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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.
| 7851. |
Sino-coso is equal 1/root2 find sino+coso |
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| 7852. |
Fdgg |
| Answer» | |
| 7853. |
solve for x and y ax/b - by/a = a+b and ax-by=2ab |
| Answer» The given equations may be written as{tex}a^2x - b^2y = a^2b + ab^2{/tex} ....... (i){tex}ax - by = 2ab{/tex} ......... (ii)Multiplying (ii) by b and subtracting the result from (i),\xa0{tex}(a^2\xa0- ab)x = a^2b - ab^2{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}(a^2\xa0- ab)x = b(a^2\xa0- ab){/tex}{tex}\\Rightarrow{/tex}{tex}x = b.{/tex}Putting {tex}x = b{/tex} in\xa0(ii), we get{tex}a b - b y = 2 a b {/tex}{tex}\\Rightarrow b y = - a b{/tex}{tex} \\Rightarrow y = \\frac { - a b } { b } = - a{/tex}Hence, x = b and y = -a. | |
| 7854. |
H.C.F of x3-3x+5 and x2-4x+3 |
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| 7855. |
Write all the trigonometry identity |
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| 7856. |
Evaluate-√30√30√30√30........infinity |
| Answer» 1 | |
| 7857. |
16*-10/*=27 |
| Answer» Thevudiya payala | |
| 7858. |
Solve for x and y (a-b)x + (a+b)y=a^2-2ab-b^2and (a+b)(x+y)=a^2+b^2 |
| Answer» Thevudiya payala | |
| 7859. |
All questions of board paper come from ncert or not for class10 |
| Answer» All the questions were come from NCERT as CBSE follow NCERT based syllabus but all the questions are NCERT but theme of questions were not. not all but some questions are value based . | |
| 7860. |
What is ming by maths |
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| 7861. |
Find the area of ring whose outer and innwr radii are respectively 23cm and 12cm |
| Answer» 1210 | |
| 7862. |
Prove that the line joining from radius of circle to tangent is 90degree |
| Answer» GIVEN\xa0: A circle C (0, r) and a tangent AB at a point P.TO PROVE :\xa0{tex}O P \\perp A B{/tex}CONSTRUCTION Take any point Q, other than P, on the tangent AB. Join OQ. Suppose OQ meets the circle at R.PROOF\xa0: We know that among all line segments joining the point O to a point on AB, the shortest one is perpendicular to AB.To prove that {tex}O P \\perp A B{/tex}, it is sufficient to prove that OP is shorter than any other segment joining O to any point of AB.Clearly,\xa0{tex}OP = OR{/tex}Now,\xa0{tex}OQ = OR + RQ{/tex}{tex}\\Rightarrow \\quad O Q > O R{/tex}{tex}\\Rightarrow \\quad O Q > O P \\quad [ \\because O P = O R ]{/tex}{tex}\\Rightarrow \\quad O P < O Q{/tex}{tex}\\therefore{/tex}\xa0{tex}O P \\perp A B{/tex} | |
| 7863. |
Find the sum of 1st n terms of an AP whose nth term is(5-6n).Hence the 15th term of this AP |
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| 7864. |
Prove curved surface area of cone |
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| 7865. |
X+yz=14Y+xz=11Z+xy=10Find the value of x, y and z. |
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| 7866. |
Explain why 13233343563715 is a composite number. |
| Answer» Bcoz 3 is the factor of this number | |
| 7867. |
17. 15. 899. 95. 6436. 45. ? |
| Answer» -729 | |
| 7868. |
Ftygttg |
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| 7869. |
Acute and obtuse angle |
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Answer» Acute angle is 0°-90°Obtuse angle is 90°-180° Acute angle is between 0 to 90 degree.and obtuse angle is between 90 to 180 degree. |
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| 7870. |
why 113/1200 is non terminating decimal? |
| Answer» It is a non terminating decimal because the prime factorisation of its denominator includes a 3.......and its factorisation is ~*2×2×2×2×5×5×3*~....???? | |
| 7871. |
Find the equation of the perpendicular bisector of the line segment joining points(7,1)&(3,5) |
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| 7872. |
factorise 3x+4x=9 |
| Answer» x(3+4)=9x(7)=9x=9/7 | |
| 7873. |
prove 1×1=2 |
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| 7874. |
New question paper provide by CBSE |
| Answer» Check Question Papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 7875. |
How the formula of sum of a.p. has derive |
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| 7876. |
Cbse 10 all book syllabus |
| Answer» This is already given in cbseguide | |
| 7877. |
a cos A -b sin A =c ,then prove a sin A+b cos A =√a^2+b^2-c^2 |
| Answer» We have, {tex}asin\\theta+bcos\\theta=c{/tex}On squaring both sides, we get{tex}(asin\\theta+bcos\\theta)^2=c^2{/tex}(a sin θ)2\xa0+ (b cos θ)2\xa0+ 2(a sin θ) (b cos θ) = c2⇒ a2\xa0sin2\xa0θ + b2\xa0cos2\xa0θ + 2ab sin θ cos θ = c2⇒ a2(1 – cos2\xa0θ) + b2\xa0(1 – sin2\xa0θ) + 2 ab sin θ cos θ = c2 {tex}[\\because sin^2\\theta+cos^2\\theta=1]{/tex}⇒ a2\xa0– a2\xa0cos2\xa0θ + b2\xa0– b2\xa0sin2\xa0θ + 2ab sin θ cos θ = c2⇒ –a2\xa0cos2\xa0θ – b2\xa0sin2\xa0θ + 2ab sin θ cos θ = c2\xa0– a2\xa0– b2 Taking Negative common,⇒ a2\xa0cos2\xa0θ + b2\xa0sin2\xa0θ – 2ab sin θ cos θ = a2\xa0+ b2\xa0– c2⇒ (a cos θ)2\xa0+ (b sin θ)2\xa0– 2(a cos θ) (b sin θ) = a2\xa0+ b2\xa0– c2⇒ {tex}(acos\\theta-bsin\\theta)^2=a^2+b^2-c^2{/tex}⇒{tex}acos\\theta-bsin\\theta{/tex} =\xa0{tex}\\pm \\sqrt { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } }{/tex}\xa0Hence proved, {tex}acos\\theta-bsin\\theta{/tex} =\xa0{tex}\\sqrt{a^2+b^2-c^2}{/tex} | |
| 7878. |
Will rd sharma questions will be there in board exam 2017 -18 |
| Answer» Yes | |
| 7879. |
What is circle? |
| Answer» A circular round figure whose boundary consists if points equidistant from a fixed point. | |
| 7880. |
The length of wire is 66m. How many circles of circumference 13.2cm can be made from this wire |
| Answer» 5 | |
| 7881. |
The athematic mean of the first three terms of an A.P is 5 . Find the middle term of this A.P. |
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| 7882. |
What is the difference between two concitive positive integer? |
| Answer» It is always -1 Eg concitive no are 12345........etc 1-2=-1 | |
| 7883. |
2x+y |
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| 7884. |
3+ 1 |
| Answer» 4 | |
| 7885. |
root 4x-1/4x+1-root 4x+1/4x-1=8/3 |
| Answer» X=+_root82/896 | |
| 7886. |
523698+5 |
| Answer» 523703 | |
| 7887. |
Trigonometry identies |
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| 7888. |
Sin°+cos°=?? |
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| 7889. |
if alfa and beta are the roots ofequation x |
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| 7890. |
In an equilateral triangle ABC, D is a point on side BC such that BD=1/3BC. Prove that 9AD2=7AB2. |
| Answer» We have {tex}BD = \\frac{1}{3}BC{/tex} Draw AP \u200b{tex} \\bot {/tex}\u200b BCIn \u200b{tex}\\triangle {/tex}\u200bAPB,AB2 = AP2 + BP2 = AP2 + (BD + DP)2= AP2 + BD2 + DP2 + 2BD.DP= (AP2 + DP2) + BD2 + 2BD.DP=AD2 + DB2 + 2BD.DP [AP2 + DP2 = AD2]=AD2+{tex}{\\left( {\\frac{1}{3}BC} \\right)^2} + 2\\left( {\\frac{1}{3}BC} \\right)(BP - BD){/tex}={tex}AD^2 + \\frac{1}{9}BC + \\frac{2}{3}BC\\left( {\\frac{1}{2}BC - \\frac{1}{3}BC} \\right){/tex}{tex}\\left[ {BP = \\frac{1}{2}BC,BD = \\frac{1}{3}BC} \\right]{/tex}={tex}A{D^2} + \\frac{1}{9}A{B^2} + \\frac{2}{3}AB\\left( {\\frac{1}{2}AB - \\frac{1}{3}AB} \\right){/tex} [BC = AB, Sides of an equilateral{tex}\\vartriangle {/tex}\u200b ]={tex}A{D^2} + \\frac{1}{9}A{B^2} + \\frac{1}{3}A{B^2} - \\frac{2}{9}A{B^2}{/tex}={tex}A{D^2} + \\frac{1}{9}A{B^2} + \\frac{1}{9}A{B^2}{/tex}\u200b{tex}\\Rightarrow {/tex}\u200b {tex}A{B^2} = A{D^2} + \\frac{2}{9}A{B^2}{/tex}\u200b{tex}\\Rightarrow {/tex}\u200b 9AB2 = 9AD2 + 2AB2\u200b{tex}\\Rightarrow {/tex}\u200b9AB2 = 9AB2 + 2AD2\u200b{tex}\\Rightarrow {/tex}\u200b 7AB2 = 9AD2\u200b{tex}\\Rightarrow {/tex}\u200b 9AD2 = 7AB2 Proved | |
| 7891. |
A³+b³ |
| Answer» (a+b)^2-2ab | |
| 7892. |
From where i shall study Book name |
| Answer» | |
| 7893. |
What is rectangle |
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Answer» A quadrilatral which have 4 side and 2 opposite side are =and// Rectangle is quadrilateral in which opposite sides are equal and parallel and each angle is of 90°. |
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| 7894. |
What is rectanl |
| Answer» | |
| 7895. |
How to become good in maths |
| Answer» By practice | |
| 7896. |
Please share the class 10 syllabus for cbse board. |
| Answer» Check syllabus section in this app . | |
| 7897. |
Explain why 13233343563715 is a composite number? |
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Answer» Find the perpendicular distance of A(10,12) from the y-axis It\'s unit digit us 5, so this no. Is divisible by 5 and hence it is composite |
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| 7898. |
What is the formula of a+b cube |
| Answer» acube+3.a square.b+3.a.bsquare+b cube | |
| 7899. |
Which type of question will come in exam of class 10 of 2017-18 |
| Answer» Check papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 7900. |
If sec theta +tan theta =p then find the value of cosec theta |
| Answer» {tex}sec\\ \\theta+ tan\\ \\theta = p{/tex} ...(i)Also {tex}sec^2\xa0\\theta - tan^2 \\theta = 1{/tex}{tex}\\Rightarrow{/tex}\xa0(sec\xa0{tex}\\theta{/tex}\xa0- tan\xa0{tex}\\theta{/tex}) (sec\xa0{tex}\\theta{/tex}\xa0+ tan\xa0{tex}\\theta{/tex}) = 1{tex}\\Rightarrow{/tex}\xa0p(sec\xa0{tex}\\theta{/tex}\xa0-\xa0tan\xa0{tex}\\theta{/tex}) = 1[using equation (i)]{tex}\\Rightarrow{/tex}\xa0sec\xa0{tex}\\theta{/tex}\xa0-\xa0tan\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{1}{p}{/tex}\xa0...(ii)(ii) - (i) we get{tex}-2 tan{/tex}\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{1-p^{2}}{p}{/tex}{tex}\\Rightarrow{/tex}- tan\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{1-p^{2}}{2 p}{/tex}{tex}\\Rightarrow{/tex}- cot\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{2 p}{1-p^{2}}{/tex}cot\xa0{tex}\\theta{/tex}\xa0{tex}=\\left(\\frac{2 p}{1-p^{2}}\\right)^{2}{/tex}{tex}=\\frac{-4 p^{2}}{\\left(1-p^{2}\\right)^{2}}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}cosec^2{/tex}\xa0{tex}\\theta{/tex}\xa0- 1\xa0{tex}=\\frac{-4 p^{2}}{\\left(1-p^{2}\\right)^{2}}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}cosec^2{/tex}\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{-4 p^{2}}{\\left(1-p^{2}\\right)^{2}}+1=\\frac{-4 p^{2}+\\left(1-p^{2}\\right)^{2}}{\\left(1-p^{2}\\right)^{2}}{/tex}{tex}cosec^2{/tex}\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{-4 p^{2}+1+p^{4}-2 p^{2}}{\\left(1-p^{2}\\right)^{2}}{/tex}{tex}=\\frac{\\left(1+p^{2}\\right)^{2}}{\\left(1-p^{2}\\right)^{2}}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}cosec\\\xa0\\theta{/tex}\xa0{tex}=\\frac{1+p^{2}}{1-p^{2}}{/tex} | |