Lösning 2.5:8 - Versionshistorik

Louwah den 4 maj 2018 kl. 13.42

Louwah — Fri, 04 May 2018 13:42:42 GMT

← Äldre version		Versionen från 4 maj 2018 kl. 13.42
Rad 1:		Rad 1:
	<math>\displaystyle \mathbf{r_G} = \frac{4\textrm{kg}(0,5;\,0,5)\textrm{m}+4\textrm{kg}(0,5;\,-0,5)\textrm{m}-1\textrm{kg}(0,75;\,-0,75)\textrm{m}}{4\, \textrm{kg} + 4\,\textrm{kg} - 1\,\textrm{kg}} </math>		<math>\displaystyle \mathbf{r_G} = \frac{4\textrm{kg}(0,5;\,0,5)\textrm{m}+4\textrm{kg}(0,5;\,-0,5)\textrm{m}-1\textrm{kg}(0,75;\,-0,75)\textrm{m}}{4\, \textrm{kg} + 4\,\textrm{kg} - 1\,\textrm{kg}} </math>

-	<math>\displaystyle = \frac{(3,25;\,0,75)}{7}\textrm{m} = (0,46;\,0,11)\,\textrm{m} ~~</math>~~ </math>	+	<math>\displaystyle = \frac{(3,25;\,0,75)}{7}\textrm{m} = (0,46;\,0,11)\,\textrm{m} </math>

Louwah — Wed, 20 Dec 2017 15:53:55 GMT

← Äldre version		Versionen från 20 december 2017 kl. 15.53
Rad 1:		Rad 1:
-	<math>r_G=\frac{~~4kg~~(0,5;0,5)m+~~4kg~~(0,5;-0,5)m-~~1kg~~(0,75;-0,75)m}{~~4kg~~+~~4kg~~-~~1kg~~}</math~~> <br\~~>	+	<math>\displaystyle \mathbf{r_G} = \frac{4\textrm{kg}(0,5;\,0,5)\textrm{m}+4\textrm{kg}(0,5;\,-0,5)\textrm{m}-1\textrm{kg}(0,75;\,-0,75)\textrm{m}}{4\, \textrm{kg} + 4\,\textrm{kg} - 1\,\textrm{kg}} </math>

-		+	<math>\displaystyle = \frac{(3,25;\,0,75)}{7}\textrm{m} = (0,46;\,0,11)\,\textrm{m} </math> </math>
-	<math>=\frac{(3,25;0,75)}{7}m=(0,46;0,11)m</math>	+

Lena Chytraeus — Wed, 16 Dec 2009 15:34:04 GMT

← Äldre version		Versionen från 16 december 2009 kl. 15.34
Rad 1:		Rad 1:
	<math>r_G=\frac{4kg(0,5;0,5)m+4kg(0,5;-0,5)m-1kg(0,75;-0,75)m}{4kg+4kg-1kg}</math> <br\>		<math>r_G=\frac{4kg(0,5;0,5)m+4kg(0,5;-0,5)m-1kg(0,75;-0,75)m}{4kg+4kg-1kg}</math> <br\>
		+

	<math>=\frac{(3,25;0,75)}{7}m=(0,46;0,11)m</math>		<math>=\frac{(3,25;0,75)}{7}m=(0,46;0,11)m</math>

Lena Chytraeus — Wed, 16 Dec 2009 13:54:23 GMT

Ny sida: <math>r_G=\frac{4kg(0,5;0,5)m+4kg(0,5;-0,5)m-1kg(0,75;-0,75)m}{4kg+4kg-1kg}</math> <br\> <math>=\frac{(3,25;0,75)}{7}m=(0,46;0,11)m</math>

Ny sida