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27801.

If a cosA +b sinA=m and a sinA-b cosA=n prove that a^2+b^2=m2+n2

Answer»
27802.

Tommorow will be math exam

Answer» Yes
27803.

Find a and d if a-3d , a-2d, a-d ,a+d, a+2d, a+3d are in Ap

Answer» Here, a = a - 3d.............. d = a - 2d - ( a - 3d)........... d = a - 2d - a + 3d........... d = 1d...
27804.

A solide

Answer» A solide ye kya hai
27805.

aCosA + bSinA =m and aSinA - bCosA =n prove that a2+b2= m2+n2

Answer» m2+n2=(acos+bcos)^2+(asin-bcos)^2 =a2cos2+a2sin2+2absincos+a2sin2+b2cos2 - 2absincos =a2(cos2+sin2)+b2(cos2+sin2) =a2(1)+b2(1) =a2+b2
27806.

(5+x/5-x)-(5-x/5+x)=15/4

Answer»
27807.

Important ouestions

Answer» Which
27808.

In an A.P , if the common different (d)=_4, and the seventh term (a7) is 4 then find the first term.

Answer» d=-4 , a7=4 a+6d=4a+6(-4)=4a=28
Given 7th term of ap = 4 Common difference= -4 If n =7 an = a+(n-1)d4=a+(4-1)-44=a+3*-44=a-124+12= a16=a
27809.

What is an conical acvity.

Answer» Conical cavity means a conical hole or cave
27810.

The sum and the product of zeros of a quadrilateral polynomial PX and 7 and 10 February then find PX

Answer»
27811.

1÷cosecA-cotA+1÷cosecB-cotB+1÷cosecC-cotC=cosecA+cosecB+cosecC+cotA+cotB+cotC

Answer» Y
27812.

Kl maths ka paper easy hogaPlease telll

Answer» It depends upon our preparation.
27813.

How find the value of middle term

Answer» By using the formula= N/2 for even term of n or N+1/2 for odd term of N
27814.

(SINA+COSA) (SECA+COSECA)= 2+SEC. COSECA

Answer»
27815.

15/2=8+p/2

Answer» 15/2=8+p/2
-1
P=-2
P=7
-1
27816.

If S29=99,S99=29Then find S128.

Answer»
27817.

Bpt theorum

Answer» Refer to your ncert book pg 124 theorem 6.1
27818.

Calculate the area of dedigned region common between the two quadrants of circle of radius 8cm each

Answer» Draw a diagonal joining opp vertices of square, Area design=2(area quadrant - area triangle.)
27819.

Please explain example 3 of chapter circles

Answer» Apply Pythagoras theorem in OPT&PRT and subract the equations u will get ur answer
27820.

How to determine hcf or lcm in word problems

Answer» For hcf - the words greatest or maximum are used and for lcm least and minimum words are used
Whenever in the question if something has to be divided into rows or colums that means we have to find the HCF......And if in the question their is given that two object have started from a position with different speeds and we have to find the least time when they both will meet again so in this case we have to find LCM...
27821.

Pls send me pythagores theorm

Answer» square of hypotenuese is equal to the sum of square of its base and perpendicular
27822.

Some importent tips for remember wile doing paper

Answer» Ensure that you have enough rough work space........and do rough work calculations neatly to avoid calculation errors (silly mistakes)
Don\'t use black pens during examinationIts good to write the whole paper using only one pen
If possible then do solutions in series
Be fully prepared and calm during the exam
Don\'t use whitner
27823.

Any expecred question for cbse maths 2018

Answer» Rs aggarwal ka saara example questions padh lo..... I guarantee you will score 70 marks... But when if you get it by your heart...
27824.

If ? ABC similar to ? PQR, ar(? ABC):ar(? PQR) =9:4,AB=18cm and BC=15cm,then find the value of QR

Answer» 10 cm
In this we have used thereom the ratio of ares of 2 similar is equal to the square of there corresponding side
Ar?ABC\\Ar?PQR = 9/49/4=(15^2/x^2)9*225=4*x^22025= 4*x^22025/4=x^2506.25=x^222.5=xSo QR =22.5
Qr is 10
10cm
27825.

Irrational number proof

Answer» Assume kro ki wo no. rational h aur us no. Ko 2 composite no. Ka factor proof kr do wo irrational ho jaega
27826.

If the pth term of AP is q and qth term is p.Then show that its nth is ( p+q-n).

Answer» ap= a+(q-1)d.......(i) and aq=a+(p-1)d.....(ii) now, solve eq. i and ii yu will get, d=-1..... substitute the value in i or ii yu will get a=p+q-1..... now an=a+(n-1)d... put values of a and d
27827.

X^3-4x^2+x+6 is a polynomial and its two roots are in ratio 2:3 find the roots of polynomial

Answer»
27828.

Show that7—√5 is irrational give that √5 is irrational.

Answer»
27829.

Prove that the tangent forms the angle of 90 at the circumference

Answer»
27830.

In triangle ABC, AB=AC .side BC is produced to D .prove that (AD²-AC²)=BD×CD

Answer»
27831.

Areaofsarcal

Answer» πr^2
Pi r^2
Circle???
27832.

sinA(1+tanA)+cosA(1+cosA)=secA+cosecA

Answer»
27833.

prove pythagoras theoram?

Answer»
27834.

Sir where all theorems

Answer» Book me
27835.

How many solutions does the pair of equations y= 0 and y= 5 has?

Answer» it has only 2 solutions
27836.

how to split the term plz mention detailed stepsx2 + 400x +960000

Answer» By middle term splitting we can write it as,\xa0x2+1200x-800x-960000\xa0x(x+1200)-800(x+1200)\xa0(x-800)(x+1200)\xa0Equating\xa0x-800=0\xa0x=800\xa0x+1200=0\xa0x=-1200\xa0
27837.

For any posutve integer n Euclid division lemma to prove that n cube-n by 6

Answer» Let :-a = n³ - n= n(n² - 1)\xa0= n(n - 1)(n + 1)= (n - 1)n (n + 1)1) Now out of three (n - 1),n and (n + 1) one must be even so a is divisible by 22) Also (n - 1),n and (n + 1) are three consecutive integers thus as proved a must be divisible by 3\xa0From (1) and (2)a must be divisible by 2 × 3 = 6Hence n³ - n is divisible by 6 for any positive integer n
27838.

Sin^-cos6

Answer» Yrr ques pura likho
27839.

The H.C.f of 96 and 404

Answer» HCF\xa0of 96 and 404 by prime factorisation method:-Since, 96 =\xa02\xa0× 2\xa0× 2\xa0× 2\xa0× 2\xa0× 3and, 404 = 2\xa0× 2\xa0× 101So, HCF of 96 and 404 = Product of common prime factors =\xa02\xa0× 2 =\xa04
H.C.F will be 4
27840.

If cosine ©+cosine^2©= 1 then prove that........

Answer» Median me median cls kaise find krte h??Kahi h n/2 krke CF me dekhte h usse jyada.Or kahi median allready diya hua h to usko cls interval me dekhte h??
27841.

X²+9x+20

Answer» Xsq + 5x + 4x + 20 = 0 x( x + 5 ) +4 ( x + 5 ) = 0( x + 5 ) ( x + 4 ) = 0x = -5 & x = -4??
27842.

Theorams

Answer» Sb kr le bhai
Konsi bhai
27843.

If x=3 is one root of the quadratic equation x² – 2kx – 6 = 0, then find the value of k.

Answer» K =5/2
27844.

What is the value of (cos² 67° – sin² 23°)?

Answer» How 0??
0
27845.

Sin. Tan

Answer»
27846.

TanA/1+secA-tanA/1-sec=2cosecA

Answer»
27847.

CosecA + cot A = pThen prove that cos A= p2 - 1/p2 + 1

Answer»
27848.

2 ka 1st...x-y= -2 and 2x-y= 1

Answer» 3rd part..., x+y= 9and 8y-x=0
Are mene aapse apne aap hi solve krne ke lie kha h baaki mujhe aa gye
2nd part ....x-3y=-10 and x-2y=10
No and thanks
27849.

Trick to solve completing the square method

Answer»
27850.

Prove that one and only one out of n , n+2, n+4 is divisible by 3.

Answer» Let the number be (3q + r){tex}n = 3 q + r \\quad 0 \\leq r < 3{/tex}{tex}\\text { or } 3 q , 3 q + 1,3 q + 2{/tex}{tex}\\text { If } n = 3 q \\text { then, numbers are } 3 q , ( 3 q + 1 ) , ( 3 q + 2 ){/tex}{tex}3 q \\text { is divisible by } 3{/tex}.{tex}\\text { If } n = 3 q + 1 \\text { then, numbers are } ( 3 q + 1 ) , ( 3 q + 3 ) , ( 3 q + 4 ){/tex}{tex}( 3 q + 3 ) \\text { is divisible by } 3{/tex}.{tex}\\text { If } n = 3 q + 2 \\text { then, numbers are } ( 3 q + 2 ) , ( 3 q + 4 ) , ( 3 q + 6 ){/tex}{tex}( 3 q + 6 ) \\text { is divisible by } 3{/tex}.{tex}\\therefore \\text { out of } n , ( n + 2 ) \\text { and } ( n + 4 ) \\text { only one is divisible by } 3{/tex}.