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CLGQKBC" The pattern followed is - I - skip one letter then the next one is used(K) N - the previous letter is used(M) D -skip one letter then the next one is used(F) I -the previous letter is used(H) A -skip one letter then the next one is used(C) Hope it helps you...

4Thank You

Tanisha Chawla 5 years ago

Hello I think the answer would be - "CLGQKBC" The pattern followed is - I - skip one letter then the next one is used(K) N - the previous letter is used(M) D -skip one letter then the next one is used(F) I -the previous letter is used(H) A -skip one letter then the next one is used(C)

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ANSWER

cos8A. cos5A_cos12A.cos9A/sin8A.cos5A+cos12A.sin9A=tan4A

Posted by Sujitha Venkatesan 5 years ago

CBSE > Class 11 > Mathematics

1 answers

Ankush Raj Pathak 5 years ago

Ans

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ANSWER

Sin 5x - 2Sin3x +Sinx ------------------------------------ Cos 5x - Cos x

Posted by Amogh Srivastav 5 years ago

CBSE > Class 11 > Mathematics

0 answers

ANSWER

Derivative of sinx from first principle

Posted by Ritik Kumar 5 years ago

CBSE > Class 11 > Mathematics

1 answers

Neev Garg 5 years ago

Cosx

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ANSWER

Limits and derivatives all formula

Posted by Ranger King 5 years ago

CBSE > Class 11 > Mathematics

1 answers

Shailendra Dhukia 5 years ago

d/dx(sinx)=cosx d/dx(cosx)=-sinx d/dx(tanx)= sec••x ••= square

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ANSWER

Value of tan20tan40tan80

Posted by Lucky Sharma 5 years ago

CBSE > Class 11 > Mathematics

2 answers

Neelesh Shukla 5 years ago

This answer is wrong

0Thank You

Lucky Sharma 5 years ago

Tan60°

3Thank You

ANSWER

how to differentiate sec x by first principle

Posted by Shanu Shanu 5 years ago

CBSE > Class 11 > Mathematics

1 answers

Shailendra Dhukia 5 years ago

d/dx(secx )= secx.tanx

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ANSWER

Find the image to the point (3,8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror .

Posted by Archi Mittal 5 years ago

CBSE > Class 11 > Mathematics

0 answers

ANSWER

Write the radiant measure of 5 degrees 37'30''

Posted by Lavish Chauhan 5 years, 1 month ago

CBSE > Class 11 > Mathematics

1 answers

Mahima Pattanayak 5 years ago

Answer: 5 degree 37 minute 30 second=0.098 Radians.

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ANSWER

Write the degree measure of 5degree 37' 30''

Posted by Lavish Chauhan 5 years, 1 month ago

CBSE > Class 11 > Mathematics

0 answers

ANSWER

Show that (2,-3,5),(1,2,3) and (7,0,-1) are collinear

Posted by Rajesh Singh 5 years, 1 month ago

CBSE > Class 11 > Mathematics

0 answers

ANSWER

Insert 6 numbers between 3 and 23 such that the resulting sequence is an AP

Posted by Abhishek Gupta 5 years, 1 month ago

CBSE > Class 11 > Mathematics

1 answers

Jagdish 5 years ago

Add 20/7 to 3 to get first number then add it again to obtain next number Common difference = (last term - first term)/no. of numbers to be inserted +1 = (23-3)/6+1 = 20/7

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ANSWER

miscellaneous exercise is important for school examination tell me I should solve miscellaneous exercise.

Posted by Kashi Sh Shakya Kashish Shakya 5 years, 1 month ago

CBSE > Class 11 > Mathematics

3 answers

Nishant Dhakare 5 years ago

Yes not for exams also for you intelligence improvement , in miscellaneous exercises question level os very suprb

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Vivek Pandey 5 years, 1 month ago

Yes v imp

1Thank You

Sameer Tanwar 5 years, 1 month ago

Yes miscellaneous example and exercise both are important as it covers nice questions which improves your understanding the topic

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ANSWER

4?2−1(?−?)2

Posted by Ram Panda 5 years, 1 month ago

CBSE > Class 11 > Mathematics

0 answers

ANSWER

prove that sin 10*sin 30*sin 50 *sin 70= 1/16

Posted by Adarsh Mund 5 years, 1 month ago

CBSE > Class 11 > Mathematics

1 answers

Sameer Tanwar 5 years, 1 month ago

We know that i) sin 2A = 2sinAcosA ii) sin(90-A)= cosA iii) sin30=1/2 LHS = sin 10 sin30 sin50sin 70 =sin 10 × 1/2 × sin (90-40)× sin(90-20) =1/2[sin 10 ×cos 40×cos 20] We multiply with[2 cos 10/2cos 10] =(1/2×1/2cos10)[2sin10cos10×cos20×cos40] =(1/4cos10)[sin20cos20×cos40] =(1/8cos10)[2sin20cos20×cos40] =(1/8cos10)[sin40cos40] =(1/16cos10)[2sin40co40] =(1/16cos10)×sin80 =(1/16cos10)×sin(90-10) =(1/16cos10)×cos10 =1/16 = RHS

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ANSWER

Limit X tends to zero sin a x + b x by x + sin BX

Posted by Deepansu Mehar 5 years, 1 month ago

CBSE > Class 11 > Mathematics

0 answers

ANSWER

Exercise 3.3 question 10

Posted by Priyanka Ghulati 5 years, 1 month ago

CBSE > Class 11 > Mathematics

2 answers

Vineeta . 5 years ago

U may search on YouTube u will get better explanation

0Thank You

Manvi Gupta 5 years, 1 month ago

Of NCERT or any other reference book...

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ANSWER

Xdy/dx-2y=e^x.x^3

Posted by Rishi Ranjan 5 years, 1 month ago

CBSE > Class 11 > Mathematics

0 answers

ANSWER

d/dx (3√x)

Posted by Harishy Dharwad 5 years, 1 month ago

CBSE > Class 11 > Mathematics

3 answers

Rudransh Bhargava 5 years, 1 month ago

7568zc fyi

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Deepanshu Tyagi 5 years, 1 month ago

d/dx(3√x)=3/2√x

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Swastik . P 5 years, 1 month ago

d/dx(3x^1/2) It will be. 3/2 x^-1/2 It can further written as 3/2√x

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(3-x)=|x-3| solve for x.

Posted by Priyanshu 11 5 years, 1 month ago

CBSE > Class 11 > Mathematics

5 answers

Deepanshu Tyagi 5 years, 1 month ago

(3-x)=|x-3| Remove modulus- ±(3-x)=x-3 3-x=x-3 or -(3-x)=x-3 x=9/2 or -3+x=x-3 x=9/2 or 0=0 So,x=9/2

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Swastik . P 5 years, 1 month ago

X=9/2

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Malavika Manoj 5 years, 1 month ago

X= 3

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Aditya Kumar 5 years, 1 month ago

The value of x will be 3

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Raunak Chauhan 5 years, 1 month ago

The value o x is 3

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ANSWER

Find the eccentricity of the ellipse 5 x square + 9 y square is equal to 1

Posted by Aman Kumar 5 years, 1 month ago

CBSE > Class 11 > Mathematics

1 answers

Aditya Kumar 5 years, 1 month ago

Eccentricity of the question is 2/3

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ANSWER

Project work

Posted by Siyaram Gupta 5 years, 1 month ago

CBSE > Class 11 > Mathematics

0 answers

ANSWER

Find the equation of the circle concentrated with x^2 +y^2 -8x +6y -5 =0 and passing through (-2,-7)

Posted by Tiyas Dutta 5 years, 1 month ago

CBSE > Class 11 > Mathematics

1 answers

Aditya Chandel 5 years, 1 month ago

x²+y²-8x+6y-35=0

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In the triangle ABC with vertices A(2,3), B (4,1) and (1,2) find the equation and length of altitude from the vertex A.

Posted by Pakhi Shah 5 years, 1 month ago

CBSE > Class 11 > Mathematics

1 answers

Raunak Chauhan 5 years, 1 month ago

First we have to find the mid point of b and c Mid point of b and c is (5/2,3/2) then length of altitude is √(2-5/2)²+(3-3/2)² √1/4+9/4 √5/2 is the length of altitude from vertex A on the side BC We can find equation by using formula two point form page number 214 ncert math

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If p is the lenght of perpendicular from the origin to the line whose intercept on the aces are a and b, then show that 1/p^2= 1/a^2+1/b^2???

Posted by Pakhi Shah 5 years, 1 month ago

CBSE > Class 11 > Mathematics