Lösung 4.1:7b - Versionsgeschichte

Tek: Replaced figure with metapost figure

2009-08-20T08:37:24Z

Replaced figure with metapost figure

← Nächstältere Version		Version vom 08:37, 20. Aug. 2009
Zeile 10:		Zeile 10:


-	<center> ~~[[Image~~:~~4_1_7_b~~.~~gif]]~~ </center>	+	<center>{{:4.1.7b - Solution - The circle x² + y² + 4y = 0}}</center>

2009-06-16T13:53:45Z

Sprache und Formulierung

← Nächstältere Version		Version vom 13:53, 16. Jun. 2009
Zeile 1:		Zeile 1:
-	Die Gleichung ~~steht~~ fast in der Standardform. Wir müssen nur ~~quadratische Ergänzung für~~ die ''y''-Terme ~~benutzen.~~	+	Die Gleichung ist fast in der Standardform gegeben. Wir müssen nur die ''y''-Terme quadratisch ergänzen:

	{{Abgesetzte Formel\|\|<math>y^2 + 4y = (y+2)^2 - 2^2\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>y^2 + 4y = (y+2)^2 - 2^2\,\textrm{.}</math>}}
Zeile 7:		Zeile 7:
	{{Abgesetzte Formel\|\|<math>x^2 + (y+2)^2 = 4</math>}}		{{Abgesetzte Formel\|\|<math>x^2 + (y+2)^2 = 4</math>}}

-	und sehen dass sie einen Kreis mit dem Mittelpunkt (0,-2) und Radius <math>\sqrt{4}=2\,</math> beschreibt.	+	und sehen, dass sie einen Kreis mit dem Mittelpunkt (0,-2) und Radius <math>\sqrt{4}=2\,</math> beschreibt.


	<center> [[Image:4_1_7_b.gif]] </center>		<center> [[Image:4_1_7_b.gif]] </center>

2009-04-02T18:20:09Z

← Nächstältere Version		Version vom 18:20, 2. Apr. 2009
Zeile 1:		Zeile 1:
-	~~The equation is almost~~ in ~~the standard form for a circle; all that is needed is for us to collect together the~~ ''y''²- ~~and ''y''-terms into a quadratic term by completing the square~~	+	Die Gleichung steht fast in der Standardform. Wir müssen nur quadratische Ergänzung für die ''y''-Terme benutzen.

	{{Abgesetzte Formel\|\|<math>y^2 + 4y = (y+2)^2 - 2^2\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>y^2 + 4y = (y+2)^2 - 2^2\,\textrm{.}</math>}}

-	~~After rewriting, the equation is~~	+	Wir erhalten so die Gleichung

	{{Abgesetzte Formel\|\|<math>x^2 + (y+2)^2 = 4</math>}}		{{Abgesetzte Formel\|\|<math>x^2 + (y+2)^2 = 4</math>}}

-	~~and we see that the equation describes a circle having its centre at~~ (0,-2) ~~and radius~~ <math>\sqrt{4}=2\,</math>.	+	und sehen dass sie einen Kreis mit dem Mittelpunkt (0,-2) und Radius <math>\sqrt{4}=2\,</math> beschreibt.


	<center> [[Image:4_1_7_b.gif]] </center>		<center> [[Image:4_1_7_b.gif]] </center>

2008-10-22T14:48:24Z

hat „Solution 4.1:7b“ nach „Lösung 4.1:7b“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:48, 22. Okt. 2008

2008-10-22T08:49:18Z

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

← Nächstältere Version		Version vom 08:49, 22. Okt. 2008
Zeile 1:		Zeile 1:
	The equation is almost in the standard form for a circle; all that is needed is for us to collect together the ''y''²- and ''y''-terms into a quadratic term by completing the square		The equation is almost in the standard form for a circle; all that is needed is for us to collect together the ''y''²- and ''y''-terms into a quadratic term by completing the square

-	{{~~Displayed math~~\|\|<math>y^2 + 4y = (y+2)^2 - 2^2\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>y^2 + 4y = (y+2)^2 - 2^2\,\textrm{.}</math>}}

	After rewriting, the equation is		After rewriting, the equation is

-	{{~~Displayed math~~\|\|<math>x^2 + (y+2)^2 = 4</math>}}	+	{{Abgesetzte Formel\|\|<math>x^2 + (y+2)^2 = 4</math>}}

	and we see that the equation describes a circle having its centre at (0,-2) and radius <math>\sqrt{4}=2\,</math>.		and we see that the equation describes a circle having its centre at (0,-2) and radius <math>\sqrt{4}=2\,</math>.

2008-10-08T11:23:46Z

← Nächstältere Version		Version vom 11:23, 8. Okt. 2008
Zeile 1:		Zeile 1:
-	The equation is almost in the standard form for a circle; all that is needed is for us to collect together the	+	The equation is almost in the standard form for a circle; all that is needed is for us to collect together the ''y''²- and ''y''-terms into a quadratic term by completing the square
-	~~<math>~~y~~^{\text{2}}</math>~~	+
-	- and	+
-	~~<math>~~y~~</math>~~	+
-	-terms into a quadratic term by completing the square	+
-		+
-		+
-	~~<math>y^{2}+4y=\left( y+2 \right)^{2}-2^{2}</math>~~	+

		+	{{Displayed math\|\|<math>y^2 + 4y = (y+2)^2 - 2^2\,\textrm{.}</math>}}

	After rewriting, the equation is		After rewriting, the equation is

		+	{{Displayed math\|\|<math>x^2 + (y+2)^2 = 4</math>}}

-	<math>x^{2}~~+\left( y+~~2 \~~right)^{2}=4~~</math>	+	and we see that the equation describes a circle having its centre at (0,-2) and radius <math>\sqrt{4}=2\,</math>.


-	and we see that the equation describes a circle having its centre at
-	<math>\left( 0 \right.,\left. -2 \right)</math>
-	and radius
-	<math>\sqrt{4}=2</math>.
-
-
-	{{NAVCONTENT_START}}
	<center> [[Image:4_1_7_b.gif]] </center>		<center> [[Image:4_1_7_b.gif]] </center>
-
-
-	{{NAVCONTENT_STOP}}

2008-09-27T12:00:45Z

← Nächstältere Version		Version vom 12:00, 27. Sep. 2008
Zeile 1:		Zeile 1:
		+	The equation is almost in the standard form for a circle; all that is needed is for us to collect together the
		+	<math>y^{\text{2}}</math>
		+	- and
		+	<math>y</math>
		+	-terms into a quadratic term by completing the square
		+
		+
		+	<math>y^{2}+4y=\left( y+2 \right)^{2}-2^{2}</math>
		+
		+
		+	After rewriting, the equation is
		+
		+
		+	<math>x^{2}+\left( y+2 \right)^{2}=4</math>
		+
		+
		+	and we see that the equation describes a circle having its centre at
		+	<math>\left( 0 \right.,\left. -2 \right)</math>
		+	and radius
		+	<math>\sqrt{4}=2</math>.
		+
		+
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
	<center> [[Image:4_1_7_b.gif]] </center>		<center> [[Image:4_1_7_b.gif]] </center>

-	~~<center> [[Image:4_1_7b.gif]] </center>~~	+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

2008-09-09T09:38:03Z

← Nächstältere Version

Version vom 09:38, 9. Sep. 2008

2008-08-21T07:06:09Z

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

2008-08-15T14:22:20Z