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Assignment No. 1

                        MTH 101 (Spring 2013)


                                                                      Maximum Marks: 15

                                                                      Due Date: 6th May, 2013  


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Question#1                                                                                         Marks4+4=8


Solve the following Absolute Value Problems and write their solutions in both inequality and interval notation.





Question#2                                                                                                  Marks 7



 Evaluate the following limit.






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Replies to This Discussion

Please Discuss here about this assignment.Thanks

Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

Assalamu Alaikum kindly please meri help kerdein maine in teenon k solutions kerliye hain per microsoft office main mjhse equations nai horhien plzz batadein k main word main kese maths ki equations likhun

I solved Q no 1 and got the ans
Please share urs ansss

plz come and discuss here.

yeah mare ans b yahe han

@shahzab khan, it is not the correct one, and @Saireen Yousaf if you have the same then both of your answers are not correct.

12/5 ≤ x ≤ 6/5 1st Q ka ans 

2nd Q ka

4  <  x < 8/3

koi baty ga k interval mein qasy likhty hen jvb baqi ya jvb bilkul correct hen mein ny khud keya hen

Limit waly ka ZERO ans ha 


@Zeeshan, don't you think that 2nd one should be in form of x>a or x<-a ? How it can be in the form of 4<x<8/3? (Just to guess)

And in the scenario if you punch in the answers of 1st one in calculator you will get(according to your solution) 2.4≤x≤1.2 (How it can be possible in real world?)

ok thank u zeeshan i solved it again and got the answers 

6/5 >= x >=12/5

and interval form is 


and the second one is 


interval is 

(8/3, 4)

frnd aur jo limit wala question hy uska answer tu btain evalute limit kaisy solved krain gain



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