CBSE > Class 11 > Mathematics | CBSE Board | Homework Help

64^(1/3)=4 because the cube root of 64 is 4 125^(1/3)=5 So, [9(4+5)^3]^(1/4) =[9(9)^3]^(1/4) =[9(729)^(1/4) =[6561]^(1/4) The forth root of 6561 is 9 The answer is 9

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A is a subset of B then prove that B compliment is a subset of A compliment

Posted by Sarthak Avhad 10 months, 2 weeks ago

CBSE > Class 11 > Mathematics

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Find the degree measures corresponding to the following radian measures

Posted by Vidhisha Khanal 10 months, 3 weeks ago

CBSE > Class 11 > Mathematics

1 answers

Srishti Thakur 10 months, 3 weeks ago

180°=π radian

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Genral solution of sinx= 1/^2

Posted by Yadunandan Goswami 10 months, 3 weeks ago

CBSE > Class 11 > Mathematics

1 answers

Inayat Kashyap 10 months, 3 weeks ago

sinx=12 ⇒sinx=sinπ6 Thus, one principal solution of the equation is π6

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Write in set builder form{1,5,10,15}

Posted by Simrandeep Kaur 10 months, 3 weeks ago

CBSE > Class 11 > Mathematics

3 answers

Ankit Class 9 months, 2 weeks ago

{ x:x is 5n&1 , n belong to N ( n=1to3)}

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Harsh Deep 10 months, 2 weeks ago

{ x:x is a 1+n where n is a multiple of 5 1to4 and n belongs to natural no.}

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Kanishk Mahajan 10 months, 3 weeks ago

{x : x is a natural number multiple of 5 and x = 1}

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Let A = { a,b } and B { x : x £ N, x is prime less than 7}. Find A X B

Posted by Vishal Gupta 10 months, 3 weeks ago

CBSE > Class 11 > Mathematics

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Let F be the set of all parallelograms, F_{2} be the set of rectangles, F_{1} be the set of rhombuses, F_{4} be the set of squares and F_{5} be the set of trapeziums in a plane then F_{1} =

Posted by Tamil Gamers 10 months, 3 weeks ago

CBSE > Class 11 > Mathematics

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Prove: (sin A + cos A) (1 - sin A cos A) = sin^3 A + cos^3 A

Posted by Jahnavi Suthar 11 months ago

CBSE > Class 11 > Mathematics

1 answers

Snehitha Snehitha 10 months, 3 weeks ago

Answer

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If U = {1,2,3,4,……..,10} is the universal set for the sets A = {2,3,4,5} and B = {1,2,3,4,5,6}, then verify that (A∪B)'' = A''∩B''.

Posted by Yaisk Jwlw 11 months ago

CBSE > Class 11 > Mathematics

2 answers

Inayat Kashyap 10 months, 3 weeks ago

, U={1,2,3,4,5,6,7} A={1,2,5,7} B={3,4,5,6} (A∪B)′=U−(A∪B) ={1,2,3,4,5,6,7}−{1,2,3,4,5,6,7} =ϕ A′∩B′=(U−A)n(U−B) ={3,4,6}n{1,2,7} =ϕ Hence (A∪B)′=A′∩B′.

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Himanshu Singh Chauhan 1221 11 months ago

7,8,9,10=7,8,9,10

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Workout for the day of week on 18th October 2100

Posted by Arav Jain 11 months ago

CBSE > Class 11 > Mathematics

1 answers

Yashi Agrawal 9 months, 2 weeks ago

Monday

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A={2,3} , B={x:x is a solution of x^2+5x+6=0}

Posted by Ritu Sinha 11 months, 1 week ago

CBSE > Class 11 > Mathematics

2 answers

Himanshu Singh Chauhan 1221 11 months ago

B=(1,-6)

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Arav Jain 11 months ago

X= teri mak ki chu

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Write the following sets in the roster form. a) A = {x : x2 + x , x ∈ R }

Posted by Daksh Kumar 11 months, 1 week ago

CBSE > Class 11 > Mathematics

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Aditi Tiwari 11 months, 1 week ago

In roster form :- {2,6,12,16,20,.........}

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If f( x) is a quadratic function, p(l)=a , p (m) = b and p(n) = c , then find p(x)

Posted by Bittu Shaw 11 months, 2 weeks ago

CBSE > Class 11 > Mathematics

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Chapter 8 binomial theorem (200-1/2)⁵

Posted by Kishanvir Kishanvir 11 months, 2 weeks ago

CBSE > Class 11 > Mathematics

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(200-1/2)⁵

Posted by Kishanvir Kishanvir 11 months, 2 weeks ago

CBSE > Class 11 > Mathematics

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Inayat Kashyap 10 months, 3 weeks ago

9752487531.22

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Find the modulus (1-i/1+i)^100

Posted by Gunjan Kashyap 11 months, 2 weeks ago

CBSE > Class 11 > Mathematics

2 answers

Aditi Tiwari 11 months, 1 week ago

√1 =1 hence modulus of (1-i/1+i)¹⁰⁰ is 1

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Aditi Tiwari 11 months, 1 week ago

(1-i/1+i)¹⁰⁰ Multiple with conjugate of denominator =(1-i*1-i/1+i*1-i)¹⁰⁰ =(1-i-i+i²/1-(i)²)¹⁰⁰ =(1-2i+(-1)/1-(-1))¹⁰⁰ =(-2i/2)¹⁰⁰ =(-i)¹⁰⁰ =(-i)⁴*²⁵+⁰ =(1)¹⁰⁰ =1 In complex form =1+0i |z|= √1²+0² |z|=√1

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To make the project to analyze y= x² graph, specifying possible domain and range

Posted by Bittu Shaw 11 months, 2 weeks ago

CBSE > Class 11 > Mathematics