Lösung 3.1:1e - Versionsgeschichte

Martin um 08:43, 12. Mai 2009

2009-05-12T08:43:17Z

← Nächstältere Version		Version vom 08:43, 12. Mai 2009
Zeile 1:		Zeile 1:
-	~~A suitable first step can be to work out the square term,~~ <math>(2-i)^2</math>~~, by expanding it,~~	+	Zuerst berechnen wir die Quadrate <math>(2-i)^2</math> durch die binomische Formel

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

-	~~After that, we calculate the remaining product~~,	+	Jetzt multiplizieren wir die beiden Faktoren,

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

2009-03-11T10:29:55Z

← Nächstältere Version

Version vom 10:29, 11. Mär. 2009

2009-03-10T13:05:24Z

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

← Nächstältere Version		Version vom 13:05, 10. Mär. 2009
Zeile 1:		Zeile 1:
	A suitable first step can be to work out the square term, <math>(2-i)^2</math>, by expanding it,		A suitable first step can be to work out the square term, <math>(2-i)^2</math>, by expanding it,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	(2-i)^2 &= 2^2 - 2\cdot 2i + i^2\\[5pt]		(2-i)^2 &= 2^2 - 2\cdot 2i + i^2\\[5pt]
	&= 4-4i+i^2\\[5pt]		&= 4-4i+i^2\\[5pt]
Zeile 10:		Zeile 10:
	After that, we calculate the remaining product,		After that, we calculate the remaining product,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	(1+i)(3-4i) &= 1\cdot3 - 1\cdot 4i + i\cdot 3 - i\cdot 4i\\[5pt]		(1+i)(3-4i) &= 1\cdot3 - 1\cdot 4i + i\cdot 3 - i\cdot 4i\\[5pt]
	&= 3-4i+3i-4i^2\\[5pt]		&= 3-4i+3i-4i^2\\[5pt]

2008-10-29T15:00:31Z

← Nächstältere Version		Version vom 15:00, 29. Okt. 2008
Zeile 1:		Zeile 1:
-	~~{{NAVCONTENT_START}}~~	+	A suitable first step can be to work out the square term, <math>(2-i)^2</math>, by expanding it,
-	A suitable first step can be to work out the square term, <math>(2-i)^2</math>, ~~with the help of the square rule:~~	+
-	~~{{NAVCONTENT_STEP}}~~	+
-	~~<math>\begin{align}(2-i)^2&=2^2-2\cdot 2i + i^2=4-4i+i^2\\~~	+
-	~~&=4-4i-1=3-4i\end{align}</math>~~	+

-	{{~~NAVCONTENT_STEP~~}}	+	{{Displayed math\|\|<math>\begin{align}
-	After that, we calculate the remaining product:	+	(2-i)^2 &= 2^2 - 2\cdot 2i + i^2\\[5pt]
-	{{~~NAVCONTENT_STEP}}~~	+	&= 4-4i+i^2\\[5pt]
-	<math>\begin{align}(1+i)(3-4i)&=1\cdot3-1\cdot 4i +i \cdot 3 - i\cdot 4i\\	+	&= 4-4i-1\\[5pt]
-	&=3-4i+3i-4i^2\\	+	&= 3-4i\,\textrm{.}
-	&=3+(-4+3)i-4\cdot (-1)\\	+	\end{align}</math>}}
-	&=3-i+4\\	+
-	&=7-i.\end{align}</math>	+	After that, we calculate the remaining product,
-	~~{{NAVCONTENT_STOP~~}}	+
		+	{{Displayed math\|\|<math>\begin{align}
		+	(1+i)(3-4i) &= 1\cdot3 - 1\cdot 4i + i\cdot 3 - i\cdot 4i\\[5pt]
		+	&= 3-4i+3i-4i^2\\[5pt]
		+	&= 3+(-4+3)i-4\cdot (-1)\\[5pt]
		+	&= 3-i+4\\[5pt]
		+	&= 7-i\,\textrm{.}
		+	\end{align}</math>}}

2008-09-18T11:57:18Z

← Nächstältere Version		Version vom 11:57, 18. Sep. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<~~center~~> ~~[[Image~~:~~3_1_1e~~.~~gif]]~~ </~~center~~>	+	A suitable first step can be to work out the square term, <math>(2-i)^2</math>, with the help of the square rule:
		+	{{NAVCONTENT_STEP}}
		+	<math>\begin{align}(2-i)^2&=2^2-2\cdot 2i + i^2=4-4i+i^2\\
		+	&=4-4i-1=3-4i\end{align}</math>
		+
		+	{{NAVCONTENT_STEP}}
		+	After that, we calculate the remaining product:
		+	{{NAVCONTENT_STEP}}
		+	<math>\begin{align}(1+i)(3-4i)&=1\cdot3-1\cdot 4i +i \cdot 3 - i\cdot 4i\\
		+	&=3-4i+3i-4i^2\\
		+	&=3+(-4+3)i-4\cdot (-1)\\
		+	&=3-i+4\\
		+	&=7-i.\end{align}</math>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

2008-09-16T14:50:21Z

← Nächstältere Version

Version vom 14:50, 16. Sep. 2008

2008-08-21T08:04:54Z

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

2008-04-07T10:42:15Z

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