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6351.

Start with b and ends with t how many words are there?

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6352.

Find n if. -: (n+2)! (2n+1)! / (2n-1)! (n+3)! 72/7

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6353.

22x22

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6354.

Find n if-: n!/2!(n-2)! : n!/4!(n-4)!

Answer» n=2 or n=3
6355.

Show that-: n!(n+2)=n!+(n+1)!

Answer» R.H.S=n!+(n+1)! =n! +(n+1)n! =n! [1+(n+1)] =n! [1+n+1] =n! (n+2) R.H.S=L. H. S
6356.

Prove that Cos^2y-cos^2x =sin (x+y)sin (x-y)

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6357.

X≥3

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6358.

(√3 +√2) ^6 - (√3-√2) ^6

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6359.

domain

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6360.

In how many ways can 9examination paper be arranged so that the best and worst never come together

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6361.

How to find the value of inverse trigonometric values such as tan-1

Answer» U can find this on net....and in examination u will have the value in last of the question..after ? In (tan -1= ....)
6362.

Solve x(2)+2=0

Answer» x(2) +2=02x+2=02x= -2x= -2/2x = -1 Ans. x= -1
6363.

Find the value of sin x/2 and if value of sin X is 1/4.

Answer» Is anwer is 1/6...?
6364.

Value of tan 18

Answer» 0.32492
6365.

√3cosec20 - sec20

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6366.

Tan13pi/12

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6367.

n!/(n-r)! (1): n=6,r=2

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6368.

Cos square 45°sin square15°=squareroot 3/2

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6369.

how to learn identities of trignometry

Answer» By writing them daily atleast 3 times
6370.

In an isosceles trapezium ABCD ,if AB is parallel to CD ,show that angle ADC+ angle ABC =180°

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6371.

Lim [x³-4x²+4x]X->2

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6372.

2!+3!

Answer» 2*1+3*2*1=8
6373.

In a triangle ABC, tanA/2 = 5/6, tanB/2 = 20/37, then tanC/2 is equal to what?

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6374.

S a huge d

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6375.

How put k+1 in pmi

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6376.

|x-1|+|x-2|>=4

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6377.

Formula for centroid of triangle

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6378.

Rules for integrarion

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6379.

What is standard position of an angle

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6380.

What is dot

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6381.

Locus andits equation

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6382.

Subset mean

Answer» A set B is said to be a subset of set A if every elements of B are in set A.for example:- let A={1,2,3}\xa0then B= {1,2} is subset of set A as every element of B i.e.1 and 2 is also element of set A.Similarly C={1} is also a subset of set A as element of C i.e.1 is in set A.
6383.

a+

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6384.

1+2+3+-------+n=n(n+1)DIVIDED BY 2

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6385.

What is a subset

Answer» "A" is considered to be a subset of ,"B" if all the elements of A is in present in B.Eg:A= {1,2,3,4}B={2,3}In the above case, B is considered to be a subset of A
6386.

If A={1,3,5,7,11} then what is the power set of A

Answer» The set of all the subset Of A is the powerset of A
6387.

Tan70=tan20+2tan50

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6388.

Area is scalar or vector quantity.Give example

Answer» Area is a scalar quantiti..because there ia no need of dimension to define area..
6389.

Write in set builder form{1,-1,i,-i}

Answer» Set builder form of {1,-1,i,-i} isA={x: x is the solution of eqaution (x2\xa0-1)(x2\xa0+1)=0} ={x: x is the solution of equation x4\xa0- 1=0} [by using identity (a-b)(a+b)=a2\xa0- b2]
6390.

if A={-1,1},find A×A×A

Answer» A={-1,1}A×A={-1,1}×{-1,1} = {(-1,-1),(-1,1),(1,-1),(1,1)}Now A×A×A={(-1,-1),(-1,1),(1,-1),(1,1)}×{-1,1} = {(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1)}
6391.

1+11+111+1111+11111+111111

Answer» 123456
6392.

If two complex number z1 and z2 are such that |z1|=|z2| is then necessary that z1=z2

Answer» No, its not necessary.For example:- let z1= 3+4i and z2= 4+3iThen |z1| ={tex}{\\sqrt {3^2 +4^2}}={\\sqrt {9+16}}={\\sqrt {25}}=5{/tex}Also |z2| ={tex}{\\sqrt {4^2+3^2}}={\\sqrt {16+9}}={\\sqrt {25}}=5{/tex}So clearly |z1| = |z2| =5But z1\xa0done not equal to z2\xa0as real part\xa0and imaginary part\xa0of z1\xa0is not equal to real part\xa0and imaginary part\xa0of z2
6393.

Find √i+√-i

Answer» Solve:\xa0{tex} √i+√(-i){/tex}{tex}=√(e^(i π/2) )+√(〖-e〗^(i π/2) ){/tex}{tex}=〖e〗^(i π/4)+ie^(i π/4){/tex}{tex}=〖e〗^(i π/4) (1+i){/tex}{tex}=〖e〗^(i π/4) (√2 e^(i π/4)){/tex}{tex}=√2 e^(i π/2){/tex}{tex}=i√2{/tex}
6394.

Find the value of tan1°tan2°tan3........tan89°.

Answer» tan1°tan2°tan3°.........tan89°=tan(90°-89°)tan(90°-88°)tan(90°-87°)...............tan(90°-46°)tan45°tan46°.....tan89° = cot89°cot88°cot87°..........cot46°tan45°tan46°......tan89° [by using identity tan(90°-A)= cotA] ={tex}1\\over tan89°{/tex}×{tex}1\\over tan88°{/tex}×.........{tex}1\\over tan46°{/tex}×tan45°×tan46°×......tan88°×tan89° = tan45° =1
6395.

How Rutherford knew that Alfa particles returned opposite direction?

Answer» With teh help of analyser\xa0
6396.

|x-2|-|x-6|

Answer» 4
6397.

Find the value of sin16

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6398.

Find value of cos (-1710°)

Answer» = -cos (1710)°=-cos (180*10-90)°= -cos (-90)°= cos (90)°= 0
5×360°-90° {360-Q}Cos90°{cos90°=0}0
Cos(1800+(-1710))Cos (90)0
6399.

Draw Venn diagram of A\'UB

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6400.

If |z+4| is greater than equal to 3,then maximum value of |z+1| is

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