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  • 5 answers

Vignesh Babu 4 years, 9 months ago

No need to get topper..... Ok But I am topper ?

Pooja . 4 years, 9 months ago

Study till u can!!

Dhino Kevin 4 years, 9 months ago

try smart work if you failed to do hardwork

Kai Pocha Lofer 4 years, 9 months ago

Go and study raa,DO OR TRY

Prachi Maithil 4 years, 9 months ago

Study consistently... This will surely make you topper...
  • 1 answers

Kai Pocha Lofer 4 years, 9 months ago

#1 Assume s=1 X=2-2(s) y=3+(s) z=4-5(s) X=2-2(1) y=3+(1) z=4-5(1) X=0 y=4 z=-1 #2 assume s=0 Do the step as same
  • 1 answers

Gaurav Seth 4 years, 9 months ago

Q u e s t i o n : What is the principal value of cos-1 ( cos 2π/3) + sin-1 ( sin 2π/3) ?

A n s w e r :

cos-1 ( cos 2π/3) + sin-1 ( sin 2π/3)

  • 1 answers

Dhino Kevin 4 years, 9 months ago

123456
  • 0 answers
  • 5 answers

Tanish Saini 4 years, 9 months ago

3/2

Rahul Yadav 4 years, 9 months ago

Sure this ans is right

Priya Singh 4 years, 9 months ago

dx/ dt= 2t, dy/dt= 3t² dy/dx= 3t.t/2t=3t/2 d2y/dx²=3/2 ans

Rahul Yadav 4 years, 9 months ago

Hii

Rahul Yadav 4 years, 9 months ago

3/5*t
  • 3 answers

Dhino Kevin 4 years, 9 months ago

prepare a handbook

Rahul Yadav 4 years, 9 months ago

Firstly you complete all chapters

Pragya Jain 4 years, 9 months ago

Try and study ncert :)
  • 2 answers

Sumit Kumar 4 years, 9 months ago

Hii

Yogita Ingle 4 years, 9 months ago

Set A={0,1,2,3,4,5} 

R be the equivalence relation on A.

Set R={(a,b)2divides(a−b)}

We have to find equivalnce class [0]

To find equivalence class {0},put b=0

⇒a−0 is multiple of 2 . 

⇒ a is multiple of 2. 

Multiples of 2 in given set are 0,2 and 4 . 

Hence equivalence class {0}={0,2,4}

  • 1 answers

Ardra T Aji 4 years, 9 months ago

X^2(x-1) + (x-1) ÷( x-1) .dx (x-1)[x^2+1]÷(x-1) .dx (x-1) in the numerator and denominator cancels and u r left with integral (x^2+1).dx =X^3÷3 + x = x^3 +3xwhole divided by x. + Constant
  • 0 answers
  • 1 answers

Sia ? 4 years, 5 months ago

https://mycbseguide.com/dashboard/content/19696

  • 1 answers

Pragya Jain 4 years, 9 months ago

(Sin²2x)/4
  • 1 answers

Student Of The Year 4 years, 9 months ago

Solution of this question is also available in this app.
  • 1 answers

Gaurav Seth 4 years, 9 months ago

On July 7, HRD Minister Ramesh Pokhriyal announced a major CBSE syllabus reduction with 30% of the syllabus slashed for the year 2020-21 for classes 9 to 12 because of the reduction in classroom teaching time due to the Covid-19 pandemic and lockdown.

CBSE has rationalized the syllabus with the help of suggestions from NCERT and the same has been notified by a new CBSE notification as well.

Deleted syllabus of CBSE Class 12 Mathematics

 

  • 4 answers

Pooja Jaat 4 years, 9 months ago

I think rd sharma

Kush Mehra 4 years, 9 months ago

TOGETHERWITH (EAD)

Khush Preet 4 years, 9 months ago

10 years

Student Of The Year 4 years, 9 months ago

I think oswaal
  • 2 answers

Kashish Agrawal 4 years, 9 months ago

Complete the question

Kashish Agrawal 4 years, 9 months ago

Didn't understood the question
  • 1 answers

Dev'S Educare 4 years, 9 months ago

Ans:-64

(-2w)^6  + (-2w2)^6

=64w^6  +  64 w^12

=64w^6(1+w^2)

=64(-w^3)

=-64

  • 2 answers

Chiranjeevi Kaushal 4 years, 9 months ago

Given , Function, y = e^4log(x) Differentiaing wrt x , we get ➡Dy/Dx = D[e^4log(x)]/Dx ➡Dy/Dx = e^4log(x) × ➡D(4log(x))/Dx ➡Dy/Dx = e^4log(x) × 4(1/x) ➡Dy/Dx = 4e^4log(x)/x Remmember : D(e^x)/Dx = e^x D(log(x))/Dx = 1/x

Rishu Singla 4 years, 9 months ago

-1/3x^3
  • 1 answers

Class 10 4 years, 9 months ago

Bahi kisa na bera ho to bataiyo Send karyo
  • 1 answers

Shubham Hatwal 4 years, 5 months ago

Put x = tant Then use formula (2 tant ÷[1-tan^2t ]= tan2t) Then tan^-1 ( tan2t)= 2t = 2 tan^-1(x ) = then do differentiation of it then answer= 2÷ { 1+x^2}
  • 3 answers

Aditya Kumar 4 years, 9 months ago

May anyone provide me the pic of its contents

Vaibhav Pandey 4 years, 9 months ago

Yes

Rishu Singla 4 years, 9 months ago

Yes
  • 0 answers
  • 2 answers

Chiranjeevi Kaushal 4 years, 9 months ago

The function is said to be "one one" if every element of set A have unique image in set B One one function is also known as" injective " The function is said to be "many one" if two or more elements of set A has same image in set B The function is said to be "onto" if every element of set B has pre image in set A It is also known as "surjective" If function f is both one one and onto then the function is called " bijective "

Nishant Pandey 4 years, 9 months ago

In 1 -1 function, single pre image have single image. In other words, different elements of domain are related with different elements of codomain.
  • 0 answers

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