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34451.

What must be added or subtracted from the polynomial is exactly divisible by x squared +2x-3.

Answer» f(x) = x4 + 2x3 - 2x2 + x - 1Remainder = -x + 2Therefore, we should add –(–x+ 2) to make it divisible exactly by x2 + 2x – 3.Thus, we should add x – 2 to x4 + 2x3 – 2x2 + x – 1 .
34452.

Detrmine wheather the value of x is zero of given polynomial or not x2+x+1,x=-1

Answer» Xsquare+x+1Substitute -1in place of x(-1)square +(-1)+11-1+1=1No the value of x is not zero of the given polynomial
34453.

Paper kitne no ka aayenga 70 ya 80

Answer» Last year, it was 80.
80 ka
I also think 80 marks but I think it will come 70 marks because this marks is on examination master hands so anything can come.
80 marks ka
80
34454.

If x^y + y^x =2xyAnd (1/x) ^1/y+(1/y)^1/x =2underrot x /yFind x and y

Answer» {tex}\\huge Do \\space you\\space think\\space this \\space is\\space valid\\space \\space :x^y\\:+\\:y^x\\:=2xy,\\:x^{\\left(-1/y\\right)}\\:+\\:y^{\\left(-1/x\\right)}=2\\cdot \\:sqrt\\left(x/y\\right){/tex}
34455.

6x+3y=6xy----(1)2x+4y=5xy----(2)Slove it ?

Answer» 6x+3y=6xy------(1)2x+4y=5xy------(2)
34456.

if x×x+x+1=0 then x×x×x=?

Answer» 3x
34457.

Find the hcf of 960 and 1575 by using Euclid algorithm

Answer» It is very long process
34458.

What do people say about the ending of the world?

Answer» Ask these kinds of question\xa0on quora.com , not here
34459.

Exercise 6.2 2nd question of triangles lesson

Answer» This app has all the NCERT solutions of Mathematics. So, you can see the solution of your question in this app.?
34460.

Best book for science,math,sst,english,hindi ...? Plz tell me ..

Answer» U like is the best book of all subjects
U like is the best book
34461.

Splitting middle term 2×^+4×-16=0

Answer» 2x square +8x-4x -16 2x(x-4)+4(x-4)(2x+4)(x-4)X=-2,X=4
34462.

When will improvement exam forum be filled

Answer»
34463.

X3-3 x+5 x-3

Answer»
34464.

9(secA)²–9(tanA)²=

Answer»
34465.

What is the difference between sinA and cosA.

Answer»
34466.

2 year

Answer» What 2 year
What
34467.

Prove that square root 3 + square root 5 is irrational

Answer»
34468.

Chapter 2 polonomials give the notes

Answer» Yess
34469.

Find the sum of the 40 positive integer divisible by 6

Answer» We know that the first 40 positive integers divisible by 6 are 6,12,18,....This is an AP with a = 6 and d = 6S40 = n/2 [2a + (n - 1)d]S40 = 20[2(6) +(40-1)6]= 20[12+234]= 4920
34470.

If a and B are zeroes of the polynomial x2-7x+10.find a quadratic polynomial whose zeros are

Answer» x2-7x+10=x2-5x-2x+10=x(x-5)-2(x-5)=(x-2)(x-5)=> x=2/5
34471.

Find five rational no between 6/5 and 7/3

Answer» 19/15,20/15,21/15,22/15,23/15
34472.

State wether it is terminating or non terminating6/15

Answer» It is terminating
Hlo
This one is terminating
Divide numerator and denomintor both by 3, we get 2/5. Denominator is in the form of 5^m, so the given no. is terminating.?
34473.

Find the zero of the quadratic polynomials 7yy-11/3y-2/3

Answer» p(y) = 7y2 -\xa0{tex}\\frac{11}{3}{/tex}y -\xa0{tex}\\frac{2}{3}{/tex}\xa0=\xa0{tex}\\frac{1}{3}{/tex}\xa0(21y2 - 11y - 2)=\xa0{tex}\\frac{1}{3}{/tex}[(7y + 1)(3y - 2)]{tex}\\therefore{/tex}\xa0Zeroes are\xa0{tex}\\frac{2}{3}{/tex}, -\xa0{tex}\\frac{1}{7}{/tex}Sum of Zeroes =\xa0{tex}\\frac{2}{3}{/tex}\xa0-\xa0{tex}\\frac{1}{7}{/tex}\xa0=\xa0{tex}\\frac{11}{21}{/tex}{tex}\\frac{-b}{a}{/tex}\xa0=\xa0{tex}\\frac{11}{21}{/tex}\xa0{tex}\\therefore{/tex}\xa0sum of zeroes =\xa0{tex}\\frac{-b}{a}{/tex}Product of Zeroes = ({tex}\\frac{2}{3}{/tex})(-{tex}\\frac{-1}{7}{/tex}) = -{tex}\\frac{2}{21}{/tex}{tex}\\frac{c}{a}{/tex}\xa0= -\xa0{tex}\\frac{2}{3}{/tex}({tex}\\frac{1}{7}{/tex}) = -{tex}\\frac{2}{21}{/tex}{tex}\\therefore{/tex}\xa0Product =\xa0{tex}\\frac{c}{a}{/tex}
34474.

100÷5

Answer» Can you use your sense of humor please. Its answer is 20
20
20 ans Ask question of sense pl
20
20
20
34475.

Find the value of k for which the system3x+y=1Kx+y=1Has (1) unique solution and (2) no solution

Answer» For no solution k=3
34476.

Find the HCF of the polynomials 2x²-x-1and 4x²+8x+3.

Answer» 2x²-x-1=2x²-2x+x-1=2x(x-1)+(x-1)=(x-1)((2x+1)and 4x²+8x+3.=4x²+6x+2x+3=2x(2x+3)+(2x+3)=(2x+3)(2x+1)Here common factor=2x+1hence\xa0HCF=2x+1\xa0
34477.

in the given figure De parallel to OQ and DE parallel to OR prove that EF parallel to OQ

Answer» In {tex}\\triangle {/tex}OQP,DE || OQ{tex}\\frac{{PE}}{{EQ}} = \\frac{{PD}}{{DO}}{/tex} .....(i)In {tex}\\triangle {/tex}OPR,DF {tex} \\bot {/tex} OR{tex}\\frac{{PD}}{{DO}} = \\frac{{PF}}{{FR}}{/tex} .....(ii)From (i) and (ii) , we get{tex}\\frac{{PE}}{{EQ}} = \\frac{{PF}}{{FR}}{/tex}{tex}\\therefore {/tex} From {tex}\\triangle {/tex}PQREF {tex} \\bot {/tex} OR
34478.

Check whether the given equation is quadratic equation:(2x-1) (x-3)=(x+5)

Answer» (2x-1) (x-3)=(x+5)2x2-x-6x+3=x+52x2-8x-2=0x2-4x-1=0this is a quadratic equation
34479.

Convert maths all chapter in hindi medium

Answer» Buy hindi medium book
34480.

5/x - 3/y= 2

Answer» 5/x - 3/y = 2(5y - 3x)/xy = 25y - 3x = 2xy
34481.

Using Euclid\'s algorithm. Find the HCF of 144 and 198

Answer» Prime factorization:{tex}144 = 2 ^ { 4 } \\times 3 ^ { 2 }{/tex}{tex}198 = 2 \\times 3 ^ { 2 } \\times 11{/tex}HCF = product of smallest power of each common prime factor in the numbers {tex}= 2 \\times 3 ^ { 2 } = 18{/tex}
HCF(144,198) 198 = 144×1+54144 = 54×2+3654 = 36 ×1 + 18 36 = 18 ×2 +0Therefore HCF (144,198) =18
34482.

If p(x)=3 x^ -2x`+6x-5,find p(2)

Answer» Hahaha, oh my god!!!, don\'t you think this is too much, you are not studying properly{tex}\\large p(x):x^3 -x^2+6x-5\\\\putting, \\space x=2\\\\p(2):(2)^3-(2)^2+6\\times2-5\\\\p(2):8-4+12-5\\\\p(2):\\boxed {11}{/tex}You just only have to replace \'{tex}\\large x{/tex}\' by the value given and perform the arithmetic operations.Now, after seeing this, don\'t forget it.\xa0And please write your qeustions properly while asking.
34483.

a+)

Answer» What is this bro type the question first
34484.

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.
34485.

Pythagoras theorum 9.1

Answer»
34486.

Why the( √2) in square under root cut

Answer» {tex}\\begin{array}{l}\\sqrt2=2^{1/2}\\\\(\\sqrt2)^2=(2^\\frac12)^2\\\\=2^{\\frac12\\times2}\\\\=2^1\\\\=2\\\\Hence\\;the\\;root\\;is\\;removed\\;\\end{array}{/tex}\xa0
34487.

Ap questions

Answer»
34488.

ABCD is a trapezium with ABparallel DC

Answer» What we have to do now we all know that one pair of trapeziym is parallel abd other one is non parallel
34489.

X_y ka factorise

Answer» Question dhaag se bhj bhai
34490.

1/cosecA-cotA-1/sinA=1/sinA-1/cosecA-citA

Answer»
34491.

Solve equation0.2x+0.3y=1.30.4x+0.5y=2.3

Answer» 0.2 x + 0.3 y = 1.3 ; 0.4 x + 0.5 y = 2.3The given system of linear equations is:0.2 x + 0.3 y = 1.3..............(1)0.4 x + 0.5 y = 2.3...................(2)From equation (1),\xa00.3 y = 1.3 - 0.2 x{tex}\\Rightarrow \\quad y = \\frac { 1.3 - 0.2 x } { 0.3 }{/tex}.........................(3)Substituting this value of y in equation(2), we get{tex}0.4 x + 0.5 \\left( \\frac { 1.3 - 0.2 x } { 0.3 } \\right) = 2.3{/tex}{tex}\\Rightarrow{/tex}0.12 x + 0.65 - 0.1 x = 0.69{tex}\\Rightarrow{/tex}0.12 x - 0.1 x = 0.69 - 0.65{tex}\\Rightarrow{/tex}0.02 x = 0.04{tex}\\Rightarrow{/tex}{tex}\\mathrm { x } = \\frac { 0.04 } { 0.02 } = 2{/tex}Substituting this value of x in equation(3), we get{tex}y = \\frac { 1.3 - 0.2 ( 2 ) } { 0.3 } = \\frac { 1.3 - 0.4 } { 0.3 } = \\frac { 0.9 } { 0.3 } = 3{/tex}Therefore, the solution is x = 2, y = 3, we find that both equation (1) and (2) are satisfied as shown below:0.2 x + 0.3 y = ( 0.2 )( 2 )+( 0.3)( 3 ) =\xa00.4 + 0.9 = 1.30.4 x + 0.5 y= ( 0.4 )( 2 ) + ( 0.5 )( 3 ) }\xa0= 0.8 + 1.5 = 2.3This verifies the solution.\u200b\u200b\u200b\u200b\u200b\u200b\u200b
0.2x+0.3y = 1.3 ....... (i)eq (i) ×102x+3y=130.4x+0.5y=2.3 ...... (ii)eq(ii) ×104x+5y=23from eq (i)x =13-3y/2putting value of x in eq (ii)4(13-3y/2)+5y=2326-6y+5y=2326-y=23y= 3putting value of y in eq (i)2x+3(3)=132x=13-9x=4/2x=2\xa0
34492.

Show that the polynomial x^2-6x+12 has no zeroes

Answer» x2\xa0- 6x + 12Here, a = 1, b = -6 and c = 12D = b2\xa0- 4acD = (-6)2\xa0- 4(1)(12)D = 36 - 48D = -12Hence, D < 0Therefore, there exists no real root of the equation.
34493.

RD YA DESK WORK SE QUESTIONS AATE HA KY

Answer» Ha aate hai...
WHAT SAY.??
YES..
Board exam me??
34494.

If 5cos theta=7sin theta find 7sin theta+ 5cos theta÷ 5sin theta + 7 cos theta

Answer» {tex}\\begin{array}{l}\\frac{7\\sin\\;\\theta+\\;5\\cos\\theta}{\\;5\\sin\\theta\\;+\\;7\\;\\cos\\;\\theta}=\\frac{7\\sin\\;\\theta+7\\sin\\;\\theta}{7\\sin\\;\\theta+{\\displaystyle\\frac{\\;7\\times5}5}\\cos\\theta}\\\\=\\frac{\\displaystyle14\\sin\\;\\theta}{7\\sin\\;\\theta+{\\displaystyle\\frac75}\\times7\\sin\\;\\theta}=\\frac{70\\sin\\;\\theta}{7\\sin\\;\\theta+49\\sin\\;\\theta}\\\\=\\frac{\\displaystyle70\\sin\\;\\theta}{\\displaystyle56\\sin\\;\\theta}=\\frac{70}{56}=\\frac54\\end{array}{/tex}
34495.

Find factor tree on 874944

Answer»
34496.

Find all the zeroes of polynomial 5x^2-15x^2-3x+9, if two of its zeroes are √3/5 and -√3/5

Answer»
34497.

Compartment exam paper 2019 cbse

Answer» Please send me the details of the compartment exam paper 2019 cbse
34498.

How can we consider this examination pattern 2020 as per previous years questions or or the latest

Answer»
34499.

Write the relationship between zeros of polynomial ax^3+bx^2+cx+d if zreoes are alpha,beta, gamma.

Answer» {tex}\\begin{array}{l}\\alpha+\\beta+\\gamma=-\\frac ba\\\\\\alpha\\beta\\gamma=-\\frac ca\\\\\\alpha\\beta+\\beta\\gamma+\\gamma\\alpha=\\frac da\\\\\\\\\\end{array}{/tex}
34500.

Sin power 4 theta minus Cos square theta

Answer» Question is incomplete