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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.
| 34601. |
Hcf of 328 and 678 |
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Answer» I was just checking this feature 2 |
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| 34602. |
For the p(x)=ax2+bx+c if α and β are zeros then find. 1. α3+β3 and 2. 1/α3+1/β3 |
| Answer» It is given that {tex} \\alpha{/tex}\xa0and {tex} \\beta{/tex}\xa0are the zeros of the quadratic polynomial\xa0{tex}f(x)=ax^2+bx+c{/tex}\xa0{tex} \\therefore \\quad \\alpha + \\beta = - \\frac { b } { a } \\text { and } \\alpha \\beta = \\frac { c } { a }{/tex}Now,\t{tex}\\alpha^3 + \\beta ^3{/tex}\t= {tex}(\\alpha + \\beta )(\\alpha^2 -\\alpha\\beta + \\beta ^2){/tex}\t=\xa0{tex}(\\alpha + \\beta)( \\alpha^2+\\beta^2-\\alpha\\beta+2\\alpha\\beta-2\\alpha\\beta){/tex}\t=\xa0{tex} (\\alpha + \\beta)[(\\alpha + \\beta)^2-3\\alpha \\beta]{/tex}\t=\xa0{tex}(\\alpha + \\beta)^3-3\\alpha \\beta(\\alpha + \\beta){/tex}\t=\xa0{tex} \\frac { - b ^ { 3 } +3abc} { a ^ { 3 } }{/tex}\t{tex}=\\frac{-b^3+3abc}{a^3}{/tex}\xa0\t\t{tex} \\frac { 1 } { \\alpha ^ { 3 } } + \\frac { 1 } { \\beta ^ { 3 } } = \\frac { \\alpha ^ { 3 } + \\beta ^ { 3 } } { ( \\alpha \\beta ) ^ { 3 } }{/tex}\t{tex}=\\frac { (\\alpha + \\beta)( \\alpha^2+\\beta^2-\\alpha\\beta) } { ( \\alpha \\beta ) ^ { 3 } }{/tex}\t{tex}=\\frac { (\\alpha + \\beta)( \\alpha^2+\\beta^2-\\alpha\\beta+2\\alpha\\beta-2\\alpha\\beta) } { ( \\alpha \\beta ) ^ { 3 } }{/tex}\t{tex}=\\frac { (\\alpha + \\beta)[(\\alpha + \\beta)^2-3\\alpha \\beta] } { ( \\alpha \\beta ) ^ { 3 } }{/tex}\t{tex}=\\frac { [(\\alpha + \\beta)^3-3\\alpha \\beta(\\alpha + \\beta)] } { ( \\alpha \\beta ) ^ { 3 } }{/tex}\t{tex} = \\frac { \\frac { - b ^ { 3 } +3abc} { a ^ { 3 } } } { \\left( \\frac { c } { a } \\right) ^ { 3 } } {/tex}\t{tex}=\\frac{-b^3+3abc}{a^3}\\;\\times\\;\\frac{a^3}{c^3}{/tex}\t{tex}= \\frac { 3 a b c - b ^ { 3 } } { c ^ { 3 } }{/tex}\t | |
| 34603. |
Express 280 in least common factor |
| Answer» | |
| 34604. |
3x/2-5y/3=-2. ,X/3+y/2=13/6 |
| Answer» The given system of equations may be written as{tex}9x -10y + 12 = 0{/tex} ...(i){tex}2x + 3y - 13 = 0{/tex}.... (ii)From (ii), we get\xa0{tex} y = \\frac { 13 - 2 x } { 3 }{/tex}Substituting\xa0{tex} y = \\frac { 13 - 2 x } { 3 }{/tex}\xa0in (i), we get{tex} 9 x - \\frac { 10 ( 13 - 2 x ) } { 3 } + 12 = 0{/tex}{tex} \\Rightarrow{/tex}\xa0{tex}27x - 10(13 - 2x) + 36 = 0{/tex}{tex} \\Rightarrow{/tex}\xa0{tex}27x -130 + 20x + 36 = 0{/tex}{tex} \\Rightarrow{/tex}\xa0{tex}47x - 94 = 0{/tex}{tex} \\Rightarrow{/tex}\xa0{tex}47x = 94{/tex}{tex} \\Rightarrow x = \\frac { 94 } { 47 } = 2{/tex}Substituting {tex}x = 2{/tex} in (i), we get9\xa0{tex} \\times{/tex} 2 - 10y + 12 = 0\xa0{tex} \\Rightarrow 10 y = 30 {/tex}{tex}\\Rightarrow y = \\frac { 30 } { 10 } = 3{/tex}Hence, x = 2 and y = 3 is the required solution. | |
| 34605. |
Check if 3q divided by p and 2q are the zeroes of the polynomial px2 + (4p2 - 3q) x - 12pq |
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| 34606. |
The middle term splitting of 2 X square + 3x - 90 = 0 |
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Answer» 2x^2 +3x - 90 =02x^2 +15x -12x - 90 =02x^2 - 12x +15x - 90 =02x(x-6) +15(x-6) =0(2x+15)(x-6) =0 so, x= -15/2 or x= 6 You solve solve it using any other method bcoz there are some equations which are difficult to be solved with middle splitting method. |
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| 34607. |
Find the HCF of 56,96, and 324 by Euclid\'s algorithms |
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Answer» Euclid division lemma:-a=bq+r324 = 96(3) + 3696 = 36(2) + 2436 = 24(1) + 1224 = 12(2) + 0hcf of 324 and 96 is 12.Now find the hcf of 12 and 56,56 = 12(4) + 812 = 8(1) + 48 = 4(2)+0hcf of 12 and 56 is 4.Therefore, hcf of 56.96 and 324 is 4 4 |
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| 34608. |
Matrix A= [cos x -sinx and A**32 =[0 -1 sin x cosx]. 1 0] then x=? |
| Answer» | |
| 34609. |
math board paper came from ncert book or from some where else??? |
| Answer» Both from book and reference book.. | |
| 34610. |
Find largest four digit number which when divided 4 , 7, 13 leaves remainder 3 in each case |
| Answer» 367 Is the required number | |
| 34611. |
Equal to -1 and B equal to 1 by 2 |
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Answer» What kya nikalna h To kay |
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| 34612. |
A=-1,d=1÷2 |
| Answer» To kya | |
| 34613. |
Proof of BPT theorem? |
| Answer» Given :\xa0In {tex}\\triangle A B C{/tex}, DE || BC and intersects AB in D and AC in E.\xa0Prove that :\xa0{tex}\\frac{AD}{DB} = \\frac{AE}{EC}{/tex}Construction:\xa0Join BC, CD and draw EF {tex}\\perp{/tex} BA and DG {tex}\\perp{/tex} CA. Now from the given figure we have,EF {tex}\\perp{/tex} BA (Construction)EF is the height of ∆ADE and ∆DBE (Definition of perpendicular)Area({tex}\\triangle{/tex}ADE) ={tex}\\frac{AD.EF}{2}{/tex} .....(1)Area({tex}\\triangle{/tex}DBE) = {tex}\\frac{DB.EF}{2}{/tex} ....(2)Divide the two equations we have{tex}\\frac{Area \\triangle ADE}{Area \\triangle DBE} = \\frac{AD}{DB}{/tex} .....(3){tex}\\frac{Area \\triangle ADE}{Area \\triangle DEC} = \\frac{AE}{EC}{/tex} .....(4)Therefore, {tex}\\triangle \\mathrm{DBE} \\sim \\triangle \\mathrm{DEC}{/tex} (Both the ∆s are on the same base and\xa0between the same || lines).....(5)Area({tex}\\triangle{/tex}DBE) = Area({tex}\\triangle{/tex}DEC) (If the two triangles are similar their\xa0areas are equal){tex}\\frac{AD}{DB} = \\frac{AE}{EC}{/tex}\xa0[from equation 3,4 and 5]Hence proved. | |
| 34614. |
Kx+y=ksquare and x+ky=1 have infinitely many solutions |
| Answer» {tex}k x + y = k ^ { 2 } \\text { and } x + k y = 1{/tex}{tex}a_1 = k , b_1 = 1, c_1= k^2{/tex}and {tex}a_2 = 1, b_2 = k , c_2 = 1{/tex}{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { k } { 1 } , \\frac { b _ { 1 } } { b _ { 2 } } = \\frac { 1 } { k } , \\frac { c _ { 1 } } { c _ { 2 } } = \\frac { k ^ { 2 } } { 1 }{/tex}For infinitely many solution{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { b _ { 1 } } { b _ { 2 } } = \\frac { c _ { 1 } } { c _ { 2 } }{/tex}{tex}\\frac { k } { 1 } = \\frac { 1 } { k } = \\frac { k ^ { 2 } } { 1 } {/tex}{tex}\\frac { k } { 1 } = \\frac { 1 } { k }{/tex}{tex} k ^ { 2 } = 1{/tex}{tex}k = \\pm 1{/tex} | |
| 34615. |
Euclid |
| Answer» Great mathematician ? | |
| 34616. |
If a and b are two prime numbers then find hcf(a,b). |
| Answer» 1 | |
| 34617. |
√cosecx+1/cosecx-1=cosx/1-sinx=cotx/cosecx-1 |
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| 34618. |
What is the HCF of Smallest prime number and smallest composite number |
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Answer» Prime number is 2Composite number is 1 Answer is 2 because smallest prime number is 2 and smallest composite number is 4 . Smallest prime number = 2Smallest composite number = 4Factors of 2 are 1 and 2Factors of 4 are 1, 2 and 4Highest common factor (HCF) = 2 1 2 29 |
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| 34619. |
What is geomitry |
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Answer» Geometry is the\xa0branch of mathematics concerned with the properties and relations of points, lines, surfaces, solids, and higher dimensional analogues. 230-220×(0.5)=? |
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| 34620. |
What is meant by a prime and Real number |
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Answer» Real numbers means in which the number is divided by 1and itself prime nos are nos which comes in 1 and its own table |
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| 34621. |
IF A(5,2) B(2,-2) and C(-2,t) are the vertices of a rightangled triangle with |
| Answer» we are given that A (5, 2), B (2, - 2) and C (-2, t) are the vertices of a right-angled triangle with {tex}\\angle B{/tex} = 90°.AB2 = ( 2 - 5)2 + ( - 2 - 2)2 =9 + 16 = 25BC2 = (- 2 - 2)2 + (t + 2)2 = 16 + (t + 2)2AC2 = (5 + 2)2\xa0+ (2 - t)2 = 49 + (2 - t)2Since {tex}\\triangle{/tex}ABC is a right angled triangle{tex}\\therefore{/tex}\xa0AC2 = AB2 + BC2or, 49 + (2 - t)2 = 25 + 16 + (t + 2)2or, 49 + 4 - 4t + t2 = 41 + t2\xa0+ 4t + 4or, 53 - 4t = 45 + 4tor, 8t = 8{tex}\\therefore{/tex}\xa0t =1 | |
| 34622. |
if the point p(x,y) is equidistant from the point A(a+b,b-a) and B(a-b,a+b) prove that bx=ay |
| Answer» |PQ| = |PR{tex}\\begin{aligned} \\sqrt { [ x - ( a + b ) ] ^ { 2 } + [ y - ( b - a ) ] ^ { 2 } } = \\sqrt { [ x - ( a - b ) ] ^ { 2 } + [ y - ( b + a ) ] ^ { 2 } } \\end{aligned}{/tex}Squaring, we get[x - (a + b)]2 + [y - (b\xa0- a)]2\xa0= [x - (a - b)]2 + [y - (a + b)]2or, [x - (a + b)]2 - [x - a + b]2\xa0= (y - a - b)2 - (y - b + a)2or, (x - a - b + x - a + b) ( x - a - b - x + a - b)= (y - a - b + y - b + a)(y - a - b - y + b - a)or, (2x - 2a) (- 2b) = (2y - 2b) (- 2a)or, (x - a)b = (y - b)aor, bx = ay.Hence Proved. | |
| 34623. |
Tan 48 ÷ cot 64 |
| Answer» Illogical | |
| 34624. |
HCF of smallest composite number and smallest prime number. |
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Answer» 2 2 bcoz smallest composite no. Is 4 and smallest prime no. Is2 1 |
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| 34625. |
What are the projects for AP??? |
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| 34626. |
Prove that 2 root 3 -1 is an irrational number |
| Answer» Let us assume that {tex}2\\sqrt 3 - 1{/tex}\xa0is a rational.\xa0numberThen, there exist positive co-primes a and b such that{tex}2\\sqrt 3 - 1 = \\frac{a}{b}{/tex}{tex}2\\sqrt 3 = \\frac{a}{b} + 1{/tex}{tex}\\begin{array}{l}2\\sqrt3=\\frac{\\mathrm a+\\mathrm b}{\\mathrm b}\\\\\\end{array}{/tex}{tex}\\sqrt 3 = \\frac{{a + b}}{{2b}}{/tex}Here\xa0{tex}\\begin{array}{l}\\frac{\\mathrm a+\\mathrm b}{\\mathrm b}\\\\\\end{array}{/tex}\xa0is a rational number ,so {tex}\\sqrt3{/tex}\xa0is a rational numberThis contradicts the fact that {tex}\\sqrt 3{/tex}\xa0is an irrational numberHence {tex}2\\sqrt 3 - 1{/tex}\xa0is irrational | |
| 34627. |
Find the quadratic equation by factorisation of 2 x square minus x + 1 by 8 is equal to zero |
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| 34628. |
Find quadratic equations by factorisation of2x-x1/8=0 |
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| 34629. |
tan A- 2 cos A tan A +2cos A-1=0, then find the value of A . |
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| 34630. |
∆ABC~∆DEF, If DE = 2AB and BC= 3cm then EF is equal to |
| Answer» Given that\xa0{tex}\\triangle{/tex}ABC ~ {tex}\\triangle{/tex}DEFWe know that when two triangles are similar, then the ratios of the lengths of their corresponding sides are equal.\xa0{tex}\\Rightarrow \\frac { A B } { D E } = \\frac { B C } { E F }{/tex}{tex}\\Rightarrow \\frac { 1 } { 2 } = \\frac { 3 } { E F }{/tex}{tex}\\Rightarrow{/tex}\xa0EF = {tex}\\frac 32{/tex} cm | |
| 34631. |
2 (ax-by)+(a+4b)=02(bx+ay)+(b-4a)=0This is question of pair of linear equation in two variable... |
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Answer» I have done.. Thnx for suggestion... Bracket open krke subtract n sumplify.... |
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| 34632. |
Check whether 14n can end with the digit zero for any natural number n. Explain |
| Answer» No 14n cannot be end with digit 0 becoz it does not have 5 in its prime factors. | |
| 34633. |
9x2-9(p+q)x+(2p2+5pq+2q2)solve this for x |
| Answer» Given{tex}9 x ^ { 2 } - 9 ( a + b ) x + 2 a ^ { 2 } + 5 a b + 2 b ^ { 2 } = 0{/tex}Now,\xa0{tex}2 a ^ { 2 } + 5 a b + 2 b ^ { 2 } = 2 a ^ { 2 } + 4 a b + a b + 2 b ^ { 2 }{/tex}{tex}= 2 a [ a + 2 b ] + b [ a + 2 b ]{/tex}= (a + 2b) (2a + b)Hence the given equation becomes,{tex}9 x ^ { 2 } - 9 ( a + b ) x + ( a + 2 b ) ( 2 a + b ) = 0{/tex}or,\xa0{tex}9 x ^ { 2 } - 3 [ 3 a + 3 b ] x + ( a + 2 b ) ( 2 a + b ) = 0{/tex}{tex}9x^2-3(a+2b)x-3(2a+b)x+(a+2b)(2a+b)=0{/tex}{tex}3x[3x-(a+2b)]-(2a+b)[3x-(a+2b)]=0{/tex}or,\xa0{tex}[ 3 x - ( a + 2 b ) ] [ 3 x - ( 2 a + b ) ] = 0{/tex}{tex}\\Rightarrow\\ either\\ 3x=a+2b\\ or\\ 3x=2a+b{/tex}{tex}\\Rightarrow\\ either\\ x=\\frac{a+2b}{3}\\ or\\ x=\\frac{2a+b}{3}{/tex}Hence, the roots =\xa0{tex}\\frac { a + 2 b } { 3 } , \\frac { 2 a + b } { 3 }{/tex} | |
| 34634. |
Comparment paper 2019 |
| Answer» Tumhe dena hai kya compartment | |
| 34635. |
P(x)=x²-2x-8 splitting middle term |
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Answer» P(x) = x² - 2x - 8 = x² - 4x + 2x - 8 = x(x - 4) + 2(x - 4) = (x - 4) (x - 2) P(x) = x² - 2x - 8= x2 - 4x + 2x - 8= x(x - 4) + 2 ( x - 4)= (x - 4) ( x + 2) |
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| 34636. |
x-y=8. 3x-3y=16 by table method |
| Answer» x - y = 8.................(1)3 x - 3 y = 16.............(2)Here,\xa0{tex}a _ { 1 } = 1 , b _ { 1 } = - 1 , c _ { 1 } = - 8{/tex}{tex}a _ { 2 } = 3 , b _ { 2 } = - 3 , c _ { 2 } = - 16{/tex}We see that\xa0{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { b _ { 1 } } { b _ { 2 } } \\neq \\frac { c _ { 1 } } { c _ { 2 } }{/tex}Hence, the lines represented by the equations(1) and (2) are parallel.Therefore, equations (1) and ( 2 ) have no solution, i.e., the given pair of linear equation is inconsistent. | |
| 34637. |
How the pi value is calculated? |
| Answer» The circumference of a circle is found with the formulaC= π*d = 2*π*rThus pi equals a circle\'s circumference divided by its diameter. | |
| 34638. |
2x_3y=6 |
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| Answer» According to question given equation is{tex}2x - 3y = 6{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}-3y = 6 - 2x{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}3y = 2x - 6{/tex}{tex}\\Rightarrow{/tex}\xa0y =\xa0{tex}\\frac{2x - 6}{3}{/tex}The solution table of the equation {tex}2x - 3y = 6{/tex} is\xa0 | x | y | \t03-3-20-4\tx - 6 = 0{tex}\\Rightarrow{/tex}\xa0x = 6Plotting the points on graph, we getthe unique solution is x = 6, y = 2.|
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| 34639. |
a2x2_3abx+2b2 |
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| 34640. |
2-(-6+3) =? |
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Answer» 5 5 is correct answer 2-(-6+3)= 2 - (-3)= 2 + 3= 5 5 Sorry its 5 not 55 typed by mistake 55 5 |
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| 34641. |
1/7x+1/6y=3 and 1/2x-1/3y=5 |
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Answer» Please solve this X=1/14 and Y=1/6 |
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| 34642. |
Is 2√x+5(√x)ka whole cube a polynomial?Give reason |
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| 34643. |
Easy way how to square numbers |
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| 34644. |
What is cross multiplication |
| Answer» The below equation is solved by cross multiplication method:Given 5x + 4y - 4 = 0 .........(i)x - 12y - 20 = 0 ........(ii)Here, a\u200b\u200b\u200b\u200b\u200b\u200b1\u200b = 5, b1\u200b\u200b\u200b\u200b\u200b= 4, c\u200b\u200b\u200b1=\xa0-4a\u200b\u200b\u200b\u200b\u200b\u200b2\u200b= 1, b\u200b\u200b\u200b\u200b\u200b\u200b2\xa0= - 12, c\u200b\u200b2\xa0= - 40By cross-multiplication method,{tex}\\frac { x } { b _ { 2 } c _ { 1 } - b _ { 1 } c _ { 2 } } = \\frac { y } { c _ { 1 } a _ { 2 } - c _ { 2 } a _ { 1 } } = \\frac { 1 } { a _ { 1 } b _ { 2 } - a _ { 2 } b _ { 1 } }{/tex}{tex}\\frac { x } { - 80 - 48 } = \\frac { y } { - 4 + 100 } = \\frac { 1 } { - 60 - 4 }{/tex}{tex}\\frac { x } { - 128 } = \\frac { y } { 96 } = \\frac { 1 } { 64 }{/tex}{tex}\\frac { x } { - 128 } = \\frac { 1 } { - 64 } \\text { and } \\frac { y } { 96 } = \\frac { 1 } { - 64 }{/tex}Hence,\xa0{tex}x = 2 \\text { and } y = \\frac { - 3 } { 2 }{/tex} | |
| 34645. |
What number squared= 12345678987654321? |
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Answer» ☒⛝ ?? No.thanks.....no sorry.......... Right answer bestie....thnx? 111111111 |
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| 34646. |
A man sold a chair and a table together for rs1520 thereby making a profit of 25 percent |
| Answer» He makes a profit of rs 304 | |
| 34647. |
135and225 find the HCF |
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Answer» 135)225(1 - 135 ----------- 90)135(1 - 90 -------- 45)90(2 - 90 --------- 0 ---------: hcf of 135 and 225 is 45 135 and 225Since 225 > 135, we apply the division lemma to 225 and 135 to obtain225 = 135 × 1 + 90Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain135 = 90 × 1 + 45We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain90 = 2 × 45 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 45,Therefore, the HCF of 135 and 225 is 45. |
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| 34648. |
route 3 sin x is equal to cos x |
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Answer» {tex}\\sqrt 3{/tex} sin x = cos xsin x / cos x = 1 / {tex}\\sqrt 3{/tex}tan x = 1/{tex}\\sqrt 3{/tex}tan 30 = 1 /{tex}\\sqrt 3{/tex}Hence x = 30 Cot is equal to 30 |
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| 34649. |
Cos 90 - sin 90 |
| Answer» 0 - 1 = -1 | |
| 34650. |
(A+b )2 |
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Answer» (a+b)²=a²+b²+2ab a2+b2+2ab. a2 + 2ab + b2 |
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