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InterviewSolution
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.
| 34801. |
Find the value of 1+2+3+.......120? |
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Answer» Hi jamin yu r in class 10 if you r in class 10 so can yu make my partner Right answer is 21255 Hi this question is not correct 7260 |
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| 34802. |
Ksusns |
| Answer» | |
| 34803. |
Two unbaised coin are tossed simultaneously find the probability of getting1 two heads |
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Answer» Probability of getting two heads =1/4 1/4 |
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| 34804. |
if p(e)=0.05, what is the probability of not E |
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Answer» 0.95 P(E)=1-P(e)So, 1- 0.05= 0.95 |
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| 34805. |
12/2x²+x-3 |
| Answer» | |
| 34806. |
Express sin a and cos a in terms of cot a |
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Answer» When we devide sina / cos a We will get cot a Edl |
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| 34807. |
State uclids division lemma |
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Answer» a=bq+r A =bq + r |
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| 34808. |
Sec$ + tan$=pFind cosec$ |
| Answer» 59 2018 5 | |
| 34809. |
Find the value of K, for which one root of the quadratic equation kx^2-14x+8=0 is 2. |
| Answer» K=5 | |
| 34810. |
6 c.m radius wale ek circle ke ek trijyakhand ka area gyat kijiye. Jiska angle 60 degree hai |
| Answer» We know that Area of sector ={tex}\\frac { \\theta } { 360 } \\pi r ^ { 2 }{/tex}\xa0Here,\xa0{tex} \\theta = 60 , r = 6{/tex}{tex}\\therefore{/tex} Required area {tex}= \\frac { 60 } { 360 } \\times \\frac { 22 } { 7 } \\times ( 6 ) ^ { 2 }{/tex}{tex}= \\frac { 1 } { 6 } \\times \\frac { 22 } { 7 } \\times 36{/tex}{tex}= \\frac { 22 \\times 6 } { 7 }{/tex}{tex}= \\frac { 132 } { 7 } = 18 \\frac { 6 } { 7 } c m ^ { 2 }{/tex} | |
| 34811. |
Sin A/√1-Sin^2 A |
| Answer» Sin/1-sin^2 Sin/cos^2. ( Square will get cancled)Sin/cos Tan | |
| 34812. |
Find the value of k for which one root of quadratic equation kx2^ -14x+8=0is2 |
| Answer» Put the value of one root that\'s given to in the place of X and solve it you will get the answer k is equal to 5 | |
| 34813. |
What does tan 45 and 60 stands for |
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Answer» Tan 45= 1 and tan 60= root3 Tangent of angle 45 and tangent of angle 60 tan 45 =1 & tan 60=√3 |
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| 34814. |
If the radius is given 4 cm so we are also take 4cm or not take 2cm means half we give me ans |
| Answer» In chapter=11 exercise =1.2 in circles | |
| 34815. |
Find the value of $ cos20°/sin70°+cos70°/sin20°-8(sin*sin)30°$ |
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Answer» 1+1-8=-6 I got my answer |
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| 34816. |
r2 h3 |
| Answer» Closer AT Poles nur wider AT Other places | |
| 34817. |
If sin x- 2cos x=1 then prove that 2sin x-cos x =2 |
| Answer» | |
| 34818. |
Two secants PAB and PAC are drawn as shown. If PA =3cm PC =2 cm cd=4cm ab? |
| Answer» | |
| 34819. |
What is composite numbee |
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Answer» Thank you And they have also more than two ✌ factors. . Examples r 4, 10 etc Composite number is a positive integer that can be formed by multiplying two smaller postive integers |
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| 34820. |
1 sin |
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Answer» 0.841471 57.3 degree |
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| 34821. |
If secant theta + tan theta is equals to p then find the value of cosecant |
| Answer» Sec + tan = p......(1)Sec ^2-tan^2 =1(Sec-tan)(sec +tan)=1(Sec-tan)(p)=1Sec-tan =1/p......(2)Adding (1) and (2)Sec+tan+sec-tan =p+1÷p 2sec = p^2+1/p .......(3)Subtract (2)from (1)Sec+tan-sec+tan=p-1/p2tan=p^2-1÷p ......(4)Divide (3)and(4)2sec÷2tan =p^2+1÷p÷p^2-1÷p 1/sin=p^2+1/p^2-1 Cosec = p^2-1÷p^2+1 | |
| 34822. |
6x2-√2x-2=0 find the two roots by prime factorization method |
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Answer» 6x2\xa0- √2 x - 2 = 06x2 - 3 √2 x- 2\xa0√2 x-2=03\xa0√2x(\xa0√2x-1) - 2(\xa0√2x-1)=0(\xa0√2x-1)(3\xa0√2x-2)=0hence x=\xa0√2/3 and x= 1/ 6x2 - 3under root 2 x- 2under root 2x -2 now factors come 6x2 - |
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| 34823. |
Define thal\'s theorum |
| Answer» Fftft | |
| 34824. |
When will be exam start? |
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Answer» Last week of feb. For Hindi students it will start 6th March and on 27th March for Sanskrit students it will starts 12 March and ends on 2 April..refer cbse q paper.. Boards....... Noooooo it will start after 20th February.... A/q to cbse It starts on 6th march and ends on 27th march March 13 i think |
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| 34825. |
Teach theorem of ch6 |
| Answer» | |
| 34826. |
cotA+cosecA-1*1/cotA-cosecA+1 |
| Answer» | |
| 34827. |
TanA+secA-1/tanA+secA+1=1+sinA/cosA |
| Answer» | |
| 34828. |
Prove 2+√5 is irTional |
| Answer» We preassume that\xa0{tex}2 + \\sqrt 5 {/tex}\xa0is a rational number.So it can be written as\xa0{tex}\\frac{a}{b}{/tex}, where a, b are co-prime integers and b is not zero.The new equation will be as below:\xa0{tex}\\implies 2 + \\sqrt 5 = \\frac{a}{b}{/tex}So, we will get\xa0{tex}\\sqrt 5 = \\frac{a}{b} - 2{/tex}As we know that 2 and {tex}\\frac{a}{b}{/tex}\xa0are rational numbers, their difference will be rational number only.on the other hand,\xa0{tex}\\sqrt 5 {/tex}\xa0is an irrational number and it can not be written as\xa0{tex}\\frac{a}{b}{/tex}.So, this contradicts the fact that\xa0{tex}\\sqrt 5 {/tex} is a rational number.\xa0Therefore our assumption is wrong and {tex}2 + \\sqrt 5 {/tex}\xa0is irrational. | |
| 34829. |
Which publication sample paper is most helpful for cbse maths class 10???? |
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Answer» Only NCERT Oswaal publications |
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| 34830. |
Bpt theorem kya hai |
| Answer» Tere ko BPT theorem nahi pata hai kya.. thik hai koi baat nahi.. main aapko help Karta Hoon Pehle Aap Ko books se refer karna padega.. theorem 6.1 pe h..there is written clearly if you don\'t know then ask me.. I will explain one by one.. Tere ko statement likhna Padega.. then prove and proof.. | |
| 34831. |
If cos 3 theta = sin 2 theta where theta is an acute angle then find the value of theta. |
| Answer» Cos 3 theta= sin2 theta Cos 3 theta= cos(90°- 2theta) 3 theta= 90°-2 theta 90°=(3+2)theta Theta =90°/5 =18° | |
| 34832. |
If x^2 +4=0 Then what will be the value of x |
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Answer» Sorry you both are wrong because the square of 2 or -2is 4 not-4 X^2+4=0 X^2=-4 X=2 -2 |
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| 34833. |
Find the zeroes of the polynomial of x²-4x-2 |
| Answer» 4+root24 upon2 and 4-root24upon2 | |
| 34834. |
Exercise 14.3 Q no.4 |
| Answer» | |
| 34835. |
Is Circle coming in exam |
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Answer» where u r dreaming.? in dhami wonderland Yes Yes Yea |
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| 34836. |
Prove that √5+2 is irrational |
| Answer» search from r.s. aggarwal u will definitely get | |
| 34837. |
How many balls each of radius 1 cm can be made from a solid sphere of lead of radius 8 cm |
| Answer» 512 | |
| 34838. |
Important theoram is lession 6 and 10 |
| Answer» BPT and Pythagoras in ch 6 because they come most of time as 4 mark question and in the CBSE sample paper the question is of Pythagoras theorem | |
| 34839. |
Cant we click a pic of question and send it ? There is no option for camera. |
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Answer» Use brainly for so No.....?? Noooo |
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| 34840. |
Express 0.25454545.... as fraction |
| Answer» 14 upon 55 | |
| 34841. |
3 cos^2 75 +4\\3 cot^30 + 9tan^2 30 .cos^2 15 +3cosec^2 60 |
| Answer» Yugg | |
| 34842. |
AB=24,BC=7,AC=? |
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Answer» AC will by equal to 25 by applying pythagoras theorem AC=25 |
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| 34843. |
How to get lvm 256 |
| Answer» | |
| 34844. |
How to prove root 5 is irrational number |
| Answer» let root 5 be rational then it must in the form of p/q [q is not equal to 0][p and q are co-prime] root 5=p/q => root 5 * q = p squaring on both sides => 5*q*q = p*p ------> 1 p*p is divisible by 5 p is divisible by 5 p = 5c [c is a positive integer] [squaring on both sides ] p*p = 25c*c --------- > 2 sub p*p in 1 5*q*q = 25*c*c q*q = 5*c*c => q is divisble by 5 thus q and p have a common factor 5 there is a contradiction as our assumsion p &q are co prime but it has a common factor so √5 is an irrational | |
| 34845. |
HCF of two numbers is 369569 . Find the numbers |
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Answer» What is the LCM ?? very tough........ ?? I am not able to do........... |
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| 34846. |
Is ncert examplar a good book for boards |
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Answer» I think ? so because most of the toppers also recommend it.. Yes because this book is prescribed by cbse and see if u are weak in maths then firstly do ncert and then start this exemplar it will definitely help u☺️ Nah... Not much good |
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| 34847. |
Find probability of getting a 52 Sunday in a leap year |
| Answer» 2/7 | |
| 34848. |
Please give solution of last board paper 2016;16;18 urgently |
| Answer» Why.???? | |
| 34849. |
Theorem 10.1 and 10.2 solution |
| Answer» Theorem no. 10.2In triangle PQO and PROOQ=OR (radius of same circle)Op=op (common)Angle Q = angleR (each 90)∆PQO conqurent ∆PROPQ=PR (cpct) | |
| 34850. |
Integration of( theta/ sin(theta))² |
| Answer» | |