdot 8}+\frac{1}{5\cdot 8}-\frac{1}{2\cdot 8}\,</math>.}}

-	~~Hence we do not need to take 8 as a factor when we multiply the top and bottom of each fraction by the product of the other fractions' denominators, but instead we~~	+	Also erweitern wir die Brüche nur mit den nicht-gemeinsamen Faktoren, 2, 3 und 5.
-	~~obtain the lowest common denominator by multiplying top and bottom by the other factors~~, 2, 3 ~~and~~ 5,	+

	{{Abgesetzte Formel\|\|<math>\frac{1\cdot 2\cdot 5}{3\cdot 8\cdot 2\cdot 5}+\frac{1\cdot 2\cdot 3}{5\cdot 8\cdot 2\cdot 3}-\frac{1\cdot 3\cdot 5}{2\cdot 8\cdot 3\cdot 5}=\frac{10}{240}+\frac{6}{240}-\frac{15}{240}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{1\cdot 2\cdot 5}{3\cdot 8\cdot 2\cdot 5}+\frac{1\cdot 2\cdot 3}{5\cdot 8\cdot 2\cdot 3}-\frac{1\cdot 3\cdot 5}{2\cdot 8\cdot 3\cdot 5}=\frac{10}{240}+\frac{6}{240}-\frac{15}{240}\,</math>.}}

-	~~The LCD is~~ 240 ~~and the answer is~~	+	Der kleinster gemeinsamer Nenner ist also 240, und der Ausdruck ist

	{{Abgesetzte Formel\|\|<math>\frac{10}{240}+\frac{6}{240}-\frac{15}{240}=\frac{10+6-15}{240}=\frac{1}{240}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{10}{240}+\frac{6}{240}-\frac{15}{240}=\frac{10+6-15}{240}=\frac{1}{240}\,</math>.}}

Tekbot: hat „Solution 1.2:3b“ nach „Lösung 1.2:3b“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 13:28:25 GMT

hat „Solution 1.2:3b“ nach „Lösung 1.2:3b“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 13:28, 22. Okt. 2008

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

Tekbot — Wed, 22 Oct 2008 08:14:58 GMT

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

← Nächstältere Version		Version vom 08:14, 22. Okt. 2008
Zeile 1:		Zeile 1:
	If we divide the denominators in succession by 2, we see that		If we divide the denominators in succession by 2, we see that

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	24&=2\cdot 2\cdot 2\cdot 3\,,\\		24&=2\cdot 2\cdot 2\cdot 3\,,\\
	40&=2\cdot 2\cdot 2\cdot 5\,,\\		40&=2\cdot 2\cdot 2\cdot 5\,,\\
Zeile 9:		Zeile 9:
	i.e. they all have a factor <math>2\cdot 2\cdot 2=8</math> in common,		i.e. they all have a factor <math>2\cdot 2\cdot 2=8</math> in common,

-	{{~~Displayed math~~\|\|<math>\frac{1}{3\cdot 8}+\frac{1}{5\cdot 8}-\frac{1}{2\cdot 8}\,</math>.}}	+	{{Abgesetzte Formel\|\|<math>\frac{1}{3\cdot 8}+\frac{1}{5\cdot 8}-\frac{1}{2\cdot 8}\,</math>.}}

	Hence we do not need to take 8 as a factor when we multiply the top and bottom of each fraction by the product of the other fractions' denominators, but instead we		Hence we do not need to take 8 as a factor when we multiply the top and bottom of each fraction by the product of the other fractions' denominators, but instead we
	obtain the lowest common denominator by multiplying top and bottom by the other factors, 2, 3 and 5,		obtain the lowest common denominator by multiplying top and bottom by the other factors, 2, 3 and 5,

-	{{~~Displayed math~~\|\|<math>\frac{1\cdot 2\cdot 5}{3\cdot 8\cdot 2\cdot 5}+\frac{1\cdot 2\cdot 3}{5\cdot 8\cdot 2\cdot 3}-\frac{1\cdot 3\cdot 5}{2\cdot 8\cdot 3\cdot 5}=\frac{10}{240}+\frac{6}{240}-\frac{15}{240}\,</math>.}}	+	{{Abgesetzte Formel\|\|<math>\frac{1\cdot 2\cdot 5}{3\cdot 8\cdot 2\cdot 5}+\frac{1\cdot 2\cdot 3}{5\cdot 8\cdot 2\cdot 3}-\frac{1\cdot 3\cdot 5}{2\cdot 8\cdot 3\cdot 5}=\frac{10}{240}+\frac{6}{240}-\frac{15}{240}\,</math>.}}

	The LCD is 240 and the answer is		The LCD is 240 and the answer is

-	{{~~Displayed math~~\|\|<math>\frac{10}{240}+\frac{6}{240}-\frac{15}{240}=\frac{10+6-15}{240}=\frac{1}{240}\,</math>.}}	+	{{Abgesetzte Formel\|\|<math>\frac{10}{240}+\frac{6}{240}-\frac{15}{240}=\frac{10+6-15}{240}=\frac{1}{240}\,</math>.}}

Tek um 08:42, 19. Sep. 2008

Tek — Fri, 19 Sep 2008 08:42:59 GMT

← Nächstältere Version		Version vom 08:42, 19. Sep. 2008
Zeile 1:		Zeile 1:
-	If we divide the denominators in succession by	+	If we divide the denominators in succession by 2, we see that
-	~~<math>~~2~~</math>~~	+
-	, we see that	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	24&=2\cdot 2\cdot 2\cdot 3\,,\\
		+	40&=2\cdot 2\cdot 2\cdot 5\,,\\
		+	16&=2\cdot 2\cdot 2\cdot 2\,,\\
		+	\end{align}</math>}}

-	<math>~~\begin{align}~~	+	i.e. they all have a factor <math>2\cdot 2\cdot 2=8</math> in common,
-	~~& 24=~~2\~~centerdot~~ 2\~~centerdot~~ 2~~\centerdot 3 \\~~	+
-	~~& 40~~=~~2\centerdot 2\centerdot \centerdot 5 \\~~	+
-	~~& 16=2\centerdot 2\centerdot 2\centerdot 2 \\~~	+
-	~~\end{align}~~</math>	+

		+	{{Displayed math\|\|<math>\frac{1}{3\cdot 8}+\frac{1}{5\cdot 8}-\frac{1}{2\cdot 8}\,</math>.}}

-	~~i.e. they all have~~ a factor	+	Hence we do not need to take 8 as a factor when we multiply the top and bottom of each fraction by the product of the other fractions' denominators, but instead we
-	~~<math>~~2~~\centerdot 2\centerdot 2=8</math>~~	+	obtain the lowest common denominator by multiplying top and bottom by the other factors, 2, 3 and 5,
-	~~in common~~,	+

		+	{{Displayed math\|\|<math>\frac{1\cdot 2\cdot 5}{3\cdot 8\cdot 2\cdot 5}+\frac{1\cdot 2\cdot 3}{5\cdot 8\cdot 2\cdot 3}-\frac{1\cdot 3\cdot 5}{2\cdot 8\cdot 3\cdot 5}=\frac{10}{240}+\frac{6}{240}-\frac{15}{240}\,</math>.}}

		+	The LCD is 240 and the answer is

-	~~<math>\frac~~{1}{~~3\centerdot 8}+\frac{1}{5\centerdot 8}-\frac{1}{2\centerdot 8}</~~math>	+	{{Displayed math\|\|<math>\frac{10}{240}+\frac{6}{240}-\frac{15}{240}=\frac{10+6-15}{240}=\frac{1}{240}\,</math>.}}
-	,	+
-		+
-	~~and hence we do not need to take~~	+
-	~~<math>8</math>~~	+
-	~~as a factor when we multiply the top and bottom of each fraction by the product of the other fractions' denominators, but instead we~~	+
-	~~obtain the lowest common denominator by multiplying top and bottom by the other factors,~~	+
-	~~<math>2,\ 3</math>~~	+
-	~~and~~	+
-	~~<math>5</math>~~	+
-	:	+
-		+
-		+
-	<math>\frac{1\centerdot 2\centerdot 5}{3\centerdot 8\centerdot 2\centerdot 5}+\frac{1\centerdot 2\centerdot 3}{5\centerdot 8\centerdot 2\centerdot 3}-\frac{1\centerdot 3\centerdot 5}{2\centerdot 8\centerdot 3\centerdot 5}=\frac{10}{240}+\frac{6}{240}-\frac{15}{240}</math>	+
-		+
-		+
-	~~The LCD is~~	+
-	~~<math>240</math>~~	+
-	~~and the answer is~~	+
-		+
-		+
-	<math>\frac{10}{240}+\frac{6}{240}-\frac{15}{240}=\frac{10+6-15}{240}=\frac{1}{240}</math>	+

Ian um 13:10, 11. Sep. 2008

Ian — Thu, 11 Sep 2008 13:10:46 GMT

← Nächstältere Version		Version vom 13:10, 11. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	If we divide the denominators in succession by
-	<~~center~~> ~~[[Image~~:~~1_2_3b.gif]]~~ </~~center~~>	+	<math>2</math>
-	{{~~NAVCONTENT_STOP~~}}	+	, we see that
		+
		+
		+	<math>\begin{align}
		+	& 24=2\centerdot 2\centerdot 2\centerdot 3 \\
		+	& 40=2\centerdot 2\centerdot \centerdot 5 \\
		+	& 16=2\centerdot 2\centerdot 2\centerdot 2 \\
		+	\end{align}</math>
		+
		+
		+	i.e. they all have a factor
		+	<math>2\centerdot 2\centerdot 2=8</math>
		+	in common,
		+
		+
		+
		+	<math>\frac{1}{3\centerdot 8}+\frac{1}{5\centerdot 8}-\frac{1}{2\centerdot 8}</math>
		+	,
		+
		+	and hence we do not need to take
		+	<math>8</math>
		+	as a factor when we multiply the top and bottom of each fraction by the product of the other fractions' denominators, but instead we
		+	obtain the lowest common denominator by multiplying top and bottom by the other factors,
		+	<math>2,\ 3</math>
		+	and
		+	<math>5</math>
		+	:
		+
		+
		+	<math>\frac{1\centerdot 2\centerdot 5}{3\centerdot 8\centerdot 2\centerdot 5}+\frac{1\centerdot 2\centerdot 3}{5\centerdot 8\centerdot 2\centerdot 3}-\frac{1\centerdot 3\centerdot 5}{2\centerdot 8\centerdot 3\centerdot 5}=\frac{10}{240}+\frac{6}{240}-\frac{15}{240}</math>
		+
		+
		+	The LCD is
		+	<math>240</math>
		+	and the answer is
		+
		+
		+	<math>\frac{10}{240}+\frac{6}{240}-\frac{15}{240}=\frac{10+6-15}{240}=\frac{1}{240}</math>

Tekbot: Lösning 1.2:3b moved to Solution 1.2:3b: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:18:03 GMT

Lösning 1.2:3b moved to Solution 1.2:3b: Robot: moved page

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Version vom 08:18, 9. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 06:26:09 GMT

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

← Nächstältere Version		Version vom 06:26, 21. Aug. 2008
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-	<center> [[~~Bild~~:1_2_3b.gif]] </center>	+	<center> [[Image:1_2_3b.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

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Lina — Wed, 26 Mar 2008 10:00:51 GMT

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