Lösung 3.1:2b - Versionsgeschichte

Moni um 11:56, 22. Aug. 2009

Moni — Sat, 22 Aug 2009 11:56:25 GMT

← Nächstältere Version		Version vom 11:56, 22. Aug. 2009
Zeile 1:		Zeile 1:
-	Nachdem die beiden Brüche nicht denselben Nenenr haben, können wir sie nicht direkt addieren. Am einfachsten wird es wenn wir beide Brüche zuerst berechnen, und dann addieren.	+	Nachdem die beiden Brüche nicht denselben Nenenr haben, können wir sie nicht direkt addieren. Am einfachsten wird es wenn wir beide Brüche zuerst berechnen und dann addieren.

-	Wir erweitern beide Brüche mit den konjugiert komplexen Nenner ~~von~~ jeweiligen Bruch,	+	Wir erweitern beide Brüche mit den konjugiert komplexen Nenner vom jeweiligen Bruch,

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

Martin um 08:50, 12. Mai 2009

Martin — Tue, 12 May 2009 08:50:42 GMT

← Nächstältere Version		Version vom 08:50, 12. Mai 2009
Zeile 1:		Zeile 1:
-	~~The two terms do not have the same denominator~~, ~~so it is not possible to subtract them directly~~. ~~It is perhaps simplest to calculate each quotient individually and then subtract the result~~.	+	Nachdem die beiden Brüche nicht denselben Nenenr haben, können wir sie nicht direkt addieren. Am einfachsten wird es wenn wir beide Brüche zuerst berechnen, und dann addieren.

-	~~We multiply the top and bottom of each fraction by the complex conjugate of its denominator~~,	+	Wir erweitern beide Brüche mit den konjugiert komplexen Nenner von jeweiligen Bruch,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 12:		Zeile 12:
	\end{align}</math>}}		\end{align}</math>}}

-	~~Then, we multiply the top and bottom of the last fraction by~~ 4, ~~so as to give make both fractions have the same denominator, and after that we subtract the numerators~~,	+	Jetzt erweitern wir den letzten Bruch mit 4, sodass beide Brüche denselben Nenner bekommen,

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

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

Tekbot — Wed, 11 Mar 2009 10:30:55 GMT

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

← Nächstältere Version

Version vom 10:30, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 13:05:54 GMT

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

← Nächstältere Version		Version vom 13:05, 10. Mär. 2009
Zeile 3:		Zeile 3:
	We multiply the top and bottom of each fraction by the complex conjugate of its denominator,		We multiply the top and bottom of each fraction by the complex conjugate of its denominator,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{3i}{4-6i}-\frac{1+i}{3+2i}		\frac{3i}{4-6i}-\frac{1+i}{3+2i}
	&= \frac{3i(4+6i)}{(4-6i)(4+6i)}-\frac{(1+i)(3-2i)}{(3+2i)(3-2i)}\\[5pt]		&= \frac{3i(4+6i)}{(4-6i)(4+6i)}-\frac{(1+i)(3-2i)}{(3+2i)(3-2i)}\\[5pt]
Zeile 14:		Zeile 14:
	Then, we multiply the top and bottom of the last fraction by 4, so as to give make both fractions have the same denominator, and after that we subtract the numerators,		Then, we multiply the top and bottom of the last fraction by 4, so as to give make both fractions have the same denominator, and after that we subtract the numerators,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{-18+12i}{52}-\frac{(5 +i)\cdot 4}{13\cdot 4}&=\frac{-18+12i}{52}-\frac{20+4i}{52}\\[5pt]		\frac{-18+12i}{52}-\frac{(5 +i)\cdot 4}{13\cdot 4}&=\frac{-18+12i}{52}-\frac{20+4i}{52}\\[5pt]
	&= \frac{-18+12i-20-4i}{52}\\[5pt]		&= \frac{-18+12i-20-4i}{52}\\[5pt]

Tek um 15:21, 29. Okt. 2008

Tek — Wed, 29 Oct 2008 15:21:51 GMT

← Nächstältere Version		Version vom 15:21, 29. Okt. 2008
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}
	The two terms do not have the same denominator, so it is not possible to subtract them directly. It is perhaps simplest to calculate each quotient individually and then subtract the result.		The two terms do not have the same denominator, so it is not possible to subtract them directly. It is perhaps simplest to calculate each quotient individually and then subtract the result.

-	We multiply the top and bottom of each fraction by the complex conjugate of its denominator:	+	We multiply the top and bottom of each fraction by the complex conjugate of its denominator,

-	<math>\begin{align}\frac{3i}{4-6i}-\frac{1+i}{3+2i} &= \frac{3i(4+6i)}{(4-6i)(4+6i)}-\frac{(1+i)(3-2i)}{(3+2i)(3-2i)}\\	+	{{Displayed math\|\|<math>\begin{align}
-	&= \frac{3i\cdot 4 + 3i\cdot 6i}{(4^2-(6i)^2}-\frac{1\cdot 3 - 1\cdot 2i + i \cdot 3 - i\cdot 2i}{3^2-(2i)^2}\\	+	\frac{3i}{4-6i}-\frac{1+i}{3+2i}
-	&= \frac{12i + 18i^2}{16+36}-\frac{3 - 2i + 3i - 2i^2}{9+4}\\	+	&= \frac{3i(4+6i)}{(4-6i)(4+6i)}-\frac{(1+i)(3-2i)}{(3+2i)(3-2i)}\\[5pt]
-	&= \frac{-18+12i}{52}-\frac{3 +(-2 + 3)i + 2}{13}\\	+	&= \frac{3i\cdot 4 + 3i\cdot 6i}{4^2-(6i)^2}-\frac{1\cdot 3 - 1\cdot 2i + i \cdot 3 - i\cdot 2i}{3^2-(2i)^2}\\[5pt]
-	&= \frac{-18+12i}{52}-\frac{5 +i}{13}\end{align}</math>	+	&= \frac{12i + 18i^2}{16+36}-\frac{3 - 2i + 3i - 2i^2}{9+4}\\[5pt]
		+	&= \frac{-18+12i}{52}-\frac{3 +(-2 + 3)i + 2}{13}\\[5pt]
		+	&= \frac{-18+12i}{52}-\frac{5 +i}{13}\,\textrm{.}
		+	\end{align}</math>}}

-	~~{{NAVCONTENT_STEP}}~~	+	Then, we multiply the top and bottom of the last fraction by 4, so as to give make both fractions have the same denominator, and after that we subtract the numerators,

-		+	{{Displayed math\|\|<math>\begin{align}
-	~~Then, we multiply the top and bottom of the last fraction by 4, so as to give make both fractions have the same denominator, and after that we subtract the numerators:~~	+	\frac{-18+12i}{52}-\frac{(5 +i)\cdot 4}{13\cdot 4}&=\frac{-18+12i}{52}-\frac{20+4i}{52}\\[5pt]
-		+	&= \frac{-18+12i-20-4i}{52}\\[5pt]
-	<math>\begin{align}\frac{-18+12i}{52}-\frac{(5 +i)\cdot 4}{13\cdot 4}&=\frac{-18+12i}{52}-\frac{20+4i}{52}\\	+	&= \frac{-38+8i}{52}\\[5pt]
-	&=\frac{-18+12i-20-4i}{52}\\	+	&= -\frac{19}{26}+\frac{2}{13}\,i\,\textrm{.}
-	&=\frac{-38+8i}{52}\\	+	\end{align}</math>}}
-	&=-\frac{19}{26}+\frac{2}{13}i.\end{align}</math>	+
-		+
-		+
-	~~{{NAVCONTENT_STOP~~}}	+

Jamesjmnewton um 09:53, 23. Sep. 2008

Jamesjmnewton — Tue, 23 Sep 2008 09:53:48 GMT

← Nächstältere Version		Version vom 09:53, 23. Sep. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<~~center~~> ~~[[Image:3_1_2b~~-1(2)~~.gif]] </center>~~	+	The two terms do not have the same denominator, so it is not possible to subtract them directly. It is perhaps simplest to calculate each quotient individually and then subtract the result.
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	We multiply the top and bottom of each fraction by the complex conjugate of its denominator:
-	<~~center~~> ~~[[Image~~:~~3_1_2b~~-2(2).~~gif]]~~ </~~center~~>	+
		+	<math>\begin{align}\frac{3i}{4-6i}-\frac{1+i}{3+2i} &= \frac{3i(4+6i)}{(4-6i)(4+6i)}-\frac{(1+i)(3-2i)}{(3+2i)(3-2i)}\\
		+	&= \frac{3i\cdot 4 + 3i\cdot 6i}{(4^2-(6i)^2}-\frac{1\cdot 3 - 1\cdot 2i + i \cdot 3 - i\cdot 2i}{3^2-(2i)^2}\\
		+	&= \frac{12i + 18i^2}{16+36}-\frac{3 - 2i + 3i - 2i^2}{9+4}\\
		+	&= \frac{-18+12i}{52}-\frac{3 +(-2 + 3)i + 2}{13}\\
		+	&= \frac{-18+12i}{52}-\frac{5 +i}{13}\end{align}</math>
		+
		+	{{NAVCONTENT_STEP}}
		+
		+
		+	Then, we multiply the top and bottom of the last fraction by 4, so as to give make both fractions have the same denominator, and after that we subtract the numerators:
		+
		+	<math>\begin{align}\frac{-18+12i}{52}-\frac{(5 +i)\cdot 4}{13\cdot 4}&=\frac{-18+12i}{52}-\frac{20+4i}{52}\\
		+	&=\frac{-18+12i-20-4i}{52}\\
		+	&=\frac{-38+8i}{52}\\
		+	&=-\frac{19}{26}+\frac{2}{13}i.\end{align}</math>
		+
		+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

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

Tekbot — Tue, 16 Sep 2008 14:51:21 GMT

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

← Nächstältere Version

Version vom 14:51, 16. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 08:05:24 GMT

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

← Nächstältere Version		Version vom 08:05, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_1_2b-1(2).gif]] </center>	+	<center> [[Image:3_1_2b-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_1_2b-2(2).gif]] </center>	+	<center> [[Image:3_1_2b-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_1_2b-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:3_1_2b-2(2).gif {{NAVCONTENT_STOP}}

Oskar — Mon, 07 Apr 2008 10:44:30 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:3_1_2b-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:3_1_2b-2(2).gif </center> {{NAVCONTENT_STOP}}

Neue Seite

{{NAVCONTENT_START}}
<center> [[Bild:3_1_2b-1(2).gif]] </center>
{{NAVCONTENT_STOP}}
{{NAVCONTENT_START}}
<center> [[Bild:3_1_2b-2(2).gif]] </center>
{{NAVCONTENT_STOP}}