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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.
| 23551. |
Class 10 maths all formua |
| Answer» It\'s very hard task .. sry .. ?? see any mathematics book .. | |
| 23552. |
Class 10 maths all chapter formule |
| Answer» Get formulae in notes :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 23553. |
Show that n²-3 is divisible by 8 If n is odd +ve integer |
| Answer» Let n = 4q + 1 (an odd integer){tex}\\therefore \\quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \\quad \\text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}{tex}= 16{q^2} + 8q{/tex}{tex}= 8 \\left( 2 q ^ { 2 } + q \\right){/tex}= 8m, which is divisible by 8. | |
| 23554. |
Class10.maths all chapter formule |
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Answer» Himanshu is right Very hard task ..??sry.. see any maths book ... |
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| 23555. |
2x+3=4 |
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Answer» 2x=4-3X=1/2 X=1/2 |
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| 23556. |
Prove that /3 is irrational number |
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Answer» Let us assume that |
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| 23557. |
500+646556+494767-94656565656 |
| Answer» -94655423833 | |
| 23558. |
An equilateral triangle of side 6 has been inscribed n a circle find area of shaded region |
| Answer» | |
| 23559. |
Tell which term forms an AP-1 -3 -5 or 4 10 16 22 |
| Answer» Only there is one possible answer | |
| 23560. |
If 2x=sec A and 2/x find the value of 2(xsq -1/xsq |
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| 23561. |
One point is called tangent Two point is called secantThree point is called ? |
| Answer» A line cannot intersect a circle at three points. | |
| 23562. |
Completing the square formula2x(square)-7x+3=0 |
| Answer» We have 2x2 - 7x + 3 = 0{tex}\\implies2( x^2 - {7 \\over 2}x + {3\\over 2}) = 0{/tex}{tex}\\implies\u200b\u200b x^2 - {7 \\over 2}x + {49 \\over 16} = {-3 \\over 2} +{ 49 \\over 16}{/tex} (Adding 49/16 to both sides){tex}\\implies x^2 -2 \\times x \\times {7 \\over 4} + ({7 \\over 4})^2 = {-24 +49 \\over 16}{/tex}{tex}\\implies (x-{7\\over4})^2 = {25 \\over 16}{/tex}{tex}\\implies x-{7\\over 4}= \\pm \\sqrt({25 \\over 16}){/tex}{tex}\\implies x={7\\over 4} \\pm {5 \\over 4}{/tex}{tex}\\implies x={7\\over 4} + {5 \\over 4}\\, and \\,x={7\\over 4} - {5 \\over 4}{/tex}{tex}\\implies x=3\\, and \\,{1\\over 2}{/tex}{tex}\\therefore{/tex}the roots of the given equation are {tex}3{/tex} and {tex}1\\over 2{/tex}. | |
| 23563. |
(tanA÷!-cotA ) +(cotA÷1-tanA) =1+secA×cosecA |
| Answer» We have,{tex}\\mathrm { LHS } = \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan A } { 1 - \\frac { 1 } { \\tan A } } + \\frac { \\frac { 1 } { \\tan A } } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan A } { \\frac { \\tan A -1 } { \\tan A } } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } - \\frac { 1 } { \\tan A ( \\tan A - 1 ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 3 } A - 1 } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[Taking LCM]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { ( \\tan A - 1 ) \\left( \\tan ^ { 2 } A + \\tan A + 1 \\right) } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[{tex}\\because{/tex}\xa0a3\xa0- b3\xa0= ( a - b )(a2\xa0+ ab + b2)]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A + \\tan A + 1 } { \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A } + \\frac { \\tan A } { \\tan A } + \\frac { 1 } { \\tan A }{/tex}{tex}\\Rightarrow{/tex}\xa0LHS = tanA + 1 + cotA [ since (1/tanA) =cotA ].= (1 + tanA + cotA){tex}\\therefore \\quad \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA ...........(1)Now, 1 + tanA + cotA = 1 +\xa0{tex}\\frac { \\sin A } { \\cos A } + \\frac { \\cos A } { \\sin A }{/tex}\xa0= 1 +\xa0{tex}\\frac { \\sin ^ { 2 } A + \\cos ^ { 2 } A } { \\sin A \\cos A }{/tex} = 1 +\xa0{tex}\\frac { 1 } { \\sin A \\cos A }{/tex}\xa0[{tex}\\because{/tex}Sin2A + Cos2 A = 1 ]\xa0= 1 + cosecAsecASo, 1 + tanA + cotA = 1+ cosecAsecA.......(2)From (1) and (2), we obtain{tex}\\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA = 1 + cosecAsecA | |
| 23564. |
Using euclids division Leema.find the H.C.F. of 96,144 |
| Answer» 144 = 96 × 1 + 4896 = 48 × 2 + 0 • • • HCF OF 96 AND 144 IS 48 | |
| 23565. |
On which date class 10 board exam will start |
| Answer» | |
| 23566. |
If tan theta=cot(60+ theta) then find the value of theta |
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Answer» Tan theta =cot (60+theta) ? Cot (90-theta) =cot (60+theta) ? 90-theta = 60 + theta ?90-60=2 theta ? 30 = 2 theta ? theta = 15 What is correct answer 650 15 30 |
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| 23567. |
3 3 |
| Answer» | |
| 23568. |
Formula of ring |
| Answer» V\xa0= (π²/4)•(a+b)•(b-a)²where:\tV is the\xa0volume of the torus\ta is the inner diameter\tb is the outer diameter | |
| 23569. |
If the perimeter of a sector of a circle of radius 5.6 cm is 17.2 cm, find the area of the sector |
| Answer» Let OAB be the given sectorLet arc AB =\xa0{tex}l{/tex}Perimeter of sector OAB = 27.2m{tex} \\Rightarrow {/tex}\xa0OA + OB + ARC AB = 27.2{tex} \\Rightarrow {/tex}\xa0{tex}l{/tex} = 27.2 - (5.7 + 5.7) = 15.8MNow, length of arc\xa0{tex} = \\frac{\\theta }{{360}} \\times 2\\pi r{/tex}{tex}\\frac{{15.8}}{{2\\pi r}} = \\frac{\\theta }{{360}}{/tex}Area of sector AOB\xa0{tex} = \\frac{\\theta }{{360}} \\times \\pi {r^2}{/tex}{tex} = \\frac{{5.8}}{{2\\pi r}} \\times \\pi {r^2}{/tex}= 7.9r{tex} = 7.9 \\times 5.7{/tex}= 45.03m2 | |
| 23570. |
Find value of K for which the equation kx^2+2x+1=0 has real and distinct roots. |
| Answer» We have, {tex}kx^2+2x+1=0{/tex}here, a=k, b=2, c=1{tex}\\therefore D=b^2-4c=(2)^2-4(k)(1)=4-4k{/tex}The given equation will have real and distinct roots,{tex}D>0\\implies4-4k>0\\implies k<1{/tex} | |
| 23571. |
Fundamental theorm |
| Answer» | |
| 23572. |
2xsquare +kx+3=0 find the value of K |
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Answer» Incomplete question Question is incomplete |
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| 23573. |
Prove that root2 + root5 is irrational number |
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| 23574. |
Prove that CosA/(1-sinA)+cos/(1+sinA)=2secA |
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| 23575. |
Find the number of two digit numbers divisible by 3 |
| Answer» The two -digit numbers divisible by 3 start from 12,15,18,21,...,99Here,\xa0{tex}a=12{/tex}{tex}d=3{/tex}{tex}a_n=a+(n-1)d{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}99=12+(n-1)(3){/tex}{tex}\\Rightarrow{/tex}{tex}\xa099=12+3n-3{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}90=3n{/tex}{tex}\\Rightarrow{/tex}\xa0n=30Thus, 30 two-digit numbers are divisible by 3. | |
| 23576. |
Quardric equation |
| Answer» Two x square+kx+3=0 give me solution | |
| 23577. |
Root 2 + root 5 is an irrational number |
| Answer» | |
| 23578. |
Write the roots of x^2 |
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| 23579. |
If sina- cosa=0 then find the value if sin^4+cos^4 |
| Answer» {tex}\\frac{1}{2}{/tex} | |
| 23580. |
18+18=? |
| Answer» 36 | |
| 23581. |
Support material 2017-2018 |
| Answer» Get revision notes from here :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 23582. |
Find the mean missing two frequency |
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| 23583. |
Volume of frustum formula derive |
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| 23584. |
A+B+B+B+80+90=______ |
| Answer» Wrong its 100 | |
| 23585. |
What is common diffrence |
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Answer» Constant value The difference b/ w two succesive numbers .. whenever no. Is an Ap .. Common difference is the difference between two term of AP a2-a1 example 2,4,6...........D=a2-a1=4-2=2 Same difference |
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| 23586. |
Solve for x: 4x^2 + 4bx - (a^2 - b^2) =0 |
| Answer» Ok done | |
| 23587. |
TanA=1/2 &cotB=3 what is the value of A+ B |
| Answer» TanA+ cot B = 7/2.. | |
| 23588. |
A wheel of a cart makes 4 revolutions per second . If the diameter of wheel is 84cm , find its speed |
| Answer» SPEED = 1056m/s.. | |
| 23589. |
9x+12y=0 |
| Answer» 3x =-4y.. | |
| 23590. |
4 drigat samikaran swadyay 4 |
| Answer» | |
| 23591. |
Basic proportionality thrm |
| Answer» Basic Proportionality Theorem : If a line is drawn parallel to one side of a triangle intersecting other two sides, then it divides the two sides in the same ratio. | |
| 23592. |
Math paper ka print kaise nikale |
| Answer» | |
| 23593. |
Find the area of a sector of a circle with radius 6 cm if angle of sector is 60°. |
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Answer» 132/7 cm sq 132/7 |
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| 23594. |
Hello...koun koun jage ho Abhi bhi???.. |
| Answer» | |
| 23595. |
Find √2and √3 |
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| 23596. |
Trigo last ex 9 |
| Answer» Get NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 23597. |
If sum of n numbers in an AP is 3n*n÷2+5n÷2 then find an25 |
| Answer» | |
| 23598. |
How do we take out the value of sin ,cos ,tan ,cosec,sec,cot more than90° |
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Answer» What is ASTP By using ASTP |
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| 23599. |
Split 207 into three parts such that these are in AP and the product of two smaller parts is 4623. |
| Answer» Let the four parts be (a - d), a and (a + d).{tex}\\therefore{/tex}\xa0a - d + a + a + d = 207{tex}\\Rightarrow{/tex}\xa03a = 207{tex}\\Rightarrow{/tex}\xa0a = 69According to given information,{tex}\\Rightarrow ( a - d ) \\times a = 4623{/tex}{tex}\\Rightarrow ( 69 - d ) \\times 69 = 4623{/tex}{tex}\\Rightarrow{/tex}\xa069 - d = 67{tex}\\Rightarrow{/tex} d = 2Thus, the three parts are a - d, a, a+ d i.e., 67, 69, 71. | |
| 23600. |
Mera un bhai bandu se sadar anurod ha ki ?????batte km kre answer bhi de. |
| Answer» | |