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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.
| 18751. |
What is the formula of Pythagoras theorm |
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Answer» (H)^2=(P)^2+(B)^2 Sum of square of small side is = square of big side H ka squre = B ka squre ✖ P ka squre |
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| 18752. |
3x-k√2x+4=0 |
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Answer» Iska answer +-4 hai Yeh answer nahi hai 3x-k√2x+4=0 =>a=3,b=k√2,c=4=> b square-5ac=> (√2k)^2-4×3×4=o =>2k2-48=o=> 2k^2=48=> k^2=48/2=>k^2=24=>k=√24 =>k=2√6 K= 2√6 |
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| 18753. |
The ratio between the sum of two Ap is (3 n +8): (7 n +15) find the ratio of 12 the term |
| Answer» Let a1, a2 and d1, d2 are the first term and common difference of two A. P. S respectively.A T Q {tex}\\frac{{\\frac{n}{2}}}{{\\frac{n}{2}}}\\frac{{\\left[ {2{a_1} + \\left( {n - 1} \\right){d_1}} \\right]}}{{\\left[ {2{a_2} + \\left( {n - 1} \\right){d_2}} \\right]}} = \\frac{{3n + 8}}{{7n + 15}}{/tex}{tex}\\frac { 12 \\text { th term of Ist } \\mathrm { A } . \\mathrm { P } } { 12 \\mathrm { th } \\text { term of } 2 \\mathrm { ndA.P } } = \\frac { a _ { 1 } + 11 d _ { 1 } } { a _ { 2 } + 11 d _ { 2 } }{/tex}put n = 23 in eq (i){tex}\\frac { 2 a _ { 1 } + 22 d _ { 1 } } { 2 a _ { 2 } + 22 d _ { 2 } } = \\frac { 3 \\times 23 + 8 } { 7 \\times 23 + 15 }{/tex}{tex}\\frac { a _ { 1 } + 11 d _ { 1 } } { a _ { 2 } + 11 d _ { 2 } } = \\frac { 7 } { 16 }{/tex} | |
| 18754. |
What is thelse theoram |
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Answer» Ncert chap. 6 ka theorem 6.1 hi h Ncert ..dekho ...mil ..jayga If a line drawns parallel to one side of a triangle to intersect the other two sides in distinct points the other two sides are divided in the same ratio In a triangle a line parrallel to the third side...divide the other two sides into equal ratio.. Line drawn ll to one side intersect the other then it divide the sides in same ratio Thales theoram also known as basic propratinally theoram It is BPT theorem |
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| 18755. |
2×3×58+2÷334 |
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Answer» 1410(approx) The answer will be 348.0059(approx) Is my answer is right Hey ...swaggy ...tell ans 515 348.005 Hey bhai 1350 Bodmas se solve kr lo Hey... |
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| 18756. |
Is there any marks for holiday homeworks |
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Answer» No No Nope Noooooooo I dont think so |
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| 18757. |
Cosec(65-thithA)-sec(25-thitha)-tan(55-thitha)+cot(35+thitha) |
| Answer» I hate tregnometry.. | |
| 18758. |
A tower ab is is 20 m high and be. It\'s shadow on the ground is20m long find the sun s. Altitude |
| Answer» Let the tower AB = 20m and shadow BC {tex}= 20\\sqrt3 m{/tex}{tex}\\angle A C B = \\theta{/tex}{tex}\\tan \\theta = \\frac { A B } { B C }{/tex}{tex}\\tan \\theta = \\frac { 20 } { 20 \\sqrt { 3 } } = \\frac { 1 } { \\sqrt { 3 } } = \\tan 30 ^ { \\circ }{/tex}\xa0{tex}\\Rightarrow \\quad \\theta = 30 ^ { \\circ }{/tex}Therefore ,the Sun’s altitude is 30°. | |
| 18759. |
Is completing the square method is in our course |
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Answer» Yes Ur welcome...?? Ok thanks Plz somebody help me U can learn by seeing videos in YouTube ...☺☺ Its not so difficult ranjan jii Now what I can do??? But actually Meri teacher n kaha h ki nhi ayegaOr isliye pdaya bhi nhi yes |
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| 18760. |
Surface area and volumes |
| Answer» Get NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 18761. |
Find the 15th term of ap with 2nd term 11 d is 9 |
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Answer» a15 = 128 Khud hi answer de diye A15 =128 and a=2 |
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| 18762. |
Solve by cross multiplication method Px + qy =p-qQx - py =p+q |
| Answer» Sorry yrrr | |
| 18763. |
2x⅔+2y=5 |
| Answer» | |
| 18764. |
Y+x+x=15root3(root3+1)+15(root3+1) |
| Answer» y+2x=60+30(root)3 | |
| 18765. |
Ap sub mujha trigo ka important formula batao |
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Answer» Most important q All are important Sb imp.hi h |
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| 18766. |
Bhai agr koi ladka 19 saal ka hai toh uss ne 12 th konsi age me complete kra hoga |
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Answer» me bhaut pareshan likin vo 5,6 saal phele 12th pass out ho chuka hai batao abbb me kya kruuuy Aur mujgse kaha ki voh19 saal ka hai Age km krwa lo date of birth me Kyu AGR ukg lkg na padhi ho to 16me complete kr lega Kyu |
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| 18767. |
Find the sum of 1st 15 multiple of 8 |
| Answer» The required AP : 8, 16,24 ,32 ......,120.First term = 8 Common difference=8 | |
| 18768. |
TanA+SecA-1/TanA-SecA+1=1+SinA/CosA |
| Answer» | |
| 18769. |
If m time s the mth term of an A P is equal to n Times its nth term find (m+n)th term of the AP |
| Answer» Let a be the first term and d be the common difference of the given AP. Then, in general the mth and nth terms can be written as\xa0Tm = a + (m - 1) d and Tn = a + (n - 1)d respectively.According to the question,we are given that,\xa0(m.Tm) = (n.Tn){tex}\\Rightarrow{/tex}\xa0m.{a + (m - 1)d} = n.{a + (n - 1)d}{tex}\\Rightarrow{/tex}\xa0a.(m - n) + {(m2 - n2) - (m - n)} . d = 0{tex}\\Rightarrow{/tex}\xa0(m - n).{a + (m + n - 1)}d.{tex}\\Rightarrow{/tex}\xa0(m - n).Tm+n = 0{tex}\\Rightarrow{/tex}Tm+n = 0 [{tex}\\because{/tex}\xa0(m-n){tex}\\neq{/tex}0].Hence, the (m + n)th term is zero. | |
| 18770. |
If 9th term of an AP is 0 , prove that its29 th term is double of its 19th term |
| Answer» We have,a9\xa0= 0{tex}\\Rightarrow{/tex}\xa0a + (9 - 1)d = 0{tex}\\Rightarrow{/tex}\xa0a + 8d = 0{tex}\\Rightarrow{/tex}\xa0a = -8dTo prove: a29 = 2a19Proof:LHS = a29= a + (29 - 1)d= a + 28d= -8d + 28d= 20dRHS = 2a19= 2 a + (19 - 1)d]= 2[ -8d + 18d]= 2\xa0{tex}\\times{/tex}10d= 20d{tex}\\therefore{/tex} LHS = RHSHence, 29th\xa0term is double the 19th term. | |
| 18771. |
Which term of the AP 19,18 1/5,17 2/5..... is the first negative term |
| Answer» Easy,,see rd Sharma example | |
| 18772. |
anyone from vindhyanagar |
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Answer» kaun ho bhai near it Yes near it |
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| 18773. |
check whether 301 is a term of the list of numbers 5,11,17,23.....? |
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Answer» A = 5 d=6 an= 301An=a+(n-1)d301= 5+(n-1)66(n-1)= 301-56(n-1)= 296n-1= 296/6No.it can never be a list of these number No it is not |
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| 18774. |
TanA/1+Tan^2A+cotA/1+cot^2A=sinAcosA prove it |
| Answer» Not possible because their are two units | |
| 18775. |
If one zero of a polynomial 3 x square - 8 x + 2 k + 1 is 7 times the other find the value of k |
| Answer» Let\xa0{tex}\\alpha{/tex}\xa0and\xa0{tex}\\beta{/tex}\xa0be the zeroes of the polynomial,\xa0{tex}3x^2-8x+2k+1{/tex}Given,\xa0{tex}\\beta = 7 \\alpha{/tex}{tex}\\therefore \\quad \\alpha + 7 \\alpha = 8 \\alpha = - \\left( - \\frac { 8 } { 3 } \\right)=\\frac{8}{3}{/tex}So,\xa0{tex}\\alpha = \\frac { 1 } { 3 }{/tex}and\xa0{tex}\\alpha \\times 7 \\alpha = \\frac { 2 k + 1 } { 3 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}7 \\alpha ^ { 2 } = \\frac { 2 k + 1 } { 3 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}7 \\left( \\frac { 1 } { 3 } \\right) ^ { 2 } = \\frac { 2 k + 1 } { 3 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}7 \\times \\frac { 1 } { 9 } = \\frac { 2 k + 1 } { 3 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { 7 } { 3 } - 1 = 2 k \\quad \\Rightarrow 2 k = \\frac { 4 } { 3 }{/tex}{tex}\\therefore k=\\frac{2}{3}{/tex} | |
| 18776. |
Chapter of the maths in the class of10 |
| Answer» No | |
| 18777. |
If one zero of a polynomial 3x_8x+2k+1is seven times the other ,find the value of k |
| Answer» Substitute the value of x | |
| 18778. |
root 3 +root2 square is irra |
| Answer» | |
| 18779. |
The difference of two numbers is 5 and the difference of their reciprocals is 1/10.Find the numbers. |
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Answer» Let the nos be x & yx=5+y (1)1/x-1/y=1/10 (2)Putting (1) in (2) we will get the answer Wrong This question is weong |
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| 18780. |
Hiii amrit I sent the question..? |
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Answer» But thank you...? Leave i got the answer...?? Did u get It?? ?? Rd sharma Rd Sharma, foundation, together with,ncrt. Which years book do you have..?? |
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| 18781. |
If 190344 is written in the form of ( 2^a 3^b 5^c 7^d 11^e...) the sum of ( a +b+c+d+e...) will be? |
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Answer» Qnswer to bata... mera galat ara hai... Just find prime factorization of that number ... |
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| 18782. |
show that one and only one out of n,n+2 or n+4 is divisible by 3 where n€N |
| Answer» Check out answer of this question in RD | |
| 18783. |
How I get all the CBSE sample papers |
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Answer» Aap search box me dalo......cbseacademics.in class 10 sample paper....hope u will get But Maine to whi se download kiye h Nahi mil Raha hai cbseacademics.in Bhai Ashu net se Kahaani par From net |
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| 18784. |
Math kisne banayi |
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Answer» Ku question samajna hai Mne Arybhatt discovered 0? Jab mile to mujhe bhi batana......punish karna h Mne |
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| 18785. |
Hello everybody |
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Answer» Hlo Hiì Hii |
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| 18786. |
Find the values of P for which the equation x2+7px+9=0 has no real roots |
| Answer» X2+7px+9=0 than b2-4ac so ,49p2-36=0 p=6/7 | |
| 18787. |
If α and β are the zeros of the polynomial 2x |
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Answer» Give complete questions Complete this question plzz InComplete question |
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| 18788. |
If ab parallel pq parallel CD ,ab=x units ,CD=y units,and pq=z units ,prove 1/x +1/y =1/z |
| Answer» In [common angle] [Each 90o] [AA criteria]Then, ....(i)Similarly, ...(ii)On adding Eq(i) and Eq(ii), we get\xa0 | |
| 18789. |
21b-32+7b-20b |
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Answer» 8(b-4) 8b-32 8b -32 ???? |
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| 18790. |
In an AP:given l=4 ,d=2 sum of number=-14 |
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Answer» If we r supposed to calculate \'a\' then and is -8 or if n then ans is 7 Pls complete the question ??? |
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| 18791. |
How many 1.maximum 2.minimum zeros can a quadratic polynomial have? |
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Answer» Max-more than 2 , min-2 2 0 zeroes. when D is lesS than 0 Sory when 0 Max=2 and min...=2 |
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| 18792. |
How to prove sum of all angles of triangle= 180° |
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Answer» Class 7th ka triangle chapter khol lo Kya linear pair= 180° ka concept use hoga Phir tum hase?? ?? By drawing a parallel lines to any sides of a triangle then use ur mind |
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| 18793. |
How to find square roots |
| Answer» By using formula | |
| 18794. |
I LOVE ? |
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Answer» Thank Whom kis ko |
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| 18795. |
Jajzbskjsns |
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Answer» Stupeet kya kys |
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| 18796. |
Find the distance between the following pairs of points(2,3),(4,1) |
| Answer» 2 root 2 | |
| 18797. |
a cosΦ +b sinΦ=ma sin Φ -b cosΦ=n |
| Answer» Given,R.H.S = m2 +n2= (a cos{tex}\\theta{/tex}\xa0+ b sin{tex}\\theta{/tex})2 + (a sin{tex}\\theta{/tex}\xa0- b cos{tex}\\theta{/tex})2 {tex}\\left[\\sin ce,\\;m\\;=\\;a\\;\\cos\\theta\\;+\\;b\\;\\sin\\theta\\;and\\;n\\;=\\;a\\;\\sin\\theta\\;-\\;b\\;\\cos\\theta\\right]{/tex}\xa0= (a2cos2{tex}\\theta{/tex}\xa0+ b2sin2{tex}\\theta{/tex}\xa0+\xa02ab cos{tex}\\theta{/tex}\xa0sin{tex}\\theta{/tex}) + (a2sin2{tex}\\theta{/tex}\xa0+ b2cos2{tex}\\theta{/tex}\xa0- 2ab sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}) {tex}\\left[\\because\\left(a\\pm b\\right)^2=a^2+b^2\\pm2ab\\right]{/tex}= a2 (cos2{tex}\\theta{/tex}\xa0+ sin2{tex}\\theta{/tex}) + b2 (sin2{tex}\\theta{/tex}\xa0+ cos2{tex}\\theta{/tex}) = a2 + b2 = L.H.S {tex}\\left[\\because\\sin^2\\theta+\\cos^2\\theta=1\\right]{/tex}therefore, {tex}m^2\\;+n^2=a^2+b^2{/tex}Hence proved. | |
| 18798. |
Hlo guys whats going on |
| Answer» study | |
| 18799. |
CosA=3/5 find value of sinA-1/tanA/2tanA |
| Answer» Sin A_1 =5/4_1 =1/4 Tan A=4/3 2Tan A=8/3 | |
| 18800. |
How to Calculate compound itntrest? |
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Answer» Right? Annually A=p+(1+ r/n)n Konsa nikalana hai quarterly, annualy ya priodicaly. First clear it |
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