Lösung 4.1:4a - Versionsgeschichte

Tek: Replaced figures with metapost figures

2009-08-20T07:30:15Z

Replaced figures with metapost figures

← Nächstältere Version		Version vom 07:30, 20. Aug. 2009
Zeile 1:		Zeile 1:
	Wir zeichnen die Punkte und sehen, dass der Abstand zwischen den Punkten die Hypotenuse des Dreiecks ist, wobei die Katheten parallel zu den Koordinatenachsen sind.		Wir zeichnen die Punkte und sehen, dass der Abstand zwischen den Punkten die Hypotenuse des Dreiecks ist, wobei die Katheten parallel zu den Koordinatenachsen sind.

-	~~[[Image~~:~~4_1_4_a~~-1(2)~~_1.gif\|~~center]]	+	<center>{{:4.1.4a - Solution - A line segment between the points (1,1) and (5,4)}}</center>

	Die Länge der Katheten erhalten wir einfach als den Unterschied des ''x''- und ''y''-Wertes von den beiden Punkten.		Die Länge der Katheten erhalten wir einfach als den Unterschied des ''x''- und ''y''-Wertes von den beiden Punkten.
Zeile 7:		Zeile 7:

	{\| align="center"		{\| align="center"
-	\|align="center"\|~~[[Image~~:~~4_1_4_a~~-1(2)~~_2.gif\|center]]~~	+	\|align="center"\|{{:4.1.4a - Solution - A triangle with the line segment between the points (1,1) and (5,4) as the hypotenuse}}
	\|-		\|-
-	\|align="center"\|<small>∆''x'' = 5 - 1 = 4  ~~and~~  ∆''y'' = 4 - 1 = 3</small>	+	\|align="center"\|<small>∆''x'' = 5 - 1 = 4  und  ∆''y'' = 4 - 1 = 3</small>
	\|}		\|}

Moni um 11:05, 9. Aug. 2009

2009-08-09T11:05:17Z

← Nächstältere Version		Version vom 11:05, 9. Aug. 2009
Zeile 12:		Zeile 12:
	\|}		\|}

-	Mit dem Gesetz des Pythagoras erhalten wir die Hypotenuse, und so auch den Abstand zwischen den Punkten,	+	Mit dem Gesetz des Pythagoras erhalten wir die Hypotenuse und so auch den Abstand zwischen den Punkten,

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

Markus.bez um 13:42, 16. Jun. 2009

2009-06-16T13:42:33Z

← Nächstältere Version		Version vom 13:42, 16. Jun. 2009
Zeile 1:		Zeile 1:
-	Wir zeichnen die Punkte, und sehen dass der Abstand zwischen den Punkten die Hypotenuse des Dreiecks ist, wo die Katheten parallel ~~mit~~ den Koordinatenachsen sind.	+	Wir zeichnen die Punkte und sehen, dass der Abstand zwischen den Punkten die Hypotenuse des Dreiecks ist, wobei die Katheten parallel zu den Koordinatenachsen sind.

	[[Image:4_1_4_a-1(2)_1.gif\|center]]		[[Image:4_1_4_a-1(2)_1.gif\|center]]
Zeile 20:		Zeile 20:


-	Hinweis: Allgemein ist der Abstand d, zwischen den Punkten <math>(x,y)</math> und <math>(a,b)</math>	+	Hinweis: Allgemein ist der Abstand d zwischen den Punkten <math>(x,y)</math> und <math>(a,b)</math>

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

Martin um 17:08, 2. Apr. 2009

2009-04-02T17:08:10Z

← Nächstältere Version		Version vom 17:08, 2. Apr. 2009
Zeile 1:		Zeile 1:
-	~~If we draw in the points in a coordinate system~~, ~~we can see the line between the points as the hypotenuse in an imaginary right-angled triangle~~, ~~where the opposite and adjacent are~~ parallel ~~with the ''x''- and ''y''-axes, respectively~~.	+	Wir zeichnen die Punkte, und sehen dass der Abstand zwischen den Punkten die Hypotenuse des Dreiecks ist, wo die Katheten parallel mit den Koordinatenachsen sind.
-		+

	[[Image:4_1_4_a-1(2)_1.gif\|center]]		[[Image:4_1_4_a-1(2)_1.gif\|center]]

-		+	Die Länge der Katheten erhalten wir einfach als den Unterschied des ''x''- und ''y''-Wertes von den beiden Punkten.
-	~~In this triangle, it is easy to measure the lengths of the opposite and the adjacent, which are simply the distances between the points in the~~ ''x''- ~~and~~ ''y''-~~directions, respectively~~.	+


Zeile 14:		Zeile 12:
	\|}		\|}

-		+	Mit dem Gesetz des Pythagoras erhalten wir die Hypotenuse, und so auch den Abstand zwischen den Punkten,
-	~~Using the Pythagorean theorem~~, ~~we can then calculate the length of the hypotenuse~~, ~~which is also the distance between the points:~~	+

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 23:		Zeile 20:


-	~~Note~~: ~~In general~~, ~~the distance between two points~~ <math>(x,y)</math> ~~and~~ <math>(a,b)</math> ~~is given by the formula~~	+	Hinweis: Allgemein ist der Abstand d, zwischen den Punkten <math>(x,y)</math> und <math>(a,b)</math>

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

Tekbot: hat „Solution 4.1:4a“ nach „Lösung 4.1:4a“ verschoben: Robot: moved page

2008-10-22T14:45:24Z

hat „Solution 4.1:4a“ nach „Lösung 4.1:4a“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:45, 22. Okt. 2008

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

2008-10-22T08:47:58Z

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

← Nächstältere Version		Version vom 08:47, 22. Okt. 2008
Zeile 17:		Zeile 17:
	Using the Pythagorean theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points:		Using the Pythagorean theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points:

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	d &= \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{4^2+3^2}\\[5pt]		d &= \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{4^2+3^2}\\[5pt]
	&= \sqrt{16+9} = \sqrt{25} = 5\,\textrm{.}		&= \sqrt{16+9} = \sqrt{25} = 5\,\textrm{.}
Zeile 25:		Zeile 25:
	Note: In general, the distance between two points <math>(x,y)</math> and <math>(a,b)</math> is given by the formula		Note: In general, the distance between two points <math>(x,y)</math> and <math>(a,b)</math> is given by the formula

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

Tek um 10:40, 3. Okt. 2008

2008-10-03T10:40:50Z

← Nächstältere Version		Version vom 10:40, 3. Okt. 2008
Zeile 1:		Zeile 1:
-	If we draw in the points in a coordinate system, we can see the line between the points as the hypotenuse in an imaginary right-angled triangle, where the opposite and adjacent are parallel with the	+	If we draw in the points in a coordinate system, we can see the line between the points as the hypotenuse in an imaginary right-angled triangle, where the opposite and adjacent are parallel with the ''x''- and ''y''-axes, respectively.
-	~~<math>~~x~~</math>~~	+
-	- and	+
-	~~<math>~~y~~</math>~~	+
-	-axes.	+

-	{{NAVCONTENT_START}}

	[[Image:4_1_4_a-1(2)_1.gif\|center]]		[[Image:4_1_4_a-1(2)_1.gif\|center]]
-	In this triangle, it is easy to measure the lengths of the opposite and the adjacent, which are simply the distances between the points in the
-	<math>x</math>
-	-
-	<math>y</math>
-	-directions.

-	[[Image:4_1_4_a-1(2)_2.gif\|center]]

		+	In this triangle, it is easy to measure the lengths of the opposite and the adjacent, which are simply the distances between the points in the ''x''- and ''y''-directions, respectively.

-	{{NAVCONTENT_STOP}}

-	~~Using Pythagoras' theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points~~:	+	{\| align="center"
		+	\|align="center"\|[[Image:4_1_4_a-1(2)_2.gif\|center]]
		+	\|-
		+	\|align="center"\|<small>∆''x'' = 5 - 1 = 4  and  ∆''y'' = 4 - 1 = 3</small>
		+	\|}


-	~~<math>\begin{align}~~	+	Using the Pythagorean theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points:
-	~~& d=\sqrt{\left( \Delta x \right)^{2}+\left( \Delta y \right)^{2}}=\sqrt{4^{2}+3^{2}} \\~~	+
-	~~& =\sqrt{16+9}=\sqrt{25}=5 \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	d &= \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{4^2+3^2}\\[5pt]
		+	&= \sqrt{16+9} = \sqrt{25} = 5\,\textrm{.}
		+	\end{align}</math>}}

-	NOTE: In general, the distance between two points
-	<math>\left( x \right.,\left. y \right)</math>
-	and
-	<math>\left( a \right.,\left. b \right)</math>
-	is given by the formula

		+	Note: In general, the distance between two points <math>(x,y)</math> and <math>(a,b)</math> is given by the formula

-	<math>d=\sqrt{~~\left~~( x-a ~~\right~~)^{2}+~~\left~~( y-b ~~\right~~)^{2}}</math>	+	{{Displayed math\|\|<math>d = \sqrt{(x-a)^2 + (y-b)^2}\,\textrm{.}</math>}}

Ian um 09:56, 27. Sep. 2008

2008-09-27T09:56:10Z

← Nächstältere Version		Version vom 09:56, 27. Sep. 2008
Zeile 1:		Zeile 1:
		+	If we draw in the points in a coordinate system, we can see the line between the points as the hypotenuse in an imaginary right-angled triangle, where the opposite and adjacent are parallel with the
		+	<math>x</math>
		+	- and
		+	<math>y</math>
		+	-axes.
		+
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
		+
	[[Image:4_1_4_a-1(2)_1.gif\|center]]		[[Image:4_1_4_a-1(2)_1.gif\|center]]
		+	In this triangle, it is easy to measure the lengths of the opposite and the adjacent, which are simply the distances between the points in the
		+	<math>x</math>
		+	-
		+	<math>y</math>
		+	-directions.

	[[Image:4_1_4_a-1(2)_2.gif\|center]]		[[Image:4_1_4_a-1(2)_2.gif\|center]]

-	~~<center> [[Image:4_1_4a-1(2).gif]] </center>~~	+
-	~~{{NAVCONTENT_STOP}}~~	+
-	~~{{NAVCONTENT_START}}~~	+
-	~~<center> [[Image:4_1_4a-2(2).gif]] </center>~~	+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+
		+	Using Pythagoras' theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points:
		+
		+
		+	<math>\begin{align}
		+	& d=\sqrt{\left( \Delta x \right)^{2}+\left( \Delta y \right)^{2}}=\sqrt{4^{2}+3^{2}} \\
		+	& =\sqrt{16+9}=\sqrt{25}=5 \\
		+	\end{align}</math>
		+
		+
		+	NOTE: In general, the distance between two points
		+	<math>\left( x \right.,\left. y \right)</math>
		+	and
		+	<math>\left( a \right.,\left. b \right)</math>
		+	is given by the formula
		+
		+
		+	<math>d=\sqrt{\left( x-a \right)^{2}+\left( y-b \right)^{2}}</math>

Tekbot: Lösning 4.1:4a moved to Solution 4.1:4a: Robot: moved page

2008-09-09T09:35:03Z

Lösning 4.1:4a moved to Solution 4.1:4a: Robot: moved page

← Nächstältere Version

Version vom 09:35, 9. Sep. 2008

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

2008-08-21T07:04:39Z

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

← Nächstältere Version		Version vom 07:04, 21. Aug. 2008
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-	[[~~Bild~~:4_1_4_a-1(2)_1.gif\|center]]	+	[[Image:4_1_4_a-1(2)_1.gif\|center]]

-	[[~~Bild~~:4_1_4_a-1(2)_2.gif\|center]]	+	[[Image:4_1_4_a-1(2)_2.gif\|center]]

-	<center> [[~~Bild~~:4_1_4a-1(2).gif]] </center>	+	<center> [[Image:4_1_4a-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_1_4a-2(2).gif]] </center>	+	<center> [[Image:4_1_4a-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}