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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.
| 12801. |
Define similar triangles |
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Answer» Jo dikhne me 1 jese hote h ♦ Hihighiizhz |
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| 12802. |
Chapter no.4 quadratic equations Exercise 4.3 Question:-9 I can\'t understand |
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Answer» Yes :-D May I help u |
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| 12803. |
Derivation of any formula can come in exam ? Mainly of trigonometric ratio and frustum formulas ? |
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Answer» I dont think so it will come Derivation of theorem can come |
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| 12804. |
Chapter 14 example 9 |
| Answer» 17. 5 | |
| 12805. |
The roots of the equation is xsq +x-p(p+1)=0 where p is constant ,are |
| Answer» {tex}x^2 + x - p(p+1) = 0{/tex}{tex}x^2 + (p+1)x - px - p(p+1) = 0{/tex}{tex}x(x+ p + 1) -p(x + p + 1) = 0{/tex}(x + p + 1)(x - p) = 0\xa0x = - p - 1, p | |
| 12806. |
If x=3 is one root of the quadratic of x square-2kr - 6= 0 the find the value of k |
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Answer» 3^2-2kr-6=09-6-2kr=03-2kr=0-2kr=-3K=3/2r K= 1/2 0 |
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| 12807. |
Name the mathematician who finding the sum of first 100 natural number? |
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Answer» Carl friedrich gauss Ye nhi aayega |
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| 12808. |
Passing marks ? |
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Answer» 13 because 20 marks toh school s mil jyege n 13 marks board s Milne p pass 33 33% Aage Apna Dekh Lena 14 |
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| 12809. |
In trigonometry questions should we the draw diagrams complesary?? |
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Answer» But only in height n distance...not in proving Yes Yes Any one is there?? |
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| 12810. |
How many terms of AP 18,16,14- - - taken so that there sum is zero |
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Answer» 19 Put sn=0 |
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| 12811. |
prove that underroot 3+2underroot3 is a irrational? |
| Answer» Do squaring | |
| 12812. |
R u ready for maths exam??? |
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Answer» I am always ready to give maths exam ?????? Nope Some left Nope....... I am also ready Yes?? |
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| 12813. |
Important quation |
| Answer» In CBSE | |
| 12814. |
Sector of circle |
| Answer» | |
| 12815. |
PROOF OF BPT THEOREM |
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Answer» Ncert textbook Theorm 6.1 Ncert textbook Theorm6.1 Check in your ncert book page no. 124 |
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| 12816. |
7x_15y=2 , x+2y=3 solve bythe subsitution method |
| Answer» y=19/29 & x=49/29 | |
| 12817. |
7x_15y=2 , x+2y=3 |
| Answer» x=49/29 & y=19/29 | |
| 12818. |
If tanø+sino=m and tanø-sinø=n. Show that m²-n²=4√mn. |
| Answer» | |
| 12819. |
What is the value of /3 |
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Answer» 1.732 1.73 1.73 1.732 |
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| 12820. |
Cos 3 theta +sin 3 theta/cos theta +sin theta |
| Answer» answer is costheta.sintheta | |
| 12821. |
From a point 100 m above the lake |
| Answer» Let AB be the surface of the lake and P be the point of observation such that {tex}AP = 100 m{/tex}. Let C be the position of the helicopter and C\' be its reflection in the lake.Then,\xa0{tex}CB = C\'B.{/tex}Let PM be perpendicular from P on CB.Then, {tex}\\angle{/tex}{tex}CPM = 30\xa0^\\circ{/tex}and {tex}\\angle{/tex}C\'PM =\xa0{tex}60^\\circ{/tex}Then,\xa0{tex}CM = h, \\ CB = h + 100.{/tex}In right {tex}\\triangle{/tex}CMP{tex}\\tan 30^{\\circ}=\\frac{\\mathrm{C} \\mathrm{M}}{\\mathrm{PM}} \\Rightarrow{\\frac{1}{ \\sqrt{3}}}=\\frac{h}{\\mathrm{PM}}{/tex}{tex}\\Rightarrow PM =\\sqrt3 h{/tex}...(i)In right {tex}\\triangle{/tex}PMC\'{tex}\\tan 60^{\\circ}=\\frac{\\mathrm{C}^{\\prime} \\mathrm{M}}{\\mathrm{PM}} \\Rightarrow \\sqrt{3}=\\frac{\\mathrm{C}^{\\prime} \\mathrm{B}+\\mathrm{BM}}{\\mathrm{PM}}{/tex}{tex}\\Rightarrow \\sqrt{3}=\\frac{h+100+100}{P M}{/tex}{tex}\\Rightarrow P M=\\frac{h+200}{\\sqrt{3}}{/tex}\xa0...(ii)From (i) and (ii), we get\xa0{tex} \\sqrt{3} h=\\frac{h+200}{\\sqrt{3}}{/tex}{tex} 3h = h + 200\xa0\\\\ 2h = 200\\\\ h = 100{/tex}Now,{tex}CB = CM + MB\\\\ = h + 100\\\\ = 100 + 100\\\\ = 200\\\\{/tex}\xa0Hence, the height of the helicopter from the surface of the lake = 200 m | |
| 12822. |
How much percentage of Ncert Material will be there in the final Examination Of maths |
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Answer» 60% 70% only 0% 90 % 85% |
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| 12823. |
What chapters are there |
| Answer» | |
| 12824. |
How to find median class |
| Answer» M=[n/2-cf]÷f*h+l | |
| 12825. |
How to draw perpendicular to the hypotenuse from the vertiex |
| Answer» | |
| 12826. |
(cosecA -sinA)(secA-cosA)=1/tanA+cotA |
| Answer» LHS: (1/sinA-sinA) (1/cosA-cosA) (1-sin sq A/sinA)(1-cos sq A/cosA) ( cos sq A/sinA)( sin sq A/cosA) = cosAsinARHS: 1/sinA/cosA+cosA/sinA Sin sq A + cos sq A/sinAcosA 1/1/sinAcosA = sinAcosA LHS =RHS HENCE PROVED | |
| 12827. |
If (x/a sin A - y/b cosA)=1 and ( x/a cosA + y/b sinA)=1, then find the value of (x^2/a^2+y^2/b^2). |
| Answer» Given: {tex}\\frac { x } { a } \\cos \\theta + \\frac { y } { b } \\sin \\theta = 1 \\text { and } \\frac { x } { a } \\sin \\theta - \\frac { y } { b } \\cos \\theta = 1 , {/tex}To prove: {tex}\\frac { x ^ { 2 } } { a ^ { 2 } } + \\frac { y ^ { 2 } } { b ^ { 2 } } = 2{/tex}Now, {tex}{\\left( {\\frac{x}{a}\\cos \\theta + \\frac{y}{b}\\sin \\theta } \\right)^2} + {\\left( {\\frac{x}{a}\\sin \\theta - \\frac{y}{b}\\cos \\theta } \\right)^2} = {(1)^2} + {(1)^2}{/tex}{tex} = \\frac{{{x^2}}}{{{a^2}}}{\\cos ^2}\\theta + \\frac{{{y^2}}}{{{b^2}}}{\\sin ^2}\\theta + 2\\frac{x}{a}\\cos \\theta \\frac{y}{b}\\sin \\theta + \\frac{{{x^2}}}{{{a^2}}}{\\sin ^2}\\theta + \\frac{{{y^2}}}{{{b^2}}}{\\cos ^2}\\theta - 2\\frac{x}{a}\\sin \\theta \\frac{y}{b}\\cos \\theta = 1 + 1{/tex}{tex} = \\frac{{{x^2}}}{{{a^2}}}{\\cos ^2}\\theta + \\frac{{{x^2}}}{{{a^2}}}{\\sin ^2}\\theta + \\frac{{{y^2}}}{{{b^2}}}{\\sin ^2}\\theta + \\frac{{{y^2}}}{{{b^2}}}{\\cos ^2}\\theta = 2{/tex}{tex} = \\frac{{{x^2}}}{{{a^2}}}\\left( {{{\\cos }^2}\\theta + {{\\sin }^2}\\theta } \\right) + \\frac{{{y^2}}}{{{b^2}}}\\left( {{{\\sin }^2}\\theta + {{\\cos }^2}\\theta } \\right) = 2{/tex}{tex} = \\frac{{{x^2}}}{{{a^2}}} + \\frac{{{y^2}}}{{{b^2}}} = 2{/tex}\xa0{tex}\\left[ {\\because {{\\cos }^2}\\theta + {{\\sin }^2}\\theta = 1} \\right]{/tex}Hence proved. | |
| 12828. |
Find the sum of first 8 multiplies of 3 |
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Answer» 108 108? Is it 108? 3+6+9+12+15+18+21+24=108 48 ..? |
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| 12829. |
mode missing frequency |
| Answer» What | |
| 12830. |
If 12th term AP is-13 and sum of its first four term is 24, what is the sum of its first 10th term |
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Answer» a=9,d=-2,a10=-9 -9 0...? 0 |
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| 12831. |
Sun of zeroes =1/4Product of zeroes = -1Find the quadratic polynomial |
| Answer» Formula for finding quadratic polynomial is= x2 - (sum of zeroes)x + (product of zeroes)Polynomial= x2 - 1/4x + (-1) =x2 - 1/4x - 1 | |
| 12832. |
Best of luck for ur last xam everyone |
| Answer» Same to u | |
| 12833. |
How much questions will be asked from ncert with same values in board? |
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Answer» Hi Hi |
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| 12834. |
Check whether the polynomal tsqure-3 is a factor of the polynomal 2t |
| Answer» | |
| 12835. |
1/cosec = cos 1/sec =sin |
| Answer» | |
| 12836. |
Tell me the making scheme for class 10 |
| Answer» Are you stupid there is no making scheme what you want cbse will make toys ??there is marking scheme stupid boy | |
| 12837. |
Is optional exercise coming boards because it is mentioned in page that it is not for exam purpose |
| Answer» Yes of course read it you will pass in this | |
| 12838. |
Find the prime factor of 2×7×11×17×23+23 |
| Answer» 23 (2×7×11×1+1)23 (14×11+1)23 (154+1)23 (155)Will find prime number of 1555 15531 31 1So, we get23×5×31There for is prime factor bcz 23,5,and31 are prime facter. By Dorser | |
| 12839. |
Which type of questions generally comes in paper ? |
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Answer» ?? Bhai sun math k paper me math ke type ke question he ayngae What type of questions ?????? |
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| 12840. |
All the best friends for last exam ?? |
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Answer» Thanku....... and same 2 u...??? Thanks and same to you same to you From 1 year it comes to just 24 hrs Same to you |
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| 12841. |
Please tell me the weightage of all chapter please |
| Answer» | |
| 12842. |
Hi friends math ki tayari hogyi sab ki |
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Answer» Ha almost Bas 2 chapter left On the way!! Yaar Ha bas hohi gaya last chapter statistics. Kar raha hu Haaa Chal rahi hai ? tumhara kya haal hai? |
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| 12843. |
How to factorise by synthetic method |
| Answer» | |
| 12844. |
TanA /1-cotA+cotA / 1-tanA= 1+secA cosecA |
| Answer» Change tanA and cotA into terms of sinA and cosA and then solve it | |
| 12845. |
Find a distance of a origin (5,6) from origin |
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Answer» Yes thats the answer Route 61 Op=√x2+y2 |
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| 12846. |
A triangle ABC is an acute angle of triangle and AD PERPENDICULAR BC PROVE THAT AC 2=AB2+BC2-2BC2×BD |
| Answer» Optional ka hai app mai dekh lo ncert solutions | |
| 12847. |
How to divide |
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Answer» Which one Kya |
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| 12848. |
How to become a topper |
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Answer» Mehnat kihiye to aap tooper ban payega Only concepts ko clear kijiye in 11 class Only practice Practice |
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| 12849. |
Should we text book |
| Answer» | |
| 12850. |
Find the least number which is divisible by all numbers from 1 to 10 |
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Answer» You have to find the LCM 1to 10 2520 |
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