Lösung 3.1:2c - Versionsgeschichte

Moni um 12:01, 22. Aug. 2009

Moni — Sat, 22 Aug 2009 12:01:14 GMT

← Nächstältere Version		Version vom 12:01, 22. Aug. 2009
Zeile 8:		Zeile 8:
	\end{align}</math>}}		\end{align}</math>}}

-	Danach erweitern wir den Bruch mit ~~den~~ konjugiert komplexen ~~nenner~~,	+	Danach erweitern wir den Bruch mit dem konjugiert komplexen Nenner,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}

Martin um 08:53, 12. Mai 2009

Martin — Tue, 12 May 2009 08:53:46 GMT

← Nächstältere Version		Version vom 08:53, 12. Mai 2009
Zeile 1:		Zeile 1:
-	~~We start by expanding the quadratic in the numerator~~,	+	Wir erweitern zuerst die Quadrate mit der binomischen Formel,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 8:		Zeile 8:
	\end{align}</math>}}		\end{align}</math>}}

-	~~Then, we multiply top and bottom by the complex conjugate of the numerator~~,	+	Danach erweitern wir den Bruch mit den konjugiert komplexen nenner,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}

Tekbot: Solution 3.1:2c moved to Lösung 3.1:2c: Robot: moved page

Tekbot — Wed, 11 Mar 2009 10:31:15 GMT

Solution 3.1:2c moved to Lösung 3.1:2c: Robot: moved page

← Nächstältere Version

Version vom 10:31, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 13:06:04 GMT

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

← Nächstältere Version		Version vom 13:06, 10. Mär. 2009
Zeile 1:		Zeile 1:
	We start by expanding the quadratic in the numerator,		We start by expanding the quadratic in the numerator,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{(2-i\sqrt{3})^2}{1+i\sqrt{3}}		\frac{(2-i\sqrt{3})^2}{1+i\sqrt{3}}
	&= \frac{2^2-2\cdot 2\cdot i\sqrt{3}+(i\sqrt{3})^2}{1+i\sqrt{3}}\\[5pt]		&= \frac{2^2-2\cdot 2\cdot i\sqrt{3}+(i\sqrt{3})^2}{1+i\sqrt{3}}\\[5pt]
Zeile 10:		Zeile 10:
	Then, we multiply top and bottom by the complex conjugate of the numerator,		Then, we multiply top and bottom by the complex conjugate of the numerator,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{1-4\sqrt{3}i}{1+i\sqrt{3}}		\frac{1-4\sqrt{3}i}{1+i\sqrt{3}}
	&= \frac{(1-4\sqrt{3}i)(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})}\\[5pt]		&= \frac{(1-4\sqrt{3}i)(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})}\\[5pt]

Tek um 15:32, 29. Okt. 2008

Tek — Wed, 29 Oct 2008 15:32:08 GMT

← Nächstältere Version		Version vom 15:32, 29. Okt. 2008
Zeile 1:		Zeile 1:
-	~~{{NAVCONTENT_START}}~~	+	We start by expanding the quadratic in the numerator,
-	We start by expanding the quadratic in the numerator ~~with the square rule:~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{(2-i\sqrt{3})^2}{1+i\sqrt{3}}
		+	&= \frac{2^2-2\cdot 2\cdot i\sqrt{3}+(i\sqrt{3})^2}{1+i\sqrt{3}}\\[5pt]
		+	&= \frac{4-4\sqrt{3}i-3}{1+i\sqrt{3}}\\[5pt]
		+	&= \frac{1-4\sqrt{3}i}{1+i\sqrt{3}}\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}\frac{(2-i\sqrt{3})^2}{1+i\sqrt{3}}&= \frac{2^2-2\cdot 2\cdot i\sqrt{3}+(i\sqrt{3})^2}{1+i\sqrt{3}}\\~~	+	Then, we multiply top and bottom by the complex conjugate of the numerator,
-	~~&=\frac{4-4\sqrt{3}i-3}{1+i\sqrt{3}}\\~~	+
-	~~&=\frac{1-4\sqrt{3}i}{1+i\sqrt{3}}\end{align}</math>~~	+

-		+	{{Displayed math\|\|<math>\begin{align}
-	{{~~NAVCONTENT_STEP}}~~	+	\frac{1-4\sqrt{3}i}{1+i\sqrt{3}}
-	~~Then, we multiply top and bottom by the complex conjugate of the numerator:~~	+	&= \frac{(1-4\sqrt{3}i)(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})}\\[5pt]
-		+	&= \frac{1\cdot 1-1\cdot i\sqrt{3}-4\sqrt{3}i\cdot 1+ 4\sqrt{3}i\cdot i \sqrt{3}}{1^2-(i\sqrt{3})^2}\\[5pt]
-		+	&= \frac{1-i\sqrt{3}-4\sqrt{3}i+4(\sqrt{3})^2i^2}{1+(\sqrt{3})^2}\\[5pt]
-	<math>\begin{align}\frac{1-4\sqrt{3}i}{1+i\sqrt{3}}&=\frac{(1-4\sqrt{3}i)(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})}\\	+	&= \frac{1-(\sqrt{3}+4\sqrt{3})i-4\cdot 3}{1+3}\\[5pt]
-	&=\frac{1\cdot 1-1\cdot i\sqrt{3}-~~1\cdot~~ 4\sqrt{3}i+ 4\sqrt{3}i\cdot i \sqrt{3}}{1^2-(i\sqrt{3})^2}\\	+	&= \frac{1-12-(1+4)\sqrt{3}i}{4}\\[5pt]
-	&=\frac{1-i\sqrt{3}-4\sqrt{3}i+4(\sqrt{3})^2i^2}{1+(\sqrt{3})^2)}\\	+	&= -\frac{11}{4}-\frac{5\sqrt{3}}{4}i\,\textrm{.}
-	&=\frac{1-(\sqrt{3}+4\sqrt{3})i-4\cdot 3}{1+3}\\	+	\end{align}</math>}}
-	&=\frac{1-12-(1+4)\sqrt{3}i}{4}\\	+
-	&=-\frac{11}{4}-\frac{5\sqrt{3}}{4}i.\end{align}</math>	+
-		+
-	~~{{NAVCONTENT_STOP~~}}	+

Jamesjmnewton um 10:07, 23. Sep. 2008

Jamesjmnewton — Tue, 23 Sep 2008 10:07:00 GMT

← Nächstältere Version		Version vom 10:07, 23. Sep. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<~~center~~> ~~[[Image~~:~~3_1_2c~~.~~gif]]~~ </~~center~~>	+	We start by expanding the quadratic in the numerator with the square rule:
		+
		+
		+	<math>\begin{align}\frac{(2-i\sqrt{3})^2}{1+i\sqrt{3}}&= \frac{2^2-2\cdot 2\cdot i\sqrt{3}+(i\sqrt{3})^2}{1+i\sqrt{3}}\\
		+	&=\frac{4-4\sqrt{3}i-3}{1+i\sqrt{3}}\\
		+	&=\frac{1-4\sqrt{3}i}{1+i\sqrt{3}}\end{align}</math>
		+
		+
		+	{{NAVCONTENT_STEP}}
		+	Then, we multiply top and bottom by the complex conjugate of the numerator:
		+
		+
		+	<math>\begin{align}\frac{1-4\sqrt{3}i}{1+i\sqrt{3}}&=\frac{(1-4\sqrt{3}i)(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})}\\
		+	&=\frac{1\cdot 1-1\cdot i\sqrt{3}-1\cdot 4\sqrt{3}i+ 4\sqrt{3}i\cdot i \sqrt{3}}{1^2-(i\sqrt{3})^2}\\
		+	&=\frac{1-i\sqrt{3}-4\sqrt{3}i+4(\sqrt{3})^2i^2}{1+(\sqrt{3})^2)}\\
		+	&=\frac{1-(\sqrt{3}+4\sqrt{3})i-4\cdot 3}{1+3}\\
		+	&=\frac{1-12-(1+4)\sqrt{3}i}{4}\\
		+	&=-\frac{11}{4}-\frac{5\sqrt{3}}{4}i.\end{align}</math>
		+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Tekbot: Lösning 3.1:2c moved to Solution 3.1:2c: Robot: moved page

Tekbot — Tue, 16 Sep 2008 14:51:41 GMT

Lösning 3.1:2c moved to Solution 3.1:2c: Robot: moved page

← Nächstältere Version

Version vom 14:51, 16. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 08:05:34 GMT

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

← Nächstältere Version		Version vom 08:05, 21. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_1_2c.gif]] </center>	+	<center> [[Image:3_1_2c.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_1_2c.gif {{NAVCONTENT_STOP}}

Oskar — Mon, 07 Apr 2008 10:45:37 GMT

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