Lösung 4.1:5b - Versionsgeschichte

Tek: Replaced figure with metapost figure

2009-08-20T07:54:33Z

Replaced figure with metapost figure

← Nächstältere Version		Version vom 07:54, 20. Aug. 2009
Zeile 14:		Zeile 14:


-	~~[[Image~~:~~4_1_5_b-~~1(2)~~.gif\|~~center]]	+	<center>{{:4.1.5b - Solution - The circle (x - 2)² + (y + 1)² = 13}}</center>

	Hinweis: Ein Kreis mit den Mittelpunkt (''a'',''b'') und Radius ''r'' hat die Gleichung		Hinweis: Ein Kreis mit den Mittelpunkt (''a'',''b'') und Radius ''r'' hat die Gleichung

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

Markus.bez um 13:46, 16. Jun. 2009

2009-06-16T13:46:22Z

← Nächstältere Version		Version vom 13:46, 16. Jun. 2009
Zeile 1:		Zeile 1:
-	Nachdem der Punkt (-1,1) auf den Kreis liegen soll, muss der Radius des ~~kreises~~ der Abstand zwischen den Punkten (-1,1) und (2,-1) sein. Wir berechnen zuerst diesen Abstand,	+	Nachdem der Punkt (-1,1) auf den Kreis liegen soll, muss der Radius des Kreises der Abstand zwischen den Punkten (-1,1) und (2,-1) sein. Wir berechnen zuerst diesen Abstand:

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 5:		Zeile 5:
	\end{align}</math>}}		\end{align}</math>}}

-	Jetzt ~~wissen~~ wir den Mittelpunkt und den Radius des Kreises, und können die Gleichung ~~erhalten,~~	+	Jetzt kennen wir den Mittelpunkt und den Radius des Kreises und können die Gleichung aufschreiben:

-	{{Abgesetzte Formel\|\|<math>(x-2)^2 + (y-(-1))^2 = (\sqrt{13})^{2}</math>}}	+	{{Abgesetzte Formel\|\|<math>(x-2)^2 + (y-(-1))^2 = (\sqrt{13})^{2}\,,</math>}}

-	oder	+	oder auch

	{{Abgesetzte Formel\|\|<math>(x-2)^{2} + (y+1)^2 = 13\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>(x-2)^{2} + (y+1)^2 = 13\,\textrm{.}</math>}}

Martin um 17:24, 2. Apr. 2009

2009-04-02T17:24:20Z

← Nächstältere Version		Version vom 17:24, 2. Apr. 2009
Zeile 1:		Zeile 1:
-	~~If the circle is to contain the point~~ (-1,1), ~~then that point's distance away from the centre (2,-1) must equal the circle's radius, ''r''. Thus, we can obtain the circle's radius by calculating the distance between~~ (-1,1) ~~and~~ (2,-1) ~~using the distance formula~~,	+	Nachdem der Punkt (-1,1) auf den Kreis liegen soll, muss der Radius des kreises der Abstand zwischen den Punkten (-1,1) und (2,-1) sein. Wir berechnen zuerst diesen Abstand,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 5:		Zeile 5:
	\end{align}</math>}}		\end{align}</math>}}

-	~~When we know the circle's centre and its radius~~, ~~we can write the equation of the circle~~,	+	Jetzt wissen wir den Mittelpunkt und den Radius des Kreises, und können die Gleichung erhalten,

	{{Abgesetzte Formel\|\|<math>(x-2)^2 + (y-(-1))^2 = (\sqrt{13})^{2}</math>}}		{{Abgesetzte Formel\|\|<math>(x-2)^2 + (y-(-1))^2 = (\sqrt{13})^{2}</math>}}

-	~~which the same as~~	+	oder

	{{Abgesetzte Formel\|\|<math>(x-2)^{2} + (y+1)^2 = 13\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>(x-2)^{2} + (y+1)^2 = 13\,\textrm{.}</math>}}
Zeile 16:		Zeile 16:
	[[Image:4_1_5_b-1(2).gif\|center]]		[[Image:4_1_5_b-1(2).gif\|center]]

-		+	Hinweis: Ein Kreis mit den Mittelpunkt (''a'',''b'') und Radius ''r'' hat die Gleichung
-	~~Note~~: ~~A circle having its centre at~~ (''a'',''b'') ~~and radius~~ ''r'' ~~has the equation~~	+

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

Tekbot: hat „Solution 4.1:5b“ nach „Lösung 4.1:5b“ verschoben: Robot: moved page

2008-10-22T14:46:44Z

hat „Solution 4.1:5b“ nach „Lösung 4.1:5b“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:46, 22. Okt. 2008

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

2008-10-22T08:48:38Z

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

← Nächstältere Version		Version vom 08:48, 22. Okt. 2008
Zeile 1:		Zeile 1:
	If the circle is to contain the point (-1,1), then that point's distance away from the centre (2,-1) must equal the circle's radius, ''r''. Thus, we can obtain the circle's radius by calculating the distance between (-1,1) and (2,-1) using the distance formula,		If the circle is to contain the point (-1,1), then that point's distance away from the centre (2,-1) must equal the circle's radius, ''r''. Thus, we can obtain the circle's radius by calculating the distance between (-1,1) and (2,-1) using the distance formula,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	r &= \sqrt{(2-(-1))^2+(-1-1)^2} = \sqrt{3^2+(-2)^2} = \sqrt{9+4} = \sqrt{13}\,\textrm{.}		r &= \sqrt{(2-(-1))^2+(-1-1)^2} = \sqrt{3^2+(-2)^2} = \sqrt{9+4} = \sqrt{13}\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}
Zeile 7:		Zeile 7:
	When we know the circle's centre and its radius, we can write the equation of the circle,		When we know the circle's centre and its radius, we can write the equation of the circle,

-	{{~~Displayed math~~\|\|<math>(x-2)^2 + (y-(-1))^2 = (\sqrt{13})^{2}</math>}}	+	{{Abgesetzte Formel\|\|<math>(x-2)^2 + (y-(-1))^2 = (\sqrt{13})^{2}</math>}}

	which the same as		which the same as

-	{{~~Displayed math~~\|\|<math>(x-2)^{2} + (y+1)^2 = 13\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>(x-2)^{2} + (y+1)^2 = 13\,\textrm{.}</math>}}


Zeile 19:		Zeile 19:
	Note: A circle having its centre at (''a'',''b'') and radius ''r'' has the equation		Note: A circle having its centre at (''a'',''b'') and radius ''r'' has the equation

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

Tek um 10:53, 7. Okt. 2008

2008-10-07T10:53:17Z

← Nächstältere Version		Version vom 10:53, 7. Okt. 2008
Zeile 1:		Zeile 1:
-	If the circle is to contain the point	+	If the circle is to contain the point (-1,1), then that point's distance away from the centre (2,-1) must equal the circle's radius, ''r''. Thus, we can obtain the circle's radius by calculating the distance between (-1,1) and (2,-1) using the distance formula,
-	~~<math>\left~~( -1 ~~\right.~~,~~\left.~~ 1 ~~\right~~)~~</math>~~, then that point's distance away from the centre	+
-	~~<math>\left~~( 2 ~~\right.~~,~~\left.~~ -1 ~~\right~~)~~</math>~~	+
-	must equal the circle's radius,	+
-	~~<math>~~r~~</math>~~. Thus, we can obtain the circle's radius by calculating the distance between	+
-	~~<math>\left~~( -1 ~~\right.~~,~~\left.~~ 1 ~~\right~~)~~</math>~~	+
-	and	+
-	~~<math>\left~~( 2 ~~\right.~~,~~\left.~~ -1 ~~\right~~)~~</math>~~	+
-	using the distance formula:	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& r=\sqrt{\left( 2-\left( -1 \right) \right)^{2}+\left( -1-1 \right)^{2}}=\sqrt{3^{2}+\left( -2 \right)^{2}} \\~~	+
-	~~& =\sqrt{9+4}=\sqrt{13} \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	r &= \sqrt{(2-(-1))^2+(-1-1)^2} = \sqrt{3^2+(-2)^2} = \sqrt{9+4} = \sqrt{13}\,\textrm{.}
		+	\end{align}</math>}}

	When we know the circle's centre and its radius, we can write the equation of the circle,		When we know the circle's centre and its radius, we can write the equation of the circle,

-		+	{{Displayed math\|\|<math>(x-2)^2 + (y-(-1))^2 = (\sqrt{13})^{2}</math>}}
-	<math>~~\left~~( x-2 ~~\right~~)^{2}+~~\left~~( y-~~\left~~( -1 ~~\right~~) ~~\right~~)^{2}=~~\left~~( \sqrt{13} ~~\right~~)^{2}</math>	+
-		+

	which the same as		which the same as

		+	{{Displayed math\|\|<math>(x-2)^{2} + (y+1)^2 = 13\,\textrm{.}</math>}}

-	<math>\left( x-2 \right)^{2}+\left( y+1 \right)^{2}=13</math>

-	{{NAVCONTENT_START}}
	[[Image:4_1_5_b-1(2).gif\|center]]		[[Image:4_1_5_b-1(2).gif\|center]]

-	{{NAVCONTENT_STOP}}
-
-	NOTE: A circle having its centre at
-	<math>\left( a \right.,\left. b \right)</math>
-	and radius
-	<math>r</math>
-	has the equation

		+	Note: A circle having its centre at (''a'',''b'') and radius ''r'' has the equation

-	<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\,\textrm{.}</math>}}

Ian um 11:20, 27. Sep. 2008

2008-09-27T11:20:33Z

← Nächstältere Version		Version vom 11:20, 27. Sep. 2008
Zeile 1:		Zeile 1:
		+	If the circle is to contain the point
		+	<math>\left( -1 \right.,\left. 1 \right)</math>, then that point's distance away from the centre
		+	<math>\left( 2 \right.,\left. -1 \right)</math>
		+	must equal the circle's radius,
		+	<math>r</math>. Thus, we can obtain the circle's radius by calculating the distance between
		+	<math>\left( -1 \right.,\left. 1 \right)</math>
		+	and
		+	<math>\left( 2 \right.,\left. -1 \right)</math>
		+	using the distance formula:
		+
		+
		+	<math>\begin{align}
		+	& r=\sqrt{\left( 2-\left( -1 \right) \right)^{2}+\left( -1-1 \right)^{2}}=\sqrt{3^{2}+\left( -2 \right)^{2}} \\
		+	& =\sqrt{9+4}=\sqrt{13} \\
		+	\end{align}</math>
		+
		+
		+	When we know the circle's centre and its radius, we can write the equation of the circle,
		+
		+
		+	<math>\left( x-2 \right)^{2}+\left( y-\left( -1 \right) \right)^{2}=\left( \sqrt{13} \right)^{2}</math>
		+
		+
		+	which the same as
		+
		+
		+	<math>\left( x-2 \right)^{2}+\left( y+1 \right)^{2}=13</math>
		+
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
	[[Image:4_1_5_b-1(2).gif\|center]]		[[Image:4_1_5_b-1(2).gif\|center]]

-	<center> [[Image:4_1_5b-1(2).gif]] </center>
-	{{NAVCONTENT_STOP}}
-	{{NAVCONTENT_START}}
-	<center> [[Image:4_1_5b-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+
		+	NOTE: A circle having its centre at
		+	<math>\left( a \right.,\left. b \right)</math>
		+	and radius
		+	<math>r</math>
		+	has the equation
		+
		+
		+	<math>\left( x-a \right)^{2}+\left( y-b \right)2=r^{2}</math>

Tekbot: Lösning 4.1:5b moved to Solution 4.1:5b: Robot: moved page

2008-09-09T09:36:23Z

Lösning 4.1:5b moved to Solution 4.1:5b: Robot: moved page

← Nächstältere Version

Version vom 09:36, 9. Sep. 2008

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

2008-08-21T07:05:19Z

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

← Nächstältere Version		Version vom 07:05, 21. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	[[~~Bild~~:4_1_5_b-1(2).gif\|center]]	+	[[Image:4_1_5_b-1(2).gif\|center]]

-	<center> [[~~Bild~~:4_1_5b-1(2).gif]] </center>	+	<center> [[Image:4_1_5b-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_1_5b-2(2).gif]] </center>	+	<center> [[Image:4_1_5b-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Martin um 14:11, 15. Aug. 2008

2008-08-15T14:11:55Z

← Nächstältere Version		Version vom 14:11, 15. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
		+	[[Bild:4_1_5_b-1(2).gif\|center]]
		+
	<center> [[Bild:4_1_5b-1(2).gif]] </center>		<center> [[Bild:4_1_5b-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}