Lösung 4.1:7a - Versionsgeschichte

Tek: Replaced figure with metapost figure

Tek — Thu, 20 Aug 2009 08:28:14 GMT

Replaced figure with metapost figure

← Nächstältere Version		Version vom 08:28, 20. Aug. 2009
Zeile 23:		Zeile 23:


-	<center> ~~[[Image~~:~~4_1_7a-2(2)~~.~~gif]]~~ </center>	+	<center>{{:4.1.7a - Solution - The circle x² + 2x + y² - 2y = 1}}</center>

Markus.bez um 13:52, 16. Jun. 2009

Markus.bez — Tue, 16 Jun 2009 13:52:49 GMT

← Nächstältere Version		Version vom 13:52, 16. Jun. 2009
Zeile 5:		Zeile 5:
	zu bringen, wo wir den Mittelpunkt und Radius ablesen können.		zu bringen, wo wir den Mittelpunkt und Radius ablesen können.

-	Wir verwenden uns von quadratischer Ergänzung für jeweils die ''x''- und ''y''-Terme, und erhalten	+	Wir verwenden uns von quadratischer Ergänzung für jeweils die ''x''- und ''y''-Terme und erhalten

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
-	x^2 + 2x &= (x+1)^2-1^2\,,\\[5pt]	+	x^2 + 2x &= (x+1)^2-1^2\,\text{ und}\\[5pt]
-	y^2 - 2y &= (y-1)^2-1^2\,,	+	y^2 - 2y &= (y-1)^2-1^2\,.
	\end{align}</math>}}		\end{align}</math>}}

Markus.bez um 13:52, 16. Jun. 2009

Markus.bez — Tue, 16 Jun 2009 13:52:00 GMT

← Nächstältere Version		Version vom 13:52, 16. Jun. 2009
Zeile 1:		Zeile 1:
-	In dieser Form ist es schwierig den Mittelpunkt und Radius des Kreises abzulesen. Daher benuzen wir quadratische Ergänzung um die Gleichung auf die Standardform	+	In dieser Form, ist es schwierig den Mittelpunkt und Radius des Kreises abzulesen. Daher benuzen wir quadratische Ergänzung, um die Gleichung auf die Standardform

	{{Abgesetzte Formel\|\|<math>(x-a)^2 + (y-b)^2 = r^2\,,</math>}}		{{Abgesetzte Formel\|\|<math>(x-a)^2 + (y-b)^2 = r^2\,,</math>}}

Martin um 18:15, 2. Apr. 2009

Martin — Thu, 02 Apr 2009 18:15:00 GMT

← Nächstältere Version		Version vom 18:15, 2. Apr. 2009
Zeile 1:		Zeile 1:
-	As the equation stands, it is difficult directly to know anything about the circle, but if we complete the square and combine ''x''- and ''y''-terms together in their own respective square terms, then we will have the equation in the standard form	+	In dieser Form ist es schwierig den Mittelpunkt und Radius des Kreises abzulesen. Daher benuzen wir quadratische Ergänzung um die Gleichung auf die Standardform

	{{Abgesetzte Formel\|\|<math>(x-a)^2 + (y-b)^2 = r^2\,,</math>}}		{{Abgesetzte Formel\|\|<math>(x-a)^2 + (y-b)^2 = r^2\,,</math>}}

-	~~and we will then be able to read off the circle's centre and radius~~.	+	zu bringen, wo wir den Mittelpunkt und Radius ablesen können.

-	~~If we take the~~ ''x''- ~~and~~ ''y''-~~terms on the left-hand side and complete the square~~, ~~we get~~	+	Wir verwenden uns von quadratischer Ergänzung für jeweils die ''x''- und ''y''-Terme, und erhalten

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

-	~~and then the whole equation can be written as~~	+	Also ist die Gleichung

	{{Abgesetzte Formel\|\|<math>(x+1)^2 - 1^2 + (y-1)^2 - 1^2 = 1\,,</math>}}		{{Abgesetzte Formel\|\|<math>(x+1)^2 - 1^2 + (y-1)^2 - 1^2 = 1\,,</math>}}

-	~~or, with the constants moved to the right-hand side,~~	+	oder auch

	{{Abgesetzte Formel\|\|<math>(x+1)^2 + (y-1)^2 = 3\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>(x+1)^2 + (y-1)^2 = 3\,\textrm{.}</math>}}

-	~~This is a circle having its centre at~~ (-1,1) ~~and radius~~ <math>\sqrt{3}\,</math>.	+	Dies ist die Gleichung eines Kreises mit dem Mittelpunkt (-1,1) und dem Radius <math>\sqrt{3}\,</math>.


	<center> [[Image:4_1_7a-2(2).gif]] </center>		<center> [[Image:4_1_7a-2(2).gif]] </center>

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

Tekbot — Wed, 22 Oct 2008 14:48:04 GMT

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

← Nächstältere Version

Version vom 14:48, 22. Okt. 2008

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

Tekbot — Wed, 22 Oct 2008 08:49:08 GMT

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

Tek um 11:19, 8. Okt. 2008

Tek — Wed, 08 Oct 2008 11:19:41 GMT

← Nächstältere Version		Version vom 11:19, 8. Okt. 2008
Zeile 1:		Zeile 1:
-	As the equation stands, it is difficult directly to know anything about the circle, but if we complete the square and combine	+	As the equation stands, it is difficult directly to know anything about the circle, but if we complete the square and combine ''x''- and ''y''-terms together in their own respective square terms, then we will have the equation in the standard form
-	~~<math>~~x~~</math>~~	+
-	- and	+
-	~~<math>~~y~~</math>~~	+
-	- terms together in their own respective square terms, then we will have the equation in the standard form,	+
-		+
-		+
-	~~<math>\left( x-a \right)^{2}+\left( y-b \right)^{2}=r^{2}</math>~~	+

		+	{{Displayed math\|\|<math>(x-a)^2 + (y-b)^2 = r^2\,,</math>}}

	and we will then be able to read off the circle's centre and radius.		and we will then be able to read off the circle's centre and radius.

-	If we take the	+	If we take the ''x''- and ''y''-terms on the left-hand side and complete the square, we get
-	~~<math>~~x~~</math>~~	+
-	- and	+
-	~~<math>~~y~~</math>~~	+
-	- terms on the left-hand side and complete the square, we get	+

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>x^{2}+2x=~~\left~~( x+1 ~~\right~~)^{2}-1^{2~~}</math>~~	+	x^2 + 2x &= (x+1)^2-1^2\,,\\[5pt]
-		+	y^2 - 2y &= (y-1)^2-1^2\,,
-		+	\end{align}</math>}}
-	~~<math>~~y^{2}-2y=~~\left~~( y-1 ~~\right~~)^{2}-1^{2}</math>	+

	and then the whole equation can be written as		and then the whole equation can be written as

-		+	{{Displayed math\|\|<math>(x+1)^2 - 1^2 + (y-1)^2 - 1^2 = 1\,,</math>}}
-	<math>~~\left~~( x+1 ~~\right~~)^{2}-1^{2}+~~\left~~( y-1 ~~\right~~)^{2}-1^{2}=1</math>	+
-		+

	or, with the constants moved to the right-hand side,		or, with the constants moved to the right-hand side,

		+	{{Displayed math\|\|<math>(x+1)^2 + (y-1)^2 = 3\,\textrm{.}</math>}}

-	<math>\~~left( x+1 \right)^~~{2}+\~~left( y-1 \right)^{2}=3~~</math>	+	This is a circle having its centre at (-1,1) and radius <math>\sqrt{3}\,</math>.
-		+

-	This is a circle having its centre at
-	<math>\left( -1 \right.,\left. 1 \right)</math>
-	and radius
-	<math>\sqrt{3}</math>.

-	{{NAVCONTENT_START}}
	<center> [[Image:4_1_7a-2(2).gif]] </center>		<center> [[Image:4_1_7a-2(2).gif]] </center>
-	{{NAVCONTENT_STOP}}

Ian um 11:52, 27. Sep. 2008

Ian — Sat, 27 Sep 2008 11:52:39 GMT

← Nächstältere Version		Version vom 11:52, 27. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	As the equation stands, it is difficult directly to know anything about the circle, but if we complete the square and combine
-	<~~center~~> ~~[[Image:4_1_7a~~-1(2)~~.gif]]~~ </~~center~~>	+	<math>x</math>
-	{{~~NAVCONTENT_STOP~~}}	+	- and
		+	<math>y</math>
		+	- terms together in their own respective square terms, then we will have the equation in the standard form,
		+
		+
		+	<math>\left( x-a \right)^{2}+\left( y-b \right)^{2}=r^{2}</math>
		+
		+
		+	and we will then be able to read off the circle's centre and radius.
		+
		+	If we take the
		+	<math>x</math>
		+	- and
		+	<math>y</math>
		+	- terms on the left-hand side and complete the square, we get
		+
		+
		+	<math>x^{2}+2x=\left( x+1 \right)^{2}-1^{2}</math>
		+
		+
		+	<math>y^{2}-2y=\left( y-1 \right)^{2}-1^{2}</math>
		+
		+	and then the whole equation can be written as
		+
		+
		+	<math>\left( x+1 \right)^{2}-1^{2}+\left( y-1 \right)^{2}-1^{2}=1</math>
		+
		+
		+	or, with the constants moved to the right-hand side,
		+
		+
		+	<math>\left( x+1 \right)^{2}+\left( y-1 \right)^{2}=3</math>
		+
		+
		+	This is a circle having its centre at
		+	<math>\left( -1 \right.,\left. 1 \right)</math>
		+	and radius
		+	<math>\sqrt{3}</math>.
		+
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
	<center> [[Image:4_1_7a-2(2).gif]] </center>		<center> [[Image:4_1_7a-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Tekbot: Lösning 4.1:7a moved to Solution 4.1:7a: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:37:43 GMT

Lösning 4.1:7a moved to Solution 4.1:7a: Robot: moved page

← Nächstältere Version

Version vom 09:37, 9. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 07:05:59 GMT

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

← Nächstältere Version		Version vom 07:05, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_1_7a-1(2).gif]] </center>	+	<center> [[Image:4_1_7a-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_1_7a-2(2).gif]] </center>	+	<center> [[Image:4_1_7a-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

← Nächstältere Version		Version vom 08:49, 22. Okt. 2008
Zeile 1:		Zeile 1:
	As the equation stands, it is difficult directly to know anything about the circle, but if we complete the square and combine ''x''- and ''y''-terms together in their own respective square terms, then we will have the equation in the standard form		As the equation stands, it is difficult directly to know anything about the circle, but if we complete the square and combine ''x''- and ''y''-terms together in their own respective square terms, then we will have the equation in the standard form

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

	and we will then be able to read off the circle's centre and radius.		and we will then be able to read off the circle's centre and radius.
Zeile 7:		Zeile 7:
	If we take the ''x''- and ''y''-terms on the left-hand side and complete the square, we get		If we take the ''x''- and ''y''-terms on the left-hand side and complete the square, we get

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	x^2 + 2x &= (x+1)^2-1^2\,,\\[5pt]		x^2 + 2x &= (x+1)^2-1^2\,,\\[5pt]
	y^2 - 2y &= (y-1)^2-1^2\,,		y^2 - 2y &= (y-1)^2-1^2\,,
Zeile 14:		Zeile 14:
	and then the whole equation can be written as		and then the whole equation can be written as

-	{{~~Displayed math~~\|\|<math>(x+1)^2 - 1^2 + (y-1)^2 - 1^2 = 1\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>(x+1)^2 - 1^2 + (y-1)^2 - 1^2 = 1\,,</math>}}

	or, with the constants moved to the right-hand side,		or, with the constants moved to the right-hand side,

-	{{~~Displayed math~~\|\|<math>(x+1)^2 + (y-1)^2 = 3\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>(x+1)^2 + (y-1)^2 = 3\,\textrm{.}</math>}}

	This is a circle having its centre at (-1,1) and radius <math>\sqrt{3}\,</math>.		This is a circle having its centre at (-1,1) and radius <math>\sqrt{3}\,</math>.