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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.
| 28201. |
sec theta +tan theta =p |
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| 28202. |
Which sample paper is best for 10 only 11th student please |
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Answer» Haa funny Geeta or bible any one will be best |
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| 28203. |
Cot\'4A-1=COSEC\'4A-2cosec2A |
| Answer» cosec^4A - 2cosec^2A =cosec^2A( cosec^2A - 2 ) =cosec^2A( cosec^2A - 1 - 1 ). =cosec^2A( cot^2A - 1 ) =( cot^2A + 1 )( cot^2A - 1)=cot^4A - 1 | |
| 28204. |
How to find the roots of the quadratic equation bby factorisation |
| Answer» By splitting their mid term!! | |
| 28205. |
A student prepares a poster on "save water" on a circular sheet |
| Answer» Ha fir | |
| 28206. |
What is linear equations mean |
| Answer» Equation having a straight line graph is called linear equation. | |
| 28207. |
If perimeter of circle and area of circle are equal then find the radius. |
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Answer» Perimeter of circle= area of circle2πr=πr×r2=r 2. Units |
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| 28208. |
Pak ka aek f 16 tod dia hamare desh nae .. PROUD TO BE INDIAN |
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Answer» Jai hind..... Jai hind..... ???????? Jai hind Ji hind ...???? |
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| 28209. |
Find the 10th term of the AP :2,7,12.... |
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Answer» 47 47 |
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| 28210. |
Explain me ex 13.3 question no 8 |
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Answer» Take the canal as a cuboid of dimensions:6,1.5 and convert 10km/hr into meter/minute we will get=1500.. NOW ACCORDING TO QUESTION 8cm STANDING WATER IS REQ. SO=1500*30*100/8(WATER FLOW FOR 30 MINUTE SO *30 AND 1M=100CM) SO WE WILL GET REWURE ANSWER THAT IS 525000m^2 |
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| 28211. |
What is zero? |
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Answer» Zero of any polynomial f(x) is the value of x for which f(x)=0e.g. for f(x)=x2-3x+2x=1 and 2 are zerothen f(1)=0 and f(2)=0 Zero is the only no. which has no sign ,neither +ve nor -ve Zero is the middle most no. of all integers. Friend zero ke chakkar mein na pado apni study karo Zero is 0 |
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| 28212. |
Exam pad is allow to carry in examination hall?????? |
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Answer» Yes but transparent one. Nope Yes Na |
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| 28213. |
NCERT 6.1 question number 5 |
| Answer» There is only up to Q no 3. .... | |
| 28214. |
In the adjoining figure angle B equal to 90 theta, angle BAC equal to theta |
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| 28215. |
X-y=3 |
| Answer» What is the que | |
| 28216. |
1/x+4- 1/x-5= 1/30 find the root |
| Answer» x=-10,1.... | |
| 28217. |
(CosΦ+secΦ)²+(sinΦ+cosecΦ)²=7+tan²Φ+cot²Φ |
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Answer» It is in ncert so easy Dono ko pahle a+b ke square par thod do answer aa jayega |
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| 28218. |
Ex. 13.1 que. 5 |
| Answer» 1/4l^2(π-24) it can be found by adding TSA of cube and hemisphere | |
| 28219. |
If sec + tan = p then find the value of cosec |
| Answer» Cosec (1+p^2) (1-p^2) | |
| 28220. |
Consttruct a more than comulative frequency distribution table ci 50-60 60-70 |
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Answer» Aur hoga ti sirf ye banega, more than 50 ,more than 60 and so on.. How to construct here......??? |
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| 28221. |
Important topics as per latest cbse syllabus |
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| 28222. |
Prove that the length of two tangents drawn from an external point to a circle are equal |
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Answer» Given = QP and PR are the external tangentProve = QP = PRConstruction = joint OP , OR &OQproof = ?POQ & ?OPR OQ =OR (radii) OP =OP (common) angle Q = angle R (90) ?POQ congruant ?OPR (R.H.S) PQ = OR (C.P.C.T ) THEOREM 10.2 (very IMP) It\'s theorem 10.2 in ncert |
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| 28223. |
how many terms of Ap 9,17,25 must be taken to give a sum of 636? |
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Answer» n=12 4n square +5n-636Split this |
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| 28224. |
(A+b+c) ×2 |
| Answer» 2a + 2b + 2c | |
| 28225. |
Solv for x :1/x+1 +2/x+2=4/x+4 |
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Answer» 2(1-√3) and 2(1+√3) First take Lcm then cross multiply them u will get the answer Put X=-1,-2,-4 |
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| 28226. |
For what value of p are 2p+1, 13 and 5p-3 three consecutive term of an Ap |
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Answer» to find consecutive term of an Ap we have a formula (A+C)=2Bhere we have a=(2p+1) b=13 c=(5p-3)by applying this in the formula.a+c=2b2p+1+5p-3=2×137p-2=267p=28p=28/7 p=4 .. thetefore value of p is 4 Terms-2p+1,13,5p-313-2p-1=5p-3-1312-2p=5p-1612+16=5p+2p28=7pp=4 Ans |
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| 28227. |
Which term of an Ap 21,18,15..........…..............is 0 |
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Answer» 21,18,15,............,na=21d=18-21 =-3an=0an=a+(n-1)d0=21+(n-1)-3-21=(n-1)-37=n-1n=88th term of an Appointment is 0 .Ans An=0 apply the formula of an. a+(n-1)d =0. 21+(n-1) (--3) = 0. 21-3n +3 =0 . 24-3n =0 -3n =-24 . n= 8. So the 8th term of the ap will be 0. |
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| 28228. |
What are the possible values of remainder r when a positive integer a is divided by 3 |
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Answer» Possible values r will be 0, 1,2 Possible values of r is 0,1,2. Ans |
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| 28229. |
(1+ Cos A + tan A)(sin A-cos A )= |
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| 28230. |
The factorisation of 3x²-15x+28 |
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Answer» L.C.M. of 3×28=74;3 742 282 14 7Therefore, 3x^2-15x+28 =>3x^2-12x-7x+28 =>3x(x-4)-7(x-4) =>(x-4)(3x-7)Then, (x-4)=0 and (3x-7)=0Theefore, x=4 and x=7/3Now,4 and 7/3 are the roots of the given equation 3x^2-15x+28 If we find discriminant it will be( -111) which means that it does not have real roots No real roots |
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| 28231. |
Saloni ji Oswal me dekho mil jayega nai to teachers se puchna |
| Answer» And Oswal nahi h mere pas... | |
| 28232. |
sorry 1 ki jagah PE 10 lika Gaya mind mat karna Anushka ji |
| Answer» | |
| 28233. |
Anushka ek question hai batao cosA=2/5 find 4plustanA value only for Anushka |
| Answer» Question sahi to hai na... Khair iska answer to 11/5 aa rha hai shayad | |
| 28234. |
If mtth term of a.p. is 1/n and nth therm finf sm of mn ted |
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| 28235. |
padlo 7 Ko paper hai |
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Answer» Marathi ka paper 5 ko hai bhai |
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| 28236. |
CotA+cosecA -1/cotA -cosecA +1=1+cosA / sinA |
| Answer» First convert cotA + cosecA -1/cotA-cosecA+1 in sinA, cosA and 1 Then rationalise it u will get the answer | |
| 28237. |
If the (1,5),(p,1),and (4,11)and power. Find the value of P |
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| 28238. |
cos A — sinA + 1 ÷ cosA + sinA — 1 = cosecA + cotA. Prove using cosec square A = 1+cot square A |
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Answer» =[CosA—SinA +1/CosA+Sin—1]×[CosA +SinA +1/CosA +SinA +1]=[Cos²A+CosA.SinA+CosA—CosA.SinA—Sin²A—SinA+CosA+SinA+1]/(CosA+SinA)²—(1)²=[Cos²A+2CosA+1—Sin²A]/[Cos²A+Sin²A+2SinACosA—1]----[:CosA.SinA—CosA.SinA=0, SinA—SinA=0]----=[2Cos²A+2CosA]/[1—1+2SinA.CosA]----[:1—Sin²A=Cos²A and Cos²A+Sin²A=1]=[2CosA(CosA+1)/[2SinA.CosA]=CosA+1/SinA--------[2CosA/2CosA=1]=CosA/SinA+1/SinA=CotA+CosecA, Hence Prooved It\'s N.C.E.R.T ques |
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| 28239. |
1+sin^2A = 3sinAcosA then prove that tanA=1 or 1/2 |
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| 28240. |
Which term of A.P. 92, 88, 84, 80 ................. is 0 ? |
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| 28241. |
Triangle ex 6.5 15 and 16 question |
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| 28242. |
Find the quadaratic poly. whose zerros are (5+2√3)&(5-2√3) |
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Answer» How X^2_ 10x+13 |
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| 28243. |
Find the distances of a point P(x,y) from the origin |
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Answer» Thanks √x*2+y*2.. Think so |
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| 28244. |
The mean of 6,6+2x,5 and 8+3x is 20.find the value of x |
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Answer» Bhai formula nhi pata kya Mean = Sum of all obs / Total No of obs. Puja ji pura solve karo x=11 |
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| 28245. |
1/(cosec x+cot x)-1/sin x=1/sin x-1/(cosec x-cot x) |
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| 28246. |
1+Cos a°+sin a° \\ 1 + cos a° -sin a° |
| Answer» Answer is 1. We can do this using identity | |
| 28247. |
Find the zeroes of the following polynomial 5√5x*x+30x+8√5 |
| Answer» -2root5÷5......n....-4÷root5........actually i cn use the proper sym for root so i \'d spell it....hope it will help uhhhhh.... | |
| 28248. |
Find the root of 4x^2+3x+5 =0 by the method of completing the square. |
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Answer» Bro competing square is easy but my way of doing is little bit diff. Is answr=3+-√11/4 |
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| 28249. |
The 4 th term of an AP is 0.prove that the 25th term of an AP is 3 times its 11th term. |
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Answer» a+3d=0 (eq..1) given, a=-3d.(eq 2)Now ,A+24d=3(a+10d) to prove.. Substitute value of a, -3d+24d=-9d+30 , 21d=21d .. Therefore proved.. NOT SURE SO PLEASE ANY TEACHER CHECK MY WAY. Anusree r u a malayali it has been given that a4=0 => a+3d=0 => a= -3d ------- (1)a11= a+10d = -3d+10d = 7d [ from (1)]a25 = a+24d = -3d+24d = 21d [ from (1)]therefore, comparing 7d and 21d we get, 21d = 3 ( 7d ) i.e. a25 = 3(a11) Hence, proved.Hope it helps. |
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| 28250. |
X^ 2 +2x + k is a factor of 2x^4+ x- 14x^2 5x 6 |
| Answer» Your question is not clear ....Please write it clearly....?? | |