Lösung 2.2:5d - Versionsgeschichte

Tek: Replaced figure with metapost figure

2009-08-18T11:40:11Z

Replaced figure with metapost figure

← Nächstältere Version		Version vom 11:40, 18. Aug. 2009
Zeile 18:		Zeile 18:


-	<center>~~[[Image~~:~~2_2_5d-~~2(2)~~_de.gif]]~~</center>	+	<center>{{:2.2.5d - Solution - The line y = -½x + 5 through the point (2,4)}}</center>

Moni: Wortwahl

2009-08-05T09:30:16Z

Wortwahl

← Nächstältere Version		Version vom 09:30, 5. Aug. 2009
Zeile 1:		Zeile 1:
-	Wenn zwei nicht-senkrechte Geraden ~~winkelrecht~~ sind, gilt für deren Steigungen <math>k_{1}</math> und <math>k_{2}</math>, dass <math>k_{1}k_{2}=-1</math>, und daher erhalten wir die Steigung von unser Geraden als	+	Wenn zwei nicht-senkrechte Geraden rechtwinkelig zueinander sind, gilt für deren Steigungen <math>k_{1}</math> und <math>k_{2}</math>, dass <math>k_{1}k_{2}=-1</math>, und daher erhalten wir die Steigung von unser Geraden als

	{{Abgesetzte Formel\|\|<math>k_{2} = -\frac{1}{k_{1}} = -\frac{1}{2}</math>}}		{{Abgesetzte Formel\|\|<math>k_{2} = -\frac{1}{k_{1}} = -\frac{1}{2}</math>}}

Bayerer um 15:03, 30. Jul. 2009

2009-07-30T15:03:17Z

← Nächstältere Version		Version vom 15:03, 30. Jul. 2009
Zeile 18:		Zeile 18:


-	<center>[[Image:2_2_5d-2(2).gif]]</center>	+	<center>[[Image:2_2_5d-2(2)_de.gif]]</center>

Bayerer um 15:01, 30. Jul. 2009

2009-07-30T15:01:39Z

← Nächstältere Version		Version vom 15:01, 30. Jul. 2009
Zeile 18:		Zeile 18:


-	<center>[[Image:2_2_5d-2(2)~~_de~~.gif]]</center>	+	<center>[[Image:2_2_5d-2(2).gif]]</center>

Bayerer um 14:59, 30. Jul. 2009

2009-07-30T14:59:40Z

← Nächstältere Version		Version vom 14:59, 30. Jul. 2009
Zeile 18:		Zeile 18:


-	<center>[[Image:2_2_5d-2(2).gif]]</center>	+	<center>[[Image:2_2_5d-2(2)_de.gif]]</center>

Markus.bez: Sprache und Formulierung

2009-06-09T09:41:00Z

Sprache und Formulierung

← Nächstältere Version		Version vom 09:41, 9. Jun. 2009
Zeile 3:		Zeile 3:
	{{Abgesetzte Formel\|\|<math>k_{2} = -\frac{1}{k_{1}} = -\frac{1}{2}</math>}}		{{Abgesetzte Formel\|\|<math>k_{2} = -\frac{1}{k_{1}} = -\frac{1}{2}</math>}}

-	~~Nachdem~~ die Gerade <math>y=2x+5</math> die Steigung <math>k_{1}=2</math> ~~hat~~.	+	Also hat die Gerade <math>y=2x+5</math> die Steigung <math>k_{1}=2</math>.

	Also ist die Gleichung unserer Gerade		Also ist die Gleichung unserer Gerade

Martin um 11:12, 13. Mär. 2009

2009-03-13T11:12:12Z

← Nächstältere Version		Version vom 11:12, 13. Mär. 2009
Zeile 1:		Zeile 1:
-	~~If two non~~-~~vertical lines are perpendicular to each other~~, ~~their slopes~~ <math>k_{1}</math> ~~and~~ <math>k_{2}</math> ~~satisfy the relation~~ <math>k_{1}k_{2}=-1</math>, ~~and from this we have that the line we are looking for must have a slope that is given by~~	+	Wenn zwei nicht-senkrechte Geraden winkelrecht sind, gilt für deren Steigungen <math>k_{1}</math> und <math>k_{2}</math>, dass <math>k_{1}k_{2}=-1</math>, und daher erhalten wir die Steigung von unser Geraden als

	{{Abgesetzte Formel\|\|<math>k_{2} = -\frac{1}{k_{1}} = -\frac{1}{2}</math>}}		{{Abgesetzte Formel\|\|<math>k_{2} = -\frac{1}{k_{1}} = -\frac{1}{2}</math>}}

-	~~since the line~~ <math>y=2x+5</math> ~~has a slope~~ <math>k_{1}=2</math>	+	Nachdem die Gerade <math>y=2x+5</math> die Steigung <math>k_{1}=2</math> hat.
-	~~(the coefficient in front of ''x'')~~.	+

-	~~The line we are looking for can thus be written in the form~~	+	Also ist die Gleichung unserer Gerade

	{{Abgesetzte Formel\|\|<math>y=-\frac{1}{2}x+m</math>}}		{{Abgesetzte Formel\|\|<math>y=-\frac{1}{2}x+m</math>}}

-	~~with~~ ''m'' ~~as an unknown constant~~.	+	wo ''m'' noch bestimmt werden muss.

-	~~Because the point~~ (2,4) ~~should lie on the line~~, (2,4) ~~must satisfy the equation of the line,~~	+	Nachdem der Punkt (2,4) auf der Gerade liegt, muss der Punkt (2,4) die Gleichung der Gerade erfüllen

	{{Abgesetzte Formel\|\|<math>4=-\frac{1}{2}\cdot 2+m\,,</math>}}		{{Abgesetzte Formel\|\|<math>4=-\frac{1}{2}\cdot 2+m\,,</math>}}

-	~~i.e.~~ <math>m=5</math>. ~~The equation of the line is~~ <math>y=-\frac{1}{2}x+5</math>.	+	Also ist <math>m=5</math>. Die Gleichung der Gerade ist daher <math>y=-\frac{1}{2}x+5</math>.


	<center>[[Image:2_2_5d-2(2).gif]]</center>		<center>[[Image:2_2_5d-2(2).gif]]</center>

Tekbot: hat „Solution 2.2:5d“ nach „Lösung 2.2:5d“ verschoben: Robot: moved page

2008-10-22T13:58:25Z

hat „Solution 2.2:5d“ nach „Lösung 2.2:5d“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 13:58, 22. Okt. 2008

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

2008-10-22T08:28:58Z

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

← Nächstältere Version		Version vom 08:28, 22. Okt. 2008
Zeile 1:		Zeile 1:
	If two non-vertical lines are perpendicular to each other, their slopes <math>k_{1}</math> and <math>k_{2}</math> satisfy the relation <math>k_{1}k_{2}=-1</math>, and from this we have that the line we are looking for must have a slope that is given by		If two non-vertical lines are perpendicular to each other, their slopes <math>k_{1}</math> and <math>k_{2}</math> satisfy the relation <math>k_{1}k_{2}=-1</math>, and from this we have that the line we are looking for must have a slope that is given by

-	{{~~Displayed math~~\|\|<math>k_{2} = -\frac{1}{k_{1}} = -\frac{1}{2}</math>}}	+	{{Abgesetzte Formel\|\|<math>k_{2} = -\frac{1}{k_{1}} = -\frac{1}{2}</math>}}

	since the line <math>y=2x+5</math> has a slope <math>k_{1}=2</math>		since the line <math>y=2x+5</math> has a slope <math>k_{1}=2</math>
Zeile 8:		Zeile 8:
	The line we are looking for can thus be written in the form		The line we are looking for can thus be written in the form

-	{{~~Displayed math~~\|\|<math>y=-\frac{1}{2}x+m</math>}}	+	{{Abgesetzte Formel\|\|<math>y=-\frac{1}{2}x+m</math>}}

	with ''m'' as an unknown constant.		with ''m'' as an unknown constant.
Zeile 14:		Zeile 14:
	Because the point (2,4) should lie on the line, (2,4) must satisfy the equation of the line,		Because the point (2,4) should lie on the line, (2,4) must satisfy the equation of the line,

-	{{~~Displayed math~~\|\|<math>4=-\frac{1}{2}\cdot 2+m\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>4=-\frac{1}{2}\cdot 2+m\,,</math>}}

	i.e. <math>m=5</math>. The equation of the line is <math>y=-\frac{1}{2}x+5</math>.		i.e. <math>m=5</math>. The equation of the line is <math>y=-\frac{1}{2}x+5</math>.

Tek um 12:44, 24. Sep. 2008

2008-09-24T12:44:43Z

← Nächstältere Version		Version vom 12:44, 24. Sep. 2008
Zeile 1:		Zeile 1:
-	If two non-vertical lines are perpendicular to each other, their ~~gradients~~	+	If two non-vertical lines are perpendicular to each other, their slopes <math>k_{1}</math> and <math>k_{2}</math> satisfy the relation <math>k_{1}k_{2}=-1</math>, and from this we have that the line we are looking for must have a slope that is given by
-	<math>k_{1}</math>	+
-	and	+
-	<math>k_{2}</math>	+
-	satisfy the relation	+

-	<math>k_{1}k_{2}=-1</math>~~, and from this we have that the line we are looking for must have a gradient that is given by~~	+	{{Displayed math\|\|<math>k_{2} = -\frac{1}{k_{1}} = -\frac{1}{2}</math>}}

-		+	since the line <math>y=2x+5</math> has a slope <math>k_{1}=2</math>
-	~~<math>k_{2}=-\frac{1}{k_{1}}=-\frac{1}{2}</math>~~	+	(the coefficient in front of ''x'').
-		+
-		+
-	since the line	+
-	<math>y=2x+5</math>	+
-	has a ~~gradient~~	+
-	<math>k_{1}=2</math>	+
-	(the coefficient in front of	+
-	~~<math>~~x~~</math>~~	+
-	).	+

	The line we are looking for can thus be written in the form		The line we are looking for can thus be written in the form

		+	{{Displayed math\|\|<math>y=-\frac{1}{2}x+m</math>}}

-	~~<math>y=-\frac{1}{2}x+m</math>~~	+	with ''m'' as an unknown constant.
-		+
-		+
-	with	+
-	~~<math>~~m~~</math>~~	+
-	as an unknown constant.	+
-		+
-	~~Because the point~~	+
-	~~<math>\left( 2 \right.,\left. 4 \right)</math>~~	+
-	~~should lie on the line,~~	+
-	~~<math>\left( 2 \right.,\left. 4 \right)</math>~~	+
-	~~must satisfy the equation of the line,~~	+
-		+
-		+
-	~~<math>4=-\frac{1}{2}\centerdot 2+m</math>~~	+
-		+
-		+
-	~~i.e.~~	+
-	~~<math>m=5</math>. The equation of the line is .~~	+
-		+

-	~~<math>y=-\frac{1}{~~2~~}x+5</math>~~	+	Because the point (2,4) should lie on the line, (2,4) must satisfy the equation of the line,

		+	{{Displayed math\|\|<math>4=-\frac{1}{2}\cdot 2+m\,,</math>}}

		+	i.e. <math>m=5</math>. The equation of the line is <math>y=-\frac{1}{2}x+5</math>.


-	~~{{NAVCONTENT_START}}~~	+	<center>[[Image:2_2_5d-2(2).gif]]</center>
-	<center> [[Image:2_2_5d-2(2).gif]] </center>	+