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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.
| 33701. |
85÷_=8500 |
| Answer» 0.01 | |
| 33702. |
Find K so that (x^2+3x+k) is a factor of (x^4 -5x^2+4 ). |
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Answer» Hi hiiii |
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| 33703. |
Find four solutin for eqn. 4x - 2y = 5 |
| Answer» (1.25,0) ,(0.75,-1) , (1,-0.5) ,(0,-2.5) | |
| 33704. |
If the mean of 1,,3,4,5,6,7,4 is (m) and mean of 3,2,2,4,3,3 (p) is (m-1) and median is then p+q= |
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| 33705. |
X³+13x²+32x+20 |
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Answer» The factors are (x+1)(x+2)(x+10)And the zeroes are - 1, - 2 & - 10 Kya krna h iska |
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| 33706. |
Value of sin 30 |
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Answer» Value of sin 30 is 1/2 Under root 3upon 2 |
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| 33707. |
Find HCF and Lcm of 18 and 24 by the prime factorisation method |
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Answer» Love you 24=18 (1)+618=6(3)+0HCF=6 |
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| 33708. |
Form a cubic polynomial with zeroes 3, 2 and -1 |
| Answer» X cube - (alpha + bita + gama)x square+ ( alpha bita + bita gama + gama alpha)x+( alpha bota gama)Where alphabita ,gama is zeros of this polynomials | |
| 33709. |
Tanx=3cotx find the value of x |
| Answer» x = 60 | |
| 33710. |
Factories by splitting Middle term 2x²+5√3+6 |
| Answer» 2x^2+root3x+4root3x+6X(2x+root3)+2root3(2x+root3)(2x+root3)(x+2root3) | |
| 33711. |
Verification of pythagorus theorum |
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| 33712. |
What is realtion between zero and cofficient of cubic polynomial not quadratic |
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| 33713. |
find zeros of cubic equation 2x3+x2-5x+2 |
| Answer» | |
| 33714. |
Sin60°•cos30° + sin30°•cos60 |
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Answer» *Shivam Mishra* ?? 1 1 Thanks for ur help shivam mishra???? Gghhygg |
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| 33715. |
2.2-2.2 |
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Answer» 1-1 0 0 |
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| 33716. |
If sn=6n_5nsquare ,find the 20th term |
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| 33717. |
If HCF (a,b)=9 ,and (a×b)=1800,then LCM (a,b)=200 .justify |
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Answer» Yes It is a formula that product of HCF and LCM is=product of the two numbers..It is only applicable for two numbers |
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| 33718. |
Find the zeroes of the polynomial x square + x - p (p+1) |
| Answer» (x)2+x-p(p+1)(X)2+x(p+1)_px-p(p+1)X(x+p+1)_p{x+(p+1)}(X-p)(x+p+1)X-p=pX+p+1=0X=_p_1 =_(p+1) | |
| 33719. |
These question I ask for you to explain in detail not directly answer okk and thanks |
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| 33720. |
If x=5-√21/2, then prove (x^3+1/x^3)-5(x^2+1/x^2)+(x+1/x)=0 |
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| 33721. |
Solvr 2+3/1999998 |
| Answer» | |
| 33722. |
Cube root of avagadro no |
| Answer» 84 446885.4 | |
| 33723. |
Xy/x |
| Answer» | |
| 33724. |
The sum of two digit no is 16 and sum of their reciprocal is1/3. Find the no. |
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Answer» 12 and 4 12 and 4 |
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| 33725. |
What must be subtracted from the polynomial f(x) =x*4+2x*3-13x*2-12x+21 |
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Answer» Ya where is gx Where is g(x)......... |
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| 33726. |
d=-3,n=16,an=-5,then a |
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Answer» a=40 40 a=40 |
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| 33727. |
First four terms of the A.P.whose first term is -3 and common difference is 1/2 are |
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| 33728. |
In a A.P,a=10,An=60,Sn=140 then n is |
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| 33729. |
Condition for the system of linear equations ax+by=c;px+qy=r to have unique solution with solution |
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| 33730. |
If the sum of zeros of a given polynomial is x3-3kx2-x-30 if 6.Find k |
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Answer» 2 2 Check the question and type it again. |
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| 33731. |
Find the sum of 10 terms of the AP 3√5, √5, 7/√5 |
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| 33732. |
Is there any other method for finding compete square of a number other than given in this app? |
| Answer» Yes by multipling the given number with given number Syntex:- (a)^2 = a * a | |
| 33733. |
Cbse class 10th ka result kb aa rha h frnds |
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Answer» 30th May 28th may |
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| 33734. |
Show that 1. Tan 48 tan 23 tan 42 tan 67=1 |
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| 33735. |
Explain why 7×11×13+1+13 is a composite number |
| Answer» Composit number :- a number whose factor 1 and itself , is called composit number.Here, 13 is common factor which is other than 1 and itself so , it is a composit number. | |
| 33736. |
Find the number nearest to 110000 but graeter than 100000 which is exactly divisible by each 8,5,21 |
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Answer» Can anyone solve this question Please solve this question |
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| 33737. |
3x +5y =2 2x - y = 3 , solve this by substitution method |
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Answer» Y =34-39/13-5/13 3x +5y= 2 ~equation 12x-y=3. Or 2x-3=y ~equ 2By putting the value of y in equation 2 we get3x+5×(2x-3)=23x+10x-15=213x=2+15X=17/13Then putting he value of x in equation 2 we get the value of y2(17/13)-3=yY=(34/13)-334-39/3Y=-5/3 Let, 3x+5y=2. is eq1 2x - y =3. is eq2In eq1, 3x + 5y =2 5y=2-3x y= 2 /5- 3x/5From eq1 we get, y= 2/5 - 3x/5 Now, We substitute y\'s value in eq2 2x - (2/5-3x/5)=3 2x -2/5+3x/5=3 10x-2+3x/5=3 13x-2=15 13X =15+2 X. =17/13Now, we find x=17/13 In eq1 we put x=17/13 3(17/13)+5y=2 51/13+5y=2 5y= 2-51/13 5y=26-51/13 5y=-25/13 Y= -25/13 ×5 Y=-5/13 Hence,we find x=17/13 and y= -5/13. |
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| 33738. |
If two zeros of the polynomial x³-3x²+x+1and a-b, a, a+b find a and b |
| Answer» Given polynomial is f(x) = x3\xa0- 3x2\xa0+ x + 1Let\xa0{tex} \\alpha{/tex}\xa0= (a - b),\xa0{tex} \\beta{/tex}\xa0= a and\xa0{tex} \\gamma{/tex}\xa0= (a + b)Now,\xa0{tex} \\alpha + \\beta + \\gamma{/tex}\xa0=\xa0{tex} - \\frac { ( - 3 ) } { 1 }{/tex}⇒\xa0(a - b) + a + ( a + b ) = 3⇒ a - b + a + a+ b = 3⇒ a + a + a = 3⇒\xa03a = 3⇒ a = 3/3⇒\xa0a = 1Also,\xa0{tex} \\alpha \\beta + \\beta y + \\gamma \\alpha = \\frac { 1 } { 1 }{/tex}⇒\xa0(a - b)a + a (a + b) + (a + b)(a - b) = 1\xa0⇒\xa0a2\xa0- ab + a2\xa0+ab + a2\xa0- b2\xa0= 1⇒\xa03a2\xa0- b2\xa0= 1 ( ∵ a = 1)⇒\xa03(1)2\xa0- b2\xa0= 1( ∵ a = 1)⇒ 3 - b2 = 1⇒\xa0b2\xa0= 2⇒\xa0b =\xa0{tex} \\pm \\sqrt{2}{/tex}Hence, a = 1 and b =\xa0{tex} \\pm \\sqrt{2}{/tex} | |
| 33739. |
Prove that cube root 3 is irrational |
| Answer» | |
| 33740. |
Show that any positive odd integer is of the form 4q+ 3 where q is some whole number |
| Answer» By Euclid\'s division algorithm,a = bq + r = 4q + rTake b = 4.Since, 0 {tex}\\leqslant{/tex}\xa0r < 4, r = 0,1, 2, 3{tex} a=4q,4q+1,4q+2 ,4q+3{/tex}Clearly, a =4q=2(2q) and\xa04q+2=2×(2q+1)So 4q and 4q+2 are evenTherefore 4q + 1, 4q + 3 are odd, as they are proceeding numbers of even numbers 4q and 4q+2.{tex}\\therefore{/tex}\xa0Any positive odd integer is of form 4q+1 or 4q+3 .Where q is a positive integer. | |
| 33741. |
If x(x)+2x+k is completely divisible by(x_1),then find the value of k |
| Answer» 2 | |
| 33742. |
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. | |
| 33743. |
Solve the equation by cross multiplication method x+y=75x+12y=7 |
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Answer» Sorry x+y=7 5x+12y=7 |
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| 33744. |
Show that(5+2root3)ka square is irrational number |
| Answer» (5+2√3)² =5²+(2√3)²+2.2√325+12+4√3=37+4√3That\'s an irrational number | |
| 33745. |
What is the syllabus of math first SA |
| Answer» | |
| 33746. |
Solve equation graphicaly method equation..5x+7y=50,7x+5y=46 |
| Answer» Khud kar | |
| 33747. |
1 + cos theta + sin theta / 1 + cos theta - sin theta = 1 + sin theta / cos theta |
| Answer» LHS\xa0{tex} \\frac{{1 + \\cos \\theta + \\sin \\theta }}{{1 + \\cos \\theta - \\sin \\theta }}{/tex}Dividing numerator and denominator by cos{tex} \\theta {/tex}{tex}= \\frac{{\\frac{1}{{\\cos \\theta }} + \\frac{{\\cos \\theta }}{{\\cos \\theta }} + \\frac{{\\sin \\theta }}{{\\cos \\theta }}}}{{\\frac{1}{{\\cos \\theta }} + \\frac{{\\cos \\theta }}{{\\cos \\theta }} - \\frac{{\\sin \\theta }}{{\\cos \\theta }}}}{/tex}{tex}= \\frac{{\\sec \\theta + 1 + \\tan \\theta }}{{\\sec \\theta + 1 - \\tan \\theta }}{/tex}Multiplying and dividing by\xa0{tex} \\sec \\theta + 1 + \\tan \\theta {/tex}{tex}= \\frac{{\\sec \\theta + 1 + \\tan \\theta }}{{\\sec \\theta + 1 - \\tan \\theta }} \\times \\frac{{\\sec \\theta + 1 + \\tan \\theta }}{{\\sec \\theta + 1 + \\tan \\theta }}{/tex}{tex}= \\frac{{{{(\\sec \\theta + 1 + \\tan \\theta )}^2}}}{{{{(\\sec \\theta + 1)}^2} - {{\\tan }^2}\\theta }}{/tex}{tex}= \\frac{{{{\\sec }^2}\\theta + 1 + {{\\tan }^2}\\theta + 2\\sec \\theta + 2\\tan \\theta + 2\\sec \\theta \\tan \\theta }}{{1 + {{\\sec }^2}\\theta + 2\\sec \\theta - {{\\tan }^2}\\theta }}{/tex}Now,\xa0{tex} 1 + {\\tan ^2}\\theta = {\\sec ^2}\\theta {/tex}{tex}= \\frac{{2{{\\sec }^2}\\theta + 2\\sec \\theta + 2\\tan \\theta + 2\\sec \\theta \\tan \\theta }}{{2 + 2\\sec \\theta }}{/tex}{tex} = \\frac{{2\\left[ {\\sec \\theta (\\sec \\theta + 1) + \\tan \\theta (1 + \\sec \\theta } \\right]}}{{2(1 + \\sec \\theta )}}{/tex}{tex} = \\frac{{(\\sec + \\tan \\theta )(\\sec \\theta + 1)}}{{(1 + \\sec \\theta )}}{/tex}{tex} = \\sec \\theta + \\tan \\theta = \\frac{1}{{\\cos \\theta }} + \\frac{{\\sin \\theta }}{{\\cos \\theta }}{/tex}{tex}= \\frac{{1 + \\sin \\theta }}{{\\cos \\theta }} = RHS{/tex} | |
| 33748. |
1/x-1/x-2=3. (Quadratic eq solve) |
| Answer» Ye bhi nhi aata | |
| 33749. |
((x/a)sin c-(y/b)cos c)=1 and ((x/a)cos c+(y/b)sin c)=1....prove that : (x^2/a^2)+(y^2/b^2)=2 |
| Answer» Hii | |
| 33750. |
If tan theta=3\\4 then cos theta is |
| Answer» 1/4 | |