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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.
| 26151. |
ax |
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Answer» because he is satisfied with his work ???????? |
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| 26152. |
If sin + cosec =2 then sin6 + cosec =? |
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Answer» Ncert se kar lo Equayion banake put kar do. |
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| 26153. |
Tt |
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Answer» What is this ? |
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| 26154. |
Write the relationship among mean, median and mode. If the mean of data is 27 and median is 33. |
| Answer» Mode = 3median - 2mean | |
| 26155. |
7/75 Has terminating decimal expension or non terminating decimal expension |
| Answer» This has terminating decimal expension | |
| 26156. |
You don\'t have to be great to start but have to start to be great . Believe that.?? |
| Answer» | |
| 26157. |
Formula of mean, median ,mode |
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Answer» Mode= sigma find xi ÷ fi It is in book |
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| 26158. |
2xsquare +x+4=0 |
| Answer» | |
| 26159. |
If the circumfrence of a circle exceed its diameter by 30cm,find the radius of the circle. |
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Answer» Aur tha Ok Muje Raakhi bandhne kab aaogi Behna? Bro mere khne ka sense kuch Aur tha Bro nhi khoge hme Ok mt kho tumari marzi? Bro hum nhi khenge Muje log pyaar se Homo sapien kehte h ? Who are you? Bahut padti ho tum ?? |
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| 26160. |
Show that( 3-√5)square |
| Answer» Bro not possible here | |
| 26161. |
Show that 3 minus root 5 whole square is irrational |
| Answer» Not possible here refer Ncert | |
| 26162. |
a+b+c=0 then find (a^4+b^4+c^4)÷a^2b^2+b^2c^2+c^2a^2 |
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| 26163. |
What is the formula of mode |
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Answer» Thank U for correcting me Aditya bro .I wrote the answer in a wrong way.? L+f-f1/2f-f1-f2*h l+(F1-F0/2F1-F0-F2)h |
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| 26164. |
ax+by=1 bx+ay=(a+b)2÷(a+b)2 -1 |
| Answer» Given equations are{tex}ax + by - 1 = 0{/tex}....(i) and {tex}bx + ay ={/tex}\xa0{tex}\\frac { ( a + b ) ^ { 2 } - \\left( a ^ { 2 } + b ^ { 2 } \\right) } { a ^ { 2 } + b ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}{tex}bx + ay ={/tex}\xa0{tex}\\frac { 2 a b } { a ^ { 2 } + b ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}{tex}(a^2 + b^2)bx + (a^2 + b^2)ay - 2ab = 0{/tex}...(ii)Here a1 = a, b1 = b, c1 = -1a2 = (a2 + b2)b,b2 = (a2 + b2)a,{tex}c_2 = -2ab{/tex}By cross multiplying method, we have{tex}\\Rightarrow{/tex}{tex}\\frac { x } { b _ { 1 } c _ { 2 } - b _ { 2 } c _ { 1 } }{/tex}\xa0=\xa0{tex}\\frac { y } { c _ { 1 } a _ { 2 } - c _ { 2 } a _ { 1 } }{/tex}\xa0=\xa0{tex}\\frac { 1 } { a _ { 1 } b _ { 2 } - a _ { 2 } b _ { 1 } }{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac { x } { b \\times ( - 2 a b ) - \\left( a ^ { 2 } + b ^ { 2 } \\right) a \\times ( - 1 ) }{/tex}\xa0=\xa0{tex}\\frac { y } { - 1 \\times \\left( a ^ { 2 } + b ^ { 2 } \\right) b - ( - 2 a b ) \\times a }{/tex}\xa0=\xa0{tex}\\frac { 1 } { a \\left( a ^ { 2 } + b ^ { 2 } \\right) a - \\left( a ^ { 2 } + b ^ { 2 } \\right) b \\times b }{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac { x } { - 2 a b ^ { 2 } + a ^ { 3 } + b ^ { 2 } a }{/tex}\xa0=\xa0{tex}\\frac { y } { - a ^ { 2 } b - b ^ { 3 } + 2 a ^ { 2 } b }{/tex}\xa0=\xa0{tex}\\frac { 1 } { \\left( a ^ { 2 } + b ^ { 2 } \\right) a ^ { 2 } - \\left( a ^ { 2 } + b ^ { 2 } \\right) b ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac { x } { a ^ { 3 } - a b ^ { 2 } }{/tex}=\xa0{tex}\\frac { y } { a ^ { 2 } b - b ^ { 3 } }{/tex}\xa0=\xa0{tex}\\frac { 1 } { \\left( a ^ { 2 } + b ^ { 2 } \\right) \\left( a ^ { 2 } - b ^ { 2 } \\right) }{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac{x}{a}{/tex}\xa0=\xa0{tex}\\frac{y}{b}{/tex}\xa0=\xa0{tex}\\frac { 1 } { a ^ { 2 } + b ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac{x}{a}{/tex}\xa0=\xa0{tex}\\frac { 1 } { a ^ { 2 } + b ^ { 2 } }{/tex}\xa0and y =\xa0{tex}\\frac { 1 } { a ^ { 2 } + b ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}x =\xa0{tex}\\frac { a } { a ^ { 2 } + b ^ { 2 } }{/tex}\xa0and y =\xa0{tex}\\frac { b } { a ^ { 2 } + b ^ { 2 } }{/tex} | |
| 26165. |
1+sinA-cosA/1+sinA+cosA ..=tanA/2 |
| Answer» (1+sinA)2-2(1+sinA)cosA+(cosA)2/(1+sinA)2-(cos)2 | |
| 26166. |
real numbers class 10 |
| Answer» | |
| 26167. |
Cross multiplication method |
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Answer» It\'s soooooooooo easiest method we can use it in any complicated questions even It\'s so easy |
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| 26168. |
Why have not mathematics in hindi langue |
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| 26169. |
QPR is a right triangle right angled at Q If QS=SR then show that PR×PR=4PS×PS-3PQ×PQ? |
| Answer» Given: PQR is a right Triangle, right-angled at QAlso QS = SRTo prove:- {tex}P{R^2} = 4P{S^2} - 3P{Q^2}{/tex}Proof:- In right-angled triangle PQR right angled at Q.{tex}P{R^2} = P{Q^2} + Q{R^2}{/tex}[By Pythagoras theorem] ....(i)Also {tex}QS = {1 \\over 2}QR{/tex} [QS=SR] .....(ii)In right-angled triangle PQS, right angled at Q.PS2 = PQ2 + QS2{tex}\\Rightarrow {/tex}{tex}P{S^2} = P{Q^2} + {\\left( {\\frac{1}{2}QR} \\right)^2}{/tex} [From (ii)]{tex}\\Rightarrow {/tex} 4PS2 = 4PQ2 + QR2 .....(iii)From (i) and (iii), we getPR2 = PQ2 + 4PS2 - 4PQ2{tex}\\Rightarrow {/tex} PR2 = 4PS2 - 3PQ2 | |
| 26170. |
Cos(A-B)=cosAsinB-sinACosB given and |
| Answer» When {tex}A= 45°\\ and\\ B = 30°{/tex}, then\xa0{tex}cos(45° - 30°) = cos 45°.cos 30° + sin 45°.sin 30°{/tex}{tex}\\Rightarrow \\quad \\cos 15 ^ { \\circ }{/tex}{tex}= \\frac { 1 } { \\sqrt { 2 } } \\times \\frac { \\sqrt { 3 } } { 2 } + \\frac { 1 } { \\sqrt { 2 } } \\times \\frac { 1 } { 2 }{/tex}{tex}= \\frac { \\sqrt { 3 } } { 2 \\sqrt { 2 } } + \\frac { 1 } { 2 \\sqrt { 2 } }{/tex}{tex}= \\frac { \\sqrt { 3 } + 1 } { 2 \\sqrt { 2 } }{/tex}{tex}= \\frac { ( \\sqrt { 3 } + 1 ) \\times \\sqrt { 2 } } { 2 \\sqrt { 2 } \\times \\sqrt { 2 } }{/tex}{tex}= \\frac { ( \\sqrt { 3 } + 1 ) \\sqrt { 2 } } { 4 }{/tex} | |
| 26171. |
How to find missing frequency in which mode is given |
| Answer» if i had an ncrt with page no 274 ?i would have gazed at the formula and went asleep ?by doing ex 14.2? | |
| 26172. |
(a+b) whole cube |
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Answer» Acube+Bcube+3ABsq.+3Asq.B (a+b)(a²+b²-ab) a cube + b cube + 3 ab ( a+b) a cube + b cube +three a square b +three a b square |
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| 26173. |
Are you mental |
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Answer» invaded i am real steel I think so NO |
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| 26174. |
Find the value of: sin50 sin40 + cos40 cosec50 |
| Answer» 2 | |
| 26175. |
If sum of first 7 terms of an AP is 49and that of 17 terms is 289 find the sum of the first n terms |
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Answer» Sn=n2(square of n) (11+n)n |
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| 26176. |
In figure op is equal to diameter of the circle . Prove that ABP is an equilateral triangle |
| Answer» Construction : Join A to Bwe have,OP = diameter{tex}\\Rightarrow {/tex}\xa0OQ + QP = diameter{tex}\\Rightarrow {/tex}\xa0Radius + QP = diameter{tex}\\Rightarrow {/tex}\xa0OQ - PQ = radiusThus OP is the hypotenuse of right angled\xa0{tex}\\triangle {/tex}AOPSo, In\xa0{tex}\\angle A O P , \\sin \\theta = \\frac { A O } { O P } = \\frac { 1 } { 2 }{/tex}\xa0{tex}\\theta = 30 ^ { \\circ }{/tex}Hence,\xa0{tex}\\angle A P B = 60 ^ { \\circ }{/tex}Now, In {tex}\\triangle A O P{/tex}AP = ABSo,\xa0{tex}\\angle P A B = \\angle P B A = 60 ^ { \\circ }{/tex}{tex}\\therefore \\triangle A P B{/tex} is an equilateral triangle. | |
| 26177. |
For what value of K the system of linear equations 2x+5y=K,Kx+15y=18 has infinitely many solutions |
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Answer» 6 K = 6 |
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| 26178. |
What will be the sample space when 4 coins are tossed |
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Answer» WHEN 4 COINS ARE TOSSEDEACH COIN HAS 2 POSSIBILITIES EITHER HEAD OR TAIL\xa0SAMPLE SPACE IS DETERMINED BY = 2^4 = 2X2X2X2 = 16HHHH HHHT HHTH HHTT HTHH HTHT HTTH HTTT\xa0\xa0THHH THHT THTH THTT\xa0TTHH TTHT TTTH TTTT 16 |
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| 26179. |
10th class any |
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Answer» Yes What do you want to ask |
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| 26180. |
If a and b are the zeroes of the polynomial f(x)=x²-5x+k such that a-b=1,find the value of k |
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Answer» Can anyone help to understand I didn\'t itIt is the example 13 of R.D.Sharma chapter polynomials Solve in notebook |
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| 26181. |
Sum of nth them formula |
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Answer» n/2 ( 2a + ( n-1)d) or n/2 ( a+l ) Nth term= n/2[2a+(n-1)d] or =n/2(a+l) N/2 [2a+(n-1)d] N/2(2a+(n-1)d) N/2(a+(n-1)d) |
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| 26182. |
Solve the quadratic question 4x2 |
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Answer» ?? I am always correct ?? Hmm sorry.. darshana is correct -1/2 2 |
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| 26183. |
Sec2+cosec2=sec2*cosec |
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| 26184. |
D, E and F are respectively the mid point of sides AB, BC and CA of Traingle abc.. |
| Answer» | |
| 26185. |
If 1 is a zeros of the polynomial ax²-3(a-1)x-1 then find the value of a |
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Answer» Ok np ..Make joke on me !! Not negative just making mockery of you Phir se ek aur negative comments ...?? i like it too Bhai Charan sparsh karaao Tbhi google se dekhte ho ?? Bihar Ayush tum kaha pe rahate ho Negative comments ...i like it ? Methile ye ans ayush Ko pata hoga bada hoshiyar hai 1 |
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| 26186. |
find a quadratic polynomial whose sum of zeroes and product of zeroes are given below1/4, _1 |
| Answer» 4XSquare -X-4 | |
| 26187. |
We want chapter wise blue print |
| Answer» See on google | |
| 26188. |
Find the 12th term of AP with first term 9 and common difference 10. |
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Answer» 119 119 |
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| 26189. |
How to find the zeros of quadratic polynomial by completing the square method |
| Answer» | |
| 26190. |
If (1+cos A)(1- cosA) = 3/4 ,find the value of secA |
| Answer» 2 | |
| 26191. |
What are composite number |
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Answer» The number which have more than 2 factors Numbers which have factors more than two i.e. 1,itself and the other those no. r called composite numbers. Eg 4 is composite number2 is prime number Which has more than 1 factors |
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| 26192. |
Prove that COSA - SINA +1/COSA+SINA-1=cosecA +cotA |
| Answer» | |
| 26193. |
TanA/1-cotA+cotA /1-tanA=1+tanA+cotA=1+secAcosecA |
| Answer» We have,{tex}\\mathrm { LHS } = \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan A } { 1 - \\frac { 1 } { \\tan A } } + \\frac { \\frac { 1 } { \\tan A } } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan A } { \\frac { \\tan A -1 } { \\tan A } } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } - \\frac { 1 } { \\tan A ( \\tan A - 1 ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 3 } A - 1 } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[Taking LCM]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { ( \\tan A - 1 ) \\left( \\tan ^ { 2 } A + \\tan A + 1 \\right) } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[{tex}\\because{/tex}\xa0a3\xa0- b3\xa0= ( a - b )(a2\xa0+ ab + b2)]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A + \\tan A + 1 } { \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A } + \\frac { \\tan A } { \\tan A } + \\frac { 1 } { \\tan A }{/tex}{tex}\\Rightarrow{/tex}\xa0LHS = tanA + 1 + cotA [ since (1/tanA) =cotA ].= (1 + tanA + cotA){tex}\\therefore \\quad \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA ...........(1)Now, 1 + tanA + cotA = 1 +\xa0{tex}\\frac { \\sin A } { \\cos A } + \\frac { \\cos A } { \\sin A }{/tex}\xa0= 1 +\xa0{tex}\\frac { \\sin ^ { 2 } A + \\cos ^ { 2 } A } { \\sin A \\cos A }{/tex} = 1 +\xa0{tex}\\frac { 1 } { \\sin A \\cos A }{/tex}\xa0[{tex}\\because{/tex}Sin2A + Cos2 A = 1 ]\xa0= 1 + cosecAsecASo, 1 + tanA + cotA = 1+ cosecAsecA.......(2)From (1) and (2), we obtain{tex}\\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA = 1 + cosecAsecA | |
| 26194. |
How to find distinct quadratic roots |
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Answer» By using quadratic formula Please send me the answer.. |
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| 26195. |
if sec thita.cos a = √3 and tan thita.cot a =√2 then find the value of tan thita and tan a ? |
| Answer» | |
| 26196. |
If cosec A+cot A=m,show that m2-1/m2+1=cosA |
| Answer» Given: cosec A + cot A = m{tex}\\Rightarrow{/tex}\xa0(cosec A + cot A)2 = (m)2 \xa0[squaring both sides ]{tex}\\Rightarrow{/tex}\xa0cosec2A + cot2A + 2 cosec A cot A = m2\xa0\xa0.......(1)Now, LHS\xa0{tex}=\\frac{m^{2}-1}{m^{2}+1}{/tex}{tex}=\\frac{\\ cosec ^{2} A+\\cot ^{2} A+2 \\ cosec A \\cot A-1}{\\ cosec ^{2} A+\\cot ^{2} A+2 \\ cosce A \\cdot \\cot A+1}{/tex}. [ From (1) ]{tex}=\\frac{\\cot ^{2} A+\\cot ^{2} A+2 \\ cosec A \\cdot \\cot A}{\\ cosec ^{2} A+\\ cosec ^{2} A+2 \\ cosec A \\cdot \\cot A}{/tex} [Since, Cosec2A - Cot2A = 1]{tex}=\\frac{2 \\cot ^{2} A+2 \\ cosec A \\cot A}{2 \\ cosec ^{2} A+2 \\ cosec A \\cot A}{/tex}{tex}=\\frac{2 \\cot A(\\cot A+\\ cosec A)}{2 \\ cosec A(\\ cosec A+\\cot A)}{/tex}{tex}=\\frac{\\cot A}{cosec A}{/tex}{tex}=\\frac{\\frac{\\cos A}{\\sin A}}{\\frac{1}{\\sin A}}{/tex}{tex}=\\frac{\\cos A}{\\sin A} \\times \\frac{\\sin A}{1}{/tex}= cos A = RHSHence, Proved. | |
| 26197. |
Cos theta minus sin theta is equal to cos theta + sin theta prove that |
| Answer» | |
| 26198. |
sin A = 5/13 Find other trignomentric ratios |
| Answer» Since sinA = 5/13adjacent side =5 and hypotenuse = 13=> base = sqrt of 169-25implies base = 12therefore cos A = 12/13tanA=5/12 secA=13/12cosecA=13/5cotA=12/5 | |
| 26199. |
Find the x. |
| Answer» | |
| 26200. |
ad is median of triangle abc the bisector of angle adb and angle adc at e and f |
| Answer» What to prove | |