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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.
| 11651. |
Sec+tan=mSec-tan=nShow that mn=1 |
| Answer» Given secA +tanA=m and secA-tanA=nTo prove:- mn=1L.H.S. mn=(secA+tanA)(secA-tanA) = sec2A - tan2A [by using identity (a+b)(a-b)=a2\xa0- b2 ] = 1+tan2A - tan2A [by using identity sec2A=1+tan2A ] = 1=R.H.S. H.Proved | |
| 11652. |
Sin(1+tan)+cos(1+cot)=sec+cosec |
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| 11653. |
Find the root of x+1/x =3, x=o.Chapter : quadratic equation |
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| 11654. |
Find the value of k, so that the equation x2-(3k-1)+2k2+2k-11=0 |
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| 11655. |
2x+2y=123x+3y=6 |
| Answer» 2x+2y=12a1\xa0= 2 b1=2 c1= 123x+3y=6a2=3 b2= 3 c2=6a1/a2=2/3 b1/b2=2/3 c1/c2=2Therefore system of equations are inconsistent and have no solution. | |
| 11656. |
What is the least number of zeroes possible for the given quadratic polynomial : ax²+bx+c ? |
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| 11657. |
What is electrolysis? |
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| 11658. |
What is fundamental theorem of arithmetic? |
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| 11659. |
Write the short note of dental caries |
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| 11660. |
What is cubic equation? |
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| 11661. |
What is prime number? |
| Answer» Prime number is\xa0a number that is divisible only by itself and 1 (e.g. 2, 3, 5, 7, 11). | |
| 11662. |
What is rhombus ? |
| Answer» a Rhombus is a parallelogram in which a diagonal bisects an interior angleor a parallelogram in which at least two consecutive sides are equal in lengthor a quadrilateral with four sides of equal lengthor a quadrilateral in which the diagonals are perpendicular and bisect each other\xa0 | |
| 11663. |
SolveX2+x+3=0 |
| Answer» 1 | |
| 11664. |
Number of zeroes of a polynomial of degree x is |
| Answer» 0 | |
| 11665. |
33%33 |
| Answer» 33%33= 33/100 x 33= 1089/100= 10.89 | |
| 11666. |
How to prove triangle |
| Answer» If the area is not equal to zero than it is triangle or if sum of angles is180° we can say it is triangle | |
| 11667. |
Log525 |
| Answer» Log (525) = 2.72......Log10\xa0525 = Log 525 / Log 10 = 2.72... / 1 = 2.72........ .(as Log 10 = 1) | |
| 11668. |
The total number of divisors of 10500 except 1 and itself is |
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| 11669. |
Property of triangle |
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Answer» Sum of angles is 180° Tri |
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| 11670. |
Easy way to memory formulae ??? How |
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Answer» Learn by heart Revise every time |
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| 11671. |
Prove theoram 6.1 |
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Answer» I know I don\'t no |
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| 11672. |
Sin30/cos45+tan45/sec60-cos30/tan45-sin60/sin90 |
| Answer» By putting the values,1/2÷1/√2+1/2-√3/2÷1-√3/2÷1=1/√2+1/2-√3=(2+√2-2√6)/2√2 | |
| 11673. |
What type of board paper is given to class 10 |
| Answer» Check Question Papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 11674. |
How will be the Question paper type can you give some hints type Question |
| Answer» Check Question Papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 11675. |
Proof that -Sin2A +cos2A =1 |
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| 11676. |
how many tanhents can a circle habe |
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Answer» A circle can have infinite number of tangents because there are infinite number of points on a circle Only one perpendicular from radius A circle can have infinite no of tangents |
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| 11677. |
What is quadratic equations |
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| 11678. |
If the area of two similar triangles are equal then prove that the triangle are confident. |
| Answer» Given:{tex}∆ABC\\sim∆PQR{/tex}and\xa0{tex}ar∆ABC=ar∆PQR{/tex}To prove:\xa0{tex}∆ABC\\cong∆PQR{/tex}Proof:\xa0{tex}∆ABC\\sim∆PQR{/tex}\xa0Also\xa0{tex}\\operatorname { ar } ( \\Delta A B C ) = \\operatorname { ar } ( \\Delta P Q R ){/tex}\xa0(given)or,\xa0{tex}\\frac { \\operatorname { ar } ( \\Delta A B C ) } { \\operatorname { ar } ( \\Delta P Q R ) } = 1{/tex}Or\xa0{tex}\\frac{AB^2}{PQ^2}=\\frac{BC^2}{QR^2}=\\frac{CA^2}{RP^2}=1{/tex}Or\xa0{tex}\\frac{AB}{PQ}=\\frac{BC}{QR}=\\frac{CA}{RP}=1{/tex}Hence we get thatAB=PQ,BC=QR and\xa0CA=RPHence\xa0{tex}∆ABC\\cong ∆PQR{/tex} | |
| 11679. |
Solve the equation: {2x-1/x+3}2{x+3/2x-1}=5, (x not equal to -3, 1/2 |
| Answer» By the process of splitting the middle term we get x=_3,1/2 | |
| 11680. |
Rd sharma ex 8.3 question no 22 class10 |
| Answer» Firstly we solve the equation by cross multiplication then the equation becomes x^2_4x_5 =0 then we solve it by splitting the middle term then the answer became x=5,_1 | |
| 11681. |
Test papers |
| Answer» Check test papers here :\xa0https://mycbseguide.com/cbse-test-papers.html | |
| 11682. |
9a^2-56a+186=0Find a |
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| 11683. |
Find the value of p for which one root of the quadratic equation Px2 - 14x +8 is 6 times the other |
| Answer» Let\xa0{tex}\\alpha{/tex}\xa0and\xa0{tex}6\\alpha{/tex}\xa0be the roots of equation.We have, {tex}px^2-14x+8=0{/tex} where a= p, b = -14, c = 8Sum of zeroes{tex} = -\\frac ba = -\\frac{-14}{p}{/tex}{tex}\\alpha +6\\alpha=\\frac{14}{p}{/tex}{tex}7\\alpha = \\frac{14}{p}{/tex}{tex}\\alpha = \\frac2p{/tex}............(i)Also, Product of the zeroes\xa0{tex} = \\frac 8p=\\frac ca{/tex}{tex}\\alpha \\times 6\\alpha = \\frac 8p{/tex}{tex}6\\alpha^2=\\frac 8p{/tex}From (i){tex}6(\\frac{2}{p})^2=\\frac 8p{/tex}{tex}6\\times \\frac {4}{p^2}=\\frac 8p{/tex}{tex}\\frac{6}{p^2}=\\frac2p{/tex}{tex}\\frac 62=\\frac{p^2}{p}{/tex}{tex}Hence, \\ p=3{/tex} | |
| 11684. |
Why root i2 I rational no. |
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| 11685. |
Sir in board exam all mean each and every question come from ncert book only or outside the book |
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| 11686. |
Find the coordinates of the equidistant from three given points A(5,3), B(5,-5) and C(1,-5). |
| Answer» Let the required points be P(x, y), thenPA = PB = PC. The points A, B, C are (5,3), (5, -5) and (1, -5) respectively{tex} \\Rightarrow{/tex}\xa0PA2\xa0= PB2\xa0= PC2{tex}\\Rightarrow{/tex}\xa0PA2 = PB2 and PB2 = PC2PA2 = PB2{tex}\\Rightarrow{/tex}\xa0(5 - x)2\xa0+ (3 - y)2\xa0= (5 - x)2\xa0+ (-5 -y)225 + x2\xa0- 10x + 9 + y2 - 6y = 25 + x2\xa0\xa0- 10x + 25 + y2\xa0+ 10y-6y - 10y = 25 - 9\xa0{tex}\\Rightarrow{/tex}\xa0-16y = 16y = -1and PB2 = PC2{tex}\\Rightarrow{/tex}\xa0(5 - x)2\xa0+ (-5 - y)2\xa0= (1 - x) + (-5 - y)225 + x2 - 10x + 25 + y2 + 10y = 1 + x2 - 2x + 25 + y2 + 10y-10x + 2x = -24\xa0{tex}\\Rightarrow{/tex}\xa0-8x = -24{tex}x = \\frac { - 24 } { - 8 } = 3{/tex}Hence, the point P is (3, -1) | |
| 11687. |
Find distance between A(6,-2) B(3,4) |
| Answer» AB =\xa0{tex}\\sqrt { ( 3 - 6 ) ^ { 2 } + ( 4+2) ^ { 2 } } = \\sqrt{13}{/tex} | |
| 11688. |
Find the value of p if the mean of the following distribution is 18 |
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| 11689. |
TanA/1-cotA + cotA/1-tanA |
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| 11690. |
which is best book to solve |
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Answer» RD sharma RS Agarwala is best RD sharma for math |
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| 11691. |
A.P |
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| 11692. |
4x+6y=3xy,8x+9y=5xy |
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Answer» Let us try to make 8x in both equationsTherefore multiply equation with 4x by 2 to make it = 8x2 x (4x + 6y = 3xy)8 x + 12 y = 6 xy .............. i8x + 9y = 5xy ....................... iii - ii3y = xy3 = xPutting the value of x in ii24 + 9y = 5x3y24 = 15y - 9y24 = 6yy = 4x = 3\xa0 Multiply 1st equation by 2Then subtract equation 2 from equation 1 ( multiplied by 2 )We will get x=3 and y=4 |
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| 11693. |
2×5 |
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Answer» Very tough question what u asked do the sum on your own 10 10 |
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| 11694. |
Solve the equation x^2-(2-√3)x-2√3=0 by the method of completing the square |
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| 11695. |
Theorem triangle |
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| 11696. |
What is general form of polynomial |
| Answer» A polynomial is a function of the form f(x) = anxn + an−1xn−1 + ... + a2x2 + a1x + a0 | |
| 11697. |
Solve for2x+y=74x-3y=-1USING SUBSISTUTION METHOD |
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Answer» The given system of equations is2x + y = 7 ... (i)4x\xa0- 3y = -1. ...(ii)From (i), we get2x + y = 7⇒ y = (7 - 2x).Substituting y = (7 - 2x) in (ii), we get4x\xa0- 3y = -1⇒ 4x - 3 (7 - 2x) = - 1⇒ 4x - 21+ 6x = -1⇒ 10x - 21 = -1⇒ 10x = - 1 + 21⇒ 10x\xa0=\xa020⇒ x = 20/10⇒ x =\xa02.Substituting x = 2 in y = (7 - 2x), we gety = (7 - 2x)⇒ y = 7 - 2(2)⇒ y = 7 - 4⇒ y = 3.Hence, the solution is x = 2, y = 3. tq |
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| 11698. |
1+2+3+4+5 |
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Answer» 15 15 15 15 |
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| 11699. |
88+88 |
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Answer» 88+88=176 176 |
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| 11700. |
Gju |
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