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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.
| 39801. |
Express the cosA in terms of cotA |
| Answer» tan A =\xa0{tex}=\\frac{1}{\\cot A}{/tex}{tex}\\Rightarrow{/tex}\xa0tan2\xa0A =\xa0{tex}\\frac{1}{\\cot ^{2} A}{/tex}{tex}\\Rightarrow{/tex}\xa0sec2\xa0A - 1 =\xa0{tex}\\frac{1}{\\cot ^{2} A}{/tex}{tex}\\Rightarrow{/tex}\xa0sec2\xa0A =\xa0{tex}\\frac{1}{\\cot ^{2} A}{/tex}+1{tex}\\Rightarrow{/tex}\xa0sec2\xa0A =\xa0{tex}\\frac{1+\\cot ^{2} A}{\\cot ^{2} A}{/tex}{tex}\\Rightarrow{/tex}\xa0sec A =\xa0{tex}\\sqrt{\\frac{1+\\cot ^{2} A}{\\cot ^{2} A}}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{1}{\\cos A}=\\sqrt{\\frac{1+\\cot ^{2} A}{\\cot ^{2} A}}{/tex}{tex}\\Rightarrow{/tex}\xa0cos A =\xa0{tex}\\sqrt{\\frac{\\cot ^{2} A}{1+\\cot ^{2} A}}{/tex}\xa0{tex}=\\frac{\\cot A}{\\sqrt{1+\\cot ^{2} A}}{/tex}{tex}\\therefore{/tex}\xa0cos A =\xa0{tex}\\frac{\\cot A}{\\sqrt{1+\\cot ^{2} A}}{/tex} | |
| 39802. |
Length of altitude of equilateral triangle of side 8 cm |
| Answer» First of its altitude divides into two equal half aftet we use pythagoras theorem its answer is route 48 | |
| 39803. |
EvaluateSin30°+tan45°-cosec60°÷Sec30°+cos60°+cot45° |
| Answer» Ghhjgg | |
| 39804. |
Prove that root5 is irrational |
| Answer» Let root5 is a rational number where root5=a/b and b is not = 0Thenroot5 =a/bSquaring both side5=a×a/b×b5bsquare=a squarea is divisible by 55 is a factor of aLet 5c=aThen5bsquare =(5c)square5bsquare =25csquarebsquare = 5csquareThenbsquare is a factor of 5b is also a factor of 5Hence,a and b are co prime numbers Therefore,our contradict the fact that root5 is an irrational number Hence,our supposition is wrong Therefore, root5 is an irrational number Hence Proved | |
| 39805. |
For the ap:3/2,1/1,-1/1,-3/2,...,write the first term a and the common difference d. |
| Answer» a=3/2 and d= 1/1 - 3/2 =-1/2 | |
| 39806. |
3x=cosecA |
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| 39807. |
1.2 |
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| 39808. |
Bro from where i can enter coupon in wallet |
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| 39809. |
If the quadratic equation is px square -3√5px +15 has two equal roots , then find the value of p ? |
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| 39810. |
Proof of BPT |
| Answer» Check in Ncert | |
| 39811. |
If x=1/2 is a solutionof quadratic equation 3x square+ 2k x - 3 =0 find the value of k ? |
| Answer» K= 0 | |
| 39812. |
Prove that 1/3-2√5 is an irrational number |
| Answer» If possible let (1/2 - √5/3) be a rational number x1/2-√5/3 => xmaking denominator equal (3-2√5)/6 => x3-2√5 = 6x3-6x = 2√52(1-2x)/2 = √51-2x = √5since , x is rational number (by our supposition)2x is also rational 1 is rational and we know that , difference of 2 rationals is rational numberso, this tells us that √5 is also rational but , √5 is irrational thus , our supposition is wrong so x is not rational , and x = 1/2 -√5/3 so, 1/2 - √5/3 is irrational hope this helpsRead more on Brainly.in - https://brainly.in/question/1462562#readmore | |
| 39813. |
If the lines given by 3x+2ky=2 |
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| 39814. |
Perimeter of triangle?? |
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Answer» Sorry for first answer It\'s addition of all three sides 3a were "a" is sides 3(sides of the triangle) Perimeter of triangle is sum of its sides That is area 1/2× base × height |
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| 39815. |
Chapter 14.3 ques no 1 |
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Answer» Mode maximum class frequency is 20 modal class is 125-145 Mode=l+(f1+f2/2f-f0-f2)h By solving it we get Mode = 135.77 units Median = First find cumulative frequency(cf) i.e 4,9,22,42,56,64,68 Now n=68 n/2=68/2=34Cf greater than 34 is 42Median class is 125-145 Now l=125,cf=22,f=20,h=20Median =l+<(n/2-cf /f) h =125(34-22/20)20 =125+12Median =137 units |
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| 39816. |
Zeroes of polynomial of 5x2-4-8x |
| Answer» p(x) = 5x2 + 8x - 4 = 0=\xa05x2 + 10x - 2x - 4 = 0= 5x(x + 2) - 2(x + 2) = 0= (x + 2)(5x - 2) = 0Hence, zeroes are -2 and\xa0{tex}\\frac 25{/tex}Verification:Sum of zeroes =\xa0{tex}- 2 + \\frac { 2 } { 5 } = \\frac { - 8 } { 5 }{/tex}Product of zeroes =\xa0{tex}( - 2 ) \\times \\left( \\frac { 2 } { 5 } \\right) = \\frac { - 4 } { 5 }{/tex}Again sum of zeroes =\xa0{tex}- \\frac { \\text { Coeff. of } x } { \\text { Coeff. of } x ^ { 2 } } = \\frac { - 8 } { 5 }{/tex}Product of zeroes =\xa0{tex}\\frac { \\text { Constant term } } { \\text { Coeff. of } x ^ { 2 } }{/tex}\xa0=\xa0{tex}\\frac{-4}{5}{/tex}Verified. | |
| 39817. |
I want a time table for, school from 7:00 to 2:00 and tution from 4:00 to 6:30 |
| Answer» Dont follow time table bcz it wastage of time nthelse but u can follow a scheduled | |
| 39818. |
Konsi helpbook sahi h sabhi subjects k liye jishse hum above 95% la saken |
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Answer» Ncert Ncert Oswal S chand nahi sahi kya Xam idea |
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| 39819. |
X×y give me answer |
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Answer» xy It will just be xy. Not solve able |
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| 39820. |
C√b+4=x. Then find the value of x+1/x |
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| 39821. |
If one zero of the polynomial ( a2+9)x2+13×6a is the reciporcal of the other find the value of a. |
| Answer» Let {tex} \\alpha{/tex}\xa0and\xa0{tex} \\frac { 1 } { \\alpha }{/tex} be the zeros of\xa0(a2\xa0+ 9)x2\xa0+ 13x\xa0+ 6a.Then, we have{tex} \\alpha \\times \\frac { 1 } { \\alpha } = \\frac { 6 a } { a ^ { 2 } + 9 }{/tex}⇒\xa01 =\xa0{tex} \\frac { 6 a } { a ^ { 2 } + 9 }{/tex}⇒\xa0a2\xa0+ 9 = 6a⇒ a2 - 6a + 9 = 0⇒\xa0a2\xa0- 3a - 3a + 9 = 0⇒\xa0a(a - 3) - 3(a - 3) = 0⇒\xa0(a - 3) (a - 3) = 0⇒\xa0(a - 3)2\xa0= 0⇒\xa0a - 3 = 0⇒ a = 3So, the value of a in given polynomial is 3. | |
| 39822. |
Sin + cos uuuuu |
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| 39823. |
How to clear the mathematics exams in one night? |
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Answer» it is possible but practise and practies First u stop thinking that math is hard and do sums with peace of mind and keep self confidence in u and all the best ??? Impossible yarr Are u crazy??????It is not possible to clear maths in just 1 night Study only |
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| 39824. |
In a,right angled at B,AC-AB=2CM BC=8CM, find value of AC. |
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Answer» AC² = AB² + BC² ,,,,,,, =›AC²=(AC-2)²+64,,,,,,, =›AC²=AC²+4-4AC+64,,,,,,, =›AC= 17 first of all let ac be x and AB be 2-x and solve by pythagoreans theorem Put bpt theorem and put the values Oo bahan copy le aa |
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| 39825. |
Tan30° |
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Answer» ★Tan30° => 1/√3★Hope this will help u my frnd!!!?? 1/✓3 1/underroot 3 |
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| 39826. |
Which chapters are more important in class 10 mathematics book and give me tips to solve maths |
| Answer» Hey I think all but if you have to score the marks then firstly do all easy chapters first and then start with trigonometry as it comes of 12 marks and then mensuration part it\'s my advice best of luck | |
| 39827. |
In an equilateral triangle ABC,ADis an altitude drawn from Aon side BC.prove that 3/4AB^2 =AD^2 |
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| 39828. |
What Is Identities |
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Answer» Identities is a word Identities are based on its formulas In mathematics identity is an equality relation A=B ,such that A and B contain some varibles and A and B produce the same value as each other regardless of what values are substituted for the values. |
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| 39829. |
A number when divide by 61 gives 27 as quotient and 32 as remainder. Find the number. |
| Answer» The no. is 1679[61 × 27 = 16471647 + 32 = 1670] | |
| 39830. |
If the 9th term of an AP is 0,prove that its 29th term is twice its 19th term |
| Answer» Case1:a9=0 a+8d=0 a=-8d Case2:a29=a+28d =(-8d)+28d =20d And a19=a+18d =(-8d)+18d =10d Then,A29 /a19=20d/10d =2Hence,a29 is twice of a19. | |
| 39831. |
What is the value of 0/2# not defined # zero |
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Answer» 0 0 |
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| 39832. |
If nth term of an ap is 4n+2 then find its 8th term |
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Answer» 34 371 8th term is 35 35 |
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| 39833. |
Write the general form of positive even integer and positive odd integer |
| Answer» 2n for even and 2n+1 for odd | |
| 39834. |
1+4+7+10.............+x there Sn is 287find x |
| Answer» 38 | |
| 39835. |
What is types of solving euclid algorithm method |
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| 39836. |
In traingle ABC and traingle DEF ,AB/DE=BC/FD,are they similar ? |
| Answer» {tex}\\Delta FDE \\sim \\Delta CAB.{/tex} | |
| 39837. |
4marks questions |
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| 39838. |
Find all the other zeroes of the polynomial p(x) =2xraise to power 4-7xcube +19xsquare - 14x+30 |
| Answer» Let the polynomial is p(x) = 2x4+ 7x3\xa0-\xa019x2\xa0- 14x + 30and {tex} \\sqrt { 2 } \\text { and } - \\sqrt { 2 }{/tex}\xa0are zeroes of polynomial p(x).{tex}\\therefore ( x - \\sqrt { 2 } ) ( x + \\sqrt { 2 } ){/tex}\xa0= x2\xa0- 2 is a factor of given polynomial.On dividing given polynomial, we getquotient : 2x2\xa0+ 7x - 15So, the other factor is 2x2\xa0+ 7x - 15Now,\xa02x2\xa0+ 7x - 15 = 0{tex}\\Rightarrow{/tex} 2x2\xa0+ 10x - 3x - 15 = 0{tex}\\Rightarrow{/tex}2x(x + 5) - 3 (x + 5) = 0{tex}\\Rightarrow{/tex}(x + 5)(2x - 3) = 0{tex}\\therefore \\quad x = - 5 , x = \\frac { 3 } { 2 }{/tex}Hence Zeores of given polynomial are {tex}\\sqrt { 2 } , - \\sqrt { 2 } , - 5 \\text { and } \\frac { 3 } { 2 }{/tex}. | |
| 39839. |
Sum of two numbers is 9 and their differences is 3 . Find the numbers . |
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Answer» X+Y=9 . ......(1)X-Y=3...........(2)From (1) and (2)Y+3+Y=92Y=9-32Y=6Y=3Put value of Y in ....(1)From ...(1)X+3=9X=9-3X=6Hence, X=6,Y=3 Let the no.be x and yAcc. To que.X+Y=9X_Y=3Adding both we get2x=12x=6Putting value of x y=3 |
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| 39840. |
in triangle pab pa=pb and the area of the triangle is 10 sq u |
| Answer» PA =\xa0{tex}\\sqrt { ( 1 - x ) ^ { 2 } + ( 2 - y ) ^ { 2 } }{/tex}PB =\xa0{tex}\\sqrt { ( 3 - x ) ^ { 2 } + ( 8 - y ) ^ { 2 } }{/tex}{tex}\\therefore{/tex}\xa0PA = PBLet coordinates of P are (x, y){tex}\\Rightarrow{/tex}\xa0{tex} \\sqrt { ( 1 - x ) ^ { 2 } + ( 2 - y ) ^ { 2 } }{/tex}{tex}= \\sqrt { ( 3 - x ) ^ { 2 } + ( 8 - y ) ^ { 2 } }{/tex}Squaring on both sides,\xa0{tex}\\Rightarrow{/tex}\xa0{tex}(1 - x)^2 + (2 - y)^2 = (3 - x)^2 + (8 - y)^2{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}(1 - x)^2 - (3 - x)^2 = (8 - y)^2 - (2 - y)^2{/tex}{tex}\\Rightarrow{/tex}\xa0(1 - x - 3 + x) (1 - x + 3 - x) = (8 - y - 2 + y) (8 - y + 2 - y){tex}\\Rightarrow{/tex}- 2(4 - 2x) = 6(10 - 2y){tex}\\Rightarrow{/tex}\xa0{tex}- 2 \\times 2 ( 2 - x ) = 6 \\times 2 ( 5 - y ){/tex}- 2 + x = 15 - 3yx = 17 - 3y ...(i)Area of\xa0{tex}\\triangle{/tex}PAB = 10 sq units{tex}\\Rightarrow{/tex}\xa0{tex} \\frac { 1 } { 2 } | x ( 2 - 8 ) + 1 ( 8 - y ) + 3 ( y - 2 ) |{/tex}\xa0= 10{tex}\\Rightarrow{/tex}\xa0{tex}| - 6 x + 8 - y + 3 y - 6 |{/tex}\xa0= 20{tex}\\Rightarrow{/tex}\xa0{tex}| - 6 x + 2 y + 2 |{/tex}\xa0= 20{tex}\\Rightarrow{/tex}\xa0{tex}\xa0- 6x + 2y + 2 = 20 {/tex}or\xa0{tex}- 6x + 2y + 2 = -20{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}- 6(17 - 3y) + 2y + 2 = 20{/tex} or {tex}- 6(17 - 3y) + 2y + 2 = - 20 {/tex}...(ii){tex}\\Rightarrow{/tex}\xa0{tex}- 102 + 18y + 2y + 2 = 20{/tex} or\xa0{tex}- 102 + 18y + 2y + 2 = - 20{/tex}{tex}\\Rightarrow{/tex}\xa020y = 120 or 20y = 80y = 6 or y = 4When y = 6, equation (i) becomes x = 17 - 18 = - 1{tex}\\therefore{/tex}\xa0Point is (- 1, 6).When y = 4, equation (i) becomes x = 17 - 12 = 5{tex}\\therefore{/tex}\xa0Point is (5, 4). | |
| 39841. |
Write the write the 10th term of an ap who\'s and whose income is essay |
| Answer» Yes | |
| 39842. |
In an AP if d=-2 n=5 and tn=0 then find the vlaue of A. |
| Answer» Given:d = -2, n = 5 and an = 0We know that , an = a + (n - 1) d{tex}0 = a + (n - 1) d{/tex}{tex}[ a_n = 0]{/tex}{tex}0 = a + ( 5 - 1 ) (-2){/tex}0 = a + 4 {tex}\\times{/tex}\xa0- 2{tex}0 = a - 8{/tex}0 + 8 = aa = 8Hence, the value of a is 8. | |
| 39843. |
Your question paper of first term |
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| 39844. |
Given sequence is5,11,21,35,53......find nth term |
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| 39845. |
Circumference of the circle |
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Answer» Tume itna nahi ataa you are very weak C= 2πr C=2πr r CircleSolve for circumferenceC=2πrr\tRadius Hii |
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| 39846. |
Find the LCM and HCF 256 and 36 and verify that LCM×HCF =product of the given numbers |
| Answer» Factors of 256=2*2*2*2*2*2*2*2*factors of 36=2*2*3*3LCM=2304HCF= 4 | |
| 39847. |
Find a quadratic polynomial whose sum and product of the zeroes are -7and-18 |
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Answer» X2+7x-18 x2+7x-18 Sum of zeroes = -7 (S) Product of zeroes =-18(P) P(x)= k(x2-Sx +P) =k(x2+7x-18)Here K = 1 P(x) = x2+7x-18 |
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| 39848. |
Find the solution of the linear equation x/a + y/b = a+3 , x/a^2 + y/b = 2 |
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| 39849. |
Am is perpendicular to bc tan b is 3/4 and tan c is 5/2 bc is 56 cm find l of am |
| Answer» In right\xa0{tex}\\triangle {/tex}AMB,\xa0tan B =\xa0{tex}\\frac { 3 } { 4 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { A M } { B M } = \\frac { 3 } { 4 }{/tex}{tex}\\Rightarrow{/tex}\xa04AM = 3BM {tex}\\Rightarrow{/tex}BM =\xa0{tex}\\frac { 4 } { 3 }{/tex}AM ...(i)In right\xa0{tex}\\triangle{/tex}AMC,tan C =\xa0{tex}\\frac { \\mathrm { AM } } { \\mathrm { MC } }{/tex}{tex}\\Rightarrow \\frac { 5 } { 12 } = \\frac { \\mathrm { AM } } { \\mathrm { MC } }{/tex}{tex}\\Rightarrow{/tex}\xa0MC =\xa0{tex}\\frac { 12 } { 5 }{/tex}\xa0AM ...(ii)Now, BM + MC = BC{tex}\\frac { 4 } { 3 }{/tex}AM +\xa0{tex}\\frac { 12 } { 5 }{/tex}AM = 56AM\xa0{tex}\\left( \\frac { 4 } { 3 } + \\frac { 12 } { 5 } \\right){/tex}\xa0= 56AM\xa0{tex}\\left( \\frac { 20 + 36 } { 15 } \\right){/tex}\xa0= 56{tex}\\Rightarrow{/tex}\xa0AM =\xa0{tex}\\frac { 56 \\times 15 } { 56 }{/tex}= 15 cm | |
| 39850. |
What is polynomial.? |
| Answer» All the algebric expressions, have only whole numbers are the components of variable. | |