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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.
| 24751. |
Find the distance between (2,3) |
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| 24752. |
sin60° cos30° +sin30° cos60° |
| Answer» And : 1 | |
| 24753. |
Solve the following pair of linear equation |
| Answer» Get lost man | |
| 24754. |
What this aap help me in studies |
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Answer» Your mind Examfear app |
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| 24755. |
Which term of the A.P 5/7,1,7/6,...is 3? |
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| 24756. |
Stores |
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| 24757. |
Tangents |
| Answer» A tangent is a line that touches a circle at one point and at one point only. It always is perpendicular to any radius drawn to the point of tangency | |
| 24758. |
P(x)+q(y)=p-qQ(x)-py=p+q solve |
| Answer» The given pair of equations ispx + qy = p - q .....(1)qx - py = p + q ....(2)Multiplying equation (1) by p and equation (2) by q, we getp2x + pqy = p2 - pq....(3)q2x - pqy = pq + q2.....(4)Adding equation (3) and equation (4), we get(p2 + q2)x = p2 + q2{tex}\\Rightarrow \\;x = \\frac{{{p^2} + {q^2}}}{{{p^2} + {q^2}}} = 1{/tex}Substituting this value of x in equation (1), we getp(1) + qy = p - q{tex}\\Rightarrow{/tex} qy = -q{tex}\\Rightarrow \\;y = \\frac{{ - q}}{q} = - 1{/tex}So, the solution of the given pair of linear equations is x = +1, y = -1.Verification, Substituting x = 1, y = -1,We find that both the equations (1) and (2) are satisfied as shown below:px + qy = p(1) + q(-1) = p - qqx - py = q(1) - p(-1) = q + p = p + qThis verifies the solution. | |
| 24759. |
What is grouped data . |
| Answer» Mean | |
| 24760. |
what is euclids division dilemma |
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| 24761. |
How many got minimum percentage to pass class 11 ( first terminal ) |
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| 24762. |
Show that √3+√5 is an irrational number |
| Answer» Firstly assumed that √3+√5 is a rational no. √3+√5=p/q (where p&q are any positive integer &co-prime &qis not equal to 0)p/q -√3=√5 (squaring both side)(p/q-√3)^2=(√5)^2p^2/q^2-2p√3/q+3= 5(Put 3 on RHS then ,we get)p^2/q^2-2p√3/q=2{Put (2p√3/q) on RHS then ,we get}(p^2/q^2)-2=-2p√3/q[(p^2)-(2q^2)]/q^2=2p√3/q(left only √3 on RHS,then solve ,we get)[p^2 - 2q^2]/2pq = √3. (We know that √3 are irrational no. )but {[p^2 - 2q^2]/2pq} is rationalThis implies that √3 is rational no.This implies that our contradict are wrong This implies that, √3 + √5 is irrational no. | |
| 24763. |
what are the zeros of polynomial x² - 68x +450 |
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| 24764. |
3 cm is a radius of a spherical and find out the 1/8 of surface area and volume area |
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| 24765. |
Find next term of the series 0,3,7,15,24,35 |
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| 24766. |
3 is irrational how to solve |
| Answer» Firstly you take 3as rational and then consider your supposition wrong and then solve it (its very easy) | |
| 24767. |
Explain the first chapter |
| Answer» You can read notes :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 24768. |
If the sum of first n even natural nos is 420 find the value of n please solve the problem |
| Answer» n(n + 1) = 420 ...Given{tex}\\Rightarrow n ^ { 2 } + n = 420{/tex}{tex}\\Rightarrow n ^ { 2 } + n - 420 = 0{/tex}Comparing with An2 + Bn + C = 0, we getA = 1, B = 1, C = -420Using the quadratic formula, {tex}n = \\frac { - B \\pm \\sqrt { B ^ { 2 } - 4 A C } } { 2 A }{/tex}we get {tex}\\Rightarrow \\frac { - 1 \\pm \\sqrt { 1 + 1680 } } { 2 } = \\frac { - 1 \\pm \\sqrt { 1681 } } { 2 }{/tex}{tex}= \\frac { - 1 \\pm 41 } { 2 } = \\frac { - 1 + 41 } { 2 } , \\frac { - 1 - 41 } { 2 } = 20 , - 21{/tex}n = -21 is in admissible as n is the number of terms.{tex}\\therefore n = 20{/tex}Hence, the required value of n is 20. | |
| 24769. |
If a-b=3 and b-c=5, then the value of a^2+b^2+c^2-ab-bc-ca |
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| 24770. |
26÷6 |
| Answer» {tex}\\frac{26}{6} = \\frac{13}{3}{/tex} | |
| 24771. |
The sum of three numbers in AP is 27 and their product is 405 find the number |
| Answer» Let no. are a-d,a,a+da-d+a+a+d=273a=27a=9(a-d)(a)(a+d)=405(9-d)(9)(9+d)=40581-dsquare=45-dsquare=-36d=6no.area-d=9-6=3a=9a+d=9+6=15 | |
| 24772. |
log e =? |
| Answer» 0.4342944819 | |
| 24773. |
B÷a×x+a÷b×y=a×a+b×bX+y=2ab .Solve it to find the value of x &y |
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| 24774. |
y4+1/y4=47 then y4-1/y4=? |
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| 24775. |
Sir CBSE me examabhi new plan ke according kitne no. Ka hoga |
| Answer» Check exam pattern here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 24776. |
If a=8 , nth term of ap is 62 and sum of nth term of ap is 210 ,find n and d. |
| Answer» n=6 and d=54/5 | |
| 24777. |
Teeta |
| Answer» Silly | |
| 24778. |
Definition of Mathmatics |
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| 24779. |
Formula of tregnometri |
| Answer» Sin square +cos square=1 | |
| 24780. |
ABCD is a square show that AB=BC=CD=AD |
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| 24781. |
Gghhh |
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| 24782. |
2+2÷2=..............plz don\'t take this question lightly and give answer. |
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Answer» 3 by bodmas 3 |
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| 24783. |
Root x + y = 11 and x + root y = 7 find x and y |
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| 24784. |
DETERMINE THE VALUE OF x such that ( 2x, 2x+2), ( 3 ,2x+ 1), (1, x+1) lie on the same line. |
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| 24785. |
Show that the product of three consecutive positive integers is divisible by 3 |
| Answer» Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer.By Euclid’s division lemma, we havea = bq + r; 0 ≤ r < bFor a = n and b = 3, we haven = 3q + r ...(i)Where q is an integer and 0 ≤ r < 3, i.e. r = 0, 1, 2.Putting r = 0 in (i), we get{tex}n = 3q{/tex}∴ n is divisible by 3.{tex}n + 1 = 3q + 1{/tex}∴ n + 1 is not divisible by 3.{tex}n + 2 = 3q + 2{/tex}∴ n + 2 is not divisible by 3.Putting r = 1 in (i), we get{tex}n = 3q + 1{/tex}∴ n is not divisible by 3.{tex}n + 1 = 3q + 2{/tex}∴ n + 1 is not divisible by 3.{tex}n + 2 = 3q + 3 = 3(q + 1){/tex}∴ n + 2 is divisible by 3.Putting r = 2 in (i), we get{tex}n = 3q + 2{/tex}∴ n is not divisible by 3.{tex}n + 1 = 3q + 3 = 3(q + 1){/tex}∴ n + 1 is divisible by 3.{tex}n + 2 = 3q + 4{/tex}∴ n + 2 is not divisible by 3.Thus for each value of r such that 0 ≤ r < 3 only one out of n, n + 1 and n + 2 is divisible by 3. | |
| 24786. |
Sinthetha |
| Answer» Costheta | |
| 24787. |
tan (a1+a2)=tan a1 +tan a2÷1-tan a1×tana2 evalute |
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| 24788. |
What is coprime number? |
| Answer» Which do not have common factor othe than 1 | |
| 24789. |
1+1_ |
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Answer» 2 but it can be 11 also 2 |
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| 24790. |
9x-3kx+k=0 find the value of k quadratic equation |
| Answer» a = 9, b = - 3 k, c = kSince roots of the equation are equalSo,\xa0{tex}b ^ { 2 } - 4 a c = 0{/tex}{tex}( - 3 k ) ^ { 2 } - ( 4 \\times 9 \\times k ) = 0{/tex}{tex}9 k ^ { 2 } - 36 k = 0{/tex}{tex}k ^ { 2 } - 4 k = 0{/tex}k (k -4 ) = 0k = 0 or k = 4\xa0 | |
| 24791. |
What is a secant of circle |
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Answer» If the circle and line have 2 common point.we call the line a secant of the circle. A line which intersects a circle in two distinct point is called a secant to the circle. |
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| 24792. |
Value of k |
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| 24793. |
When will routine of board issue |
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| 24794. |
If m=cosA-sinA and n=cosA+sinA Then show that |
| Answer» There\'s nothing to show | |
| 24795. |
if sin^2+cos^2=9,then sin+cos=? |
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| 24796. |
Find the circumcentre of the triangle whose vertices are (-2,-3), (-1,0), (7,-6). |
| Answer» Let the coordinates of the circumcentre of the triangle be P(x, y). Then, we know that circumentre of a triangle is equidistant from each of its vertices.\xa0{tex}PA = \\sqrt {{{(x + 2)}^2} + {{(y + 3)}^2}} {/tex}{tex} \\Rightarrow P{A^2} = {x^2} + {y^2} + 4x + 6y + 13{/tex}\xa0..... (i){tex} \\Rightarrow PB = \\sqrt {{{(x + 1)}^2} + {{(y - 0)}^2}} {/tex}{tex} \\Rightarrow P{B^2} = {x^2} + {y^2} + 2x + 1{/tex}\xa0...... (ii){tex}PC = \\sqrt {{{(x - 7)}^2} + {{(y + 6)}^2}} {/tex}{tex} \\Rightarrow {/tex}\xa0PC2 = x2 + y2 - 14x + 12y + 85 ..... (iii)Now, PA2 = PB2{tex} \\Rightarrow {/tex}\xa0x2 + y2 + 4x + 6y + 13 = x2 + y2 + 2x + 1{tex} \\Rightarrow {/tex}\xa02x + 6y = -12{tex} \\Rightarrow {/tex}\xa0x + 3y = -6 ..... (iv)And PA2 = PC2x2 + y2 + 4x + 6y + 13 = x2 + y2 - 14x + 12y + 8518x - 6y = 72{tex} \\Rightarrow {/tex} 3x - y = 12 ..... (v)And PB2 = PC2{tex} \\Rightarrow {/tex}\xa0x2 + y2 + 2x + 1 = x2 + y2 - 14x + 12y + 85{tex} \\Rightarrow {/tex}\xa016x - 12y = 84\xa0{tex} \\Rightarrow {/tex}\xa04x - 3y = 21 ....... (vi)Solving (iv) and (v) we get x = 3, y = -3Hence, circumcentre of the triangle is (3, -3) | |
| 24797. |
What are All formulas used in surface area and volumes |
| Answer» Volume of cuboid=lbh . Total surface area of cuboid =2 (lb+bh+hl). Curved surface area of cuboid =2h (l+b). Volume of cube=a×a×a . T.S.A. Of cube 6a×a . C.S.A. of cube =4×a×a. T.S.A. Of cylinder=2×py×r(h+r). C.S.A. of cylinder =2×py×r×h. Volume of cylinder =py×r×r×h . T.S.A. OF cone=py×r (l+r) . C.S.A. of cone =py×r×l . VOLUME of cone =1/3×py×r×r×h. T.S.A of sphere =4×py×r×r . C.S.A. of sphere =4×py×r×r. Volume of sphere =4/3×py×r×r×r.C.S.A of hemisphere =2×py×r×r. T.S.A of hemisphere =3×py×r×r.volume of hemisphere =2/3×py×r×r×rC.S.A of frustum =py×l (r1+r2) +py (r×r1+r×r2). T.S.A of frustum =py×l (r1+r2) + py (r×r1+r×r2). Volume of frustum =1/3×py×h (r×r1+r×r2+r1×r2) | |
| 24798. |
If the mean and median of set of numbers is 8.9 and 9 respectively, then find the mode |
| Answer» 8.7 | |
| 24799. |
I have to understand the Ap statement problems |
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| 24800. |
SinA |
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