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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.
| 38601. |
Obtain the other zeroes of the polynomial 9x4-6x3-35x2+24x-4, if two of its zeroes are 2 and -2 ? |
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Answer» Is it correct question.........???????☺☺☺☺ And then factorise the quotient P(x)= 9x4-6x3-35x2+24x-4Alpha×beta =-2Alpha+beta=2(x2-2x-2)÷by p(x) |
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| 38602. |
Rd sharma questions |
| Answer» What does this mean? | |
| 38603. |
Why the value of pie is 22/7 |
| Answer» This is NTNR | |
| 38604. |
If x+a is a factor of 2x+2ax+5x+10 |
| Answer» Here the given polynomial f(x)=2x2 + 2ax + 5x +10If x+a is a factor of f(x) then x+a=0 or x = -a{tex}\\Rightarrow{/tex}f(-a) = 0{tex}\\Rightarrow{/tex}2(-a)2 + 2a.(-a) + 5(-a) + 10 =0{tex}\\Rightarrow{/tex}2a2 - 2a2 -5a +10 =0{tex}\\Rightarrow{/tex}-5a = -10{tex}\\Rightarrow{/tex}a=2. | |
| 38605. |
Given, tan theta- cot theta =7, find tan^3 theta + cot^theta |
| Answer» | |
| 38606. |
Questoin paper of last year |
| Answer» You can get it from internet | |
| 38607. |
Find quadratic polynomial whose zeroes (5+2√3)and (5-2√3) |
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Answer» Alfa=5+2root3 beta=5-2root3Alfa + beta = 5+2root3+5-2root3 +2root3 - 2root3 will be cancelledSo alfa+beta=5+5=10Alfa*beta=(5+2root3)*(5-2root3) =25-10root3+10root3-12 =13The quadratic polynomial is xsquare -(alfa+beta)x + (alfa*beta)So the polynomial be will xsquare -10x +13 Are bhai question aur answer khud he post karte ho Thanks booe de bhai x2-10x+25-4√3 X×x-10x+13 |
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| 38608. |
Identities of trigonometry |
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Answer» How many you wants? There are many identities. |
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| 38609. |
prove that 2+2= 5 |
| Answer» galti se | |
| 38610. |
All the identity of chapter introduction to trigonometry |
| Answer» There are many identities....how many you want? | |
| 38611. |
Prove that diameter of circle bisects two equal part |
| Answer» | |
| 38612. |
Are yaar tum log ques. Kyu pooch re ho ab to result bi aa gaya khush rao |
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Answer» It\'s so because we all are in class 10 Aap abhi 11 me hai lagta hai aapko pata nahi hai Kuki mey 10 th mey hu |
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| 38613. |
Prove. That _/2 _/3 is irrational no |
| Answer» Given in maths book | |
| 38614. |
Show that the square of an odd integer is of the form 4 Q + 1 for some integer q |
| Answer» Let x\xa0be an odd integer.On dividing x\xa0by 4, let\xa0q\xa0be the quotient and\xa0r\xa0be the remainder.So, by Euclid\'s division lemma, we have: x\xa0= 4q\xa0+\xa0r, where 0 <_\xa0r\xa0< 4.Therefore, x2\xa0= ( 4q + r\xa0)2\xa0. = 16q2 +\xa0r2\xa0+ 8qr.....(i).where 0 <_\xa0r\xa0<\xa04.Case l. When\xa0r\xa0= 0 Putting\xa0r\xa0= 0 in (I), we get:\xa0 x2\xa0= 16q2\xa0\xa0= 4 ( 4q2\xa0) = 4 Q, where Q = 4q2\xa0is an integer.Case ll. When\xa0r\xa0= 1\xa0 Putting\xa0r\xa0= 1 in (I), we get:\xa0 x2\xa0= ( 16q2\xa0+ 1 + 8q\xa0) = 4 ( 4q2\xa0+ 2q\xa0) + 1 = ( 4 Q + 1), where Q = ( 4q2\xa0+ 2q\xa0) is an integer.\xa0Clearly, 4Q is even and since x is odd. So x is not equal to 4Q.Hence, the square of an odd integer is of form ( 4Q + 1) for some integer Q.\xa0\xa0 | |
| 38615. |
What is the value of pie? |
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Answer» The exact value of pie in between 22/7 and 3.14 22/7 3.14or22/7 The value of pie does not give us completely 3.14 also 22/7 |
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| 38616. |
In triangle, angle B=90,ifAB=2cm and AC=3cm,than find all trigonometric ratio |
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Answer» What Cc?? Cc After finding the side BC.....(root 5)Sin theta = 2/3Cos theta = root 5/3Tan theta = 2/root 5Cot theta = root 5/2Sec theta = 3/root 5Cosec theta = 3/2 |
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| 38617. |
Playing with number |
| Answer» Dont woery u are in merit list | |
| 38618. |
Find quadric equation ax2+dx+c |
| Answer» | |
| 38619. |
Koi easy trick nahi h exercise 3 ka |
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Answer» Hey, have you changed your last name Which chapter priyanshu Konsi ch ki exercise |
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| 38620. |
5 year questions paper with solutions |
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Answer» You can take out from internet Zy. But how |
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| 38621. |
Sccfgb13566 |
| Answer» What happens | |
| 38622. |
If one zero of the polynomial (a2+9)x2+13x+6a is reciprocal 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. | |
| 38623. |
Find the zeroes of the polynomial : 1.x2+x-a (a+1) |
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Answer» How the zeroes are a and a+1 can u plz give detailed answer Zeros are:- -(a+1) and a. Is a+1 a part of the whole equation?? Plz someone answer it? |
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| 38624. |
Find the H.C.F of 65 and 117 and express it in form 65m +117n |
| Answer» First find the HCF of 65 and 117 by Using Euclid\'s division algorithm,117 = 65{tex}\\times{/tex} 1 + 5265 = 52{tex}\\times{/tex} 1 + 1352 = 13{tex}\\times{/tex} 4 + 0So, HCF of 117 and 65 = 13HCF = {tex}65m + 117n{/tex}For, {tex}m= 2{/tex} and {tex}n = -1{/tex},HCF = 65{tex}\\times{/tex} 2 + 117{tex}\\times{/tex} (-1)= 130 - 117= 13Hence, the integral values of m and\xa0n are 2 and -1 respectively and the HCF of 117 and 65 is 13. | |
| 38625. |
In 8.4 Question 4th part of 5th |
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| 38626. |
Hcf edi |
| Answer» | |
| 38627. |
How to prove proposnality their of triangles |
| Answer» What do you mean by personality? | |
| 38628. |
5437× 1003 by suitable arrangement |
| Answer» 5437 × 1001 = 5437 × (1000 + 1)= 5437 × 1000 + 5437 × 1= 54,37,000 + 5437 = 54,42,437 | |
| 38629. |
√X +y =43 upon x + √y =55 find the value X and y |
| Answer» Itna padh k kya karega thoda chill mar | |
| 38630. |
How to psss mathmatics compartment exam |
| Answer» Chaman cutiya churan phunk k | |
| 38631. |
how to apply rechecking of cbse class 10 board result from whom website |
| Answer» | |
| 38632. |
Find the zeroes of x^2+3/4x+5 |
| Answer» Guys plz answer it | |
| 38633. |
In the figure AD=4cm BD=3cm CB=12cm then find the value of cot theta |
| Answer» In right\xa0{tex}\\triangle{/tex}ABD,\xa0{tex}\\angle D = 90 ^ { \\circ }{/tex}{tex}AB^2 = AD^2 + BD^2{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}AB^2 = 4^2 + 3^2{/tex}{tex}\\Rightarrow{/tex}\xa0AB =\xa0{tex}\\pm 5{/tex}\xa0cm{tex}\\Rightarrow{/tex}\xa0{tex}AB = 5 cm{/tex} (Neglecting negative value)In right\xa0{tex}\\triangle{/tex}ABC , cot\xa0{tex}\\theta{/tex}\xa0=\xa0{tex}\\frac { \\mathrm { BC } } { \\mathrm { AB } } = \\frac { 12 } { 5 }{/tex} | |
| 38634. |
What is heron\'s formula |
| Answer» RootS(S-a)(S-b)(S-c) | |
| 38635. |
If 9×5×2 =529and 4×7×2=724then find the value of 3×9×8 |
| Answer» 983 | |
| 38636. |
What is the relation between sin snd cot |
| Answer» Cot theta == Sin theta / cos theta..So,, Sin theta==cot theta×cos theta | |
| 38637. |
Px +qy = p-q Qx-py =p+q |
| 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. | |
| 38638. |
Show that 247 And187 are co-prime. |
| Answer» | |
| 38639. |
How to calculate persentage |
| Answer» Got marks÷total marks×100 | |
| 38640. |
How to know the given frequency is cumulative frequency or simple frequency?? |
| Answer» | |
| 38641. |
Parallelogram properties |
| Answer» Opp.angle sum 180\'.Diagonals bisect .Opp sides are ll and equal.Any diagonal divide into two. congruent triangle having equal area. | |
| 38642. |
We can apply the both form of rechecking and supplyment or not |
| Answer» Yee | |
| 38643. |
What is the rechecking form date |
| Answer» | |
| 38644. |
Is it possible to have two numbers whose HCF is 18 and LCM is 760 |
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Answer» Yes correct No cuz H.C.F. is always factor of lcm |
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| 38645. |
3x2-4√3x+4=0 find roots |
| Answer» The given quadratic equation is{tex}3 x ^ { 2 } - 4 \\sqrt { 3 } x + 4 = 0{/tex}Here, a = 3, {tex}b = - 4 \\sqrt { 3 }{/tex}, c = 4{tex}\\therefore{/tex} discriminant = b2 - 4ac{tex}= ( - 4 \\sqrt { 3 } ) ^ { 2 } - 4 ( 3 ) ( 4 ){/tex}= 48 - 48 = 0Hence, the given quadratic equationhas two equal real roots.The roots are {tex}= - \\frac { b } { 2 a } , - \\frac { b } { 2 a }{/tex}{tex}\\text { i.e. } - \\frac { ( - 4 \\sqrt { 3 } ) } { 2 \\times 3 } , - \\frac { ( - 4 \\sqrt { 3 } ) } { 2 \\times 3 } , \\text { i.e. } \\frac { 2 } { \\sqrt { 3 } } , \\frac { 2 } { \\sqrt { 3 } }{/tex} | |
| 38646. |
How to prepare the whole maths book of class 10 in a month with rd shrma for board examnation |
| Answer» No it is not possible , you have to do revision after completing a chapter. | |
| 38647. |
x2 -b2=(2x-a) |
| Answer» | |
| 38648. |
For any positive integer n,prove that n3-3 is divisible by 6 |
| Answer» n3\xa0- n = n (n2\xa0- 1) = n (n - 1) (n + 1)\xa0Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.⇒ n (n – 1) (n + 1) is divisible by 3.\xa0Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q, then n is divisible by 2.If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.⇒ n (n – 1) (n + 1) is divisible by 2.Since, n (n – 1) (n + 1) is divisible by 2 and 3.∴ n (n-1) (n+1) = n3\xa0- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)\xa0 | |
| 38649. |
Difference between lemma and algorithm |
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Answer» Difference Algorithm: an algorithm is a series of well defined steps which gives a procedure for solving a type of problem.Lemma: lemma its proven statement used to prove another statement . A lemma is a proven statement used for proving another statement. An algorithm is a series of well defined steps which gives a procedure for solving a type of problem |
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| 38650. |
Factorise by splitting the middle term method -25x+156=0 |
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Answer» Sorry it can\'t be split because splitting is done with middle term but this equation has not middle term. It can be done in quadratic equation only. Nyx |
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