Lösung 3.1:4c - Versionsgeschichte

Martin um 10:29, 12. Mai 2009

Martin — Tue, 12 May 2009 10:29:11 GMT

← Nächstältere Version		Version vom 10:29, 12. Mai 2009
Zeile 1:		Zeile 1:
-	~~If we subtract~~ <math>2z</math> ~~from both sides~~,	+	Subtrahieren wir <math>2z</math> von beiden Seiten,

	{{Abgesetzte Formel\|\|<math>iz+2-2z=-3</math>}}		{{Abgesetzte Formel\|\|<math>iz+2-2z=-3</math>}}

-	~~and then subtract~~ <math>2</math> ~~from both sides~~, ~~we have~~ <math>z</math> ~~terms left only on the left~~-~~hand side~~,	+	und subtrahieren danach <math>2</math> von beiden Seiten, haben wir nur noch <math>z</math>-Terme auf der linken Seite,

	{{Abgesetzte Formel\|\|<math>iz-2z=-3-2\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>iz-2z=-3-2\,\textrm{.}</math>}}

-	~~After taking out a factor~~ <math>z</math> ~~from the left-hand side~~,	+	Ziehen wir den Faktor <math>z</math> von der lunken Seite herauf, erhalten wir

	{{Abgesetzte Formel\|\|<math>(i-2)z=-5\,,</math>}}		{{Abgesetzte Formel\|\|<math>(i-2)z=-5\,,</math>}}

-	~~we obtain, after dividing by~~ <math>-2+i</math>,	+	und dividieren wir beide Seiten durch <math>-2+i</math> erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 21:		Zeile 21:
	= 2+i\,\textrm{.}\end{align}</math>}}		= 2+i\,\textrm{.}\end{align}</math>}}

-	~~A quick check shows that~~ <math>z=2+i</math> ~~satisfies the original equation,~~	+	Wir kontrollieren zur Sicherheit ob <math>z=2+i</math> auch wirklich die Gleichung erfüllt

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
-	\text{~~LHS~~} &= iz+2 = i(2+i)+2 = 2i-1+2 = 1+2i\,,\\[5pt]	+	\text{Linke Seite} &= iz+2 = i(2+i)+2 = 2i-1+2 = 1+2i\,,\\[5pt]
-	\text{~~RHS~~} &= 2z-3 = 2(2+i)-3 = 4+2i-3 = 1+2i\,\textrm{.}	+	\text{Rechte Seite} &= 2z-3 = 2(2+i)-3 = 4+2i-3 = 1+2i\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Tekbot: Solution 3.1:4c moved to Lösung 3.1:4c: Robot: moved page

Tekbot — Wed, 11 Mar 2009 10:32:55 GMT

Solution 3.1:4c moved to Lösung 3.1:4c: Robot: moved page

← Nächstältere Version

Version vom 10:32, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 13:06:54 GMT

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

← Nächstältere Version		Version vom 13:06, 10. Mär. 2009
Zeile 1:		Zeile 1:
	If we subtract <math>2z</math> from both sides,		If we subtract <math>2z</math> from both sides,

-	{{~~Displayed math~~\|\|<math>iz+2-2z=-3</math>}}	+	{{Abgesetzte Formel\|\|<math>iz+2-2z=-3</math>}}

	and then subtract <math>2</math> from both sides, we have <math>z</math> terms left only on the left-hand side,		and then subtract <math>2</math> from both sides, we have <math>z</math> terms left only on the left-hand side,

-	{{~~Displayed math~~\|\|<math>iz-2z=-3-2\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>iz-2z=-3-2\,\textrm{.}</math>}}

	After taking out a factor <math>z</math> from the left-hand side,		After taking out a factor <math>z</math> from the left-hand side,

-	{{~~Displayed math~~\|\|<math>(i-2)z=-5\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>(i-2)z=-5\,,</math>}}

	we obtain, after dividing by <math>-2+i</math>,		we obtain, after dividing by <math>-2+i</math>,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	z &= \frac{-5}{-2+i}		z &= \frac{-5}{-2+i}
	= \frac{-5(-2-i)}{(-2+i)(-2-i)}		= \frac{-5(-2-i)}{(-2+i)(-2-i)}
Zeile 23:		Zeile 23:
	A quick check shows that <math>z=2+i</math> satisfies the original equation,		A quick check shows that <math>z=2+i</math> satisfies the original equation,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\text{LHS} &= iz+2 = i(2+i)+2 = 2i-1+2 = 1+2i\,,\\[5pt]		\text{LHS} &= iz+2 = i(2+i)+2 = 2i-1+2 = 1+2i\,,\\[5pt]
	\text{RHS} &= 2z-3 = 2(2+i)-3 = 4+2i-3 = 1+2i\,\textrm{.}		\text{RHS} &= 2z-3 = 2(2+i)-3 = 4+2i-3 = 1+2i\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Tek um 07:55, 30. Okt. 2008

Tek — Thu, 30 Oct 2008 07:55:47 GMT

← Nächstältere Version		Version vom 07:55, 30. Okt. 2008
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}
	If we subtract <math>2z</math> from both sides,		If we subtract <math>2z</math> from both sides,

-		+	{{Displayed math\|\|<math>iz+2-2z=-3</math>}}
-	<math>iz+2-2z=-3</math>	+
-		+

	and then subtract <math>2</math> from both sides, we have <math>z</math> terms left only on the left-hand side,		and then subtract <math>2</math> from both sides, we have <math>z</math> terms left only on the left-hand side,

-		+	{{Displayed math\|\|<math>iz-2z=-3-2\,\textrm{.}</math>}}
-	<math>iz-2z=-3-2</math>	+
-		+

	After taking out a factor <math>z</math> from the left-hand side,		After taking out a factor <math>z</math> from the left-hand side,

-		+	{{Displayed math\|\|<math>(i-2)z=-5\,,</math>}}
-	<math>(i-2)z=-5</math>	+
-		+

	we obtain, after dividing by <math>-2+i</math>,		we obtain, after dividing by <math>-2+i</math>,

		+	{{Displayed math\|\|<math>\begin{align}
		+	z &= \frac{-5}{-2+i}
		+	= \frac{-5(-2-i)}{(-2+i)(-2-i)}
		+	= \frac{(-5)\cdot(-2)-5\cdot(-i)}{(-2)^2-i^2}\\[5pt]
		+	&= \frac{10+5i}{4+1}
		+	= \frac{10+5i}{5}
		+	= 2+i\,\textrm{.}\end{align}</math>}}

-	~~<math>\begin{align}z&=\frac{-5}{-2+i}=\frac{-5(-2-i)}{(-2+i)(-2-i)}=\frac{(-5)\cdot(-2)-5\cdot(-i)}{(-2)^2-i^2}\\~~	+	A quick check shows that <math>z=2+i</math> satisfies the original equation,
-	~~&=\frac{10+5i}{4+1}=\frac{10+5i}{5}=2+i.\end{align}</math>~~	+
-		+
-		+
-	A quick check shows that <math>z=2+i</math> satisfies the original equation	+
-		+
-		+
-	~~<math>\begin{align}LHS &= iz+2=i(2+i)+2=2i-1+2=1+2i\\~~	+
-	~~RHS &= 2z-3 = 2(2+i)-3=4+2i-3=1+2i.\end{align}</math>~~	+

-	{{~~NAVCONTENT_STOP~~}}	+	{{Displayed math\|\|<math>\begin{align}
		+	\text{LHS} &= iz+2 = i(2+i)+2 = 2i-1+2 = 1+2i\,,\\[5pt]
		+	\text{RHS} &= 2z-3 = 2(2+i)-3 = 4+2i-3 = 1+2i\,\textrm{.}
		+	\end{align}</math>}}

Jamesjmnewton um 10:51, 23. Sep. 2008

Jamesjmnewton — Tue, 23 Sep 2008 10:51:49 GMT

← Nächstältere Version		Version vom 10:51, 23. Sep. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<~~center~~> ~~[[Image:3_1_4c~~.~~gif]]~~ </~~center~~>	+	If we subtract <math>2z</math> from both sides,
		+
		+
		+	<math>iz+2-2z=-3</math>
		+
		+
		+	and then subtract <math>2</math> from both sides, we have <math>z</math> terms left only on the left-hand side,
		+
		+
		+	<math>iz-2z=-3-2</math>
		+
		+
		+	After taking out a factor <math>z</math> from the left-hand side,
		+
		+
		+	<math>(i-2)z=-5</math>
		+
		+
		+	we obtain, after dividing by <math>-2+i</math>,
		+
		+
		+	<math>\begin{align}z&=\frac{-5}{-2+i}=\frac{-5(-2-i)}{(-2+i)(-2-i)}=\frac{(-5)\cdot(-2)-5\cdot(-i)}{(-2)^2-i^2}\\
		+	&=\frac{10+5i}{4+1}=\frac{10+5i}{5}=2+i.\end{align}</math>
		+
		+
		+	A quick check shows that <math>z=2+i</math> satisfies the original equation
		+
		+
		+	<math>\begin{align}LHS &= iz+2=i(2+i)+2=2i-1+2=1+2i\\
		+	RHS &= 2z-3 = 2(2+i)-3=4+2i-3=1+2i.\end{align}</math>
		+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Tekbot: Lösning 3.1:4c moved to Solution 3.1:4c: Robot: moved page

Tekbot — Tue, 16 Sep 2008 14:53:21 GMT

Lösning 3.1:4c moved to Solution 3.1:4c: Robot: moved page

← Nächstältere Version

Version vom 14:53, 16. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 08:06:24 GMT

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

← Nächstältere Version		Version vom 08:06, 21. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_1_4c.gif]] </center>	+	<center> [[Image:3_1_4c.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_1_4c.gif {{NAVCONTENT_STOP}}

Oskar — Mon, 07 Apr 2008 10:53:19 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:3_1_4c.gif </center> {{NAVCONTENT_STOP}}

Neue Seite

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<center> [[Bild:3_1_4c.gif]] </center>
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