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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.
| 22551. |
1÷x+3+2÷2x+1=3÷x+4 |
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| 22552. |
Prove that root 3+ roor 5 is irrational |
| Answer» Prove root 3 separately and root 5 separate | |
| 22553. |
Prove that 3 - 2 root 5 is an irrational number |
| Answer» Refer exercise 1.3, 2 problem | |
| 22554. |
Cos A minus Sin A + 1 upon Cos A + Sin A minus 1 is equal to cosec A + cot a |
| Answer» Refer exercise 8.4 ,5 main 5problem | |
| 22555. |
Derivation of formula = volume of frustum of cone |
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| 22556. |
Divide the polynomial 3x + 2x - 2 by 4x-1 |
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| 22557. |
Hfghff |
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Answer» Bol be Ha bol Anyone online |
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| 22558. |
If the points1,x,5,2 |
| Answer» Incomplete question Miss Babita ji❌ | |
| 22559. |
If x+a is factor of 2x2+2ax+5x+10, find a |
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Answer» x + a is factor x = -a2x2 + 2ax + 5x + 10 =02(-a)2 + 2 a(-a) + 5(-a) + 10 = 02a2 - 2a2 - 5a + 10 = 0- 5a + 10 = 0- 5a = -10a = 10/5a = 2 1 |
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| 22560. |
2sin€- 3 |
| Answer» We know that, Sin30° =(1/2), Cos45°=(1/√2), tan60°=√3, putting these values in the given expression, we get :-{tex}2{\\sin ^2}30^\\circ - 3{\\cos ^2}45^\\circ + {\\tan ^2}60^\\circ {/tex}{tex} = 2{\\left( {\\frac{1}{2}} \\right)^2} - 3\\left( {\\frac{1}{{\\sqrt 2 }}} \\right)^2+ {\\left( {\\sqrt 3 } \\right)^2}{/tex}{tex} = \\frac{2}{4} - \\frac{3}{2} + 3{/tex}{tex} = \\frac{{2 - 6 + 12}}{4}{/tex}{tex} = \\frac{8}{4} = 2{/tex} | |
| 22561. |
If alpha and beta are zeros of polynomial ax²+bx+c find alpha²-beta² |
| Answer» a²-b²=(a+b)²-2b²-2ab | |
| 22562. |
Prove that sin 4 theta + cos 4 theta upon 1 minus 2 sin square theta cos squared theta equal to 1 |
| Answer» Sin4°+cos4°÷1-2sin2°cos2°={sin2°}2+{cos2°}2÷1-2sin2°cos2° = (sin2°+cos2°)2-2(sin2°cos2°)÷1-2sin2°cos2°=1^2-2(sin2°cos2°)÷1-2sin^2°cos^2°=1. a^2+b^2=(a+b)^2-2ab | |
| 22563. |
Qudratic equation question no.9 ex-4.3 |
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| 22564. |
If nth term of an AP is ( 2n+1), what is the sum of its first three terms? |
| Answer» 15............. | |
| 22565. |
Tan theta + sec theta is equal to P find the value of cosec theta |
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| 22566. |
Find 8th term from the end of the AP 7,10,13..184 |
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Answer» The required A.P is 184, ....,13, 10, 7a=184 d=-3 n=8an=a+(n-1)da8=184+(8-1)-3 =184+7×-3 =184-21 =163 Shaayad 28 Nahi 140 |
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| 22567. |
Ok Bye we will meet at 9oclock at night take care |
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| 22568. |
I hve one more doubt..if cos theta +sin theta=root2costheta,prove:costheta-sin theta=root2sin theta |
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| 22569. |
(sin theta/1+cos theta )+(1+cos theta/sin theta)=2cosec teta...prove |
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Answer» Ypp Yesss Boy Na gi Zeeshan aap girl ho By cross multiple(Sin theta)² +(1+cos theta)²/sin theta (1+ cos thetha)=(Sin²+1+cos² +2cos)/(sin(1+cos){2(1+cos)}/{sin(1+cos)} =2/sin=2cossec |
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| 22570. |
Hii nd good morning everyone ?☺ |
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Answer» Lgta hai aap offline hai prachi Hii prachi aap kya kr rhe ho gud mng☺ Good morning my dear friend You too Good morning and have a nice day ?? Good morning |
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| 22571. |
Is anyone have any idea about the math paper???simple or tuff comparitively last year paper |
| Answer» Not tuff nor simple . It\'s medium | |
| 22572. |
By frnds gd night........???? |
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Answer» Kya hua Sahoo ji neend aa rhi hai Good ? sahooooo jiiiiii |
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| 22573. |
Sum of squares of two consecutive natural number is 4,2,1 find nos solve these |
| Answer» Sqare of two natural no. Hai lekin 4,2,1kaise aa gya | |
| 22574. |
Explain the theorems of chapter 6 |
| Answer» Kon sa therom | |
| 22575. |
Prove areas theorem of ch 6 |
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| 22576. |
Cos4A /sin2B+cos2B /sin4A=1 Prove it Where 4and2 are their powers |
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| 22577. |
For what value of k will k+9,2k-1and2k+7are the consecutive terms of an A.P? |
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Answer» Hii any one is online K is 18 K=18 2k-1-k-9=2k+7-2k+1 K-10=8 K=18 |
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| 22578. |
Find the value of k ,for which one root of the quadratic equation kxsquare- 14x +8 =0 is 2 |
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Answer» Put x =2 then , K(2)^2-14(2)+8=0 4k-28+8=0 K=5 The value of k is 5 K=5 |
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| 22579. |
find zeros of polynamial xdegree3_5xdeg2_2x+24 product of two zeroes is 12 |
| Answer» U have ab =12Now aby= -(d)÷a12y= -(24)/1 = -24y=-2 = x+2 will be factor of f(x)On division u will have quotient x2-7x+12Factorisation will give zeroes -2 , 3,and 4Thats ur answer | |
| 22580. |
Is x=2,y=3 a solution of linear equation 2x + 3y - 13 |
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Answer» x = 2 , y = 32x + 3y - 13= 2(2) + 3(3) - 13= 4 + 9 - 13= 13 - 13= 0 Yes 0 is answer |
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| 22581. |
Show that any positive odd integer is of form 6q +1,or,6q+3,or,6q+5,where q is some integer |
| Answer» Let a be any positive integer and b = 6∴ by Euclid’s division lemmaa = bq + r, 0≤ r and q be any integer q ≥ 0∴ a = 6q + r,where, r = 0, 1, 2, 3, 4, 5If a is even then then remainder by division of 6 is 0,2 or 4Hence r=0,2,or 4or A is of form 6q,6q+2,6q+4As, a = 6q = 2(3q), ora = 6q + 2 = 2(3q + 1), ora = 6q + 4 = 2(3q + 2).If these 3 cases a is an even integer.but if the remainder is 1,3 or 5 then r=1,3 or 5or A is of form 6q+1,,6q+3 or,6q+5Case 1:a = 6q + 1 = 2(3q) + 1 = 2n + 1,\xa0Case 2: a = 6q + 3 = 6q + 2 + 1,= 2(3q + 1) + 1 = 2n + 1,\xa0Case 3: a = 6q + 5 = 6q + 4 + 1= 2(3q + 2) + 1 = 2n + 1This shows that odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. | |
| 22582. |
Find the hcf of 65 and 117 express in the form of 65m+117n |
| Answer» Euclid\'s Division Lemma :-a = bq +r117 > 65117 = 65 × 1 + 52 ----> [ 2 ]65 = 52 x 1 + 13 -----> [1]52 = 13 x 4 + 0HCF = 1313 = 65m + 117nFrom [ 1] ,13 = 65 - 52 x 1From [2] ,52 = 117 - 65 x 1 ----> [3]Hence ,13 = 65 - [ 117 - 65 x 1 ] ------> from [3]= 65 x 2 - 117= 65 x 2 + 117 x [-1 ]m = 2n = -1 | |
| 22583. |
Find the value of k for which the points (3k-1,k-2), (k,k-7) , and (k-1,-k-2) are collinear. |
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Answer» Thanks bhai Equation will be like this 4k^2 -12k+0=0 then taking 4k as common then 4k(k-3)=0 And u will get the answer as mentioned Right answer is K=0 and K = 3 2k^2-6k-5=0,meri ye equation ban rhi hai , but k ki value nhi aa rhi Answer KYa aa rha h mera to koi square = 3k aa rha h Plz solve this problem |
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| 22584. |
Solve the equation(x/x+1)+(x+1/x)=34/15 |
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Answer» First we take LCM and the answer we will obtain is...xsq.+(x+1)sq./(x+1)x= 34/15. Then opening the brackets...2xsq.+2x+1/xsq.+x= 34/15. Now cross multiplication...30xsq.+30x+15= 34xsq.+34x. Now sloving this equation we get the values of x=-5/2 and 3/2 2x^2 +2x+1= 34x^2+34x/15Now 4x^2 +4x-15=0 (4x^2-10x)+(6x-15)=0 Taking common 2x(2x-5)+3(2x-5)=0 (2x+3)(2x-5)=0 X= -3/2 and 5/2. |
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| 22585. |
Thnks |
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Answer» But from my point of view trigonometry too Triangles in my opinion Which is the most complicated chapter of ncert? My pleasure no problem Bt kisliye |
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| 22586. |
When ab+cd =6 the wx +yz =1106 |
| Answer» Hey sulakh complete ur question | |
| 22587. |
Is anyone there to answer by below given question? |
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Answer» Sorry but it\'s not. Answered already |
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| 22588. |
After how many places of decimals will the decimal expansion of 23457/2³x5⁴ |
| Answer» Upto 4 places | |
| 22589. |
How can I prepare for final examination |
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Answer» My mam gave me the marks list of every chapter:Real no: 6mPolynomials: 3mLinear eq: 5mQuadratic eq: 5mAP: 7mTriangles: 8mCoordinate: 6mIntraduction to trignometry: 8mApplications: 4mCircles: 3mConst: 4mAreas related: 3mSurface areas: 7mStatistics: 7mProbability: 4m it depends if u hv English or social then first go through the chp n then learn one answer if cant remember then learn n write in rough By reading |
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| 22590. |
Show that any positive odd integer is the form 6q+1, or 6q+3 ,or 6q+5 , where q is some integer |
| Answer» Let a positive odd integer be ,a ,and , b =6 6q +1, 6q +3 ,6q +5 a= bq + ra = 6q + ra = 6q + 1 (r =1, 3, 5)a = 6q + 3a = 6q + 5. Proved | |
| 22591. |
Give all the formulae of trigonometry |
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| 22592. |
Sec + tan =p find cosec |
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Answer» (P^2-1)(p^2+1) Kya question itna hi hai . |
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| 22593. |
Find the middle term of the AP 6,13,20.....216 |
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Answer» 111 Riya answer is correct bhavya ne 16th term likha hai or tumne us term ko nikala hai both answers are correct but bhavya it should be 16th term..... Bhavya apna solution dijiye 16 Middle term of an AP is 111 an=a+(n-1)d216=6+(n-1)7216-6=(n-1)7210/7=n-130=n-131=nMiddle term 31+1/232 is the middle term |
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| 22594. |
Find find the roots of equation in root 2 X square + 9 is equal to 9 |
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Answer» 2x2 +9 = 92x2 = 9 -92x2 = 0x 2 = 0x = 0 0 Roots are 0 |
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| 22595. |
After how many decimal places will the decimal expansion of ଶଷଶర×ହterminate? |
| Answer» Hey are yu from odisha | |
| 22596. |
How Can I Be Good At Maths?...I dont Know Anything...Suggest some Chapter To Start First. |
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Answer» Phele un chapters ko padho jo easy h aur jyada marks ke bhi Hii Sekhar |
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| 22597. |
Mr kumar and mr sharma togetherdonated20 |
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| 22598. |
Mr kumar |
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| 22599. |
When our class 10 date sheet |
| Answer» \tDateSubjectsThursday, March 7Mathematics - 041Wednesday, March 13Science - Theory - 086\xa0 Science W/O Practical - 090Tuesday, March 19Hindi Course-A - 002\xa0 Hindi Course-B - 085Saturday, March 23English Comm. - 101\xa0 English LNG & LIT - 184Friday, March 29Social Science - 087\t | |
| 22600. |
Merry Christmas to all my friends ????????????? good ? and sweet dreams to all my friends?????? |
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Answer» Merry Christmas riahabh ji Merry Christmas Rishabh ........ Same. 2 u |
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