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Jimbo 420

Posted: Thu Jun 08, 2006 11:46 am Post subject: cetroid finder 


this program is suposed to find the centroid of a triangle i can get 2 line eqasions but can't get it to find the centroid need help
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upthescale

Posted: Thu Jun 08, 2006 7:37 pm Post subject: (No subject) 


are you asking for help? go the the main menu, and go in Turing Help, and post this same post there, because this is users submisions, not help quesitons!






Delos

Posted: Thu Jun 08, 2006 9:11 pm Post subject: (No subject) 


Moved.
And Jimbo 420, please change your sig. Statements promoting negative stereotypes, racism, or any other form of discriminatory thought are not tolerated. Even in a humorous form. I'm sure you did not mean any harm by it, but it still is really pushing the envelope.






upthescale

Posted: Thu Jun 08, 2006 10:00 pm Post subject: (No subject) 


how is that sterotype delos?






Clayton

Posted: Thu Jun 08, 2006 10:09 pm Post subject: (No subject) 


^the above post is spam^ you have been warned too many times to count and it is really annoying plz stop, it is a stereotype because it is using Jews (joo) as a point of humour (and improperly i might add)






Jimbo 420

Posted: Mon Jun 12, 2006 11:48 am Post subject: (No subject) 


any1 know how to finish this program it gives the 2 lines and the 2 lines must be solved to find the answer






TheOneTrueGod

Posted: Mon Jun 12, 2006 1:07 pm Post subject: (No subject) 


Theres two methods.
A) Substitution
L1: 2X + 4Y = 6
L2: 3X  6Y = 12
from L1: X = (64Y)/2
X = 3  2Y
sub into L2:
3(32Y)  6Y = 12
9  6Y  6Y = 12
6Y = 3
Y = 1/2
Sub back into the L1 eqn
2X + 4(1/2) = 6
2X  2 = 6
2X = 8
X = 4
so the point of intersection is (4,1/2)
B)Add the equations
L1: X + 3Y = 5
L2: X  4Y = 3
L1  L2 : XX + 3Y  (4Y) = 5  3
3Y + 4Y = 2
7Y = 2
Y = 2/7
Now continue by subbing it in to the equations.
Summary
You'd probably be better off using the first method, though you'd have to be specific as to the input.






Jimbo 420

Posted: Wed Jun 14, 2006 11:59 am Post subject: Nice 


yeh i had something like that going on thanks for the help tho im pretty sure it works good now thanks man






Jimbo 420

Posted: Wed Jun 14, 2006 12:06 pm Post subject: more help 


ok i can't seem to figure out how to code the substitution equasions? would it work easier using a procedure(which i havent used many of:)






