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BSc Tuition > Learn BSc Tuition > Anyone have doubts in BSC maths.. can contact me.....
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Anyone have doubts in BSC maths.. can contact me.. send me photo of your qus. and I will reply with the answer.

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Good... will send..
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Apptitute training / campus recruitment training /diplomo/bank po/lic entrance/rrb/bank clerical

definetly
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Math Educator for Std.11th ,12th , Engineering Entrance and Degree Level with 11+ Years Experience

IF ANYONE HAVE MATHS PROBLEMS OF 12TH STD. THEN SEND ME PIC OR SCAN OF PROBLEMS , I WILL REPLY WITH COMPLETE SOLUTIONS.. I TEACH MATHS 10TH , 12TH STD., CET AND JEE MAINS LEVEL ENGG. ENTRANCE EXAM IN BANGALORE AND ALSO ONLINE.. I AM GOLD MEDAL WINNER IN MATHS OLYMPIAD..
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Tutor

what have u written is not clear.. symbols are missing... so paste only picture of problems.
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History

12 th std 2015/12/1 apply
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Hindi

A non abelian smallest proper subgroup in this . 6. 8. 12. 24 Plz answer immediately
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12th std 2015/12/1 apply
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Outside the purview of our domain
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BTech BSc Inter Tutor Supply and Regular

show that {xn}n?N{xn}n?N is an increasing and upper bounded sequence, which shall imply that {xn}n?N{xn}n?N converges. Next we shall find the limit using the recursion relation. A. First, {xn}n?N{xn}n?N is an increasing sequence. This can be shown inductively: For k=1k=1: x1=p+x0??????>p?=x0x1=p+x0>p=x0. If...
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show that {xn}n?N{xn}n?N is an increasing and upper bounded sequence, which shall imply that {xn}n?N{xn}n?N converges. Next we shall find the limit using the recursion relation. A. First, {xn}n?N{xn}n?N is an increasing sequence. This can be shown inductively: For k=1k=1: x1=p+x0??????>p?=x0x1=p+x0>p=x0. If xk?10x?x0=p>0. Then xn+1=xn+p???????p+x??????xn+1=xn+p?p+x. But limxn+1=xlimxn+1=x, and hence x=p+x?????? x=p+x or x2=x+p, x2=x+p, or x=1±1+4p??????2. x=1±1+4p2. We choose the one with the plus sign since the other one is negative. Hence x=1+1+4p??????2. read less
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