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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.
| 27051. |
4X ki power 2+5underoot2X-3 |
| Answer» | |
| 27052. |
Solve the equation 3x_5y+1=0 by cross multiplication |
| Answer» No.2 equation kaha gya? | |
| 27053. |
x-3y =33x-9y =2This of linear equation in two variable |
| Answer» | |
| 27054. |
Hlo koi h ..m aa gyi pdh ke ... Or ab fir se jaa rhi hu pdne... |
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Answer» Byy.. Okk bye |
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| 27055. |
abx2+(b2-ac)x-bc=0 yeh jo 2 hai vo x ka square hai..or b mai bhi.. |
| Answer» {tex}abx^2 + (b^2 - ac)x - bc = 0\xa0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}abx^2 + b^2x - c(ax + b) = 0\xa0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}bx(ax + b) - c(ax + b) = 0\xa0{/tex}{tex}\\Rightarrow{/tex}{tex}(bx - c)(ax + b) = 0\xa0{/tex}x =\xa0{tex}\\frac{c} {b}{/tex},\xa0{tex}\\frac{-b}{a}{/tex} | |
| 27056. |
(CosecA -cot A)2=1-cos A÷1+cosA |
| Answer» LHS=(cosecA - cotA)2\xa0=\xa0{tex}\\left( \\frac { 1 } { \\sin A } - \\frac { \\cos A } { \\sin A } \\right) ^ { 2 }{/tex}{tex}= \\left( \\frac { 1 - \\cos A } { \\sin A } \\right) ^ { 2 }{/tex}{tex}= \\frac { ( 1 - \\cos A ) ^ { 2 } } { \\sin ^ { 2 } A }{/tex}{tex}= \\frac { ( 1 - \\cos A ) ^ { 2 } } { 1 - \\cos ^ { 2 } A }{/tex}{tex}= \\frac { ( 1 - \\cos A ) ^ { 2 } } { ( 1 - \\cos A ) ( 1 + \\cos A ) }{/tex}{tex}= \\frac { 1 - \\cos A } { 1 + \\cos A }{/tex}Hence proved | |
| 27057. |
what is the simplest form of justification |
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| 27058. |
Explaination of practice paper 3 level 1 question 5 |
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| 27059. |
Question of 3.3 of question 1 ka 3rd in Linear equation in two variable |
| Answer» Abe kutte tujhe dekh ke samajh Nahi aa raha Kya saale | |
| 27060. |
Exercise 15.1 |
| Answer» | |
| 27061. |
-x+3=5 |
| Answer» X=-3=2 | |
| 27062. |
Ye 3696 kon h bhai ... Ya bhen ... Thanks hi mil rhe h ...yaha se.. |
| Answer» | |
| 27063. |
1÷7x+1÷6y=3 and 1÷2x-1÷3y=5 solve by reducing method |
| Answer» {tex}\\frac{1}{7x}{/tex}\xa0+\xa0{tex}\\frac{1}{6y}{/tex}\xa0= 3 .......(i)and\xa0{tex}\\frac{1}{2x}{/tex}\xa0-\xa0{tex}\\frac{1}{3y}{/tex}\xa0= 5 ...........(ii)Multiplying equation (ii) by\xa0{tex}\\frac{1}{2}{/tex}, we get\xa0{tex}\\frac{1}{4x}{/tex}\xa0-\xa0{tex}\\frac{1}{6y}{/tex}\xa0=\xa0{tex}\\frac{5}{2}{/tex} ..........(iii)Adding eq. (i) and (iii), we get{tex}\\frac{1}{4x}{/tex}\xa0+\xa0{tex}\\frac{1}{7x}{/tex}\xa0=\xa0{tex}\\frac{5}{2}{/tex}\xa0+ 3{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{7 + 4}{28x}{/tex}\xa0=\xa0{tex}\\frac{11}{2}{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac{11}{28x}{/tex}\xa0=\xa0{tex}\\frac{11}{2}{/tex}{tex}\\Rightarrow{/tex}\xa028x = 2\xa0{tex}\\Rightarrow{/tex}\xa0x =\xa0{tex}\\frac{1}{14}{/tex}Putting the value of x in eq.(i), we get{tex}\\frac{1}{7(\\frac1{14})}{/tex}\xa0+\xa0{tex}\\frac{1}{6y}{/tex}\xa0= 3y =\xa0{tex}\\frac{1}{6}{/tex}Hence x = {tex}\\frac{1}{14}{/tex} and y = {tex}\\frac{1}{6}{/tex} is the solution of given system of equations. | |
| 27064. |
Byee.... thodi der mai aati hu...???????????????? |
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Answer» Ok Bilkul..bbye |
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| 27065. |
Find the largest number which divides 245 and 1245 leaving remainder 5 in each case |
| Answer» Why kharif crops cannot be grown in rabi season | |
| 27066. |
If -2 is a zero of polynomial 3x2+4x+2k,find the value of k |
| Answer» Let f(x)=3x2 + 4x +2kIf -2\xa0is zero of f(x) then f(-2)=0 then f(-2)=0{tex}\\Rightarrow{/tex}\xa03 × (-2)² + 4 × -2 + 2k = 0{tex}\\Rightarrow{/tex}\xa012 - 8 + 2k = 0{tex}\\Rightarrow{/tex}\xa04 + 2k = 0{tex}\\Rightarrow{/tex}\xa02k = -4{tex}\\Rightarrow{/tex}\xa0k =\xa0{tex}\\frac{{ - 4}}{2}{/tex}{tex}\\Rightarrow{/tex}\xa0k = -2 | |
| 27067. |
It being given that 1 is zero of the polynomial 7x-x3-6 find its other zeroes. |
| Answer» The given polynomial is -x3 + 7x - 6 and let\xa0f (x) = -x3 + 7x - 6 .Since 1 is a zero of f(x), so (x - 1) is a factor of f(x).Now we divide f(x) = -x3 + 7x - 6\xa0by (x - 1), we obtain\xa0Where quotient =\xa0(-x2\xa0- x + 6){tex}\\therefore{/tex}\xa0f(x) = (-x3\xa0+ 7x -\xa06) = (x - 1)(-x2\xa0- x + 6){tex}= - ( x - 1 ) \\left( x ^ { 2 } + x - 6 \\right) = - ( x - 1 ) \\left( x ^ { 2 } + 3 x - 2 x - 6 \\right){/tex}{tex}= ( 1 - x ) [ x ( x + 3 ) - 2 ( x + 3 ) ] = ( 1 - x ) ( x + 3 ) ( x - 2 ){/tex}{tex}\\therefore{/tex}\xa0f(x) = 0 {tex}\\Rightarrow{/tex}\xa0{tex}( 1 - x ) ( x + 3 ) ( x - 2 ){/tex} = 0{tex}\\Rightarrow{/tex}\xa0(1 - x ) = 0 or (x + 3) = 0 or (x - 2) = 0{tex}\\Rightarrow{/tex}\xa0x =\xa01 or x = -3 or x = 2.Thus, the other zeros are -3 and 2 | |
| 27068. |
Factorise X square + 11 x + 13 |
| Answer» b² - 4ac = (11)² - 4(1)(13) = 121 - 52 = 69x = -b +/- √69/2a. =. -11+/-√69/2 So x = -11 + √69/2. Or. x = -11-√69/2 | |
| 27069. |
Find the centre of the circle passing through the point A, |
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| 27070. |
Definition of polynomial |
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| 27071. |
√17640 |
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Answer» Bcoz last mei sirf 1 zero hai Perfect nhi ho skta |
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| 27072. |
10 board me kya aya hai |
| Answer» Check the syllabus here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 27073. |
M dekhu teri photo 100 100 baaaaaar kude .......❤️❤️❤️???????? |
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Answer» ? Dekhu teri photo infinity times.....munde ??? Hi...gungun...bestie.. Yeah...! |
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| 27074. |
Given that √2 is a zero of cubic polynomial 6x^3+√2x^2-10x-4√2, find its other two zeros |
| Answer» Assume f(x) = {tex}6{x^3} + \\sqrt 2 {x^2} - 10x - 4\\sqrt 2{/tex}If {tex}\\sqrt 2{/tex}\xa0is the zero of f(x), then {tex}(x - \\sqrt 2 ){/tex}\xa0will be a factor of f(x). So, by remainder theorem when f(x) is divided by {tex}(x - \\sqrt 2 ){/tex}, the quotient comes out to be quadratic.Now we divide {tex}6{x^3} + \\sqrt 2 {x^2} - 10x - 4\\sqrt 2{/tex}\xa0by\xa0{tex}(x - \\sqrt 2 ){/tex}.{tex}\\therefore \\;f(x) = (x - \\sqrt 2 )(6{x^2} + 7\\sqrt 2 x + 4){/tex}\xa0(By Euclid’s division algorithm){tex}= (x - \\sqrt 2 )(6{x^2} + 4\\sqrt 2 x + 3\\sqrt 2 x + 4){/tex}\xa0( By factorization method )For zeroes of f(x), put f(x) = 0{tex}\\therefore (x - \\sqrt 2 )(6{x^2} + 4\\sqrt 2 x + 3\\sqrt 2 x + 4) = 0{/tex}{tex}\\Rightarrow \\;(x - \\sqrt 2 )[2x(3x + 2\\sqrt 2 ){/tex}\xa0{tex}+ \\sqrt 2 (3x + 2\\sqrt 2 )] = 0{/tex}{tex}\\Rightarrow (x - \\sqrt 2 )(3x + 2\\sqrt 2 )(2x + \\sqrt 2 ) = 0{/tex}{tex}\\Rightarrow x - \\sqrt 2 = 0{/tex}\xa0or {tex}3x + 2\\sqrt 2 = 0{/tex}\xa0or {tex}2x + \\sqrt 2 = 0{/tex}{tex}\\Rightarrow x = \\sqrt 2{/tex}\xa0or {tex}x = \\frac{{ - 2\\sqrt 2 }}{3}{/tex}\xa0or {tex}x = \\frac { - \\sqrt { 2 } } { 2 }{/tex}So, other two roots are {tex}= \\frac{{ - 2\\sqrt 2 }}{3}{/tex}\xa0and {tex}\\frac{{ - \\sqrt 2 }}{2}{/tex}. | |
| 27075. |
Sin60°+cos30° |
| Answer» Sin60° + cos30°(√3/2) + (√3/2)= ( (√3+ √3)/ 2= 2√3/2= √3 | |
| 27076. |
A cube of side 5 cm is painted all slongbits faces... |
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| 27077. |
Find the zeroes of polynomial x2-6x+1 |
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| 27078. |
Using euclid algorithm to find the HCF of 595 and 252 and then express it in the form of 595m + 252n |
| Answer» Given integers,595 and 252Applying Euclid division algorithm to 595 and 252 we get,595=252×2+91.........................................1Now applying ,Euclid division algorithm to 252 and 91 we get,252=91×2+70.....….………...........................2Now applying Euclid division algorithm to 91 and 70,we get91=70×1+21…...............................................3Now applying Euclid division algorithm to 70 and 21 we get,70=21×3+7...................................................4Now applying Euclid division algorithm to 21 and 7 we get,21=7×3+0......................................................5The remainder at this stage is zero.Hence HCF of 595 and 252 is 7.eq1,eq2,eq3, and eq4 can be written as,595-(252×2)=91..........................................6252-(91×2)=70.............................................791-(70×1)=21...................................................870-(21×3)=7.....................................................9Now eq9 we have,(as we express HCF in the form of a equation that is the reason we start from eq9)7=70-(21×3)By putting eq8 in above equation we have,7=70-(91-70×1)37=70-91×3+70×37=70×4-91×3By putting eq7 in above equation we have,7=(252-91×2)4-91×37=252×4-91×8-91×37=252×4-91×11By putting eq6 in above equation we have,7=252×4-(595-252×2)117=252×4-595×11+252×227=252×26-595×117=595(-11)+252(26)7=595m+252n, where m=-11 and n=26.Hence HCF of 595 and 252 is in the form of 595m+252n,where m=-11 and n=26. | |
| 27079. |
What are the types of linear eq.in two variables? |
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| 27080. |
Identity of equation |
| Answer» Euclid\'s algorithm | |
| 27081. |
Prove that square of any positive integer is of the form 5q,5q+1,5q+2,5q+3,5q+4 for some integer |
| Answer» Any integer can be written in the form 5m , 5m+1, 5m+2 (where m is any integer)(5m)² = 25m² = 5 × 5m² = 5q (where q = 5m²)(5m+1)² = (5m)² + 2 ×5m × 1 + 1² [by using identity- (a+b)²= (a²+ 2ab +b² )] = 25m² + 10 m + 1 = 5 (5m²+ 2m) +1 = 5q +1 ( where q= 5m²+2m)(5m+2)²= (5m)² + 2× 5m × 2 +2² [by using identity- (a+b)²= (a²+ 2ab +b² )] = 25m² + 20 m +4 = 5 (5m²+4m ) + 4 = 5q+4 (where q= 5m²+4m)Hence proved | |
| 27082. |
2$53 |
| Answer» What is this | |
| 27083. |
Ch3 3.1 question 1 |
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| 27084. |
I did not understand page 30 question example 2 plus explain |
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| 27085. |
Cramers rule |
| Answer» Cramer\'s Rule is a method that uses determinants for solving systems of linear equations. To explain the method and how these determinants are generated, this lecture will first illustrate the solution to a system of linear equations by the addition/elimination method without simplifying the indicated products and sums of the coefficients. The intent is to demonstrate that these products and sums can be represented by determinants. | |
| 27086. |
What are polynomial |
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Answer» A polynomial is a mathematical expression that consists of variables and constants combined using addition, subtraction and multiplication. Variables may have non-negative integer exponents. Although division by a constant is allowed, division by a variable is not allowed.The degree of a polynomial is the exponent of the highest degree term.\tA polynomial of degree 0 is called a constant polynomial.\tA polynomial of degree 1 is called a linear polynomial.\tA polynomial of degree 2 is called a quadratic polynomial.\tA polynomial of degree 3 is called a cubic polynomial. A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. |
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| 27087. |
ex4.4 qno.8 |
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| 27088. |
Why use costita |
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| 27089. |
Find whether -1 is a zero of the polynomial p(x) =5xsquare-3x-1 |
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Answer» X=-1P(x)=5x^2-3x-1=5(-1)^2-3(-1)-1=5+3-1=8-1=7 Hence,x=-1 is not a zero of p(x) putting x=-1we observe that the remainder of equation comes 7. so it is not zeroes of polynomial |
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| 27090. |
Find the mean median and mode of 2,3,5,8,2,5,9,1,4,7,3,7,6,4,4,8 |
| Answer» Please answer friends | |
| 27091. |
2+2-2=2 |
| Answer» 0 | |
| 27092. |
Sllabus |
| Answer» | |
| 27093. |
If alpha and beta are the zeroes of polynomial 3x2-4x-4 find (alpha +1)(beta+1) |
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| 27094. |
In graphical formx+2y=702x+y=95 |
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| 27095. |
Guys what are internal exam? |
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| 27096. |
60/x+240/y=4,100/x+200/y=25/6 find x&y |
| Answer» X=0 ,y=0 | |
| 27097. |
What is the rational number between root 2 ND root3 |
| Answer» There r infinity rational no. | |
| 27098. |
Solve equation by cross mulyiplication method |
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Answer» Where\'s equation.? Equation kaha h |
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| 27099. |
(x2-y2)2 |
| Answer» (x2-y2)2= (x2)2 - 2x2y2 + (y2)2= x4 - 2x2y2 + y4 | |
| 27100. |
When class 10 result declared? |
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Answer» No info has been given Kisi v din result aa sakta hai jaise 12 ka aaya tha |
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