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InterviewSolution
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.
| 35651. |
Value of tan60 |
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Answer» Root 3=1.732 Root 3 Under root 13 _/3 _/3 Root 3 |
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| 35652. |
Pythagoures formula |
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Answer» H sq.=b sq.+p.sq Hsp=Psq+Bsq AC sq = AB sq + BC sq AC sq.=AB sq.+BC sq. |
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| 35653. |
Largest source of bauxite in india name the state |
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Answer» Odisha Oddisha Orissa is the largest source of bauxite |
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| 35654. |
√3x^2-2√2^x-2√3=0 solvethe equ by quadratic formula |
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| 35655. |
What is a section formulA |
| Answer» The section formula is X=mx2nx1/m+n and Y=my2ny1/m+n | |
| 35656. |
Who is the most important book for maths . RS aggraval ,R D Sharma. Or Ncert? |
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Answer» Examination point of view . ?%is best I agree with ncert RD Sharma because it contains all questions of ncert and RS Aggarwal Ncert Ncert Rs Aggarwal Of course ncert |
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| 35657. |
a3_b3+ab+1 solve this problem |
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Answer» Please solve this problem this is the challenge for class 10 given by teacher Really Sunidhi chauhan is in 10 class. I don\'t know if I know I can solve this is the problem |
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| 35658. |
In fig. 6. 36, QR/QS =QT/PR and.‹1=‹2.Show that triangle PQS is similar to triangle TQR |
| Answer» Given:\xa0In figure, {tex}\\frac{{QR}}{{QS}} = \\frac{{QT}}{{PR}}{/tex}and {tex}\\angle{/tex} 1 = {tex}\\angle{/tex} 2To prove: {tex}\\triangle PQS \\sim \\triangle TQR{/tex}Proof: In {tex}\\triangle {/tex} PQR {tex}\\because {/tex}{tex}\\angle{/tex} 1= {tex}\\angle{/tex} 2{tex}\\therefore {/tex} PR = QP (1).......[ {tex}\\because {/tex} sides opposite to equal angle of a triangle are equal]Now, {tex}\\frac{{QR}}{{QS}} = \\frac{{QT}}{{PR}}{/tex} ......given{tex} \\Rightarrow \\frac{{QR}}{{QS}} = \\frac{{QT}}{{QP}}{/tex} (2).......Using(1)Again in {tex}\\triangle PQS{/tex} and {tex}\\triangle TQR{/tex}{tex}\\because \\frac{{QR}}{{QS}} = \\frac{{QT}}{{QP}}...........From(2){/tex}{tex}\\therefore \\frac{{QS}}{{QR}} = \\frac{{QP}}{{QT}}{/tex} and {tex}\\angle SQP = \\angle RQT{/tex}{tex}\\therefore \\triangle PQS \\sim \\triangle TQR{/tex}...........SAS similarity criterion | |
| 35659. |
What is the sum of all natural numbers from 1 to 100 . |
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Answer» Use formula n(n+1)divided by 2 5050 AND The Answer will be 5050 |
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| 35660. |
find the least no. which is divisible by all the numbers from 1 to 10. |
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Answer» 2520 by taking l.c.m of all no 1 to 10 you will get the answer Vaishnavi 1 |
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| 35661. |
Find the value of p if the difference of square of the zeroes of polynomial x2 +px+45 is 144 |
| Answer» If the squared difference of the zeroes of the quadratic polynomial f(x) = x2 + px + 45 is equal to 144, then ,we have to find the value of p.Let {tex}\\alpha{/tex}\xa0and {tex}\\beta{/tex}\xa0be the zeroes of the given quadratic polynomial.{tex}\\therefore{/tex}\xa0{tex}\\alpha{/tex} + {tex}\\beta{/tex} = - p and {tex}\\alpha\\beta{/tex}= 45 ...(i)Given, ({tex}\\alpha{/tex} - {tex}\\beta{/tex})2 = 144or, ({tex}\\alpha{/tex} + {tex}\\beta{/tex})2- 4{tex}\\alpha{/tex}{tex}\\beta{/tex}\xa0= 144or, (-p)2\xa0- 4 {tex}\\times{/tex}\xa045 = 144 [Using (i)]p2\xa0- 180 = 144p2 = 144 + 180 = 324{tex}\\therefore{/tex}\xa0p = ± {tex}\\sqrt{324}{/tex}= ± 18Hence,\xa0the value of p is ± 18. | |
| 35662. |
9-5 |
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Answer» 4 Xxx Sususijsksk |
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| 35663. |
Show that the polynomial f(x)= x^4+4x^2 +6 has no zeros |
| Answer» f(x) = x4\xa0+ 4x2\xa0+ 6= (x2)2\xa0+ 4x2\xa0+ 6Let x2\xa0=n,Then, f(x) = n2\xa0+ 4n + 6,Here a=1,b=4,c=6The discriminant(D) = {tex}\\text{b}^2-4\\mathrm{ac}=\\;(4)^2-4\\times1\\times6=16-24=-8{/tex}Since the discriminant is negative so this polynomial has no zerosHence, f(x) = x4\xa0+ 4x2\xa0+ 6 has no zero. | |
| 35664. |
How to draw BD perpendicular to AC in right triangle ABC?? |
| Answer» Bisect angle ABC | |
| 35665. |
159%......=15900 |
| Answer» 10000 | |
| 35666. |
If √6+√6+√6+.............Infinite |
| Answer» 1 | |
| 35667. |
In a triangle abc de parallel to bc find the value of x |
| Answer» As DE\xa0{tex}\\parallel{/tex} BC{tex}\\therefore \\frac{AD}{AB}=\\frac{AE}{AC}{/tex},\xa0{tex}{/tex}.{tex}\\frac{x}{2x+1}=\\frac{x+3}{2x+8}{/tex}(x + 3)(2x + 1) = x(2x + 8)2x2 + x + 6x + 3 = 2x2 + 8x3 = 8x - 7xx = 3 | |
| 35668. |
Sin^2 20+sin^2 70- tan^2 45 |
| Answer» Sin^2 20°+sin(90°-20°)-tan^2 45°Sin^2 20°+cos^2 20°-tan^2 45°Now ,. Sin2 A + cos2 A = 1 ( identity)And tan 45 °= 1 tan2 45°= 1^21 - 1 =0. | |
| 35669. |
Find the value of k for which the system of equtionsx-2y=3 and 3x+ky=1has a unique solution |
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Answer» for unique solutions a1/a2 is not equal to b1/b2 so 1 /3 is not equal to -2/k so k is not equal to -6 so k is all values except -6 1/3=-2/k1/3×2/kK=6 |
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| 35670. |
2x+3y=√2 |
| Answer» 2x+3y=√2Squaring both side(2x+3y)²=24x²+9y²+2.2x.3y=24x²+9y²+12xy=24x²+12xy+9y²=24x²+6xy+6xy+9y²=22x(2x+3y)+3y(2x+3y)=22x+3y=22x+3y=2 | |
| 35671. |
14 lesson mai a kya hota h |
| Answer» Middle value of xi if it has two values then we can take any of them | |
| 35672. |
Prove that Sin1° × sin2°× ..............× sin89° × sin90°=√3/2 |
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| 35673. |
The sum of pth terms of AP is q and qth terms of AP is p show that p+q th term of AP is -(p+q). |
| Answer» Let a be the first term and d the common difference of the given A.P.{tex}\\therefore S_{p}=\\frac{p}{2}{/tex}\xa0[2a + (p - 1)d] = q\xa0{tex}\\Rightarrow{/tex}\xa02a + (p - 1)d\xa0{tex}=\\frac{2 q}{p}{/tex} ….(i)And\xa0{tex}S_{q}=\\frac{q}{2}{/tex}\xa0[2a + (q - 1)d] = p{tex}\\Rightarrow{/tex}\xa02a + (q - 1)d\xa0{tex}=\\frac{2 p}{q}{/tex}\xa0….(ii)Subtracting eq. (ii) from eq. (i) we get(p - q)d =\xa0{tex}\\frac{2 q}{p}-\\frac{2 p}{q}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0(p - q)d\xa0{tex}=\\frac{2\\left(q^{2}-p^{2}\\right)}{p q}{/tex}{tex}\\Rightarrow{/tex}\xa0(p - q)d\xa0{tex}=\\frac{-2}{p q}{/tex}(p2\xa0- q2){tex}\\Rightarrow{/tex}\xa0(p - q)d\xa0{tex}=\\frac{-2}{p q}{/tex}\xa0(p + q)(p - q)\xa0{tex}\\Rightarrow d=\\frac{-2}{p q}{/tex}\xa0(p + q)Substituting the value of d in eq. (i) we get2a + (p - 1)\xa0{tex}\\left[\\frac{-2(p+q)}{p q}\\right]=\\frac{2 q}{p}{/tex}{tex}\\Rightarrow 2 a=\\frac{2 q}{p}+\\frac{2(p-1)(p+q)}{p q}{/tex}{tex}\\Rightarrow a=\\frac{q}{p}+\\frac{(p-1)(p+q)}{p q}{/tex}{tex}a=\\frac{q^{2}+p^{2}+p q-p-q}{p q}{/tex}Now\xa0Sp+q\xa0{tex}=\\frac{p+q}{2}{/tex}\xa0[2a + (p + q - 1)d{tex}=\\frac{p+q}{2}\\left[\\frac{2 q^{2}+2 p^{2}+2 p q-2 q-2 q}{p q}+\\frac{(p+q-1)[-2(p+q)}{p q}\\right]{/tex}{tex}=\\frac{p+q}{2}\\left[\\frac{2q^{2} + 2p^{2} + 2pq - 2p - 2q -2p^{2} -2 p q+2 p-2 p q-2 q^{2}+2 q}{p q}\\right]{/tex}{tex}=\\frac{p+q}{2}\\left[\\frac{-2 p q}{p q}\\right]{/tex}\xa0= -(p + q)\xa0hence proved. | |
| 35674. |
x^5+a^5 is divided by x+a |
| Answer» x5 + a5 by x + aWe stop here since the remainder is zero,So,quotient\xa0{tex}= x ^ { 4 } - a x ^ { 3 } + a ^ { 2 } x ^ { 2 } - a ^ { 3 } x + a ^ { 4 }{/tex}remainder = 0Therefore,{tex}\\text { Quotient } \\times \\text { Divisor } + \\text { Remainder }{/tex}{tex}= \\left( x ^ { 4 } - a x ^ { 3 } + a ^ { 2 } x ^ { 2 } - a ^ { 3 } x + a ^ { 4 } \\right) ( x + a ) + 0{/tex}{tex}= x ^ { 5 } + a x ^ { 4 } - a x ^ { 4 } - a ^ { 2 } x ^ { 3 } + a ^ { 2 } x ^ { 3 }{/tex}{tex}+ a ^ { 3 } x ^ { 2 } - a ^ { 3 } x ^ { 2 } - a ^ { 4 } x + a ^ { 4 } x + a ^ { 5 }{/tex}{tex}= x ^ { 5 } + a ^ { 5 }{/tex}= DividendTherefore, the division algorithm is verified. | |
| 35675. |
√2is irrational no. |
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Answer» Ya √2 is an irrational number and u can prove it Yes No Yes |
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| 35676. |
What is the surface area |
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Answer» In general, the surface area is the sum of all the areas of all the shapes that cover the surface of the object. howmuch place covered by figure thats called its area |
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| 35677. |
SolveaX + bY = a - b and bX - aY = a + b |
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Answer» Ritik Chaudhary? Multiply whole equation (1) by \'b\' and multiply whole equation (2) by \'a\' , Then solve by elimination method.HOPE THIS HELP YOU! |
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| 35678. |
Dipesh I am not fail in mathematics .I am the top score In class .now I am so happy |
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Answer» Harshita sry your exam is over Harshita your is over Out of 30 ,28.5 How much marks u got .... |
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| 35679. |
Converse of bpt theorm |
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| 35680. |
The common difference of an A.P is -2.find it\'s sum,if first term is 100and last term is-10. |
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Answer» 2520 Varun pornia your answer is wrong The formula of sn is ~1/2{2a+(n-1)d} And another formula is sn~{1/2(a+l)} So\' I use second formula Given; a=100 l=10 Put value in formula; we get 1/2 (100+10) = 1/2×110 =55 is anwer 152 |
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| 35681. |
Can 6n end with zero |
| Answer» 6n =(2.3)n=2n.3n. (.means multiply) Therefore prime factorization of 6n cannot end with 0 | |
| 35682. |
Find the medianof distribution |
| Answer» Incomplete question | |
| 35683. |
In a |
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| 35684. |
Prove that 1-cos=sin |
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| 35685. |
For what value of p are 2p+1, 13,5p-3of AP find AP |
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Answer» Let a1=2p+1, a2=13, and a3=5p-3 Solve it a2-a1=a3-a2 Find, 13-(2p+1)=5p-3-(13)13-2p-1=5p-3-1312-2p=5p-1612+16=5p+2p28=7pP=28/7P=4 Math in hindi |
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| 35686. |
12×12 |
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Answer» 144 144 Are you serious , you are in grade 10 144 144 144 144 |
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| 35687. |
√2 is irrational |
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Answer» Yes Yes Please complete ur ques. |
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| 35688. |
In ΔABC, AD is perpendicular to BC such that AD^2 =BD. CD. prove that ΔABC is right angles at A |
| Answer» AD2\xa0= BD\xa0{tex}\\times{/tex}\xa0CDor,\xa0{tex}\\frac { A D } { C D } = \\frac { B D } { A D }{/tex}Therefore,\xa0{tex}\\triangle A D C \\sim \\triangle B D A{/tex}\xa0(by SAS)or,\xa0{tex}\\angle{/tex}BAD =\xa0{tex}\\angle{/tex}ACD;{tex}\\angle{/tex}DAC =\xa0{tex}\\angle{/tex}DBA\xa0(Corresponding angles of similar triangles){tex}\\angle{/tex}BAD\xa0+ {tex}\\angle{/tex}ACD + {tex}\\angle{/tex}DAC + {tex}\\angle{/tex}DBA = 180o [sum of angles of ∆]or, 2{tex}\\angle{/tex}BAD + 2{tex}\\angle{/tex}DAC = 180oor,\xa0{tex}\\angle{/tex}BAD + {tex}\\angle{/tex}DAC = 90oTherefore,\xa0{tex}\\angle{/tex}A = 90o | |
| 35689. |
Find the ratio in which p p(4,m)divides the line segment joining the point m(2,3)(6,-3) |
| Answer» Let us assume that point P divides the line segment AB in the ratio (k : 1)Using section formula,P(4, m) =\xa0{tex}\\frac{6k+2}{k+1} , {/tex}{tex}\\frac{-3k+3}{k+1}{/tex}{tex}\\therefore{/tex}\xa0{tex}\\frac{6 \\mathrm{k}+2}{\\mathrm{k}+1}{/tex}\xa0= 4\xa0{tex}\\Rightarrow 6k+2=4k+4 \\Rightarrow 2k=2{/tex}{tex}\\Rightarrow{/tex}\xa0k = 1{tex}\\therefore{/tex}\xa0The ratio is 1 : 1Now, m =\xa0{tex}\\frac{-3k+3}{k+1}{/tex}\xa0=\xa0{tex}\\frac{-3+3}{2}{/tex} = 0 | |
| 35690. |
Form a cubic polynomial with zeroes 3,2,_1 |
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| 35691. |
Prove cosec2fjfjjddjxkckicjcudnsksoz |
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Answer» What is this umang ???? Tere bap ne ye type kiya tha kya |
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| 35692. |
If tanx=sin 45° cos45°+ sin30°,then find x |
| Answer» x=45° | |
| 35693. |
What is root of 13 |
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Answer» 169 169 |
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| 35694. |
Find the area of figure formed by joining the points A(-4,-2) B(-3,-5) C(3,-2) and D(2,4) |
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| 35695. |
If the ratio of areas of 2 triangles is 81:100 then what is the ratio of their corresponding sides |
| Answer» 9:10 | |
| 35696. |
What is the Euclid\'s division lemma |
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Answer» Boring chapter This formula valid when r is equal to or greater than 0 and b must be greater than r . a= b q + r Ooo Divident = Divisor × Quotient + Remainder |
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| 35697. |
Prove that SinA - cosA + 1 / sinA + cosA -1 = 1 / secA -tanA using identity sec2A = 1 + tan2A |
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| 35698. |
Section of clockwise Revolution does the hour hand of a clock when it goes from 3 to 9 and 427 |
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| 35699. |
X2*658837x |
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| 35700. |
If Sn dwnotes the sum of an AP whose common difference is d and first term is a find Sn-2,Sn-1,Sn-2 |
| Answer» Refer the example in RS AGARWAL BOOK | |