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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.
| 2451. |
Palagom |
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| 2452. |
If radius of circle is 6cm and length of arc is12 cm.find the area of sector. |
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Answer» You need to mention which type of triangle it is Very easy yrr ...36cm2 {area of sector} |
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| 2453. |
Tan A+secA-1. Cos A---------------------- = -----------TanA - secA+1. 1-sinA |
| Answer» Convert all values in sin and cos At last see for any sort of multiplication | |
| 2454. |
Find the hcf of 81 and 237 and express it as a linear combination of 81 237 |
| Answer» Since, 237 > 81On applying Euclid\'s division algorithm, we get237 = 81 {tex}\\times{/tex}\xa02 + 75 ..........(i)81 = 75 {tex}\\times{/tex}\xa01 + 6 ..........(ii)75 = 6 {tex}\\times{/tex}\xa012 + 3 .........(iii)6 = 3 {tex}\\times{/tex}\xa02 + 0 .............(iv)Hence, HCF (81,237) = 3.In order to write 3 in the form of 81x + 237y,Now,{tex}\\style{font-family:Arial}{\\begin{array}{l}3=75-6\\times12(\\;from(iii))\\\\=75-(81-75\\times1)\\times12\\\\=75-(81\\times12-75\\times12)\\\\=75-81\\times12+75\\times12\\\\=75(1+12)-81\\times12\\\\=75\\times13-81\\times12\\\\=13\\times(237-81\\times2)-81\\times12\\;(\\;Substuting\\;75\\;from(i))\\\\=13\\times237-13\\times81\\times2-81\\times12\\\\=237\\times13-81\\times(26+12)\\\\=81(-38)+237\\times13\\\\=81x+237y\\end{array}}{/tex}Hence, x = -38 and y = 13\xa0 | |
| 2455. |
2×2 what |
| Answer» 4 | |
| 2456. |
4x+3y-6z/3x+7 |
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| 2457. |
Integration of greatest integral function |
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| 2458. |
Hey guys ....can anyone explain me example 3 chapter 10 Ncert? |
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| 2459. |
Hello ishi .. Akhir tm aa hi gyi .. |
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| 2460. |
Time table of cbse board of class10 |
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| 2461. |
Prove of BPT Theorem |
| Answer» Given :\xa0In {tex}\\triangle A B C{/tex}, DE || BC and intersects AB in D and AC in E.\xa0Prove that :\xa0{tex}\\frac{AD}{DB} = \\frac{AE}{EC}{/tex}Construction:\xa0Join BE, CD and draw EF {tex}\\perp{/tex} BA and DG {tex}\\perp{/tex} CA. Now from the given figure we have,EF {tex}\\perp{/tex} BA (Construction)EF is the height of ∆ADE and ∆DBE (Definition of perpendicular)Area({tex}\\triangle{/tex}ADE) ={tex}\\frac{AD.EF}{2}{/tex} .....(1)Area({tex}\\triangle{/tex}DBE) = {tex}\\frac{DB.EF}{2}{/tex} ....(2)Divide the two equations we have{tex}\\frac{Area \\triangle ADE}{Area \\triangle DBE} = \\frac{AD}{DB}{/tex} .....(3){tex}\\frac{Area \\triangle ADE}{Area \\triangle DEC} = \\frac{AE}{EC}{/tex} .....(4)Therefore, {tex}\\triangle \\mathrm{DBE} \\sim \\triangle \\mathrm{DEC}{/tex} (Both the ∆s are on the same base and\xa0between the same || lines).....(5)Area({tex}\\triangle{/tex}DBE) = Area({tex}\\triangle{/tex}DEC) (If the two triangles are similar their\xa0areas are equal){tex}\\frac{AD}{DB} = \\frac{AE}{EC}{/tex}\xa0[from equation 3,4 and 5]Hence proved. | |
| 2462. |
(a+b)²x²-4abx-(a-b)²=0 Find by factarisation |
| Answer» We have,{tex}(a + b)^2x^2 - 4abx - (a - b)^2 = 0{/tex}In order to factorize {tex}(a + b)^2x^2 - 4abx - (a - b)^2{/tex}, we have to find two numbers \'l\' and \'m\' such that.l + m = -4ab and lm = -(a + b)2(a - b)2Clearly, (a - b)2 + [-(a + b)2] = -4ab and\xa0lm = -(a + b)2(a - b)2l = (a - b)2 and m = -(a + b)2Now,{tex}(a + b)^2x^2 - 4abx - (a - b)^2 = 0{/tex}{tex}\\Rightarrow{/tex}{tex}(a + b)^2x^2 - (a + b)^2x + (a - b)^2x - (a - b)^2 = 0{/tex}{tex}\\Rightarrow{/tex} (a + b)2x [x - 1] + (a - b)2[x - 1] = 0{tex}\\Rightarrow{/tex} (x - 1)[(a + b)2x + (a - b)2] = 0{tex}\\Rightarrow{/tex} x - 1 = 0 or (a + b)2x + (a - b)2 = 0{tex}\\Rightarrow{/tex} x = 1 or {tex}x = - \\frac{{{{(a - b)}^2}}}{{{{(a + b)}^2}}}{/tex} | |
| 2463. |
Some plz give me some tips tommorow is my sst preboard |
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| 2464. |
find zeroes of polynomial p2-19p/6-7/6 ? |
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| 2465. |
Hard easy |
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| 2466. |
If cos A=2/5,find the value of 4+4 tan×tan A |
| Answer» Ans is 25 | |
| 2467. |
If alpha and beta are zeros of polynomial ax^2 + bx + c = 0 then evaluate alpha^2 -beta^2 |
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Answer» U\'ll get 16b^4(1-ca)/a^4 (Alpha sq -beta sq)^2 |
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| 2468. |
In an AP, the sum of n terms is 1/2 (5n^2+3n).Find its 40th term |
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| 2469. |
Vghv |
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| 2470. |
What is tan 60 |
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Answer» tan 60 =√3 root 3 Root 3 |
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| 2471. |
Fulform of mathmatics |
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Answer» The is no definite full form of math.....but people make it\'s full form to visualize about it...... It has no real full form |
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| 2472. |
How to study maths |
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Answer» Don\'t study maths, practice maths Study maths like maths |
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| 2473. |
263575 |
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| 2474. |
7 easy chapter?? |
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| 2475. |
In the adjoining figure ABCD is a trapezium |
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| 2476. |
Probability of a sure event is? |
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Answer» 1 Is it 1 ? 1 |
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| 2477. |
Is ur app really help me for getting good results in board exam |
| Answer» Ya y not | |
| 2478. |
a/x-b + b/x-a = 2 |
| Answer» x=a+b or x=(a+b)/2 | |
| 2479. |
some important tips for fiit jee ftre examination |
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| 2480. |
In a right triangle prove that 1+tan^2theta =sec^2theta |
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| 2481. |
If secA+tanA=m & secA-tanA=n.find the value of root nm |
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| 2482. |
Pq parallel to bc and ap:pb =1:2. Find ar (apq)/ar (abc). |
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| 2483. |
COT a + Tan B/cot B +tan a =CotA +TanB |
| Answer» We have,L.H.S{tex} = \\frac{{\\cot A + \\tan B}}{{\\cot B + \\tan A}}{/tex}{tex} = \\frac{{\\frac{{\\cos A}}{{\\sin A}}}}{{\\frac{{\\cos B}}{{\\sin B}}}} + \\frac{{\\frac{{\\sin B}}{{\\cos B}}}}{{\\frac{{\\sin A}}{{\\cos A}}}}{/tex}{tex} = \\frac{{\\frac{{(\\cos A\\cos B + \\sin A\\sin B}}{{\\sin A\\cos B}}}}{{\\frac{{(\\cos A\\cos B + \\sin A\\sin B)}}{{\\sin B\\cos A}}}}{/tex}{tex} = \\frac{{(\\cos A\\cos B + \\sin A\\sin B)}}{{\\sin A\\cos B}} \\times \\frac{{\\sin B\\cos A}}{{(\\cos A\\cos B + \\sin A\\sin B)}}{/tex}{tex} = \\frac{{\\cos A}}{{\\sin A}} \\cdot \\frac{{\\sin B}}{{\\cos B}}{/tex}{tex} = \\cot A \\cdot \\tan B = R.H.S{/tex}therefore,\xa0{tex}\\frac{{\\cot A + \\tan B}}{{\\cot B + \\tan A}} = \\cot A \\cdot \\tan B{/tex}Hence proved. | |
| 2484. |
The distance between the point A(0,6) B(0,2)is |
| Answer» 4 | |
| 2485. |
equation 3x*-k |
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| 2486. |
of real no chapter |
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| 2487. |
1×1000 |
| Answer» 1000 | |
| 2488. |
? ABC is circumscribing a circle. Find the length of BC |
| Answer» Given,{tex}AR = 4 \\ cm .{/tex}Also,\xa0{tex}AR = AQ{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}AQ = 4 cm\xa0{/tex}Now,\xa0{tex}QC = AC - AQ{/tex}{tex}= 11 cm - 4 cm = 7 cm{/tex}\xa0...(i)Also,\xa0{tex}BP = BR{/tex}{tex}\\therefore{/tex}\xa0BP = 3 cm and PC = QC{tex}\\therefore{/tex}\xa0PC = 7 cm [From (i)]\xa0{tex}BC = BP + PC \\\\ = 3 cm + 7 cm \\\\ = 10 cm{/tex} | |
| 2489. |
In an ap 6th term is half the 4th term |
| Answer» a6=a+5d or x÷2=a+5da4=a+3d or x=a+3dThen solve it by elemination | |
| 2490. |
Solve for x4^x - 3 ×2 ^x - 128=0 |
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| 2491. |
Formula of distance |
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Answer» Distance = Speed ---------- Time speed/ time |
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| 2492. |
What is the polynomial |
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| 2493. |
QT and RS are medians of a trianglePQR right angled at P. Prove that 4(QT2+RS2)=5OR2 |
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| 2494. |
how many terms of the AP 9, 17,25,.......must be taken to give a sum of 636 |
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| 2495. |
If tanA=a/x,find the value of x/√a^2+x^2. |
| Answer» CosA | |
| 2496. |
If √3sin◆(theeta)=cos◆Find the value of3cos square◆+2÷3cos◆+2 |
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| 2497. |
Prove that Vn-1 + Vn-1 is a irrational no. |
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| 2498. |
Divide |
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Answer» Ok Sorry trigonometry identities |
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| 2499. |
what is the value of sin tita |
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Answer» You r wrong riya usne sin thita ka value pucha hai The value of sin tita always lies between 0 and one |
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| 2500. |
If the points (x,y) is equidistant from the points (a-b, a+b) and (-a-b, a+b), prove that x-a=0 |
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