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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

27701.

Is exame pad is allowed in examination hall

Answer» Yes....but it should be transparent Suman ......All the best for ur exams ??
27702.

Please send all formula of class 10

Answer»
27703.

Find the mode of following data120 110 130 110 120 140 130 120 140 120

Answer» 120
Mode is 120
120 paji
27704.

Maths mi graph kise banye

Answer» If cbseguide would allow pictures then surely i can answer you bt...still the feature isnt introduced yet. ...call your school buddies for help
Answer me...
27705.

If Cos theta -sin theta =root 2sin theta prove that cos theta + sin theta = root 2 sin theta

Answer» Galat question hProof mei √2Costheta hoga
27706.

If the mth of an AP is 1/n & nth term is 1/m then show that its (mn)th term is 1.

Answer» Rd sharma me solutuon diya h
27707.

The length of tangent drawn from an external point at a circle are equal

Answer» Refer Ncert
27708.

Anyone here who is hoping excessively to prevent leaking of paper like yesteryear...?

Answer»
27709.

Prove that thetangents drawn at the end of a diameter of a circle are parallel

Answer» Given: PQ is a diameter of a circle with centre O. The lines AB and CD are the tangents at P and Q respectively.\xa0To prove: AB\xa0{tex}\\parallel{/tex} CDProof: AB is a tangent to the circle at P and Op is the radius through the point of contact{tex}\\because{/tex}{tex}\\angle{/tex}OPA = 90o .......(1) [The tangent at any point of a circle is perpendicular to the radius through the point of contact]{tex}\\because{/tex}\xa0CD is a tangent to the circle at Q and OQ is the radius through the point of contact.{tex}\\therefore{/tex}\xa0{tex}\\angle{/tex}OQD = 90o ........(2) [The tangent at any point of a circle is perpendicular to the radius through the point of contact]From (1) and (2),{tex}\\angle{/tex}OPA = {tex}\\angle{/tex}OQDBut these form a pair of equal alternate angles.{tex}\\because{/tex}\xa0AB\xa0{tex}\\parallel{/tex} CD
27710.

Root √5 is irrational

Answer» Let us assume that √5 is rational, then it will be =a/ba and be are coprime numbers. √5=a/bSquaring both sides5=a²/b²5b²=a²It means 5 will divide a² and also a. 5 will divide a for some integer c5c=a (squaring both sides) 25c² = a²So 25c²=5b²5c²=b²It contradicts that a and b are coprime, our assumption was wrong and √5 is irrational
How i can solve it
Yes bro
27711.

That root 2 + root 2 is irrational

Answer» Let that be equal to p/q where p q r integers .....Sqaure on b.sAnd after that u know it
27712.

All the best to all students who use this app for their maths exam

Answer» Thanq..n same to u
27713.

Csa of conical top

Answer»
27714.

Full mathematics fomulae

Answer» Plzz search on google
27715.

Find lcm

Answer» Its not full question
27716.

The ratio of area of 2 similar triangles is equal to the square of the corresponding

Answer» Sides
Dont know
27717.

area of major or minor sector

Answer» Area of sector theta /360 ×πr²Major sector area of circle - area of minor sector πr²-theta/360×πr²
27718.

What is the value at unit place value 9°

Answer»
27719.

1+sinA+cosA/1+cosA-sinA=1+sinA/cosA

Answer»
27720.

If circumference of a circle is equal to perimeter Of Square then ratios of Their areas is?

Answer» 14:11
27721.

Two zeroes of polynomial x^4- 13x^3+52x^2-64x+24 are3+ root5 and 3- root 5 find the other zeroes

Answer» 3+-root5 equals to me x rkh kr..squring krdijiye...gx nikl jaega
Rd sharma is a book
Bese ye..question RD ka..hai kya
Is question ko pd kr..lg raha hai..ki mera abhi..bohot bcha hai???
Error 305?
27722.

Converse of bpt

Answer» Line bisecting the other two sides of traingle will be parallel to third side
27723.

what are the possible values of remainder r, when a positive integer "a" is divisible by 3?

Answer» Values of r can be 0,1,2 by euclids division lemma
27724.

Find the mode of the following data 120,130,140,150,160,170

Answer» Haan ungrouped data mai...
Mode highest no. Hota hai ?
170
27725.

Itnnaa sannataa kyu hai bhai!!!???

Answer»
27726.

Slove for x and mx+y=m+n m(1\\m_n-1/m-n)x+n(1/n_m_content://media/external/file/1080221m+n

Answer»
27727.

A12 the term of ap is -3, -1/2, 2

Answer»
27728.

Find the value of k so that 2x find the value of k so that 2 X square + kx + 3 is equal to zero

Answer» 2x2 + kx +3=0We know that quadratic equation has two equal roots only when the value of discriminant is equal to zero.Comparing equation 2x 2 +kx +3=0 with general quadratic equation ax2 + bx + c =0, we get a = 2, b = k and c = 3Discriminant = b2 − 4ac = k2 - 4(2)(3) = k2 -24Putting discriminant equal to zerok2 - 24 = 0 ⇒ k2 = 24{tex}\\Rightarrow k = \\pm \\sqrt { 24 } = \\pm 2 \\sqrt { 6 } \\Rightarrow k = 2 \\sqrt { 6 } , - 2 \\sqrt { 6 }{/tex}
27729.

Formula of Area of minor segment

Answer» Area of sector - area of tirangle
Theta/360×πr^2- r^2 × sin theta/2
27730.

1. 1. 1. 1A+b+x=. A+ b+. X

Answer»
27731.

√2 is irrational no.

Answer»
27732.

Find the distance between A(a,b)&(-a,-b)

Answer»
27733.

(a+b) whole squa re is identify me agar a= 2a diya hai too a whole square me kya lenge

Answer»
27734.

If \uf020\uf044ABC ~ \uf044QRP, ar(\uf044ABC)ar(\uf044QRP) = 94, and BC = 15 cm, then find PR.

Answer»
27735.

If sec theta + tan theta=p then,find the value of cosec theta

Answer» Secθ+tanθ=p ----------------------(1)∵, sec²θ-tan²θ=1or, (secθ+tanθ)(secθ-tanθ)=1or, secθ-tanθ=1/p ----------------(2)Adding (1) and (2) we get,2secθ=p+1/por, secθ=(p²+1)/2p∴, cosθ=1/secθ=2p/(p²+1)∴, sinθ=√(1-cos²θ)=√[1-{2p/(p²+1)}²]=√[1-4p²/(p²+1)²]=√[{(p²+1)²-4p²}/(p²+1)²]=√[(p⁴+2p²+1-4p²)/(p²+1)²]=√(p⁴-2p²+1)/(p²+1)=√(p²-1)²/(p²+1)=(p²-1)/(p²+1)∴, cosecθ=1/sinθ=1/[(p²-1)/(p²+1)]=(p²+1)/(p²-1) I HOPE THIS HELPS!
27736.

The height of a cone is 120cm.A small cone is cut off at the top

Answer» Complete the question please....
27737.

1+tan

Answer»
27738.

1/a+b+x =1/a +1/b + 1/c

Answer» Question clear nhi ho paa rha hai...
27739.

Probability dice

Answer» Easy hai
27740.

Pythagoras thearm

Answer» Pythagoras TheoremStatement: In a right angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.\xa0Given: A right triangle ABC right angled at B.To prove: AC2\xa0= AB2\xa0+ BC2Construction: Draw BDACProof :We know that: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.ADBABCSo,(Sides are proportional)Or, AD.AC = AB2\xa0... (1)Also,BDCABCSo,Or, CD. AC = BC2\xa0... (2)Adding (1) and (2),AD. AC + CD. AC = AB2\xa0+ BC2AC (AD + CD) = AB2\xa0+ BC2AC.AC = AB2\xa0+ BC2AC2\xa0= AB2\xa0+ BC2Hence Proved.
27741.

Proof pythagores thorem

Answer»
27742.

Write the value of (1-sin square theta) sec square theta

Answer» Cos theta
Cos squre theta
27743.

If we do multiply with negative numbers

Answer» Ans will be positive-5 * (-5)= 25
27744.

Sec Q + tan Q = p prove that

Answer»
27745.

A dice thrown twice find the probability that 5 will not come up either time

Answer» U can search in ncert solutions in this app
25/36
27746.

Ex-3.5 -2

Answer» .
27747.

WHAT IS DIGIT AT UNIT PLACE OF NINE SQUARE AND

Answer»
27748.

Under root 5

Answer»
27749.

Find the distance of x axis from point (3,-4)?

Answer» 4 units
27750.

tan a+b=90

Answer»