Lösung 3.1:2d - Versionsgeschichte

Moni um 12:06, 22. Aug. 2009

Moni — Sat, 22 Aug 2009 12:06:45 GMT

← Nächstältere Version		Version vom 12:06, 22. Aug. 2009
Zeile 25:		Zeile 25:
	\end{align}</math>}}		\end{align}</math>}}

-	Diesen Bruch berechnen wir, indem wir den ~~bruch~~ mit den konjugiert komplezen Nenner erweitern,	+	Diesen Bruch berechnen wir, indem wir den Bruch mit den konjugiert komplezen Nenner erweitern,

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

Martin um 08:57, 12. Mai 2009

Martin — Tue, 12 May 2009 08:57:38 GMT

← Nächstältere Version		Version vom 08:57, 12. Mai 2009
Zeile 1:		Zeile 1:
-	~~Because the expression is rather large~~, ~~we work step by step~~. ~~We start by making the numerator and denominator each have the same denominator~~,	+	Nachden der Ausdruck groß und kompliziert ist, arbeiten wir in Schritten. Zuerst schreiben wir den Zähler und Nenner wie einzelne Brüche,

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

-	~~Hence~~,	+	Und also erhalten wir,

	{{Abgesetzte Formel\|\|<math>\frac{5-\dfrac{1}{1+i}}{3i+\dfrac{i}{2-3i}} =\ \frac{\dfrac{4+5i}{1+i}}{\dfrac{9+7i}{2-3i}} = \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{5-\dfrac{1}{1+i}}{3i+\dfrac{i}{2-3i}} =\ \frac{\dfrac{4+5i}{1+i}}{\dfrac{9+7i}{2-3i}} = \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}\,\textrm{.}</math>}}

-	~~We multiply out the numerator and denominator~~	+	Wir erweitern den Zähler und den Nenner,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 25:		Zeile 25:
	\end{align}</math>}}		\end{align}</math>}}

-	~~This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator~~,	+	Diesen Bruch berechnen wir, indem wir den bruch mit den konjugiert komplezen Nenner erweitern,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 35:		Zeile 35:
	\end{align}</math>}}		\end{align}</math>}}

-	~~If we divide up the numbers into factors~~,	+	Zerlegen wir alle Zahlen in ihre Primfaktoren,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 43:		Zeile 43:
	\end{align}</math>}}		\end{align}</math>}}

-	~~we can simplify the answers~~	+	können wir die Antwort weiter vereinfachen,

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

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

Tekbot — Wed, 11 Mar 2009 10:31:35 GMT

Solution 3.1:2d moved to Lösung 3.1:2d: 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 13:06:14 GMT

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

← Nächstältere Version		Version vom 13:06, 10. Mär. 2009
Zeile 1:		Zeile 1:
	Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,		Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	5-\frac{1}{1+i}		5-\frac{1}{1+i}
	&=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i}		&=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i}
Zeile 14:		Zeile 14:
	Hence,		Hence,

-	{{~~Displayed math~~\|\|<math>\frac{5-\dfrac{1}{1+i}}{3i+\dfrac{i}{2-3i}} =\ \frac{\dfrac{4+5i}{1+i}}{\dfrac{9+7i}{2-3i}} = \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{5-\dfrac{1}{1+i}}{3i+\dfrac{i}{2-3i}} =\ \frac{\dfrac{4+5i}{1+i}}{\dfrac{9+7i}{2-3i}} = \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}\,\textrm{.}</math>}}

	We multiply out the numerator and denominator		We multiply out the numerator and denominator

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{(4+5i)(2-3i)}{(9+7i)(1+i)}		\frac{(4+5i)(2-3i)}{(9+7i)(1+i)}
	&= \frac{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\[5pt]		&= \frac{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\[5pt]
Zeile 27:		Zeile 27:
	This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator,		This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{23-2i}{2+16i}		\frac{23-2i}{2+16i}
	&= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt]		&= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt]
Zeile 37:		Zeile 37:
	If we divide up the numbers into factors,		If we divide up the numbers into factors,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	14 &= 2\cdot 7\,,\\[5pt]		14 &= 2\cdot 7\,,\\[5pt]
	372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt]		372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt]
Zeile 45:		Zeile 45:
	we can simplify the answers		we can simplify the answers

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{14}{260}-\frac{372}{260}\,i		\frac{14}{260}-\frac{372}{260}\,i
	&= \frac{2\cdot 7}{2\cdot 2\cdot 5\cdot 13}-\frac{2\cdot 2\cdot 3\cdot 31}{2\cdot 2\cdot 5\cdot 13}\,i\\[5pt]		&= \frac{2\cdot 7}{2\cdot 2\cdot 5\cdot 13}-\frac{2\cdot 2\cdot 3\cdot 31}{2\cdot 2\cdot 5\cdot 13}\,i\\[5pt]

Tek um 15:41, 29. Okt. 2008

Tek — Wed, 29 Oct 2008 15:41:52 GMT

← Nächstältere Version		Version vom 15:41, 29. Okt. 2008
Zeile 1:		Zeile 1:
-	~~{{NAVCONTENT_START}}~~	+	Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,
-	Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator:	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	5-\frac{1}{1+i}
		+	&=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i}
		+	= \frac{5+5i-1}{1+i}
		+	= \frac{4+5i}{1+i}\,,\\[5pt]
		+	3i+\frac{i}{2-3i}
		+	&= \frac{3i(2-3i)}{2-3i}+\frac{i}{2-3i}
		+	= \frac{6i-9i^2+i}{2-3i}
		+	= \frac{9+7i}{2-3i}\,\textrm{.}
		+	\end{align}</math>}}

-	<math>\begin{align}5-\frac{1}{1+i}&=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i} = \frac{5+5i-1}{1+i} = \frac{4+5i}{1+i},\\
-	3i+\frac{i}{2-3i} &= \frac{3i(2-3i)}{2-3i}+\frac{i}{2-3i}=\frac{6i-9i^2+i}{2-3i}=\frac{9+7i}{2-3i}.\end{align}</math>
-
-
-	{{NAVCONTENT_STEP}}
	Hence,		Hence,

		+	{{Displayed math\|\|<math>\frac{5-\dfrac{1}{1+i}}{3i+\dfrac{i}{2-3i}} =\ \frac{\dfrac{4+5i}{1+i}}{\dfrac{9+7i}{2-3i}} = \frac{(4+5i)(2-3i)}{(9+7i)(1+i)}\,\textrm{.}</math>}}

-	<math>\frac{5-\frac{1}{1+i}}{3i+\frac{i}{2-3i}}=\frac{\frac{4+5i}{1+i}}{\frac{9+7i}{2-3i}}=\frac{(4+5i)(2-3i)}{(9+7i)(1+i)}.</math>
-
-
-	{{NAVCONTENT_STEP}}
	We multiply out the numerator and denominator		We multiply out the numerator and denominator

		+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{(4+5i)(2-3i)}{(9+7i)(1+i)}
		+	&= \frac{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\[5pt]
		+	&= \frac{8-12i+10i+15}{9+9i+7i-7}\\[5pt]
		+	&= \frac{23-2i}{2+16i}\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}\frac{(4+5i)(2-3i)}{(9+7i)(1+i)}&= \frac{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\~~	+	This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator,
-	~~&=\frac{8-12i+10i+15}{9+9i+7i-7}\\~~	+
-	~~&=\frac{23-2i}{2+16i}.\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{23-2i}{2+16i}
		+	&= \frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\[5pt]
		+	&= \frac{23\cdot 2 -23\cdot 16i -2i\cdot 2 +2i \cdot 16i}{2^2-(16i)^2}\\[5pt]
		+	&= \frac{46-368i-4i-32}{4+256}\\[5pt]
		+	&= \frac{14-372i}{260}\,\textrm{.}
		+	\end{align}</math>}}

-	{{NAVCONTENT_STEP}}
-	This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator:
-
-
-	<math>\begin{align}\frac{23-2i}{2+16i}&=\frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\
-	&=\frac{23\cdot 2 -23\cdot 16i -2i\cdot 2 +2i \cdot 16i}{2^2-(16i)^2}\\
-	&=\frac{46-368i-4i-32}{4+256}\\
-	&=\frac{14-372i}{260}\end{align}</math>
-
-
-	{{NAVCONTENT_STEP}}
	If we divide up the numbers into factors,		If we divide up the numbers into factors,

		+	{{Displayed math\|\|<math>\begin{align}
		+	14 &= 2\cdot 7\,,\\[5pt]
		+	372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\,,\\[5pt]
		+	260 &= 10\cdot 26=2\cdot 5\cdot 13\cdot 2\,,
		+	\end{align}</math>}}

-	~~<math>\begin{align}14 &= 2\cdot 7\\~~	+	we can simplify the answers
-	~~372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\\~~	+
-	~~260 &= 10\cdot 26=2\cdot 5\cdot 13\cdot 2\end{align}</math>~~	+
-		+
-		+
-	we can simplify the answers:	+
-		+
-		+
-	~~<math>\begin{align}\frac{14}{260}-\frac{372}{260}i&=\frac{2\cdot 7}{2\cdot 2\cdot 5\cdot 13}-\frac{2\cdot 2\cdot 3\cdot 31}{2\cdot 2\cdot 5\cdot 13}i\\~~	+
-	~~&=\frac{7}{2\cdot 5\cdot 13}-\frac{3\cdot 31}{5\cdot 13}i\\~~	+
-	~~&=\frac{7}{130}-\frac{93}{65}i\end{align}</math>~~	+

-	{{~~NAVCONTENT_STOP~~}}	+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{14}{260}-\frac{372}{260}\,i
		+	&= \frac{2\cdot 7}{2\cdot 2\cdot 5\cdot 13}-\frac{2\cdot 2\cdot 3\cdot 31}{2\cdot 2\cdot 5\cdot 13}\,i\\[5pt]
		+	&= \frac{7}{2\cdot 5\cdot 13}-\frac{3\cdot 31}{5\cdot 13}\,i\\[5pt]
		+	&= \frac{7}{130}-\frac{93}{65}\,i\,\textrm{.}
		+	\end{align}</math>}}

Jamesjmnewton um 10:24, 23. Sep. 2008

Jamesjmnewton — Tue, 23 Sep 2008 10:24:36 GMT

← Nächstältere Version		Version vom 10:24, 23. Sep. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<~~center~~> ~~[[Image:3_1_2d~~-1(2).~~gif]]~~ </~~center~~>	+	Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator:
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image~~:~~3_1_2d~~-2(2)~~.gif]]~~ </~~center~~>	+	<math>\begin{align}5-\frac{1}{1+i}&=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i} = \frac{5+5i-1}{1+i} = \frac{4+5i}{1+i},\\
		+	3i+\frac{i}{2-3i} &= \frac{3i(2-3i)}{2-3i}+\frac{i}{2-3i}=\frac{6i-9i^2+i}{2-3i}=\frac{9+7i}{2-3i}.\end{align}</math>
		+
		+
		+	{{NAVCONTENT_STEP}}
		+	Hence,
		+
		+
		+	<math>\frac{5-\frac{1}{1+i}}{3i+\frac{i}{2-3i}}=\frac{\frac{4+5i}{1+i}}{\frac{9+7i}{2-3i}}=\frac{(4+5i)(2-3i)}{(9+7i)(1+i)}.</math>
		+
		+
		+	{{NAVCONTENT_STEP}}
		+	We multiply out the numerator and denominator
		+
		+
		+	<math>\begin{align}\frac{(4+5i)(2-3i)}{(9+7i)(1+i)}&= \frac{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\
		+	&=\frac{8-12i+10i+15}{9+9i+7i-7}\\
		+	&=\frac{23-2i}{2+16i}.\end{align}</math>
		+
		+
		+	{{NAVCONTENT_STEP}}
		+	This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator:
		+
		+
		+	<math>\begin{align}\frac{23-2i}{2+16i}&=\frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\
		+	&=\frac{23\cdot 2 -23\cdot 16i -2i\cdot 2 +2i \cdot 16i}{2^2-(16i)^2}\\
		+	&=\frac{46-368i-4i-32}{4+256}\\
		+	&=\frac{14-372i}{260}\end{align}</math>
		+
		+
		+	{{NAVCONTENT_STEP}}
		+	If we divide up the numbers into factors,
		+
		+
		+	<math>\begin{align}14 &= 2\cdot 7\\
		+	372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\\
		+	260 &= 10\cdot 26=2\cdot 5\cdot 13\cdot 2\end{align}</math>
		+
		+
		+	we can simplify the answers:
		+
		+
		+	<math>\begin{align}\frac{14}{260}-\frac{372}{260}i&=\frac{2\cdot 7}{2\cdot 2\cdot 5\cdot 13}-\frac{2\cdot 2\cdot 3\cdot 31}{2\cdot 2\cdot 5\cdot 13}i\\
		+	&=\frac{7}{2\cdot 5\cdot 13}-\frac{3\cdot 31}{5\cdot 13}i\\
		+	&=\frac{7}{130}-\frac{93}{65}i\end{align}</math>
		+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

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

Tekbot — Tue, 16 Sep 2008 14:52:01 GMT

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

← Nächstältere Version

Version vom 14:52, 16. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 08:05:44 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_2d-1(2).gif]] </center>	+	<center> [[Image:3_1_2d-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_1_2d-2(2).gif]] </center>	+	<center> [[Image:3_1_2d-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_1_2d-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:3_1_2d-2(2).gif {{NAVCONTENT_STOP}}

Oskar — Mon, 07 Apr 2008 10:49:25 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:3_1_2d-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:3_1_2d-2(2).gif </center> {{NAVCONTENT_STOP}}

Neue Seite

{{NAVCONTENT_START}}
<center> [[Bild:3_1_2d-1(2).gif]] </center>
{{NAVCONTENT_STOP}}
{{NAVCONTENT_START}}
<center> [[Bild:3_1_2d-2(2).gif]] </center>
{{NAVCONTENT_STOP}}