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# GDB No.3 is opened and Friday, July 24, 2015

 GDB No.3 Dated: Jul 14, 15 Dear Students, A new GDB has been created and it will be open from July 15, 2015. Due date of the discussion board is July 24, 2015. Moreover a new functionality (Preview) has been added in GDB to verify that your post is correctly displaying. So before posting your comments you should first preview your post and if the equations are correctly displaying then you can post the comment. Remember once you post the comments on GDB then you cannot edit or repost your comment. Don't post your GDB reply on regular MDB.
• The purpose of this GDB is to assess your knowledge and computational skills.
• Produce you own work. Copying the text from any other student or from any website is strictly prohibited. You will get zero marks in this case.
• Write the mathematical equations/symbols in Math Type software and paste it in GDB. For the procedure watch this video based on the problems and solution of previous GDB  Problem and Solutions Video.
• Last date of this GDB is July 24, 2015. There will not any relaxation after due date. You have ample days to complete this task. Submit your answers at the earliest to avoid losing your marks.

Important Note: A new functionality (Preview) has been added in GDB to verify that your post is correctly displaying. So before posting your comments you should first preview the post and if the equations are correctly displaying then you can post the comment. Remember once you post the comments on GDB then you cannot edit or repost your comment.

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### Replies to This Discussion

Need Solution Please..... Koi Ha Laiq Bacha,

this is the sol. of gdb mth301

gdb%20sol.%20mth301.docx

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Dear Students Don’t wait for solution post your problems here and discuss ... after discussion a perfect solution will come in a result. So, Start it now, replies here give your comments according to your knowledge and understandings....

y=0 mean right side of y or upper region of quadrant or positive side of y?

y is varying from 0 to  sqrt(1-x^2)

y = sqrt(1-x^2) is a circle centered at (0,0)

for y = 0, this circle will be to the right of y axis because x is also varying from 0 to 1. Had x varied from -1 to 1, then it was going to be half circle above x axis. However, x is varying from 0 to 1 i.e. one fourth of a complete circle.

Thanks Bro i got your point..

Wow Bro that was awesome Thanks.

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