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MTH101 -Calculus and Analytical Geometry Assignment NO #1 Discussion and Solutions Spring Fall 2013 Due date:27 nov 2013

 

 

 

 

Assignment: # 01 (Fall-2013)

 

Mth101 (Calculus & Analytical Geometry)

 

Lecture: 01 – 10

Total Marks = 20

Due date:  27-11-2013

 

INSTRUCTIONS:-

 

  1. In order to attempt this assignment you should have full command on Lecture# 01 to Lecture # 10

 

  1. Try to get the concepts, consolidate your concepts and ideas from these questions which you learn in Lecture # 01 to Lecture # 10.
  2. You should concern recommended books for clarify your concepts if handouts are not sufficient.
  3. Try to make solution by yourself and protect your work from other students. If we found the solution files of some students are same then it ‘ll be rewarded zero marks to all those students.
  4. You are supposed to submit your assignment in Word format any other formats like scan images, PDF format etc will not be accepted and will be give zero marks.
  5. Assignments through e-mail will not be accepted after the due date.If there is any problem in submitting your assignment through LMS, you can send your solution file through email with in due date.
  6. You are advised to upload your assignment at least two days before Due date.

 

 

 

Q.1                                                                                                        Marks 5

 

Solve the inequality  and write the solution set in interval form.

                    

Q.2                                                                                                         Marks 5

 

Find the equation of a circle whose diameter has endpoints (4, -1) and (-6, 7).

 

 

Q.3                                                                                                         Marks 5

 

  Evaluate the following limit:

 

                                                        

Q.4                                                                                                         Marks 5

 

  If  and, then find the composite functions .

See Attachment File 

 

 

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Replies to This Discussion

ya 1st ka jvb interval mein likha ha ap ny

yes... Interval mai hay .... 

is this is true interval if not then why

yess.. its true ... 

I got x> -4/5 i want to know how to convert it to interval form please tell me the method I want to understand. thanks

LIMIT waky ja -1

what are you doing with limit

 

see this example to solve q no 3

Good works....

Thanks

|5x+11|>7
7 > 5x + 11 > -7
7 >5x+11 -7 > 5x + 11
-4 / 5 > x -18 / 5 > x
This can b writen in interval notation as ( -4/5 , -18/5 )

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Some one's answers 

mujhe question no. 1 may ye samjh nai aai hai k hum x< -18 \5 or x > -4 \5 ko line pe kaise  show kare or jawab shayad phir 2now taraf se he _ infinity may aae ga.......... plz koi help kare

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