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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.
| 16551. |
2019 ka board paper was easy so they made 2020 paper very difficult yes or no |
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Answer» Whats your insta It may be difficult because 2019 board exam finished before the given date and 2020 board classes has more time to cover there syllabus and revise it.But there is one more rule of giving 2 category of paper one is simple and 2nd one is difrficult. |
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| 16552. |
√3sin¢=cos¢ find¢ |
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| 16553. |
if 1+sin^2theta=3sin theta .cos theta than prove that ( tan theta =1or1/2). |
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| 16554. |
identity of a^3+b^3+c^3-3abc |
| Answer» (a+b+c)(a^2+b^2+c^2-ab-bc-ca) | |
| 16555. |
Factories x²+32x-273=0 ? |
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| 16556. |
Sin(B+C/2)=cos(A/2). Prove |
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Answer» A+B+C/2=180/2B+C/2=90=-A/2B+C/2=90-A/2multiplying both sides by sinA we getsin(B+C/2)=sin(90-A/2)sin(B+C/2)=cos(A/2)hence it is proved A+B+C/2=180/2 In triangle abc A+B+C=180dividing both side by 2A/2 +B/2+c/2=90b/2+c/2=90-a/2sin(b+c/2)=sin(90-a/2)sin(b+c/2)=cot(a/2) |
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| 16557. |
Given that √2 is a zero of the cubic polynomial 6x®3 |
| Answer» what is \'\'®\'\' here then i can answer | |
| 16558. |
P(k-1,2) is equidistant from point A (3,k) and B (k,5) find k |
| Answer» This question is from which chapter and which exercise of which book | |
| 16559. |
Is there any change in pattern of 10th final exam of maths |
| Answer» Yes ,in this year there are 20mcq questions | |
| 16560. |
Cosec^2 theta - cot^2 theta = 1 |
| Answer» cosec^2theta-cot^2theta =1lhs1/sin^2theta=cos^2/sin^2theta by taking lcm we get1-cos^2theta /sin^2theta{1-cos^2theta =sin^2theta}so \'sin^2theta/sin^2theta cancelout u got the answer 1 1 proved | |
| 16561. |
Solution of theorem 6.9 |
| Answer» See on this aap in maths | |
| 16562. |
Ratio and proportion for std. 7 |
| Answer» Geometre need which work | |
| 16563. |
What is combustion reaction |
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Answer» A chemical process in which a substance reacts with oxygen to give off heat and light is called combustion. Ys |
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| 16564. |
Difference of parallelogram and rectangles |
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Answer» \tA parallelogram is a quadrilateral with 2 pairs of opposite, equal and parallel sides.\tA rectangle is a quadrilateral with 2 pairs of opposite, equal and parallel sides but also\xa0forms right angles between adjacent sides. Rectangle is also also a parallelogram whose all angles are 90°. But generally paralleogram is a quadrilateral whose opposite sides are parallel. |
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| 16565. |
How to answer questions |
| Answer» You can check solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 16566. |
Chapter 13 ; ex 13. 4 questions no. 3 |
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Answer» Then I can solve Type question and send You can see all the NCERT solutions that u want in this app( My cbseguide) okk?? |
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| 16567. |
Using Euclid division find hcf of 240 and 228 |
| Answer» HCF of 240 and 228 is 12 | |
| 16568. |
Tan^4 20° |
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| 16569. |
(2352968)(50355659)585995 |
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| 16570. |
What are the main topics to revise for maths Olympiad?? |
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| 16571. |
Trigonometry chapter is very hard.so i want to cbse this chapter is remove in the board exam. |
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Answer» Trignometry GoPCat gigDJ dokgx |
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| 16572. |
7.1 P(2,-5), B(2,9) |
| Answer» Let the point of x-axis be P(x, 0)Given A(2, -5) and B(-2, 9) are equidistant from PThat is PA = PBHence PA2\xa0= PB2 → (1)Distance between two points is\xa0{tex}\\sqrt{[(x_2\xa0- x_1)^2\xa0+\xa0(y_2\xa0- y_1)^2]}{/tex}PA =\xa0{tex}\\sqrt{[(2\xa0- x)^2\xa0+\xa0(-5\xa0- 0)^2]}{/tex}PA2\xa0= 4 - 4x +x2\xa0+ 25=\xa0x2\xa0- 4x + 29Similarly,\xa0PB2\xa0=\xa0x2\xa0+ 4x + 85Equation (1) becomesx2\xa0- 4x + 29 =\xa0x2\xa0+ 4x + 85- 8x = 56x = -7Hence the point on x-axis is (-7, 0) | |
| 16573. |
a+b+c=3,ab+bc+ca=671 |
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Answer» Question incomplete Your questions is incomplete |
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| 16574. |
Which of the following Algebraic expressions are polynomial 4x_3 what is the answer |
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Answer» Nnn X=3/4 linear |
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| 16575. |
LM || AB if AL = X-3 AC = 2X BM = X-2 and BC = 2X + 3 FIND THE VALUE OF X. |
| Answer» We have, AL = x - 3, AC = 2x, BM = x - 2 and BC = 2x + 3, and we need to find the value of x.In\xa0{tex}\\Delta{/tex}ABC, we have{tex}L M \\| A B{/tex}{tex}\\therefore \\quad \\frac { A L } { L C } = \\frac { B M } { M C }{/tex}\xa0[By Thaley\'s Theorem]{tex}\\Rightarrow \\quad \\frac { A L } { A C - A L } = \\frac { B M } { B C - B M }{/tex}{tex}\\Rightarrow \\quad \\frac { x - 3 } { 2 x - ( x - 3 ) } = \\frac { x - 2 } { ( 2 x + 3 ) - ( x - 2 ) }{/tex}{tex}\\Rightarrow \\quad \\frac { x - 3 } { x + 3 } = \\frac { x - 2 } { x + 5 }{/tex}{tex} \\Rightarrow{/tex}\xa0(x - 3) (x + 5) = (x - 2) (x + 3){tex} \\Rightarrow{/tex}\xa0x2 + 2x -15 = x2 + x - 6{tex} \\Rightarrow{/tex}\xa0x = 9 | |
| 16576. |
how to do cross multiplication |
| Answer» There is a separate formula for this in NCERT | |
| 16577. |
I cant understand completing square method |
| Answer» Completing square ka matlab hamko us polynomial ko kisi na kisi ke whole square me lana hai jabardasti agar us polynomial me hame kuch add karna pade alaga se to yad rahe ki subtract bhi karna must hai | |
| 16578. |
If Sp=Sq then show that Sp+q = 0 |
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Answer» Yes So Mahi is this from Ap? |
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| 16579. |
I cannot understand substitution method |
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Answer» In substitution method first of all you have to find the value of first variable for example take a as a first variable and B A second variable let the expression be to A + 5 b is equal to 6 Now possible find the value of a is equal to-5 b so put the value of stay in another equation and their you will find the value of B and then put the value of be in this occasion so you will get the value of a subtitute ka arth hota hai ki us chiz ko hatakar usi ke jaisa kuch rakhna padega same yehi kam polynomial me bhi hoga first eqn se x ka maan nikalo aur usi ko second app me ke x ke jagah par first eqn la x rakho aapka y ka man asani se nikal jayega |
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| 16580. |
X2-(/3+1)x+/3=0 |
| Answer» Solved it by completing the square method | |
| 16581. |
Middle term split of x2 - x - 4 |
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Answer» x2-x-4=0x2-(2-2)x-4=0x2-2x+2x-4=0x(x-2)+2(x-2)=0(x+2)(x-2)=0x-2=0x=2. ANSx+2=0x=-2. NEGLECTED VALUE x2-2x+2x-4x(x-2)+2(x-2)(x-2)(x+2) Madhar chod |
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| 16582. |
What is euclid division lamma |
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Answer» For a pair of given positive integers ‘a’ and ‘b’, there exist unique integers ‘q’ and ‘r’ such thata=bq+r , where 0≤r |
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| 16583. |
Is root 2 ,root 8,root 18 ,root32 an ap? If yes then find its next two terms |
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Answer» Yes it is an ap it common difference is √2. its next terms are a+ 4d, a+5d i.e.√2+4*√2=5√2=√50 and √2+5*√2=6√2= √72 No its not in AP as the difference is not same. |
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| 16584. |
new. current. syllabus |
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Answer» Check syllabus here :\xa0https://mycbseguide.com/cbse-syllabus.html you can chek this on icon of my cbse guide |
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| 16585. |
Aparimay sankhya ka 2 example |
| Answer» aparimay means irrational so √2 ,√3 | |
| 16586. |
Find the root of equation 5x2-6x-28=0 by the method of completing the squre |
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| 16587. |
xsquare add 9x substract 5400 |
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| 16588. |
In an AP the sum of first ten terms is -150 and the sum of its next ten terms is -550.Find the AP |
| Answer» According to the question,the sum of first 10 terms of an AP is -150 and the sum of its next 10 terms is -550Let a be the first term and d be the common difference of the given AP.Then, we haveS10=-150{tex}\\Rightarrow \\frac { 10 } { 2 } [ 2 a + 9 d ] = - 150{/tex}{tex}\\Rightarrow{/tex}5[2a+9d]=-150{tex}\\Rightarrow{/tex}2a+9d=-30...(i)Clearly, the sum of first 20 terms =-150+(-550)=-700{tex}\\therefore{/tex}S20=-700{tex}\\Rightarrow \\frac { 20 } { 2 } [ 2 a + 19 d ] = - 700{/tex}{tex}\\Rightarrow{/tex}10[2a+19d]=-700{tex}\\Rightarrow{/tex}2a+19d=-70...(iii)Subtracting (i) from (ii), we get10d=-40{tex}\\Rightarrow{/tex}d=-4{tex}\\Rightarrow{/tex}2a=-30-9(-4)=-30+36=6{tex}\\Rightarrow{/tex}a=3Thus, we have\xa0First term=a=3Second term= a+d=3+2(-4)=-1Third term=a+2d=3+2(-4)=3-8=-5Fourth term=a+3d=3+3(-4)=3-12=-9Thus, the given AP is 3,-1,-5,-9,.... | |
| 16589. |
Who will win newzealand or england |
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Answer» ?????? England NewZealand |
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| 16590. |
A line AB is 8 cm in length, AB is produced to P such that BP square= AB.AP. Find length of BP |
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| 16591. |
Definition of statics |
| Answer» Statics is\xa0the branch of mechanics concerned with bodies at rest and forces in equilibrium. | |
| 16592. |
Factorize 24x2 - 65x + 21 |
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Answer» So, 24x2-56x-9+21 Then we split 504 into 56 and 9 First of all we will multiply 21 and 24x2 |
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| 16593. |
Prove that:-Tan7°tan23°tan60°tan67°tan83°=√3 |
| Answer» tan7*tan83*tan67*tan23* tan60 now,tan7 °= cot 83° ,tan67= cot 23 therfore 1*1tan60°= √3 | |
| 16594. |
(√2+√5)³ simplify |
| Answer» 2√2+5√5+2√10 | |
| 16595. |
Solution of maths exercise in class 10\xa0 |
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| 16596. |
Elimination method solve |
| Answer» write the equation in order ,eleminate a variable by change its sign(substraction)and addinge.g -- 2x+3y -- 1 ..... eq 1st 2x+2y -- 0 ..... eq 2nd (-) (-) -- (-) y -- 1 then put the value of yin eq 1nd /2nd 2x+3y -- 1 2x+3 -- 1 2x -- 1-3 x-- -1 | |
| 16597. |
Solve 3x-3y-7=0 and 3x-3y-15=0 by substitution method |
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Answer» X=31/3?? Y=8?? |
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| 16598. |
14.1 Q5 by direct method |
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| 16599. |
If cosA =12/13, verify that sinA(1- tanA)=35/156 |
| Answer» Net pe search krle | |
| 16600. |
if asinA+bcosA=c,then prove that acosA-bsinA=root of a2+b2-c2 |
| Answer» We have, {tex}asin\\theta+bcos\\theta=c{/tex}On squaring both sides, we get{tex}(asin\\theta+bcos\\theta)^2=c^2{/tex}(a sin θ)2\xa0+ (b cos θ)2\xa0+ 2(a sin θ) (b cos θ) = c2⇒ a2\xa0sin2\xa0θ + b2\xa0cos2\xa0θ + 2ab sin θ cos θ = c2⇒ a2(1 – cos2\xa0θ) + b2\xa0(1 – sin2\xa0θ) + 2 ab sin θ cos θ = c2 {tex}[\\because sin^2\\theta+cos^2\\theta=1]{/tex}⇒ a2\xa0– a2\xa0cos2\xa0θ + b2\xa0– b2\xa0sin2\xa0θ + 2ab sin θ cos θ = c2⇒ –a2\xa0cos2\xa0θ – b2\xa0sin2\xa0θ + 2ab sin θ cos θ = c2\xa0– a2\xa0– b2 Taking Negative common,⇒ a2\xa0cos2\xa0θ + b2\xa0sin2\xa0θ – 2ab sin θ cos θ = a2\xa0+ b2\xa0– c2⇒ (a cos θ)2\xa0+ (b sin θ)2\xa0– 2(a cos θ) (b sin θ) = a2\xa0+ b2\xa0– c2⇒ {tex}(acos\\theta-bsin\\theta)^2=a^2+b^2-c^2{/tex}⇒{tex}acos\\theta-bsin\\theta{/tex} =\xa0{tex}\\pm \\sqrt { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } }{/tex}\xa0Hence proved, {tex}acos\\theta-bsin\\theta{/tex} =\xa0{tex}\\sqrt{a^2+b^2-c^2}{/tex} | |