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InterviewSolution
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.
| 41001. |
Find the perimeter of triangle with vertices (0,4)(0,0) and (3,0 |
| Answer» First find length of each sides of ∆Let A( 4 , 0) B(0, 0) and C (0 , 3)use distance formula,AB =√(4²+0) =4BC= √(0+3²) = 3CA =√(4²+3²) =5Now , perimeter of ∆ = 3 + 4 + 5= 12 unit | |
| 41002. |
Got Departed from the school or not? |
| Answer» | |
| 41003. |
Lcm of18 |
| Answer» Gs | |
| 41004. |
in 2019 board exam which types questions will come in 10 cbse board exam please say me any....... ?? |
| Answer» | |
| 41005. |
Triangles chapter kise pasand hai? |
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Answer» I hate it Samjha do mujhe plz Mujhe Anyone online answer plz? |
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| 41006. |
Rationsl number between √ 2 and √3 |
| Answer» 1.2 | |
| 41007. |
Sb pagal ho gye hai sb kya ye homework help ke liye hai baat krne ke liye nhi plz concentrate ok |
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Answer» Ok sorry aradhana Yes u r right Sahi bol rahe ho ap padhai ke liye hai ye |
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| 41008. |
Prove the following identity:SecA+TanA/CosecA+CotA=1+SinA/CosA |
| Answer» | |
| 41009. |
What is known as centriod |
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Answer» Centroid is centre of a triangle where all it medians intersect each other. Center of any figure |
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| 41010. |
Write the 4th term of an ap if its nth term is 3n+2 |
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Answer» n =4Therefore,3(4) + 2 = 14Hence the 4th term of an A.P. is 14 3(4)+2=14 14 |
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| 41011. |
For what value of a the point (a,1) (1,-1) & (11,4) are collinear |
| Answer» Use the formula area of triangle with vertices | |
| 41012. |
CosA - sinA +1/cosA + sinA -1= cosecA + cotA |
| Answer» | |
| 41013. |
Triangle ABC~ tri DEF if AB=4 cm ,BC=3.5cm , CA=2.5cm,DF=7.5cm, then find the perimeter of tri DEF |
| Answer» | |
| 41014. |
Where B square minus 4 AC implies |
| Answer» Discriminant of the quadratic equation | |
| 41015. |
Prove n3(cube)-n is divsble by 3 for any positive integer n |
| Answer» | |
| 41016. |
1+secA/secA = sin^2A/1-cosA |
| Answer» | |
| 41017. |
2+2-4 |
| Answer» 0 | |
| 41018. |
What about the solution of 12th class ncert exampler |
| Answer» | |
| 41019. |
What perimeter |
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Answer» Perimeter is the meaurement of length of boundary of a 2d figure Perimeter means distance around a figure or curve. We can only measure perimeter of a closed figure/2 dimensional shape or curve as movement around a closed figure or curve is possible.Perimeter of a square = sum of all sides = 4 x sidePerimeter of Rectangle = L + L + B + B |
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| 41020. |
Tow triangle are similar so prove that they are congrueant |
| Answer» Same type of question is present in ncert please check it....... | |
| 41021. |
(sin50+threta)-cos(40-threta)+tan1tan10tan20tan70tan80tan90 |
| Answer» | |
| 41022. |
what is distance formula in chapter 7 |
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Answer» Distance formula - √(x2 - x1)+(y2 - y1) Root mein (x2 - X1) ka whole square + (Y2 - Y1) ka whole square |
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| 41023. |
Find the area of rectangular field whose side is measures 35m and diagonal 37m |
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Answer» Sorry pls Shradhanand llpps solution batao 770m sq. Shayad 420m^2 |
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| 41024. |
if sec theata + tan theata =p then find valu of cosec theata |
| Answer» | |
| 41025. |
If x=-1/2 is the solution of a quadratic equation 3x2+2kx-3=0 find the value of k |
| Answer» -9/4 | |
| 41026. |
34 divide 2 |
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Answer» 17 2/34 =0.0588235294. 2√34(2√34(17 34 0 |
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| 41027. |
A ladder of length 4m makes an angle |
| Answer» Complete ur que | |
| 41028. |
Yar method of completing squares kaise karte hai agar ata hai toh btauu muje nhi ata☹️☹️?? |
| Answer» Hy durga | |
| 41029. |
In triangle ,if AD is the median ,then showthat (AB)2+(AC)2=2(AD2+BD2) |
| Answer» | |
| 41030. |
Find a quadratic polynomial whose zeroes are -3 and -1 |
| Answer» x2+4x -3 is the quadratic equation | |
| 41031. |
22+22+56+56+56 |
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Answer» 212 212 |
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| 41032. |
AB+CD=BC+DA |
| Answer» Bhai question toh sahi dal | |
| 41033. |
Q. Sin90 + Cos 90 =0 |
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Answer» sin 90 = 1 and cos 90 = 0Sin 90 + Cos 90= 1 + 0= 1 Why |
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| 41034. |
If sum of first term n is 3n/2+13n/2 .find 25th term |
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Answer» Miss puja kya app answer de sakengi The question which i asked has came in our test The question which u are saying i have solved inxamidea this is another Please check your question................i think it should be 3n square/2 +13n/2 Please answer this question |
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| 41035. |
Cos thitha . Tan thitha =sin thitha prove this |
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Answer» Lhs = cos theta ×sinTheta/cos theta (since tan theta =sin theta /cos theta ) =now cancel cos theta and cos theta =sin thetha Therefore lhs =rhs Proved Cos theta.tan theta=sin theta. Cos theta.sin theta/cos theta{tan theta=sin/cos} so sin=sin |
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| 41036. |
Square root of 800 |
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Answer» Its 20 root 2..... 20√2 |
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| 41037. |
2x +3x |
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Answer» x(2+3)x(5)5x 5x 5x |
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| 41038. |
Wt r decimal numbers? |
| Answer» | |
| 41039. |
What is prime and composite number |
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Answer» Prime number means the number which only divisible by the number and one.Composite number means the number will divisible by one or more numbers.Have a nice day, Prime numbers are those numbers which are divisible by one and themselves only They are numbers |
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| 41040. |
Prove that (cosec theta +cot theta) square =sec theta +1/sec theta-1 |
| Answer» {(1/sin)+(cos/sin)}² ={(1+cos)/sin}² =(1+cos)²/sin²=(1+cos)²/(1-cos²)=(1+cos)(1+cos)/(1-cos)(1+cos)=(1+cos)/(1-cos)=(1+1/sec)/(1-1/sec)=(sec+1)/(sec-1) | |
| 41041. |
If SinA+2cosA=1 then prove : 2sinA-cosA=2 |
| Answer» | |
| 41042. |
Important question in chapter 8(trigonometry) |
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Answer» In trignometry ther is no important question.All are important Exercise 8.4 all questions |
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| 41043. |
Finding next three terms 1. √2,√6,√9,√12... |
| Answer» | |
| 41044. |
Rupali pandav sorry |
| Answer» | |
| 41045. |
Find the roots of the equation 5x-6x-2=0 by the method of completing the square |
| Answer» 5x2 - 6x - 2 = 0Multiplying the above equation by 1/5{tex} \\Rightarrow {x^2} - \\frac{6}{5}x - \\frac{2}{5} = 0{/tex}{tex}\\Rightarrow x ^ { 2 } - \\frac { 6 } { 5 } x + \\left( \\frac { 3 } { 5 } \\right) ^ { 2 } - \\left( \\frac { 3 } { 5 } \\right) ^ { 2 } - \\frac { 2 } { 5 } = 0{/tex}{tex}\\Rightarrow \\left( x - \\frac { 3 } { 5 } \\right) ^ { 2 } = \\frac { 9 } { 25 } + \\frac { 2 } { 5 }{/tex}{tex}\\Rightarrow \\left( x - \\frac { 3 } { 5 } \\right) ^ { 2 } = \\frac { 9 + 10 } { 25 }{/tex}{tex}\\Rightarrow \\left( x - \\frac { 3 } { 5 } \\right) ^ { 2 } = \\frac { 19 } { 25 }{/tex}{tex}\\Rightarrow x - \\frac { 3 } { 5 } = \\pm \\frac { \\sqrt { 19 } } { 5 }{/tex}{tex}\\Rightarrow x = \\frac { 3 } { 5 } \\pm \\frac { \\sqrt { 19 } } { 5 }{/tex}{tex}\\Rightarrow x = \\frac { 3 + \\sqrt { 19 } } { 5 } \\text { or } x = \\frac { 3 - \\sqrt { 19 } } { 5 }{/tex} | |
| 41046. |
Find 25th term of the A.P , -5 , -5/2 , 0 ,5/2. |
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Answer» n=25a=-5d=a2-a1d=-5/2-(-5)d=-5/2+5=-5+10/2=5/2an=a+(n-1)da25=-5+(25-1)5/2 =-5+12*5 =-5+60a25=55 Please answer me? |
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| 41047. |
Find roots of quadratic equation: x - 1/x=3 , x is not equal to 0 |
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Answer» Hlo mehak Aapne mujhe pehchana ki nahi x2-1=3xx2-3x-1=0{tex}x = {3\\pm \\sqrt{9+4} \\over 2}={3\\pm\\sqrt{13}\\over 2}{/tex} X-1/3=3x(x)-1/x=3x2-1/x=3x2-1=3xx2-3x-1=0comparing equation with ax2+bx+c=0a=1,b=-3,c=-1 answer Please answer me |
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| 41048. |
If x=a, y=b is the solution of the pair of equation x-y=2 and x +y=4 find the value of a and b |
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Answer» x=a=3y=b=1 Given= x=a and y=bThe equation are x-y=2. equation-1x+y=4. equation-2 Adding equation 1 form equation 2x+y=4x-y=22x=6x=6/2x=3Substitute the value of x in eq 2x+y=43+y=4y=4-3y=1 x = a , y = bx - y = 2a - b = 2....... (i)x + y = 4a + b = 4 ....... (ii)Add (i) and (ii)2a = 6a = 6/2 = 3But a = 3 in (ii)3 + b = 4b= 4 - 3b = 1So, a = 3 and b = 1 |
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| 41049. |
Show that for odd positive integer to be a perfect square ,it should be of the form 8k+1 |
| Answer» Since, any odd positive integer n is of the form 4m\xa0+ 1 or 4m + 3.if n = 4m + 1n2 = (4m + 1)2= 16m2 + 8m + 1= 8(2m2 + m) + 1So n2 = 8q + 1 ........... (i)) (where q = 2m2 + m is a positive integer)If n = (4m + 3)n2 = (4m + 3)2= 16m2 + 24m + 9= 8(2m2 + 3m + 1) + 1So n2 = 8q + 1 ...... (ii) (where q = 2m2 + 3m + 1 is a positive integer)From (i) and (ii) we conclude that the square of an odd positive integer is of the form 8q + 1, for some integer q. | |
| 41050. |
Q. If the points (p, q); (m, n) and (p-m, q-n) are collinear. Show that pn=qm. |
| Answer» Given points are collinear. Therefore[p {tex}\\times{/tex}\xa0n + m(q - n) + (p - m) q] - [m {tex}\\times{/tex}\xa0q + (p - m) n + p (q - n)] = 0{tex}\\Rightarrow{/tex}\xa0(pn + qm - mn + pq - mq) - (mq + pn - mn + pq - pn) = 0{tex}\\Rightarrow{/tex}\xa0(pn + p q - mn) - (mq - mn + pq) = 0{tex}\\Rightarrow{/tex}\xa0pn - mq = 0{tex}\\Rightarrow{/tex}\xa0pn = qm | |