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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.
| 30151. |
If the 7times the 7th term of an ap its equal to 11times the 11th term of an ap then find 18th term |
| Answer» 7th term = (a+6d) 11th term = (a+10d)(a+6d)7 since 7th term of the ap is 7 times and the same for 11th term(a+6d)7=(A+10d)11 (given)7a + 42d =11a +110d (on solving)7a - 11a = 110d - 42d-4a = -68da= -17dwe know 18th term = (a+17d)................................. (i)now we substitute the value of a in (i) which emplies, -17d + 17d = 0 | |
| 30152. |
How many questions are asking from chapter-Triangle in CBSE Board examination? |
| Answer» Check blue print | |
| 30153. |
Explain thelss theoram |
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Answer» If a line is drawn parallel to one side of triangle, to intersect the other side in distinct point ,the other tow sides are divided in same ratio Visit our ncert book...evrthng is nt possible to weite here Take out your rs aggarwal math book |
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| 30154. |
a=7p=6q+5=5r+4=4s+3=3t+2=2u+1,a=? |
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| 30155. |
If the probability of winning a game is 0.995 then what will be probability of losing a game?? |
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Answer» .005 0.005 Subtract 0.995 from 1 that will be the probability of losing game 0.5 0.995 is ratio\xa0 |
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| 30156. |
Define fundamental tharom ofAP |
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Answer» Fogg chal rha hai Kya chal rha hai |
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| 30157. |
Find the area bounded by the line x+y=10 and both the co-ordinate axes. |
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| 30158. |
Show that numbers 8n can never end with digit 0 for any natural number n. |
| Answer» If any numbr ends with 0, then it must be divisible by 10 .Hence it should have 5 and 2 as a factors. Now the factorization of 8n\xa0are:{tex}\\therefore 8 ^ { n } = ( 2 \\times 2 \\times 2 ) ^ { n } = 2 ^ { n } \\times 2 ^ { n } \\times 2 ^ { n }{/tex}Hence 5 is not in the factors of 8n.From the fundamental theorem of arithmetic, we know that the prime factorization of every composite number is unique.So it is clear that 8n is not divisible by 10.{tex}\\therefore{/tex}\xa08n\xa0can never end\xa0with 0. | |
| 30159. |
SecA-tanA/secA+tanA=cos^2A/(1+sinA)^2 |
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| 30160. |
Write maximum number of zeros for p (x)=x3+3×2-3×+1 |
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| 30161. |
Hii tamana and amit are you here |
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Answer» Hi friend I Himangshu Maji Ok Na Are mai hoon h Hii Hii |
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| 30162. |
Who know me???? |
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Answer» Hello Hii Aarushi ☺ Complicated life What is your name ? Nhi mai aapko nhi janti R you don\'t know me Hmm ??? |
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| 30163. |
Please help me... |
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Answer» Is someone solving that question I have posted it already yesterday Hmm Is someone here to solve my question ?? My math question Kya hua |
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| 30164. |
Hii amit |
| Answer» Hiiiiiii | |
| 30165. |
For what value of k will k+9, 2k-1, and 2k+7 are the consecutive terms of an AP? |
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Answer» 18 Use d=a2-a1= a3-a2 . Hlo |
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| 30166. |
Define "identity"? |
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| 30167. |
Exercise 5.2 |
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Answer» Question Which auestion |
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| 30168. |
Anyone want to be my friend |
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Answer» OK but ik kam Karo Thanks jashan I am ur friend Thanks ss Samaaya friend ka mtlb dost hota hai Riddhi ha Kyaa baat h bro continue........ What?? |
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| 30169. |
4root 16 -6 *3root343 +18×5root 243 -root 196Simplyfily |
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| 30170. |
What is the relationship between hcf and lcm |
| Answer» Hcf is a factor of lcm | |
| 30171. |
Tan square theta+cot square theta is equal to sec square theta cosec square theta |
| Answer» Lhs = sin sq/cos sq +cos sq/sin sq. Then take a l.c.m, then it comes sin4(power)+ cos4/ Cos2 . Sin2 power are equal in numerator then (sin2 +cos2)2/cos2sin2 = (1)sq / cos2 sin2= lhs. Rhs= 1/cos2.sin2 | |
| 30172. |
Identities all |
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Answer» ??? For what |
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| 30173. |
Amit this is a last mess to you bye forever |
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Answer» No no Use laga ki mujhe kuch nhi pata chalega par aisa kuch nhi h Waise usne tumhe pareshan to nhi kiya |
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| 30174. |
2tan30degree by 1+tan square 30 degree.plzz solve this question |
| Answer» 2×1/underoot3 upon 1+1uponunderoot3^2=2 upon underoot3 upon 1+1upon3 = 2upon underoot3 upon 6 upon 3 =2upon underoot 3 ×1upon 2 = underoot 3 value of cos30 | |
| 30175. |
Define aqua regia |
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Answer» 3 parts of HCl and 1 part of HNO3 is called aqua regia They can dissolve all of the metal even gold and platinum. It is a solution of HNO3 and Hcl and has the capability to react with gold etc. It is freshly prepared mixture of concentreted hydrochloric acid and concentreted nitric acid in the ratio of3:1.aqua regia is highly corrosive fuming liquid.......... Able to dissolve gold and platinum. A mixture of 3 part of concentrated hydrochloric acid and 1 part of concentrated nitric acid. |
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| 30176. |
Ram RAM r koi friend h kkk koi bhii |
| Answer» Kyaaa | |
| 30177. |
What is the formula of area of major and minor sector of a circle ? |
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Answer» Doston mein dil mein aata hun , samagh mein nhi .so bye .i can be your friend . Thita /360 py r2 |
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| 30178. |
Agarwal aap toh intelligent ho jaldi smj gyi |
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Answer» Tuuuu Who Agrawal?? |
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| 30179. |
R makha Kya chll ryaaa kon h riddhi or jashan |
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Answer» From bhiar I am riddhi kyo |
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| 30180. |
Hiii aahan pls apni sis sa baat karavna |
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| 30181. |
Anjali are you here |
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| 30182. |
What is equation |
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Answer» Oops this is maths sorry A chemical equation is the symblic represention of a chemical reaction |
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| 30183. |
If sin theta=cos theta,then find 2tan there+cos²theta |
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| 30184. |
x2 +2x+3 |
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| 30185. |
Hi koi tho batt krr |
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| 30186. |
Anjali are you here... |
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Answer» Yrr vo jab Aya tho tho plz yousa Bata Dena ki ma yousa yad kr ra tha Noo |
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| 30187. |
T^n = sin^n + cos^n. to prove :- T^3 - T^5 / T^1 = T^5 - T^7/T^3 |
| Answer» We know,{tex}{T_n} = {\\sin ^n}\\theta + {\\cos ^n}\\theta {/tex}{tex}\\therefore {T_3} = {\\sin ^3}\\theta + {\\cos ^3}\\theta {/tex}{tex}{T_5} = {\\sin ^5}\\theta + {\\cos ^5}\\theta {/tex}{tex}{T_1} = \\sin \\theta + \\cos \\theta {/tex}{tex}{T_7} = {\\sin ^7}\\theta + {\\cos ^7}\\theta {/tex}+ cot{tex}\\theta{/tex}\xa0= m and cosec{tex}\\theta{/tex}\xa0- cot{tex}\\theta{/tex}\xa0= n,LHS\xa0{tex} = \\frac{{{T_3} - {T_5}}}{{{T_1}}}{/tex}{tex} = \\frac{{{{\\sin }^3}\\theta + {{\\cos }^3}\\theta - \\left( {{{\\sin }^5}\\theta + {{\\cos }^5}\\theta } \\right)}}{{\\sin \\theta + \\cos \\theta }}{/tex}{tex} = \\frac{{{{\\sin }^3}\\theta + {{\\cos }^3}\\theta - {{\\sin }^5}\\theta - {{\\cos }^5}\\theta }}{{\\sin \\theta + \\cos \\theta }}{/tex}{tex} = \\frac{{{{\\sin }^3}\\theta - {{\\sin }^5}\\theta + {{\\cos }^3}\\theta - {{\\cos }^5}\\theta }}{{\\sin \\theta + \\cos \\theta }}{/tex}{tex} = \\frac{{{{\\sin }^3}\\theta (1 - {{\\sin }^2}\\theta ) + {{\\cos }^3}\\theta (1 - {{\\cos }^2}\\theta )}}{{\\sin \\theta + \\cos \\theta }}{/tex}{tex} = \\frac{{{{\\sin }^3}\\theta {{\\cos }^2}\\theta + {{\\cos }^3}\\theta {{\\sin }^2}\\theta }}{{\\sin \\theta + \\cos \\theta }}{/tex}\xa0{tex}\\left[ \\begin{gathered} \\because 1 - {\\sin ^2}\\theta = {\\cos ^2}\\theta \\hfill \\\\ 1 - {\\cos ^2}\\theta = {\\sin ^2}\\theta \\hfill \\\\ \\end{gathered} \\right]{/tex}{tex} = \\frac{{{{\\sin }^2}\\theta {{\\cos }^2}\\theta (\\sin \\theta + \\cos \\theta )}}{{(\\sin \\theta + \\cos \\theta )}}{/tex}{tex} = {\\sin ^2}\\theta {\\cos ^2}\\theta {/tex}\xa0RHS\xa0{tex} = \\frac{{{T_5} - {T_7}}}{{{T_3}}}{/tex}{tex} = \\frac{{{{\\sin }^5}\\theta + {{\\cos }^5}\\theta - ({{\\sin }^7}\\theta + {{\\cos }^7}\\theta )}}{{\\left( {{{\\sin }^3}\\theta + {{\\cos }^3}\\theta } \\right)}}{/tex}{tex} = \\frac{{{{\\sin }^5}\\theta + {{\\cos }^5}\\theta - {{\\sin }^7}\\theta - {{\\cos }^7}\\theta }}{{({{\\sin }^3}\\theta + {{\\cos }^3}\\theta )}}{/tex}{tex} = \\frac{{{{\\sin }^5}\\theta - {{\\sin }^7}\\theta + {{\\cos }^5}\\theta - {{\\cos }^7}\\theta }}{{{{\\sin }^3}\\theta + {{\\cos }^3}\\theta }}{/tex}{tex} = \\frac{{{{\\sin }^5}\\theta \\left( {1 - {{\\sin }^2}\\theta } \\right) + {{\\cos }^5}\\theta \\left( {1 - {{\\cos }^2}\\theta } \\right)}}{{{{\\sin }^3}\\theta + {{\\cos }^3}\\theta }}{/tex}{tex} = \\frac{{{{\\sin }^5}\\theta {{\\cos }^2}\\theta + {{\\cos }^5}\\theta {{\\sin }^2}\\theta }}{{{{\\sin }^3}\\theta + {{\\cos }^3}\\theta }}{/tex}{tex} = \\frac{{{{\\sin }^2}\\theta {{\\cos }^2}\\theta \\left( {{{\\sin }^3}\\theta + {{\\cos }^3}\\theta } \\right)}}{{\\left( {{{\\sin }^3}\\theta + {{\\cos }^3}\\theta } \\right)}}{/tex}{tex} = {\\sin ^2}\\theta {\\cos ^2}\\theta {/tex}LHS = RHSHence proved. | |
| 30188. |
From circle |
| Answer» | |
| 30189. |
Sintheta/1-costheta + tantheta/1+costheta = sectheta*cosectheta + cottheta |
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Answer» Yes U will get cottheta +1/costheta*sintheta Add LHS and convert tan theta to sin theta/cos theta Lengthy hai....!! |
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| 30190. |
If m times the mth term of an AP is equal to the nth term show that (m+n)th term of that AP is zero |
| 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. | |
| 30191. |
Monika mera msg k reply kro |
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Answer» ***"*"""**** Bolo No shantanu not on facebook Tum Facebook pe ho What happen? Kya hua shantanu?? |
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| 30192. |
Sin^2A+cos^2A |
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Answer» -(-1) =1 1 |
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| 30193. |
Is there ASS Congruency |
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Answer» Ji nhi No |
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| 30194. |
Molika i luv u so much ???❤❤?❤????❤?????????????????????????????? |
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Answer» Mera shantanu bhaiya Pagal ho gyi leken tujshe tabhi pyar kronga Tera bff Pati Bhaiyaaaaaa |
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| 30195. |
Kl Milta h love u |
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Answer» Nahi shantanu bhaiya Saaiya Tum n.a. galt type kiya h Bhah bhah OK bhaiya |
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| 30196. |
Molina byee |
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Answer» Bhaiya shantanu I luvu meri jaan Molina nahi molika byee bhaiya |
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| 30197. |
Meri darling |
| Answer» Mera bhai | |
| 30198. |
What is the level of maths paper in board class 10 |
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Answer» I think it will be more interesting than the other exams Average 70% from ncert |
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| 30199. |
Hello mam may I know why which book I do maths |
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Answer» Ncert mainly NCERT and RD SHARMA are must Vidya question bank is much better |
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| 30200. |
Prove that 1+ cot^2A/1+cosecA=1/sinA |
| Answer» =Cosec^2A/cosecA (1+cot^2A=cosec^2A)=cosecA=1/sinA | |