Hello Coders,
My name is Gian. I am currently writing several small programs for a side project. My time is running out and im still short of 3 programs. I am looking for a quick, easy and understanding programmer to work with. I will give detailed descriptions of the programs. I will also pay accordingly. Please contact me at Gianmbaio@yahoo.com or send me a message here on CodeCall. I hope to hear from you soon! Thanks!
9 replies to this topic
#1
Posted 15 April 2008 - 08:39 AM
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#2
Guest_Jordan_*
Posted 15 April 2008 - 09:38 AM
Guest_Jordan_*
Moved to correct forum.
#3
Posted 05 May 2008 - 01:53 AM
Maybe you can highlight more on what type of programming you need!
God is real... unless declared an integer
my blog :: http://techarraz.com/
#4
Posted 05 May 2008 - 08:00 AM
In particular, what language requirements are needed?
#5
Guest_Jaan_*
Posted 06 May 2008 - 02:52 AM
Guest_Jaan_*
It was posted on 04-15-2008, 07:39 PM And now it's 05-06-2008 ,01:52 PM
soo..
soo..
#7
Posted 05 November 2010 - 01:25 AM
programming c++ project
#8
Posted 05 November 2010 - 01:30 AM
the algorith of thi function and its program i have been stuck on it for a week now!!
write a double function, f(), with one double parameter, x, that will evaluate either f1(x)=x.xsin(x) + x/e if x<=1.0 or f2(x)=x.x.xcos(x)-log(x) if x>1.0, and return result
write a double function, f(), with one double parameter, x, that will evaluate either f1(x)=x.xsin(x) + x/e if x<=1.0 or f2(x)=x.x.xcos(x)-log(x) if x>1.0, and return result
#9
Posted 05 November 2010 - 01:33 AM
please reach me @ Hngema.1@gmail.com or jst reply her on codecall, i will really appreciate it!!
tanx
tanx
#10
Posted 06 November 2010 - 04:31 AM
sherz said:
the algorith of thi function and its program i have been stuck on it for a week now!!
write a double function, f(), with one double parameter, x, that will evaluate either f1(x)=x.xsin(x) + x/e if x<=1.0 or f2(x)=x.x.xcos(x)-log(x) if x>1.0, and return result
write a double function, f(), with one double parameter, x, that will evaluate either f1(x)=x.xsin(x) + x/e if x<=1.0 or f2(x)=x.x.xcos(x)-log(x) if x>1.0, and return result
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