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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.
| 11451. |
Find 2 consecutive natural no. The sum of the whose squares is 145 |
| Answer» Solution: let first number = xThen second number = x +1ATQ,=> x2 + (x+1)2 = 145=> x2\xa0+ x2 +2x + 1\xa0= 145=> 2x2 +2x - 144= 0divide by 2=> x2 + x - 72 = 0 => x2 + 9x - 8x +72 = 0=> x(x+9) -8(x+9) = 0=> (x+9)(x-8)= 0=> x = 8, -9as x is natural number so -9 cannot be value of x.So numbers are 8,9.\xa0\xa0 | |
| 11452. |
What is the proability of getting a composite no. When a die is thrown once and twice??????? |
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| 11453. |
How to answer the cistern questions and canal questions\xa0 |
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| 11454. |
the area and circumfrence of a circle are numerically equal . what is the radius of circle\xa0 |
| Answer» Ans. Let Radius of Circle = r\xa0Then Circumference of Circle =\xa0{tex}2\\pi r{/tex}Area of Circle =\xa0{tex}\\pi r^2{/tex}According to Ques.{tex}=> 2\\pi r = \\pi r^2 {/tex}{tex}=> r = 2{/tex}So Radius of circle = 2 unit | |
| 11455. |
In an AP the first trem is 12 and c.d is 16 .if the last term of the AP is 252 find its middle term |
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Answer» a8= a + 7d =12+7×16 =12+112 =124\xa0 The Question Should be "In an AP the first trem is 12 and c.d is 6. if the last term of the AP is 252 find its middle term"Ans.\xa0First Term a = 12\xa0Commond Difference d = 6Last term an\xa0= 252We know\xa0=> an = a + (n-1)d=> 252 = 12 + (n-1)6\xa0=> 240 = (n-1)6=> 40 = n -1\xa0=> n = 41\xa0Total Number of terms = 41\xa0Middle term =\xa0{tex}({n+1\\over 2 })^{th} term = ({41+1\\over 2 })^{th} term = 21^{th} term{/tex}a21\xa0\xa0= 12\xa0+ (21-1)6=> a21\xa0= 12 + 120 = 132 First term(a) =12 Last term(l) =252l = a+(n-1)d252 = 12+(n-1)252 = 12+16n-16256 = 16n\xa0n = 16n/2 = 16/2 = 8So, middle term is 8.\xa0 |
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| 11456. |
Give some important Questions for Mathematics class 10 sa2\xa0BE FAST PLEASE |
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| 11457. |
Find the value of\xa0k\xa0for which the points A (-1,3),B (2,k) and C (5,-1) are collinear.\xa0 |
| Answer» According to question,Area of triangle ABC = 0Given: A (-1, 3), B (2, k) and C (5, -1)=> {tex}{1 \\over 2}\\left| {\\left( { - 1} \\right)\\left( {k + 1} \\right) + 2\\left( { - 1 - 3} \\right) + 5\\left( {3 - k} \\right)} \\right| = 0{/tex}=> {tex}\\left| { - k - 1 - 8 + 15 - 5k} \\right| = 0{/tex}=> {tex}\\left| { - 6k + 6} \\right| = 0{/tex}=> {tex}-6k=-6{/tex}=> {tex}k=1{/tex} | |
| 11458. |
Find the reflection (image) of point (7,-5) in the point (-3,1). |
| Answer» Solution: A point reflection exists when a figure is built around a single point called the center of the figure, or point of reflection.For every point in the figure, there is another point found directly opposite it on the other side of the center such that the point of reflection becomes the midpoint of the segment joining the point with its image.Now for point (7,-5) around (-3,1) we will have the coordinates of the reflected point as:x coordinate = 2(-3) - 7 = -13y coordinate = 2(1) - (-5) = 7Thus (-13,7) is the required point. | |
| 11459. |
i want to know the weightage of each chapter in class x mathematics |
| Answer» No one know the weightage of each chapter, instead of that I will tell you the weightage of each unit:-UNITS NAME MARKSI Number System 04II Algebra 20III Trignometry 12IV Coordinate 08 GeometryV Geometry 16VI Mensuration 10VII Statistics and 10 Probability TOTAL= 80 | |
| 11460. |
What is zero of the polynomial. l2\xa0-15 |
| Answer» For finding the zeroes of the polynomial, we equate it with zero.I2 - 15 = 0=> I2 = 15=> I = {tex} \\pm \\sqrt {15} {/tex}Therefore, the zeroes of the given polynomial is 15 and -15. | |
| 11461. |
If hcf of 210 and 55 is expressible in the form 210×5+55p,find p |
| Answer» 210 = 2 x 3 x 5 x 755 = 5 x 11Therefore, HCF (210, 55) = 5Now, 210 x 5 + 55 x p = 5=> 1050 + 55p = 5=> 55p = 5 - 1050=> 55p = -1045=> p = -1045/55=> p = -19 | |
| 11462. |
the area of isoceles triangle EFG is 60 sq .m and EF=EG=13 m find the FG |
| Answer» Let the length of FG be x m.Using Heron\'s formula,s = {tex}{{13 + 13 + x} \\over 2} = {{26 + x} \\over 2}{/tex}Area of triangle EFG = 60 sq.m=> {tex}\\sqrt {\\left( {{{26 + x} \\over 2}} \\right)\\left( {{{26 + x} \\over 2} - 13} \\right)\\left( {{{26 + x} \\over 2} - 13} \\right)\\left( {{{26 + x} \\over 2} - x} \\right)} {/tex}\xa0= 60=> {tex}\\sqrt {\\left( {{{26 + x} \\over 2}} \\right)\\left( {{{26 + x - 26} \\over 2}} \\right)\\left( {{{26 + x - 26} \\over 2}} \\right)\\left( {{{26 + x - 2x} \\over 2}} \\right)} {/tex}\xa0= 60=> {tex}\\sqrt {\\left( {{{26 + x} \\over 2}} \\right)\\left( {{x \\over 2}} \\right)\\left( {{x \\over 2}} \\right)\\left( {{{26 - x} \\over 2}} \\right)} {/tex}\xa0= 60=> {tex}{x \\over 2}\\sqrt {\\left( {{{676 - {x^2}} \\over 4}} \\right)} = 60{/tex}=> {tex}{x \\over 4}\\sqrt {676 - {x^2}} = 60{/tex}=> {tex}x\\sqrt {676 - {x^2}} = 240{/tex}Squaring both sides,=> {tex}{x^2}\\left( {676 - {x^2}} \\right) = 57600{/tex}=> {tex}{x^4} - 676{x^2} + 57600 = 0{/tex}=> {tex}{x^4} - 576{x^2} - 100{x^2} + 57600 = 0{/tex}=> {tex}{x^2}\\left( {{x^2} - 576} \\right) - 100\\left( {{x^2} - 576} \\right) = 0{/tex}=> {tex}\\left( {{x^2} - 576} \\right)\\left( {{x^2} - 100} \\right) = 0{/tex}=> {tex}{x^2} - 576 = 0{/tex} and {tex}{x^2} - 100 = 0{/tex}=> {tex}{x^2} = 576{/tex} and {tex}{x^2} = 100{/tex}=> {tex}x=24{/tex}\xa0m and {tex}x = 10{/tex}\xa0m\xa0 | |
| 11463. |
ax + by = 1\xa0bx + ay = (a+b)^2 / a2 + b2 |
| Answer» ax+by=1...........(i)bx+ay=(a+b)2/a2+b2..........(ii)multiplying equation (i) by \'b\' and equation (ii) by \'a\'abx+b2y=b...........(iii)abx+a2y=a(a+b)2/a2+b2.......(iv)subtracting(iv) from (iii) we get,b2y-a2y=b-a(a+b)2/a2+b2y(b2-a2)=b(a2+b2)-a(a+b)2/a2+b2y(b2-a2)=a2b+b3-[a(a2+2ab+b2]/a2+b2y(b2-a2)=a2b+b3-a3-2a2b-ab2/a2+b2y(b2-a2)=b3-a3-a2b-ab2/a2+b2y(b2-a2)=(b3-a3)-ab(a+b)/a2+b2y(b+a)(b-a)=(b-a)(b2+ab+a2)/a2+b2y=(b2+ab+a2)/(a2+b2)(b+a)\xa0 | |
| 11464. |
When to do HCF in problem sums? |
| Answer» When you need to find out maximum or largest value\xa0 | |
| 11465. |
When to do LCM in problem sums? |
| Answer» When you have to find smallest least minimum number\xa0 | |
| 11466. |
What are the zeroes of quadratic polynomial x2+99x+127? |
| Answer» We need to find zeroes of x^2+99*x+127=0We have certainly got a formula for determining the zeroes of a 2 degree equation.That is called Discriminant formula or D- formula or Shri Dharacharya Method.Roots of the 2 degree equation are given by=(-b+(b^2-4*a*c)^(1/2))/2*a and (-b-(b^2-4*a*c)^(1/2))/2*a\xa0So, Zeroes of this equation is given by\xa0(-99+(99^2-4*1*127)^(1/2))/2*1 and (-99-(99^2-4*1*127)^(1/2))/2*1Hence,zeroes are -1.3 and -97.7 respectively. | |
| 11467. |
The value of\xa0 cot xcot x-cot 3x+ tan xtan x -tan 3x\xa0Will be?? |
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| 11468. |
In an isosceles triangle ABC, if AB=BC and AB2=2AC2\xa0,then angle C will be???\xa0 |
| Answer» Ans.Given: AC = BCABC is an Isosceles triangleAB2\xa0= 2 AC2To Find =\xa0{tex}\\angle C{/tex}Proof :\xa0As\xa0AB2\xa0= 2AC2=>\xa0AB2\xa0= AC2\xa0\xa0+ AC2=>\xa0\xa0AB2\xa0= AC2\xa0\xa0+ BC2 [Given AC = BC]So By Converse of Pythagoras theorem\xa0{tex}\\angle C = 90^o{/tex}\xa0 | |
| 11469. |
SecA (1+SecA)(SecA-tanA)=1\xa0 |
| Answer» secA(1-sinA)(secA+tanA)\xa0=(secA-tanA)(secA+tanA)\xa0=1... using identity | |
| 11470. |
How to get full marks in maths exam 2017-18 |
| Answer» Practise and practice. Solve as much sample questions as you can .Try to solve each type of questions related to a chapter with thorough understanding.And most importantly practice a chapter that many times until and unless you can solve the questions with minimal time.Because in exam time with precision is very important factor. | |
| 11471. |
Unit tests of class 10 pattern 2017\xa0 |
| Answer» Every school has to take three periodic tests for every subject and the best two will be considered.\xa0Periodic tests has a weightage of 10 marks.\xa02-3 chapters will be assessed in every subject.\xa0All the best. Make the most of your summer vacations!\xa0 | |
| 11472. |
What is best way of solving out an root term? |
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| 11473. |
what are properties of integers\xa0 |
| Answer» Ans.\xa0 | |
| 11474. |
If a+b=5 and 3a+2b=20, Then value of a+b is:1) 102) 153) 204) 25 |
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Answer» Ooo.But this ques. had come in chs exam Question Must be :\xa0If a + b = 5 and 3a + 2b = 20, then 3a + b will beAns.\xa04) 25a + b = 5 .............(1)3a + 2b = 20 .........(2)From (1), a = 5 - b\xa0put this value in (2), we get\xa0=> 3( 5 - b ) + 2b = 20=> 15 - 3b + 2b = 20\xa0=> - b = 5\xa0=> b = -5\xa0put values of b in (1), we get=> a - 5 = 5\xa0=> a = 10\xa0Now, 3a + b = 3(10 ) - 5= 30 - 5 = 25 |
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| 11475. |
If cos (a+b)=0, then sin (a-b)=1) cos 2b2) cos b3) sin 2a4) sin a |
| Answer» 1) cos 2bExplanation:cos (a+b) =0=> cos (a+b) = cos 90=> a+b = 90=> a = 90- bNowsin(a+b)= sin(90-b-b)= sin(90-2b)= cos 2b [as sin (90 - x) = cos x] | |
| 11476. |
find the prime factorisation of the denominator of the rational number equivalent to 8.39 |
| Answer» 8.39=839/100therefore we have to find the prime factorisation of 100.prime factorisation of 100 :- 2 * 5 * 2 * 5.........\xa0 | |
| 11477. |
Name shape of the curve of a quadratic polinomial\xa0 |
| Answer» The graph of a quadratic function is called a Parabola.A parabola is roughly shaped like the letter "U". | |
| 11478. |
best topic of maths in which i make survey report , file ,and a chart\xa0 |
| Answer» Statistics | |
| 11479. |
name the best topic in which we make survey reports,charts and a file in maths.\xa0 |
| Answer» Statistics | |
| 11480. |
If one zero of quadtric polynomial x2+(a+1)x-6 is 2 , then what is the value of a? |
| Answer» As 2 is zero of given polynomial thenp(2) = 0=> (2)2 + (a+1)2 -6 = 0=> 4 + 2a + 2 - 6 = 0=> 0 + 2a = 0=> 2a = 0=> a = 0 | |
| 11481. |
Find prime factorizatiom of denominator of rational number equivalent to 8.39\xa0 |
| Answer» We can write 8.39 as\xa0{tex}839\\over 100{/tex}So denominator is 100100 = 2*2*5*5So prime factors of 100 are 2,5\xa0 | |
| 11482. |
{tex}2√45+3√20÷2√5{/tex} |
| Answer» {tex}{2 \\over {\\sqrt {45} }} + {3 \\over {\\sqrt {20} }} \\div {2 \\over {\\sqrt 5 }}{/tex}= {tex}{2 \\over {3\\sqrt 5 }} + {3 \\over {2\\sqrt 5 }} \\times {{\\sqrt 5 } \\over 2}{/tex}= {tex}{2 \\over {3\\sqrt 5 }} + {3 \\over 4}{/tex}= {tex}{{8 + 9\\sqrt 5 } \\over {12\\sqrt 5 }}{/tex}= {tex}{{8 + 9\\sqrt 5 } \\over {12\\sqrt 5 }} \\times {{\\sqrt 5 } \\over {\\sqrt 5 }}{/tex}= {tex}{{8\\sqrt 5 + 45} \\over {60}}{/tex} | |
| 11483. |
Prove that √7 is an irrational |
| Answer» Ans. let us assume that {tex}\\sqrt 7{/tex}\xa0be rational.Then it must in the form of {tex}p \\over q {/tex}[q ≠ 0] [p and q are co-prime]{tex}\\sqrt7 = {p \\over q}{/tex}=>{tex}q \\sqrt 7 = p {/tex}Squaring on both sides{tex}=> 7q^2= p^2 {/tex} .............. (1)\xa0p2\xa0is divisible by 7p is divisible by 7p = 7c [c is a positive integer][squaring on both sides ]p2\xa0= 49 c2\xa0..............\xa0(2)Subsitute p2\xa0in eq\xa0(1) we get=> 7q2\xa0= 49c2\xa0=>q2 = 7c2=> q is divisble by 7 thus q and p have a common factor 7.There is a contradiction as our assumsion p and q are co primebut it has a common factor.So that {tex}\\sqrt 7{/tex}\xa0is an irrational. | |
| 11484. |
Show that √2 is irrational. |
| Answer» Ans.Assume{tex} \\sqrt{2}{/tex} is rational,i.e. it can be expressed as a rational fraction of the form {tex}\\frac{b}{a},{/tex}where a and b are two relatively prime integers.Now, since {tex}\\sqrt{2}=\\frac{b}{a}{/tex},squaring both sides,we have\xa0{tex} 2=\\frac{b^2}{a^2}{/tex}, or {tex}b^2=2a^2{/tex}.Since 2a2\xa0is even, b2\xa0must be even, and since b2\xa0is even, so is b.Let b=2c.We have 4c2=2a2\xa0and thus a2=2c2.Since 2c2\xa0is even, a2\xa0is even, and since a2\xa0is even, so is a.However, two even numbers cannot be Co-prime,so {tex}\\sqrt 2{/tex}\xa0cannot be expressed as a rational fraction;Hence {tex}\\sqrt 2{/tex}\xa0is irrational | |
| 11485. |
Two families?...................... Find reservation charge |
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| 11486. |
Show that there is no positive integer n for which √n+1+√n+1 is rational\xa0 |
| Answer» Let us assume that there is a positive integer n for {tex}\\sqrt{n-1}+\\sqrt{n+1}{/tex}which is rational and equal to {tex}\\frac pq{/tex}, where p and q are positive integers and (q\xa0{tex}\\neq{/tex}\xa00).{tex}\\sqrt { n - 1 } + \\sqrt { n + 1 } = \\frac { p } { q }{/tex}......(i)or,\xa0{tex}\\frac { q } { p } = \\frac { 1 } { \\sqrt { n - 1 } + \\sqrt { n + 1 } }{/tex}on multiplication of numerator and denominator by\xa0{tex}\\sqrt{n-1}-\\sqrt{n+1}{/tex}\xa0we get{tex}= \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { ( \\sqrt { n - 1 } + \\sqrt { n + 1 } ) ( \\sqrt { n - 1 } - \\sqrt { n + 1 } ) }{/tex}{tex}= \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { ( n - 1 ) - ( n + 1 ) } = \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { - 2 }{/tex}or,\xa0{tex}\\sqrt { n + 1 } - \\sqrt { n - 1 } = \\frac { 2 q } { p }{/tex} ........(ii)On adding (i) and (ii), we get{tex}2 \\sqrt { n + 1 } = \\frac { p } { q } + \\frac { 2 q } { p } = \\frac { p ^ { 2 } + 2 q ^ { 2 } } { p q }{/tex}{tex}\\sqrt{n+1}\\;=\\frac{p^2+2q^2}{2pq}{/tex}...............(iii)From (i) and (ii),{tex}\\style{font-family:Arial}{\\sqrt{n-1}\\;=\\frac{p^2-2q^2}{2pq}}{/tex}........(iv)In RHS of (iii) and (iv)\xa0{tex}\\frac{p^2+2q^2}{2pq}\\;and\\;\\frac{\\displaystyle p^2-2q^2}{\\displaystyle2pq}\\;are\\;rational\\;number\\;because\\;p\\;and\\;q\\;are\\;positive\\;integers{/tex}But it is possible only when (n + 1) and (n - 1) both are perfect squares.Now n+1-(n-1)=n+1-n+1=2Hence they differ by 2 and two perfect squares never differ by 2.So both (n + 1) and (n -1 ) cannot be perfect squares. Hence there is no positive integer n for which\xa0{tex}\\style{font-family:Arial}{\\sqrt{n-1\\;}+\\sqrt{n+1}}{/tex} is rational | |
| 11487. |
\xa0if one zero of polynomial 2x^2-(3k-1) x-9 is the\xa0negative of the other find K and the zeros. |
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Answer» Also find the zeros Ans. Polynomial : 2x2 - (3k-1) x - 9\xa0on comparing with ax2\xa0+ bx + c, we get, a = 2, b = -(3k-1) and c = -9Let one zero =\xa0{tex}\\alpha {/tex},\xa0then second zero ll be =\xa0{tex}- \\alpha {/tex}We knowSum\xa0of zeroes =\xa0{tex}-{b\\over a}{/tex}=>\xa0{tex}-\\alpha + \\alpha = {-[{-({3k-1)}}]\\over 2}{/tex}=>\xa0{tex}0 = {3k-1\\over 2}{/tex}=> 3k -1 = 0\xa0=> 3k = 1\xa0=> k =\xa0{tex}{1\\over 3}{/tex} |
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| 11488. |
Find the quadratic polynomial ,the sum of whose root is √2 and their product is 1/3\xa0 |
| Answer» Ans.Given : Sum of roots =\xa0{tex}\\sqrt 2{/tex}Product of roots =\xa0{tex}{1\\over 3}{/tex}We know,\xa0Sum of roots =\xa0{tex}-b\\over a{/tex}=>\xa0{tex}{\\sqrt 2\\over 1 } = {-b\\over a}{/tex}=>\xa0{tex}{3\\times \\sqrt 2\\over 1\\times 3 } = {-b\\over a}{/tex}=>\xa0{tex}{3 \\sqrt 2\\over 3 } = {-b\\over a}{/tex}On comparing,we get, a = 3 and b =\xa0{tex}-3\\sqrt 2{/tex}Also,\xa0Products of the root =\xa0{tex}c\\over a{/tex}=>\xa0{tex}{1\\over 3} = {c\\over a}{/tex}on comparing a = 3 and c = 1So required polynomial :\xa0{tex}3x^2 -3\\sqrt 2x+1{/tex} | |
| 11489. |
{tex}2√3x²-5x+√3{/tex} |
| Answer» {tex}2\\sqrt 3x^2-5x+\\sqrt 3 =0{/tex}{tex}=> 2\\sqrt 3x^2-3x-2x+\\sqrt 3 =0{/tex}{tex}=> \\sqrt 3x(2x-\\sqrt 3)-1(2x-\\sqrt 3) =0 {/tex}{tex}=>( \\sqrt 3x-1)(2x-\\sqrt 3) =0 {/tex}{tex}=>x= {1\\over \\sqrt 3},{\\sqrt 3\\over 2} {/tex} | |
| 11490. |
Solve for X;4x^2-2 (a^2+b^2)x+a^2b^2=0 |
| Answer» {tex}4{x^2} - 2\\left( {{a^2} + {b^2}} \\right)x + {a^2}{b^2} = 0{/tex}=> {tex}4{x^2} - 2{a^2}x - 2{b^2}x + {a^2}{b^2} = 0{/tex}=> {tex}2x\\left( {2x - {a^2}} \\right) - {b^2}\\left( {2x - {a^2}} \\right) = 0{/tex}=> {tex}\\left( {2x - {a^2}} \\right)\\left( {2x - {b^2}} \\right) = 0{/tex}=> {tex}2x - {a^2} = 0{/tex}\xa0and {tex}2x - {b^2} = 0{/tex}=> {tex}x = {{{a^2}} \\over 2}{/tex}\xa0and {tex}x = {{{b^2}} \\over 2}{/tex} | |
| 11491. |
Is 7 × 5 × 3 × 2 +3 a composite number? Justify your answer |
| Answer» {tex}7\\times 5 \\times 3 \\times 2 +3{/tex}Taking 3 common, we get{tex}3(7\\times 5 \\times 2+1){/tex}=\xa0{tex}3(70+1){/tex}=\xa0{tex}3\\times 71{/tex}Yes the number is a composite number because it has 2 factors 3 and 71 other than one and itself. | |
| 11492. |
How to do cross-multiplication in simple method without using a1 a2 b1 b2 method\xa0x-3y-3=0\xa0 |
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| 11493. |
Find the sum of first 40 positive integers divisible by 6 |
| Answer» Acc.to que.\xa040 positive integers which are divisible by 6 is can be written in the form of A.p\xa06+12+18+..........+240a= 6 , d = a2 - a1 = 12 -6 =6n= 40 and an or nth terms = 240Therefore\xa0\xa0Sn = n/2 ×(a +an)= 4o/2 ×(6+240)=20×246=4920Hence sum of 40 positive integers which r divisible by 6 is 4920\xa0 | |
| 11494. |
find the zeroes of the polynomial ( 6x2-3- 7x) |
| Answer» Let\xa0{tex}P\\left( x \\right) =6{ x }^{ 2 }-7x-3{/tex}We have the zeroes of a polynomial are the values of which satisfy {tex}P\\left( x \\right) =0{/tex}\xa0Now\xa0{tex}P\\left( x \\right) =0{/tex}{tex}\\Rightarrow 6{ x }^{ 2 }-7x-3=0\\\\ \\Rightarrow 6{ x }^{ 2 }-9x+2x-3=0\\\\ \\Rightarrow 3x\\left( 2x-3 \\right) +1\\left( 2x-3 \\right) =0\\\\ \\Rightarrow \\left( 3x+1 \\right) \\left( 2x-3 \\right) =0\\\\ \\Rightarrow 3x+1=0\\quad or\\quad 2x-3=0\\quad \\\\ \\Rightarrow x=\\frac { -1 }{ 3 } ,x=\\frac { 3 }{ 2 } {/tex}Hence zeroes of the given polynomial are\xa0{tex}\\frac { -1 }{ 3 } and\\frac { 3 }{ 2 } {/tex} | |
| 11495. |
Find the sum of all natural number that are less than 100 and divisible by 4 |
| Answer» Number which are divisible by 4 form An AP:4,8,12,16,..... 96First term a = 4Common difference d = 8-4=4Last term l = 96Total Number of terms ==> an\xa0= a\xa0+ (n-1)d=> 96 = 4 +(n-1)4=> 92 = 4n-4=> 96 = 4n=> n = 24Sum of all numbers\xa0Sn\xa0=\xa0{tex}{n\\over 2}[a+l]{/tex}=>\xa0{tex}{24\\over 2}[4+96]{/tex}= 1200 | |
| 11496. |
9,15,21,27,... .\xa0 |
| Answer» 33As it is a AP with common difference 6. | |
| 11497. |
Tan^2A/sec^2A-1=? Please and this. |
| Answer» {tex}tan^2A\\over sec^2A-1{/tex}=\xa0{tex}sec^2A-1\\over sec^2A-1{/tex}\xa0as\xa0{tex}sec^2\\theta -tan^2\\theta = 1{/tex}= 1 | |
| 11498. |
Solve this, tan2A/sec2A-1=? |
| Answer» | |
| 11499. |
Solve:(a+2b)x + (2a-b)y=2(a-2b)x + (2a+B)y =3 |
| Answer» The given system of equations are :(a + 2b)x + (2a - b)y = 2So, (a + 2b)x + (2a - b)y - 2 = 0 ............(i)And (a - 2b) x + (2a + b) y = 3So, (a - 2b)x + (2a + b)y - 3 = 0 .........(ii)The given equations is in the form ofa1x + b1y + c1 = 0and a2x + b2y + c2 = 0So, we geta1 = a + 2b, b1 = 2a - b, c1 = -(2)a2 = a - 2b, b2 = 2a + b, c2 = (-3)By cross-multiplication method:{tex} \\frac{x}{{ - 2a + 5b}} = \\frac{y}{{a + 10b}} = \\frac{1}{{10ab}}{/tex}Now, {tex}\\frac{x}{{ - 2a + 5b}} = \\frac{1}{{10ab}} {/tex}{tex} ⇒ x = \\frac{{5b - 2a}}{{10ab}}{/tex}And {tex}\\frac{y}{{a + 10b}} = \\frac{1}{{10ab}} {/tex}{tex} ⇒ y = \\frac{{a + 10b}}{{10ab}}{/tex}The solution of the system of equations are {tex}x = \\frac{{5b - 2a}}{{10ab}}{/tex}\xa0and {tex}y = \\frac{{a + 10b}}{{10ab}}{/tex}\xa0respectively. | |
| 11500. |
PYTHAGORAS THEOREM VERIFICATION. |
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Answer» Take a right triangle ABC in which | |