Hallo i am trayng to sin in to my mensager and its say that an herro has acord.and its says that i am not conect to the net mensager how can i resol this problem?

and i have windows xp and the mensanger dos not let me up dat to the new mensager.many thanks

Re: sining in to hotmail mensager

In the messenger options / tools there is an option to diagnoise and solve problems, you can use that of you can diasble any firewall you have or open a port on the firewall for messenger

the first solution is recomended

Posted on Jan 09, 2008

Cual es el modelo de la camara para buscar los drivers para ti?

Mar 03, 2011 | Operating Systems

It seems you're working with radians on whatever product or software you're using. sin(53.1rad)=0.302... while sin(53.1degrees)=0.799... which is what you said you expected. Make sure you set degrees as your mode and you'll get your expected results. Thanks for using Fixya!

Oct 08, 2010 | Operating Systems

puedes utilizar el comando "net user" y veras quienes estan conectados

saludos desde uruguay, x.lautaro@hotmail.com (software an web)

you can use the command "net user" and you see who are conected

saludos desde uruguay, x.lautaro@hotmail.com (software an web)

you can use the command "net user" and you see who are conected

Sep 02, 2010 | Microsoft Windows XP Professional

This is not Microsoft, so you have not installed one of "OUR" games.

The games you see on a new pc are just demo/trial versions.

Have you checked your "Download" folder for your game

The games you see on a new pc are just demo/trial versions.

Have you checked your "Download" folder for your game

Jul 15, 2010 | Operating Systems

The data1.cab file contains all the compressed installer files. Try downloading the software again, turn off your virus scanner and any other software that might be running and install it again!

Jul 11, 2009 | Operating Systems

plz downgrade to sp2 and check

Aug 08, 2008 | Microsoft Windows XP Professional With...

Does this help, it may be a start.

A Cos[w t - è[t]] + B è''[t]==0, where A and B are arbitrary constants?

If you expand the Cosine term, you get A Cos[w t] Cos[è[t]] + A Sin[w t] Sin[è[t]] +B è''[t] ==0, which can be approximated as:

B è''[t] + A Sin[w t] è[t]== -A Cos[w t]

So is the frequency of small oscillations just Sqrt[(A Sin[w t])/B]?

A Cos[w t - è[t]] + B è''[t]==0, where A and B are arbitrary constants?

If you expand the Cosine term, you get A Cos[w t] Cos[è[t]] + A Sin[w t] Sin[è[t]] +B è''[t] ==0, which can be approximated as:

B è''[t] + A Sin[w t] è[t]== -A Cos[w t]

So is the frequency of small oscillations just Sqrt[(A Sin[w t])/B]?

Nov 07, 2007 | Operating Systems

Oct 20, 2016 | Operating Systems

Oct 20, 2016 | Operating Systems

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i can sine in on hotmail

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