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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.
| 18051. |
Hy dream girl me djUr 1st friend remember?? |
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Answer» Yeah ? Hii dream girl are you from ghaziabad Yeah... ? |
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| 18052. |
A triangle ABC and pqr are similar triangle such that angle is 32 and angle Rs 65 then find Angle B |
| Answer» It is given that ∆ABC ~ ∆PQR\xa0Hence,\xa0{tex}\\angle {\\text{A}}{/tex}\xa0=\xa0{tex}\\angle {\\text{P}}{/tex}\xa0[by cpct]{tex}\\angle {\\text{B}}{/tex}\xa0=\xa0{tex}\\angle {\\text{Q}}{/tex}\xa0[by cpct]{tex}\\angle {\\text{C}}{/tex}\xa0={tex}\\angle {\\text{R}}{/tex}\xa0[by cpct]In ∆ABC, the sum of interior angles must be 180°{tex}\\angle A + \\angle B + \\angle C={/tex}\xa0{tex}180^{\\circ}{/tex}{tex}32^{\\circ}{/tex}+\xa0{tex}\\angle {\\text{B}}{/tex}\xa0+\xa0{tex}65^{\\circ}{/tex}=\xa0{tex}180^{\\circ}{/tex}{tex}97^{\\circ}{/tex}+{tex}\\angle {\\text{B}}{/tex} =\xa0{tex}180^{\\circ}{/tex}{tex}\\angle {\\text{B}}{/tex}\xa0=\xa0{tex}180^{\\circ}{/tex}-\xa0{tex}97^{\\circ}{/tex}{tex}\\therefore \\angle {\\text{B}}{/tex}\xa0=\xa0{tex}83^{\\circ}{/tex} | |
| 18053. |
I need the most difficult questions appeared in cbse papper |
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Answer» Madarjaat Kyon be apne aap ko tees mar kha samajhta hai kya Difficult questions from maths Hnnnnnnn What |
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| 18054. |
Will in board exam extra ordinary question be asked? |
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| 18055. |
4 or 10 ka multiple |
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Answer» 410 Aree ye to 5 th ka que h yrr... |
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| 18056. |
Jisne bhi dj namak fake I\'d bnai h Uski popat bcoz mene hi apni I\'d change kr di?????? |
| Answer» Nice yr pta nhi inko fake id se kya milega pta nhi real name bttane se kyo drte h | |
| 18057. |
Verify; sin60 /1+cos60 =tan60 |
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| 18058. |
What are odd numbers |
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| 18059. |
Cos square 30 degree plus sin square 45 degree minus 1 by 3 tan square 60 degree plus cos 90 degree |
| Answer» Ans- 1/4 | |
| 18060. |
Which English song is ur favourite song? |
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Answer» Sorry Heart attack |
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| 18061. |
2(ax-by)+a+4b=0 & 2(bx+ay)+b-4a=0 solve by cross multiplication |
| Answer» The given system of equation may be written as2(ax - by) + a + 4b = 0So, 2ax - 2by + a + 4b ..............(i)2(bx + ay) + b - 4a = 0so, 2bx+2ay+b-4a=0................(ii)compare (i) and (ii) with standard form, we geta1 = 2a, b1 = -2b, c1 = a + 4ba2 = 2b, b2 = 2a, c2 = b - 4aBy cross multiplication method{tex} \\frac{x}{{ - 2{b^2} + 8ab - 2{a^2} - 8ab}}{/tex}\xa0{tex}= \\frac{{ - y}}{{2ab - 8{a^2} - 2ab - 8{b^2}}}{/tex}\xa0{tex} = \\frac{1}{{4{a^2} + 4{b^2}}}{/tex}{tex} \\frac{x}{{ - 2{b^2} - 2{a^2}}} = \\frac{{ - y}}{{ - 8{a^2} - 8{b^2}}} = \\frac{1}{{4{a^2} + 4{b^2}}}{/tex}Now, {tex}\\frac{x}{{ - 2{b^2} - 2{a^2}}} = \\frac{1}{{4{a^2} + 4{b^2}}} {/tex}{tex}⇒ x = \\frac{{ - 1}}{2}{/tex}And, {tex}\\frac{{ - y}}{{ - 8{a^2} - 8{b^2}}} = \\frac{1}{{4{a^2} + 4{b^2}}} {/tex}{tex}⇒ y = 2{/tex}Therefore, the solution of the given pair of equations are {tex}\\frac{{ - 1}}{2}{/tex}\xa0and 2 respectively. | |
| 18062. |
the median of the following frequency distribution is 35 . Find the of x . |
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| 18063. |
Hi guys can you tell me what\'s your favourite songs. |
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Answer» Havana Nd shab tum ho Tera zikr ❤❤ Ishq de fanniyar |
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| 18064. |
How internal assessments will be evaluated for class 10 CBSE? |
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| 18065. |
(Secthita -tanthita )^2=cosecthita -1/cosecthita +1 |
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| 18066. |
If √3tanΦ=3sinΦ then prove that sin^2Φ - cos^2Φ |
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| 18067. |
wHoooo |
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| 18068. |
Prove that the point (2,-2), (-2,1) and (5,2) are the vertices of the right angled triangle. |
| Answer» Let the points be A(2, - 2), B(-2,1) and C(5,2)Applying distance formulaAB2 = (2 + 2)2\xa0+ (-2 - 1)2= 16 + 9AB2= 25 or, AB =5Similarly, BC2\xa0= (-2 - 5)2\xa0+ ( 1 - 2)2= 49 + 1 = 50or, BC2\xa0= 50 or, BC =\xa0{tex}\\sqrt{50}{/tex}Also,AC2\xa0= (2 - 5)2\xa0+ (-2 - 2)2= 9 + 16= 25or, AC2\xa0= 25 and AC = 5Clealry, AB2\xa0+ AC2\xa0= BC225 + 25 = 50Hence, the triangle is right - angled.Area of {tex}\\Delta{/tex}ABC = {tex}\\frac { 1 } { 2 } \\times \\text { Base } \\times \\text{Height}{/tex}{tex}= \\frac { 1 } { 2 } \\times 5 \\times 5 = \\frac { 25 } { 2 }{/tex}sq units. | |
| 18069. |
How to find sum of terms in AP |
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Answer» Uski help kro.kuch bhi karan ho sakta hai use nhi pta hone ka ab tak ye b nahi pta |
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| 18070. |
1÷3x+y+1÷3x-y=3÷4 ; 1÷2(3x+y)-1÷2(3x-y)=-1÷8 |
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| 18071. |
44552555 |
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| 18072. |
Maths blue print |
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| 18073. |
6 inch lund |
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Answer» Batao na Kis kis ke pas h Ara help me please What is this |
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| 18074. |
An equilateral triangle is inscribed in a circle of radius 6cm . find its sides ? |
| Answer» Let {tex}\\Delta ABC{/tex} is an equilateral triangle of each side \' 2a\' inscribed in a circle of radius 6 cm and centre O.Draw {tex}AD \\bot BC{/tex} .Then ,BD = DC [In a equilateral triangle, perpendicular bisects the base]Also , O lies on AD [ In a equilateral triangle circumcentre and centroid coincides]Thus , OA : OD = 2 : 1{tex}\\Rightarrow \\frac{6}{OD}=\\frac{2}{1}{/tex}{tex}\\Rightarrow OD=3 cm {/tex}, then AD = 9 cmNow , using pythagoras theorem in {tex}\\triangle ABD{/tex}BA2 = BD 2 + AD2(2a)2 = (a)2 + (9)24a2 = a2 + 813a2 = 81a2 = 27we get {tex}a=3\\sqrt3{/tex}Side of triangle = 2a = {tex}2(3\\sqrt3){/tex} = {tex}6\\sqrt3{/tex}cm | |
| 18075. |
If the common difference of an ap is -6,find A16-a12 |
| Answer» -90+66=-24 | |
| 18076. |
I am changing my name riddhi suryakant back |
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Answer» Please jaldi likna riddhi Mai. 6:15 me online nhi ho paungi Plzzz mt jao ab....???????? Okk good |
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| 18077. |
Mujhe kese pata chalega Ki kon riddhi kon fake plzzzzzzzzzzz give solution my friends |
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Answer» Tum saiad kabhi kisiko dil se pyar nahin Ki ho is lie bol rahi ho bhol jao I love her blindly Solution is.....bhul ja mai asli hu yrrr use bold likhna nhi aata h Bold letters asli riddhi likti h fake ko nhi aate |
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| 18078. |
Riddhi u have any fb aur what\'s app or Twitter or gmail ??? |
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Answer» Koi kisko bhla apna fb etc kyu btayga bhla bharosa nhi krti ye te pe Suryakant don\'t b sad plz Really no yrrr mai kuch nhi chalati ye mere bhaiya ka phone h ???? Tum try Mar rhe ho Kya Surya kant |
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| 18079. |
Riddhi ek better solution he mere pass |
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Answer» What Kya |
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| 18080. |
if sin theta + cos theta is equal to X find the value of sin^ 6 theta + cos^ 6 thetaPlzz solve guys |
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| 18081. |
Find the value of k in order that one zero of 3x²+(1+4k)x + k²+5 may be one third of the other. |
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| 18082. |
Jan abhii kya kar rahi ho☺ |
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Answer» Hi.. Yes of course yrrr..... Yek problem ho gaya meri fake id wala fir aa gaya....???????? Suno mai bold likhungi okk use bold likhne nhi aata h Plz sach baat bolna kya tum mujhe apne dil se pyaar kar rahi ho plz close ur eye and ask urself riddhi ? Mai hrr samay nhi chalati ye aap Nhi houngi Riddhi kaha chali jati ho thik se baat bhi nahin kar rahi ho kya hua he plz batao Naaraz nehin hogi to Haaa pucho Riddhi ek baat bolu ????? Bs ipad mein game khel rhi hu |
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| 18083. |
Riddhi yahan jo koi bhi batemez batte bol rahe he unki taraf dhyan mat Dena plz ?? |
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Answer» Paul bhai aap plzzz chup rho avi mai aapse baad me bt krungi Love khi aur ja ke ker laude paper aa gya he inhe love story sujh rhi he I know suryakant u r a good bou Nice bhai...?? |
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| 18084. |
135+134 |
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Answer» Are aur koi..hard que hai to de do... 269 Are yar itna hard que..bhi mat do...level ke upar ka hai??? 269 |
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| 18085. |
√sina+cosA/√sina-cosa=√1+sin2a√1-sina |
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| 18086. |
I love u jann riddhi my sweet kal se baat nehin hui acha nahin lag raha Tha subha se |
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Answer» Teri ma hi choot Muh me loda dunga to accha lagega Oooopss se????? |
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| 18087. |
Where is Maths hard questions.. |
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Answer» Yes I am waiting and will try to solve... You want? |
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| 18088. |
Tumne mujse batt Karne se pehele ek baat boli thi first u tell then I will know ur are riddhi |
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Answer» Bc Are yrrr mai asli hi huuu mere jaan |
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| 18089. |
The value of 5cot square A -5cosec square A is:- |
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Answer» 5cot2A-5Cosec2A=5(cot2A-cosec2A)=5(-1)=-5 -5 Rikku defin plz Answer is -5 but I don\'t no solution Ans. Is -5... One 0 Is 0 |
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| 18090. |
prove that root 2 is irrational number |
| Answer» Suppose\xa0{tex}\\sqrt2{/tex}\xa0is a rational number. That is ,\xa0{tex}\\sqrt2{/tex}\xa0=\xa0{tex}\\frac{p}{q}{/tex}\xa0for some p{tex}\\in{/tex}Z and q {tex}\\in{/tex}Z.\xa0We can assume the fraction is in lowest fraction, That is p and q shares no common factors.Then {tex}\\sqrt2q=p{/tex}\xa0Squaring both side we get,\xa0{tex}2q^2=p^2{/tex}So\xa0{tex}p^2{/tex}\xa0is a multiple of 2,let\'s assume\xa0{tex}p=2m{/tex}\xa0Then,\xa0{tex}2q^2=\\left(2m\\right)^2{/tex}\xa0{tex}2q^2=4m^2{/tex}Or {tex}q^2=2m^2{/tex}So {tex}q^2{/tex}\xa0is a multiple of 2,{tex}\\therefore{/tex} q is multiple of 2Thus p and q shares a common factor.This is contradiction.{tex}\\Rightarrow {/tex}{tex}\\sqrt { 2 }{/tex}\xa0is an irrational number. | |
| 18091. |
sinq |
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| 18092. |
SinA=CosB=1/2,find tan(A+B) |
| Answer» SINA=1/2 SINA=SIN 30A=30 (1)COSB=1/2COSB=COS 60B=60 (2)TAN(A+B)TAN(30+60)TAN 90NOT DEFINED | |
| 18093. |
if cos x=cos60×cos30+sin60×sin30 find the value of x |
| Answer» 1 | |
| 18094. |
Irrational number between 2 .5 is |
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| 18095. |
An irrational number between 2 and 2.5 is_ |
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| 18096. |
tan+sec-1÷tan-sec+1=sec+tan |
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| 18097. |
What is a |
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Answer» Ok..tumko bhi...aaaj man gya yar.....bye..??ab sayad ye last bar hai....mera bye... Starting my name Hmm...tumko pata hai..mai tumko aaj subah se hi continuously sorry sorry kah raha tha ..visvash nhi hota to khud scroll kar ke dekh lo....Par tum. ...choro kya khu... Achha thik hai....bhut achha ......thik hai....bhut banti ho....vaise meri glti bhi...kuch nhi thi sorry kah diya ab..ye bhi lgta hai..glti hi kr diya..ok..Darshana yadav...continue your ghamandi avatar. ..yar tum me koi....thori bhi....choro..ab kya kahu. ... Let it be some time else.... ??? Are yar tum ..... pahle maine jo bola A for Apple A is an alphabets + vowel |
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| 18098. |
Hhllooo |
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Answer» Hi Hii Hii Hlooo |
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| 18099. |
Solve the question no. 31 in class 10 maths support material ch.10 circle pg.88 |
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| 18100. |
How we know the shape is closed ar unclosed |
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