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  • 1 answers

Gaurav Seth 4 years, 5 months ago

We have to show that P(A ∩ B) = P(A) ∩ P(B)

LHS:

Let x ∈ P(A ∩ B)

=> x ∈ (A ∩ B)

=> x ∈ A and x ∈ B .............1

RHS:

x ∈ P(A) ∩ P(B)

=> x ∈ A and x ∈ B  ............2

From equation 1 and 2, we get

P(A ∩ B) = P(A) ∩ P(B)

Hence proved.

  • 1 answers

Roopam Sharma 4 years, 5 months ago

Hii Aadya ji
  • 0 answers
  • 1 answers
idk
  • 5 answers

Manisha Dhibar 4 years, 5 months ago

Shi kha....

Manish Kumar 4 years, 5 months ago

Bhai mzak kr rhe ho kya ?

Anshu Yadav 4 years, 5 months ago

1

Tec Om 4 years, 5 months ago

1

? S. S. ? 4 years, 5 months ago

Are u serious?!???sry to say this....but agr apko ye nii aata...to 11th ki maths kya ghnta solve krr paoge...
  • 1 answers

Gaurav Seth 4 years, 5 months ago

If u equals to {1, 2, 3, 4,.....,10} is the universal set for the two sets a={2,3,4,5} and b={1,2,3,4,5,6} then verify that (a u b)" equals to a" n b"

Given

     a = { 2, 3, 4, 5 }   and   b = { 1, 2, 3, 4, 5, 6 }

Then

     a ∪ b = [ everything in one or the other ]

               = { 1, 2, 3, 4, 5, 6 }

so

   ( a ∪ b )' = [ the complement, so everything in u that's not in a∪b ]

                  = { 7, 8, 9, 10 }.         ...(*)

Also

     a' = { 1, 6, 7, 8, 9, 10 }    and    b' = { 7, 8, 9, 10 }

so

     a' ∩ b' = [ those elements in both a' and b' ]

                 = { 7, 8, 9, 10 }.          ...(**)

The sets at (*) and (**) are the same, verifying the relationship

    ( a ∪ b )' = a' ∩ b'.

  • 2 answers

Smita Mohapatra 4 years, 5 months ago

What's the question asking Here what to find Plzz mention

Syed Amaan 4 years, 5 months ago

Ssss
  • 1 answers

Ff Sedn 4 years, 5 months ago

Hi
  • 3 answers

Ff Sedn 4 years, 5 months ago

Hi jyoti

Syed Amaan 4 years ago

What are the set

Siddarth Asati Asati 4 years, 5 months ago

Where are the sets
  • 1 answers

Siddarth Asati Asati 4 years, 5 months ago

You can saw from google
  • 2 answers

Siddarth Asati Asati 4 years, 5 months ago

A=(1,2,5) B=(6,7) X is the universal set for A and B (X=(1,2,3...10) B'=(1,2,3,4,5,6,7,8,9,10) From LHS side (1,2,3) intersection (1,2,3,4,5,6,7,8,9,10) (1,2,5)(which is equal to A) Hence it is proved

Tec Om 4 years, 5 months ago

A=(1,2,5)
B=(6,7)
X is the universal set for A and B { X =(1,2,3,4...10)} B'={1,2,3,4,5,8,9,10}
from LHS side
{1,2,5} intersection{1,2,3,4,5,8,9,10}
{1,2,5} (which is equal to A)
Hence it is proved
  • 1 answers

Monika Gautam 4 years, 5 months ago

it is a circle whose centre is (4,-3) and radius 5
  • 1 answers

Gaurav Seth 4 years, 5 months ago

 

Q : Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (-4, 0, 0) is equal to 10?

Answer:

  • 1 answers

Siddarth Asati Asati 4 years, 5 months ago

Good morning
  • 2 answers

Himanshu Singh 4 years, 5 months ago

5+1?

Akash Pandey 4 years, 5 months ago

ye konse ch me se hai bhai...
  • 3 answers

Manisha Dhibar 4 years, 5 months ago

Bt what was the question
Apna question to bolo bro.....

4 years, 5 months ago

Ok if anybody else can answer
  • 2 answers
Answer is A. ...Kyonki (A U B) mein A aur B dono ke saare elements honge... So jab iska A se intersection hoga tab A aaraam se common aa jaayega?

4 years, 5 months ago

A
  • 2 answers

Winter Wøłf Yt 4 years, 5 months ago

Tell me u got

Winter Wøłf Yt 4 years, 5 months ago

Search in rd sharma
  • 0 answers
  • 0 answers
  • 0 answers
  • 2 answers

4 years, 5 months ago

Well That person is sure now who am i . I request that person now not to reply and just dont reveal my identity

? S. S. ? 4 years, 5 months ago

Oh really...u aren't Smarakee...then why is ur id - smar****@gmail.com....?!
  • 2 answers

Ff Sedn 4 years, 5 months ago

Hi

Gaurav Seth 4 years, 5 months ago

LHS = sin²(6x) - sin²(4x)
Use the formula,
(a² - b²) = (a - b)(a + b)
= (sin6x -sin4x)(sin6x +sin4x)
Now,
Use the formula,
sinC + sinD = 2sin(C+D)/2.cos(C-D)/2
sinC - sinD = 2cos(C+D)/2.sin(C-D)/2

={2cos(6x+4x)/2.sin(6x-4x)/2}{2sin(6x+4x)/2.cos(6x-4x)/2}
={2cos5x.sinx}{2sin5x.cosx}
={2sin5x.cos5x}{2sinx.cosx}

Use the formula,
sin2A = 2sinA.cosA

= sin2(5x).sin2(x)
= sin10x.sin2x =RHS

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