Lösung 2.1:4c - Versionsgeschichte

Markus.bez um 08:16, 9. Jun. 2009

Markus.bez — Tue, 09 Jun 2009 08:16:01 GMT

← Nächstältere Version		Version vom 08:16, 9. Jun. 2009
Zeile 1:		Zeile 1:
	Anstatt alle Terme miteinander zu multiplizieren, untersuchen wir, welche Terme miteinander multipliziert <math>x</math>- und <math>x^2</math>-Terme ergeben		Anstatt alle Terme miteinander zu multiplizieren, untersuchen wir, welche Terme miteinander multipliziert <math>x</math>- und <math>x^2</math>-Terme ergeben

-	Beim <math>x</math>-Term sehen wir, dass es nur [[eine]] Kombination von Termen gibt, die multipliziert miteinander einen <math>x</math>-Term ergeben	+	Beim <math>x</math>-Term sehen wir, dass es nur ''eine'' Kombination von Termen gibt, die multipliziert miteinander einen <math>x</math>-Term ergeben

	{{Abgesetzte Formel\|\|<math>(\underline{x}-x^{3}+x^{5})(\underline{1}+3x+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 1\cdot 2} + \cdots</math>}}		{{Abgesetzte Formel\|\|<math>(\underline{x}-x^{3}+x^{5})(\underline{1}+3x+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 1\cdot 2} + \cdots</math>}}

Markus.bez um 08:15, 9. Jun. 2009

Markus.bez — Tue, 09 Jun 2009 08:15:37 GMT

← Nächstältere Version		Version vom 08:15, 9. Jun. 2009
Zeile 1:		Zeile 1:
-	Anstatt alle Terme miteinander zu multiplizieren, untersuchen wir welche Terme, multipliziert ~~miteinander,~~ <math>x</math>- und <math>x^2</math>-Terme ergeben	+	Anstatt alle Terme miteinander zu multiplizieren, untersuchen wir, welche Terme miteinander multipliziert <math>x</math>- und <math>x^2</math>-Terme ergeben

-	~~Bei den~~ <math>x</math>-Term sehen wir dass es nur eine Kombination von Termen gibt die multipliziert miteinander einen <math>x</math>-Term ergeben	+	Beim <math>x</math>-Term sehen wir, dass es nur [[eine]] Kombination von Termen gibt, die multipliziert miteinander einen <math>x</math>-Term ergeben

	{{Abgesetzte Formel\|\|<math>(\underline{x}-x^{3}+x^{5})(\underline{1}+3x+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 1\cdot 2} + \cdots</math>}}		{{Abgesetzte Formel\|\|<math>(\underline{x}-x^{3}+x^{5})(\underline{1}+3x+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 1\cdot 2} + \cdots</math>}}
	Der Koeffizient von ''x'' ist also <math>1\cdot 2 = 2\,</math>.		Der Koeffizient von ''x'' ist also <math>1\cdot 2 = 2\,</math>.

-	Auch für den <math>x^2</math>-Term gibt es nur eine Möglichkeit die Terme miteinander zu multiplizieren	+	Auch für den <math>x^2</math>-Term gibt es nur eine Möglichkeit, die Terme miteinander zu multiplizieren

	{{Abgesetzte Formel\|\|<math>(\underline{x}-x^{3}+x^{5})(1+\underline{3x}+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 3x\cdot 2} + \cdots</math>}}		{{Abgesetzte Formel\|\|<math>(\underline{x}-x^{3}+x^{5})(1+\underline{3x}+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 3x\cdot 2} + \cdots</math>}}

-	Der ~~Koefficient~~ von <math>x^2</math> ist also <math>3\cdot 2 = 6\,</math>.	+	Der Koeffizient von <math>x^2</math> ist also <math>3\cdot 2 = 6\,</math>.

Martin um 19:29, 28. Feb. 2009

Martin — Sat, 28 Feb 2009 19:29:46 GMT

← Nächstältere Version		Version vom 19:29, 28. Feb. 2009
Zeile 1:		Zeile 1:
-	~~Instead of multiplying together the whole expression~~, ~~and then reading off the coefficients~~, ~~we investigate which terms from the three brackets together give terms in ''~~x~~''¹ and ''~~x~~''².~~	+	Anstatt alle Terme miteinander zu multiplizieren, untersuchen wir welche Terme, multipliziert miteinander, <math>x</math>- und <math>x^2</math>-Terme ergeben

-	~~If we start with the term in ''~~x~~'', we see that there is only one combination of a term from each bracket which, when multiplied, gives ''~~x~~''¹,~~	+	Bei den <math>x</math>-Term sehen wir dass es nur eine Kombination von Termen gibt die multipliziert miteinander einen <math>x</math>-Term ergeben

	{{Abgesetzte Formel\|\|<math>(\underline{x}-x^{3}+x^{5})(\underline{1}+3x+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 1\cdot 2} + \cdots</math>}}		{{Abgesetzte Formel\|\|<math>(\underline{x}-x^{3}+x^{5})(\underline{1}+3x+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 1\cdot 2} + \cdots</math>}}
		+	Der Koeffizient von ''x'' ist also <math>1\cdot 2 = 2\,</math>.

-	~~so, the coefficient in front of ''x'' is~~ <math>~~1\cdot~~ 2 ~~= 2\,~~</math>.	+	Auch für den <math>x^2</math>-Term gibt es nur eine Möglichkeit die Terme miteinander zu multiplizieren
-		+
-	~~As for ''x''², we also have only one possible combination~~	+

	{{Abgesetzte Formel\|\|<math>(\underline{x}-x^{3}+x^{5})(1+\underline{3x}+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 3x\cdot 2} + \cdots</math>}}		{{Abgesetzte Formel\|\|<math>(\underline{x}-x^{3}+x^{5})(1+\underline{3x}+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 3x\cdot 2} + \cdots</math>}}

-	~~The coefficient in front of ''~~x~~''² is~~ <math>3\cdot 2 = 6\,</math>.	+	Der Koefficient von <math>x^2</math> ist also <math>3\cdot 2 = 6\,</math>.

Tekbot: hat „Solution 2.1:4c“ nach „Lösung 2.1:4c“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 13:47:45 GMT

hat „Solution 2.1:4c“ nach „Lösung 2.1:4c“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 13:47, 22. Okt. 2008

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

Tekbot — Wed, 22 Oct 2008 08:23:48 GMT

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

← Nächstältere Version		Version vom 08:23, 22. Okt. 2008
Zeile 3:		Zeile 3:
	If we start with the term in ''x'', we see that there is only one combination of a term from each bracket which, when multiplied, gives ''x''¹,		If we start with the term in ''x'', we see that there is only one combination of a term from each bracket which, when multiplied, gives ''x''¹,

-	{{~~Displayed math~~\|\|<math>(\underline{x}-x^{3}+x^{5})(\underline{1}+3x+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 1\cdot 2} + \cdots</math>}}	+	{{Abgesetzte Formel\|\|<math>(\underline{x}-x^{3}+x^{5})(\underline{1}+3x+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 1\cdot 2} + \cdots</math>}}

	so, the coefficient in front of ''x'' is <math>1\cdot 2 = 2\,</math>.		so, the coefficient in front of ''x'' is <math>1\cdot 2 = 2\,</math>.
Zeile 9:		Zeile 9:
	As for ''x''², we also have only one possible combination		As for ''x''², we also have only one possible combination

-	{{~~Displayed math~~\|\|<math>(\underline{x}-x^{3}+x^{5})(1+\underline{3x}+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 3x\cdot 2} + \cdots</math>}}	+	{{Abgesetzte Formel\|\|<math>(\underline{x}-x^{3}+x^{5})(1+\underline{3x}+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 3x\cdot 2} + \cdots</math>}}

	The coefficient in front of ''x''² is <math>3\cdot 2 = 6\,</math>.		The coefficient in front of ''x''² is <math>3\cdot 2 = 6\,</math>.

Tek um 10:26, 23. Sep. 2008

Tek — Tue, 23 Sep 2008 10:26:40 GMT

← Nächstältere Version		Version vom 10:26, 23. Sep. 2008
Zeile 1:		Zeile 1:
-	Instead of multiplying together the whole expression, and then reading off the coefficients, we investigate which terms from the three brackets together give terms in	+	Instead of multiplying together the whole expression, and then reading off the coefficients, we investigate which terms from the three brackets together give terms in ''x''¹ and ''x''².
-	~~<math>~~x~~^{1}</math>~~	+
-	and	+
-	~~<math>~~x~~^{2}</math>~~.	+

-	If we start with the term in	+	If we start with the term in ''x'', we see that there is only one combination of a term from each bracket which, when multiplied, gives ''x''¹,
-	~~<math>~~x~~</math>~~, we see that there is only one combination of a term from each bracket	+
-	which, when multiplied, gives	+
-	~~<math>~~x~~^{1}</math>~~,	+

		+	{{Displayed math\|\|<math>(\underline{x}-x^{3}+x^{5})(\underline{1}+3x+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 1\cdot 2} + \cdots</math>}}

-	<math>~~\left( x-x^{3}+x^{5} \right)\left(~~ 1~~+3x+5x^{2}~~ \~~right)\left(~~ 2~~-7x^{~~2~~}-x^{4}~~ \~~right)=...+x\centerdot 1\centerdot 2+...~~</math>	+	so, the coefficient in front of ''x'' is <math>1\cdot 2 = 2\,</math>.

-	~~so, the coefficient in front of~~	+	As for ''x''², we also have only one possible combination
-	~~<math>~~x~~</math>~~	+
-	is	+
-	~~<math>1\centerdot 2=2</math>.~~	+

-	~~As for~~	+	{{Displayed math\|\|<math>(\underline{x}-x^{3}+x^{5})(1+\underline{3x}+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 3x\cdot 2} + \cdots</math>}}
-	<math>x^{2}</math>~~, we also have only one possible combination:~~	+

-		+	The coefficient in front of ''x''² is <math>3\cdot 2 = 6\,</math>.
-	~~<math>\left( x-x^{3}+x^{5} \right)\left( 1+3x+5x^{2} \right)\left( 2-7x^{2}-x^{4} \right)=...+x\centerdot 3x\centerdot 2+...</math>~~	+
-		+
-		+
-	The coefficient in front of	+
-	~~<math>~~x~~^{2}</math>~~	+
-	is	+
-	<math>3\~~centerdot~~ 2=6</math>	+

Ian um 14:36, 15. Sep. 2008

Ian — Mon, 15 Sep 2008 14:36:02 GMT

← Nächstältere Version		Version vom 14:36, 15. Sep. 2008
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-	{{~~NAVCONTENT_START~~}}	+	Instead of multiplying together the whole expression, and then reading off the coefficients, we investigate which terms from the three brackets together give terms in
-	<~~center~~> ~~[[Image:2_1_4c~~.~~gif]]~~ </~~center~~>	+	<math>x^{1}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
		+	<math>x^{2}</math>.
		+
		+	If we start with the term in
		+	<math>x</math>, we see that there is only one combination of a term from each bracket
		+	which, when multiplied, gives
		+	<math>x^{1}</math>,
		+
		+
		+	<math>\left( x-x^{3}+x^{5} \right)\left( 1+3x+5x^{2} \right)\left( 2-7x^{2}-x^{4} \right)=...+x\centerdot 1\centerdot 2+...</math>
		+
		+	so, the coefficient in front of
		+	<math>x</math>
		+	is
		+	<math>1\centerdot 2=2</math>.
		+
		+	As for
		+	<math>x^{2}</math>, we also have only one possible combination:
		+
		+
		+	<math>\left( x-x^{3}+x^{5} \right)\left( 1+3x+5x^{2} \right)\left( 2-7x^{2}-x^{4} \right)=...+x\centerdot 3x\centerdot 2+...</math>
		+
		+
		+	The coefficient in front of
		+	<math>x^{2}</math>
		+	is
		+	<math>3\centerdot 2=6</math>

Tekbot: Lösning 2.1:4c moved to Solution 2.1:4c: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:37:23 GMT

Lösning 2.1:4c moved to Solution 2.1:4c: Robot: moved page

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Version vom 08:37, 9. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 06:35:49 GMT

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

← Nächstältere Version		Version vom 06:35, 21. Aug. 2008
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Lina — Mon, 31 Mar 2008 10:01:53 GMT

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