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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.
| 3951. |
What is nth terms. |
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Answer» Number of terms No of terms |
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| 3952. |
Is subject ke theorem kisne banaye |
| Answer» | |
| 3953. |
Equ solve karte same method ka Nam mention karna hai kya. Like \'using substitution method, etc\' |
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Answer» No Yes |
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| 3954. |
2 is a prime no. Or not? |
| Answer» Yes 2 is prime no. | |
| 3955. |
why can\'t I see the CBSE timetable on the Web? and where can I check it |
| Answer» Cbse. nice. in | |
| 3956. |
Value of x if 157 + 42 = 53x |
| Answer» | |
| 3957. |
If the mean if 1,2,3.....,x is 6x/11,find the value of x??? |
| Answer» Find n and then take sum / n = avg and find x | |
| 3958. |
Maths practical |
| Answer» No | |
| 3959. |
List Tow challenges of political parties.. |
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Answer» Corruption Dynastic succession Lack of internal democracy Lack of internal democracy Money and muscle power Give one more point if you know. Competition from other parties and growing demands of the ublic. |
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| 3960. |
Show that 3+/2 is an irrational number |
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Answer» Hii.aapke passNCERT hoge to usme dekho..usme h hiw to show irrational no.similar eg. Yes it is a irrational number |
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| 3961. |
find the value of p for which the equation is px2-px+1=0 |
| Answer» | |
| 3962. |
-2++2 |
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Answer» -2++2: -2+2: 0 Ans 0 -2++2-2+2O Itna bhi nhi pata 0 0 |
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| 3963. |
In an isosceles triangle ABC, If AC=BC and AB^2=2AC^2 then find angle C |
| Answer» 90° | |
| 3964. |
State the fundamenatl theorem of arithmatic |
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Answer» Yup No prblm Thankuuuuuu Weeeeeellllcome Thankuuuuuuuuu Pge 8 pr haa Haa wait batata hu Mtlb theory m h khi Real nos. Me hai 1st ch Kha Ncert me dekh lo .. yaad nhi rehta |
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| 3965. |
Solve the quadratic equation(x-1)²-5(x-1)-6=0 |
| Answer» x=7 and 0 | |
| 3966. |
How to find square Root |
| Answer» By long division merhod | |
| 3967. |
dogs nylon so kjc |
| Answer» ?? | |
| 3968. |
How to use componendo and divinendo |
| Answer» Not necessary in 10th class | |
| 3969. |
A hemispherical bowl of internal |
| Answer» Let the height of water in the cylindrical vessel be h cm.Given Internal radius of hemispherical bowl (r1) = 9 cm,Internal radius of the cylindrical bowl (r2) = 6 cmAs the content of hemispherical bowl has been put into the cylindrical vessel.{tex}\\Rightarrow{/tex}\xa0Volume of hemispherical bowl = Volume of cylindrical bowl{tex}\\Rightarrow \\frac { 2 } { 3 } \\pi r _ { 1 } ^ { 3 } = \\pi r _ { 2 } ^ { 2 } h{/tex}{tex}\\Rightarrow h = \\frac { 2 } { 3 } \\times \\left( \\frac { r _ { 1 } ^ { 3 } } { r _ { 2 } ^ { 2 } } \\right){/tex}{tex}\\Rightarrow h = \\frac { 2 } { 3 } \\times \\left( \\frac { 9 ^ { 3 } } { 6 ^ { 2 } } \\right){/tex}{tex}\\Rightarrow h = \\frac { 2 } { 3 } \\times \\frac { 729 } { 36 }{/tex}{tex}\\Rightarrow{/tex}\xa0h = 13.5 cmThus, the height of the water in the cylindrical vessel is 13.5 cm. | |
| 3970. |
Trigonometry exercise 8.4 Q5.5 |
| Answer» | |
| 3971. |
Maths marking scheme 2018 |
| Answer» | |
| 3972. |
X+1/x-1+x-1/x+1=5/6 |
| Answer» 17/7i | |
| 3973. |
Find the least positive integer which is divisible by first 5 natural numbers |
| Answer» | |
| 3974. |
The base of an isosceles triangle is 12cm and its perimeter is 32 cn find its area |
| Answer» 48cm sq. | |
| 3975. |
Define construction |
| Answer» | |
| 3976. |
Prove that the distance of the point (A cos x , A sin x) from the point (0,0) is independent of x |
| Answer» | |
| 3977. |
If so 4n-n find the 10th term |
| Answer» | |
| 3978. |
What is the formula of mode? |
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Answer» Mode= L+[f1-f0/2f1-f0-f2]×h Thanks l=(f1+f0)÷(2(f1+f0+f2) |
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| 3979. |
Hiii guys!Exam ki preparation kaisa chal raha hai |
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Answer» Bhai tu happy rah Achha |
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| 3980. |
Ex 11.2 |
| Answer» | |
| 3981. |
Exercise 3.2 2 part of 1 question |
| Answer» | |
| 3982. |
find the value of a so that the point (3 ,a) lies on the line represented by 2x-3y=5. |
| Answer» Since the point (3, a) lies on the line 2x - 3y = 5, we have2\xa0{tex}\\times{/tex}\xa03 - 3\xa0{tex}\\times{/tex}\xa0a = 5{tex}\\Rightarrow{/tex}\xa06 - 3a = 5{tex}\\Rightarrow{/tex}\xa03a = 1{tex}\\Rightarrow{/tex}\xa0{tex}a = \\frac { 1 } { 3 }{/tex} | |
| 3983. |
all formule of chapter 13 |
| Answer» | |
| 3984. |
What is meant by cosA |
| Answer» | |
| 3985. |
Sec A(plus) Tan A=p.Find Cosec A |
| Answer» | |
| 3986. |
TanA +secA-1÷tanA-secA+1= 1+sinA÷cosA |
| 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 | |
| 3987. |
Why can\'t I solve maths |
| Answer» Because ur not having interest in it Hmm.....?? | |
| 3988. |
Sec A |
| Answer» | |
| 3989. |
Solve for x:√3xsq-2√2x-2√3=0 |
| Answer» | |
| 3990. |
Sum of first P terms of an ap is a p square + b p find its common difference |
| Answer» We are given that sum\xa0of first p terms of this APSp= ap2 + bpLet the first term= x and common difference = d{tex}{\\mathrm S}_1=\\mathrm a(1)^2+\\mathrm b(1)=\\mathrm a+\\mathrm b{/tex}So first term x = a+bNow S2 = a(2)2 + b(2)S2 = 4a + 2b,{tex}\\mathrm{So}\\;{\\mathrm s}_2={\\mathrm a}_1+{\\mathrm a}_2=\\mathrm x+\\mathrm x+\\mathrm d=4\\mathrm a+2\\mathrm b{/tex}2x + d = 4a+2b2(a+b)+d = 4a+2b2a+2b+d=4a+2bd=4a+2b-2a-2bd = 2aTherefore, the common difference of the given AP is 2a. | |
| 3991. |
The decimal expansion of 9/125 will terminate how many places of decimals |
| Answer» 3 | |
| 3992. |
The volume of a hemisphere is 2425 1/2 cm .find the total surface area. |
| Answer» As, volume of hemisphere = 2425{tex}\\frac { 1 } { 2 } c m ^ { 3 }{/tex}{tex}\\Rightarrow \\frac { 2 } { 3 } \\pi r ^ { 3 } = 2425 \\frac { 1 } { 2 }{/tex}{tex}\\Rightarrow \\frac { 2 } { 3 } \\times \\frac { 22 } { 7 } r ^ { 3 } = \\frac { 4851 } { 2 }{/tex}{tex}\\Rightarrow r ^ { 3 } = \\frac { 4851 \\times 3 \\times 7 } { 2 \\times 2 \\times 22 }{/tex}{tex}\\Rightarrow r ^ { 3 } = \\frac { 441 \\times 3 \\times 7 } { 2 \\times 2 \\times 2 }{/tex}{tex}\\Rightarrow r ^ { 3 } = \\frac { 21 ^ { 3 } } { 2 ^ { 3 } }{/tex}{tex}\\Rightarrow r = \\frac { 21 } { 2 } c m{/tex}So, the curved surface area of the hemisphere ={tex}2 \\pi r ^ { 2 }{/tex}{tex}= 2 \\times \\frac { 22 } { 7 } \\times \\frac { 21 } { 2 } \\times \\frac { 21 } { 2 } = 693 \\mathrm { cm } ^ { 2 }{/tex} | |
| 3993. |
Sin (456) |
| Answer» | |
| 3994. |
If sec A=x+1÷4x,then prove that secA+tanA=2x or 1÷2x |
| Answer» Use the identity sec^2 theta = 1 + tan^2 to find tan theta and then put the values in the given equation..u will find the answer..? | |
| 3995. |
Ouadratic equation |
| Answer» | |
| 3996. |
Sample paper questions for 2018 |
| Answer» Check sample papers here :\xa0https://mycbseguide.com/cbse-sample-papers.html | |
| 3997. |
2÷r |
| Answer» | |
| 3998. |
Maths ncert pg 140 question 4 |
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Answer» Wlcm! I like maths Okk thnx angle 1=angle 2 given so sides PR=PQ Than PA put at PR in given question Vo to Pata Hai AA similarity criteria use or sakte hai Questions tell Easy question hai Rd hai Pura answer ni chahiye tarika Bata do koi Rs hai |
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| 3999. |
If cos is equal to 2 upon 5, find the value of 4 + 4 tan square A. |
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Answer» 25 Ans. is 25. |
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| 4000. |
If cos A=2/5 , find the value of 4+4 tan square A. |
| Answer» Ans.25 | |