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  • 1 answers

Karthik Murali 8 years, 2 months ago

For n=1 it is true. Assume for n=k it is true Then 2^k>k Prove that 2^(k+1)>k+1 LHS=2^k.2>2k=k+k>k+1 as k>1
  • 1 answers

Manash Pratim Saikia 8 years, 2 months ago

9.999999e15
  • 1 answers

Vinay Bagai 8 years, 2 months ago

1+p^2/2p
  • 0 answers
  • 1 answers

Akshay Satish 8 years, 2 months ago

y=mx
  • 1 answers

Shreenidhith Hemaraju 8 years, 2 months ago

4.6
  • 1 answers

Himanshi Verma 8 years, 2 months ago

2
  • 0 answers
  • 1 answers

Rahul Poddar 8 years, 2 months ago

(a-d)+a+(a+d)=3a=27, a=9, (a-d) a (a+d)=(9-d) 9 (9+d)=648,(81-9d)(9+d)=648,729+81d-81d-9d2=648,729-648=9 d2, 81/9=d2, d2=9, d=3 .... now a=9, d=3,..required number. 6,9,12
  • 2 answers

Srishti Rachna 8 years, 2 months ago

NO....

It's         (tan a +tan b)/(1-tan a. tan b)   :-v

Me Souhardee❤️ 8 years, 2 months ago

Tan a +tan b/1+tan a.tan b
  • 2 answers

Vasu Deekshith 8 years, 2 months ago

a=1d=2 TN =1+(n-1)2 =1+2n-2 =2n-1

Vasu Deekshith 8 years, 2 months ago

Tn=a+(n-1)d
  • 5 answers

Himanshu Shukla 8 years, 1 month ago

Correct answer is 21 if repeition is not allowed

Himanshu Shukla 8 years, 1 month ago

Answer is 21....repeuti

Dipens Rajput 8 years, 2 months ago

24 if repitetion of digit is not allowed 64 if rep. is allowed

Rahul Poddar 8 years, 2 months ago

64

Rahul Poddar 8 years, 2 months ago

6
J
  • 0 answers

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