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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.
| 38801. |
If x equal to 5 find the value of x -5 . |
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Answer» 10-10 0 0 ? 0 If x=5 Put value of x as 5X-55-5=0 So easy is thisss ☺☺☺☺?? 0 |
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| 38802. |
25%5 |
| Answer» 1.25 | |
| 38803. |
1+cosA/sinA +sinA/1+cosA |
| Answer» 2/sinA | |
| 38804. |
how to find the area of a segment? write the formula also |
| Answer» | |
| 38805. |
2×π2][ |
| Answer» 2*22/7*2 | |
| 38806. |
Can any one tell me the ans of 3 lesson 3.3 , 1 que 1 solution ans how 4+y+y =142y=10 |
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Answer» ....... Open 3.3 u will get and if am right so tell me 5 4+y+y=14......4+2y=14........4+10=14 (2y=10)............nice joke?????? Plzzz tell me if any no the ans |
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| 38807. |
What isrational number |
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Answer» Rational no. Which no. Which in the form of p/q where q does not equal to zero Numbers which can be expressed in the form of p/q are known as rational number The number of the form p/q, where p & q are integers, and q is not equal to 0 are called Rational Numbers. |
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| 38808. |
Formula to find AP |
| Answer» a+d,a+2d,a+3d | |
| 38809. |
Rolles method |
| Answer» | |
| 38810. |
How can I learn the prove question |
| Answer» Instead of cramming or learning just practice the questions more and more | |
| 38811. |
Find the value of q(x) = root 3x square + 10 X + 7 root 3 |
| Answer» 3x+3x+7x+73x+7(x+1)3x+7=0, x+1=0X=-7/3 and x=-1 | |
| 38812. |
2sin68°/cos22°-2cot15°/5tan75°-(3tan45°tan20°tan40°tan50°tan70°)/5 |
| Answer» 2sin(90°-22°)/cos22°-2cot(90°-75°)/5tan75°-(3tan45°tan20°tan40°tan50°tan70°)/5= 2cos22°/cos22°- 2tan75°/5tan75°- 3/5= 2-2/5-3/5= 8/5-3/5= 5/5=1 | |
| 38813. |
Prove that 2-root4 is irrational number? By any shortest method. |
| Answer» 2root 4 is equal to 4 because root 4is 2 | |
| 38814. |
if the zeroes of p(x) a(x)2-5x+10 are reciprocal of each other then find the value of a,b |
| Answer» Let zeroes of this polynomial be alpha and 1/alpha then alpha x 1/alpha equal c/a then the value of a is equal 10 | |
| 38815. |
Can i get class 10 2019 syllabus |
| Answer» Ya,u can get through this app | |
| 38816. |
For what value of k will the following equations have no solution.(3k+1)x+3y =2 ; (k2+1)x+(k-2)y=5 |
| Answer» Given linear equation is(3k + 1)x + 3 y - 2 = 0 .......... (i)(k2 + 1)x + (k - 2)y - 5 = 0 ............ (ii)Compare with a1x + b1y + c = 0 and a2x + b2y and c2 = 0a1 = 3k + 1 , b1 = 3 , c1 = -2and a2 = k2+ 1 , b2 = k - 2, c2 = -5The given system of equations will have no solution, if{tex} \\frac { a_1 } { a_2 } = \\frac { b_1 } { b_2} \\neq \\frac { c_1 } { c_2 }{/tex}{tex} \\frac { 3 k + 1 } { k ^ { 2 } + 1 } = \\frac { 3 } { k - 2 } \\neq \\frac { - 2 } { - 5 }{/tex}{tex}\\Rightarrow \\quad \\frac { 3 k + 1 } { k ^ { 2 } + 1 } = \\frac { 3 } { k - 2 } \\text { and } \\frac { 3 } { k - 2 } \\neq \\frac { 2 } { 5 }{/tex}Now,\xa0{tex}\\frac { 3 k + 1 } { k ^ { 2 } + 1 } = \\frac { 3 } { k - 2 }{/tex}{tex}\\Rightarrow{/tex}\xa0(3k + 1)(k - 2)=3(k2 + 1){tex}\\Rightarrow{/tex}\xa03k2 - 5k - 2 =3k2 + 3{tex}\\Rightarrow{/tex}\xa0-5k - 2 =3{tex}\\Rightarrow{/tex}\xa0-5k = 5{tex}\\Rightarrow{/tex}\xa0k = -1Clearly,\xa0{tex}\\frac { 3 } { k - 2 } \\neq \\frac { 2 } { 5 }{/tex}\xa0for k = -1.Hence, the given system of equations will have no solution for k = -1. | |
| 38817. |
ex4.2 |
| Answer» You can get thos whole exercise in this app | |
| 38818. |
A and B are two positive integer then show that root 2 is lies between a/b and a+2b/a+b |
| Answer» We do not know whether\xa0{tex} \\frac { a } { b } < \\frac { a + 2 b } { a + b } \\text { or, } \\frac { a } { b } > \\frac { a + 2 b } { a + b }{/tex}.Therefore, to compare these two numbers, let us compute\xa0{tex} \\frac { a } { b } - \\frac { a + 2 b } { a + b }{/tex}We have,{tex} \\frac { a } { b } - \\frac { a + 2 b } { a + b } = \\frac { a ( a + b ) - b ( a + 2 b ) } { b ( a + b ) }{/tex}\xa0{tex} = \\frac { a ^ { 2 } + a b - a b - 2 b ^ { 2 } } { b ( a + b ) } = \\frac { a ^ { 2 } - 2 b ^ { 2 } } { b ( a + b ) }{/tex}{tex} \\therefore \\quad \\frac { a } { b } - \\frac { a + 2 b } { a + b } > 0{/tex}{tex} \\Rightarrow \\quad \\frac { a ^ { 2 } - 2 b ^ { 2 } } { b ( a + b ) } > 0{/tex}{tex} \\Rightarrow{/tex} a2 - 2b2 > 0{tex} \\Rightarrow{/tex} a2> 2b2{tex} \\Rightarrow \\quad a > \\sqrt { 2 } b{/tex}and,\xa0{tex} \\frac { a } { b } - \\frac { a + 2 b } { a + b } < 0{/tex}{tex} \\Rightarrow \\quad \\frac { a ^ { 2 } - 2 b ^ { 2 } } { b ( a + b ) } < 0{/tex}{tex} \\Rightarrow{/tex} a2 - 2b2 < 0{tex} \\Rightarrow{/tex}a2 <2b2{tex} \\Rightarrow \\quad a < \\sqrt { 2 } b{/tex}Thus,\xa0{tex} \\frac { a } { b } > \\frac { a + 2 b } { a + b }{/tex}, if\xa0{tex}a > \\sqrt { 2 b }{/tex}\xa0and\xa0{tex} \\frac { a } { b } < \\frac { a + 2 b } { a + b }{/tex},\xa0if\xa0{tex} a < \\sqrt { 2 } b{/tex}.So, we have the following cases:CASE I When\xa0{tex} a > \\sqrt { 2 } b{/tex}In this case, we have{tex} \\frac { a } { b } > \\frac { a + 2 b } { a + b } \\text { i.e., } \\frac { a + 2 b } { a + b } < \\frac { a } { b }{/tex}We have to prove that{tex} \\frac { a + 2 b } { a + b } < \\sqrt { 2 } < \\frac { a } { b }{/tex}We have,{tex} a > \\sqrt { 2 } b{/tex}{tex} \\Rightarrow{/tex} a2> 2b2 [Adding a2 on both sides]{tex} \\Rightarrow \\quad 2 a ^ { 2 } + 2 b ^ { 2 } > \\left( a ^ { 2 } + 2 b ^ { 2 } \\right) + 2 b ^ { 2 }{/tex}\xa0[Adding\xa02b2 on both sides]{tex} \\Rightarrow \\quad 2 \\left( a ^ { 2 } + b ^ { 2 } \\right) + 4 a b > a ^ { 2 } + 4 b ^ { 2 } + 4 a b{/tex}\xa0[Adding 4ab on both sides]{tex} \\Rightarrow \\quad 2 \\left( a ^ { 2 } + 2 a b + b ^ { 2 } \\right) > a ^ { 2 } + 4 a b + 4 b ^ { 2 }{/tex}{tex} \\Rightarrow \\quad 2 ( a + b ) ^ { 2 } > ( a + 2 b ) ^ { 2 }{/tex}{tex} \\Rightarrow \\quad \\sqrt { 2 } ( a + b ) > a + 2 b{/tex}{tex} \\Rightarrow \\quad \\sqrt { 2 } > \\frac { a + 2 b } { a + b }{/tex} ........(i)Again,{tex} a > \\sqrt { 2 } b {/tex}{tex}\\Rightarrow \\frac { a } { b } > \\sqrt { 2 }{/tex} .......(ii)From (i) and (ii), we get{tex} \\frac { a + 2 b } { a + b } < \\sqrt { 2 } < \\frac { a } { b }{/tex}CASE II When\xa0{tex} a < \\sqrt { 2 } b{/tex}In this case, we have{tex} \\frac { a } { b } < \\frac { a + 2 b } { a + b }{/tex}We have to show that\xa0{tex} \\frac { a } { b } < \\sqrt { 2 } < \\frac { a + 2 b } { a + b }{/tex}We have,{tex} a < \\sqrt { 2 } b{/tex}{tex} \\Rightarrow \\quad a ^ { 2 } < 2 b ^ { 2 }{/tex}{tex} \\Rightarrow \\quad a ^ { 2 } + a ^ { 2 } < a ^ { 2 } + 2 b ^ { 2 }{/tex}\xa0[Adding a2 on both sides]{tex} \\Rightarrow \\quad 2 a ^ { 2 } + 2 b ^ { 2 } < a ^ { 2 } + 2 b ^ { 2 }+ 2 b ^ { 2 }{/tex}\xa0[Adding 2b2 on both sides]{tex}\\Rightarrow \\quad 2 a ^ { 2 } + 2 b ^ { 2 } < a ^ { 2 } + 4 b ^ { 2 }{/tex}{tex} \\Rightarrow \\quad 2 a ^ { 2 } + 4 a b + 2 b ^ { 2 } < a ^ { 2 } + 4 a b + 4 b ^ { 2 }{/tex}\xa0[Adding 4ab on both sides]{tex} \\Rightarrow \\quad 2 ( a + b ) ^ { 2 } < ( a + 2 b ) ^ { 2 }{/tex}{tex} \\Rightarrow \\sqrt { 2 } ( a + b ) < a + 2 b{/tex}{tex} \\Rightarrow \\quad \\sqrt { 2 } < \\frac { a + 2 b } { a + b }{/tex}\xa0. ...(iii){tex} \\Rightarrow \\quad a < \\sqrt { 2 } b \\Rightarrow \\frac { a } { b } < \\sqrt { 2 }{/tex} ....(iv)From (iii) and (iv), we get{tex} \\frac { a } { b } < \\sqrt { 2 } < \\frac { a + 2 b } { a + b }{/tex}Hence,\xa0{tex} \\sqrt { 2 }{/tex}\xa0lies between\xa0{tex} \\frac { a } { b }{/tex}\xa0and\xa0{tex} \\frac { a + 2 b } { a + b }{/tex}. | |
| 38819. |
Cos squared theta + sec squared 30 degree + sin square theta |
| Answer» | |
| 38820. |
Is it required in matric to remember and understand t -ratios greater than 90 degrees |
| Answer» | |
| 38821. |
If the 2 zeros of the polynomial x2-bx+care in the ratio2:3show that 6b2=25c |
| Answer» | |
| 38822. |
As terminating number has a rational or irrational number |
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Answer» Rational Terminating |
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| 38823. |
Value of 49! |
| Answer» I think question is wrong | |
| 38824. |
X + √y=7 &. √x+y=11 solve for x and y |
| Answer» | |
| 38825. |
What are the total surface area and curved surface area and volume of all shapes |
| Answer» | |
| 38826. |
Find the value of k for which 2 X + 3 is a factor of 2 x cube + 9 x square minus x minus b |
| Answer» i couldnot able to understand that you want the value of k but there is no unknown qwantity of k in this question .Here only b is unknown variable | |
| 38827. |
hcf and division method |
| Answer» | |
| 38828. |
Find the quadratic polynomial whose zeroes are 5and-5 |
| Answer» x^2-x-10 | |
| 38829. |
X=-√2is a zero of p(x) =x^-√2x+k find the value of k |
| Answer» The value of k is 0 | |
| 38830. |
Three x square - 4 root 3 x +4=0 |
| Answer» | |
| 38831. |
Pair of this linear equation consistent or not:- X+y=14 x-y=4 |
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Answer» Yes, this equations are consist .They also have a unique solution. Yes they r consisting a1/a2 not =to b1/b2~ 1/1 not=to 1/-1 Internet pr search kro 9 and 5 X+y=14X-y= 4-_+___-_____ 2y=10 y=5Puttion the value of y=5 we getX+y=14X+5=14X=9 Plz ans me fastly!!?? |
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| 38832. |
We can become good friends. |
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Answer» Why not shivani Or niche tumhara ans glt h Abhi koi ni h syd tumhari sunne wala |
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| 38833. |
Is there is RD sharma examples ? |
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Answer» Bro did u understand ur Question after writing becoz sorry i cant... If u want rd sharma example then buy the book Wt do u mean |
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| 38834. |
The sum of two consecutive number is 313 find the number |
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Answer» The two numbers are 156 and 157 .The two numbers are 12 and 13. |
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| 38835. |
Prove that product of any three consecutive integer is divisible with 6 |
| Answer» Let three consecutive numbers be x, (x + 1) and (x + 2)Let x = 6q + r 0 {tex}\\leq r < 6{/tex}{tex}\\therefore x = 6 q , 6 q + 1,6 q + 2,6 q + 3,6 q + 4,6 q + 5{/tex}{tex}\\text { product of } x ( x + 1 ) ( x + 2 ) = 6 q ( 6 q + 1 ) ( 6 q + 2 ){/tex}if x = 6q then which is divisible by 6{tex}\\text { if } x = 6 q + 1{/tex}{tex}= ( 6 q + 1 ) ( 6 q + 2 ) ( 6 q + 3 ){/tex}{tex}= 2 ( 3 q + 1 ) .3 ( 2 q + 1 ) ( 6 q + 1 ){/tex}{tex}= 6 ( 3 q + 1 ) \\cdot ( 2 q + 1 ) ( 6 q + 1 ){/tex}which is divisible by 6if x = 6q + 2{tex}= ( 6 q + 2 ) ( 6 q + 3 ) ( 6 q + 4 ){/tex}{tex}= 3 ( 2 q + 1 ) .2 ( 3 q + 1 ) ( 6 q + 4 ){/tex}{tex}= 6 ( 2 q + 1 ) \\cdot ( 3 q + 1 ) ( 6 q + 1 ){/tex}Which is divisible by 6{tex}\\text { if } x = 6 q + 3{/tex}{tex}= ( 6 q + 3 ) ( 6 q + 4 ) ( 6 q + 5 ){/tex}{tex}= 6 ( 2 q + 1 ) ( 3 q + 2 ) ( 6 q + 5 ){/tex}which is divisible by 6{tex}\\text { if } x = 6 q + 4{/tex}{tex}= ( 6 q + 4 ) ( 6 q + 5 ) ( 6 q + 6 ){/tex}{tex}= 6 ( 6 q + 4 ) ( 6 q + 5 ) ( q + 1 ){/tex}which is divisible by 6if x = 6q + 5{tex}= ( 6 q + 5 ) ( 6 q + 6 ) ( 6 q + 7 ){/tex}{tex}= 6 ( 6 q + 5 ) ( q + 1 ) ( 6 q + 7 ){/tex}which is divisible by 6{tex}\\therefore {/tex}\xa0the product of any three natural numbers is divisible by 6. | |
| 38836. |
If sin A =8/17, find other trignometric ratios of angle A |
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Answer» Cos A =15/17, Tan A =8/15, Cosec A =17/8, Sec A =17/15, Cot A =15/8 Sin a=8/17Cos a=15/17Tan a=8/15Cosec a=17/8Sec a=17/15Cot a=15/8 |
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| 38837. |
555+455 |
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Answer» Not This type of question is note for class 10 students 1010 |
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| 38838. |
How should I plan my preparation for boards |
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Answer» Varun.... I think u have got your answer... Researchers have told water boosts your mental health or performance Study Ncrt deeply then after goes to other extra books Ritu whats the role of water i know its too hot temperature but like to know what water does... In prepartion of exams Know your weakness,try to overcome it.Be thorough with the book and revise them as much as you can. Give sufficient time to studies. Practice previous year question papers. Take regular breaks during study time. Drink sufficient water...... |
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| 38839. |
Which question bank/guide is best for mathematics? |
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Answer» First R.S aggarwal then oswaal and try R.D sharma while preparing for finals. RD Sharma, 100% Success, Exam-Idea S N Upadhaya book |
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| 38840. |
Solve the following pair of linear equations graphically 2x+3y=12, 2y-1=x(2y-x=1) |
| Answer» | |
| 38841. |
If sin theta =a^2-b^2/a^2+b^2,find the value of all ratio of theta |
| Answer» Let us draw a right triangle ABC in which\xa0{tex}\\angle B A C = \\theta{/tex}{tex}\\sin \\theta = \\frac { a ^ { 2 } - b ^ { 2 } } { a ^ { 2 } + b ^ { 2 } }{/tex}\xa0...... Given{tex}\\Rightarrow \\frac { B C } { A C } = \\frac { a ^ { 2 } - b ^ { 2 } } { a ^ { 2 } + b ^ { 2 } } \\Rightarrow \\frac { B C } { a ^ { 2 } - b ^ { 2 } } = \\frac { A C } { a ^ { 2 } + b ^ { 2 } } = k ( \\text { say } ){/tex}where k is a positive number.{tex}\\Rightarrow B C = k \\left( a ^ { 2 } - b ^ { 2 } \\right){/tex}{tex}A C = k \\left( a ^ { 2 } + b ^ { 2 } \\right){/tex}In\xa0{tex}\\Delta A B C{/tex}{tex}\\because \\angle B = 90 ^ { \\circ }{/tex}{tex}\\therefore A C ^ { 2 } = A B ^ { 2 } + B C ^ { 2 } \\ldots \\ldots{/tex}\xa0By Pythagoras theorem{tex}\\Rightarrow \\mathrm { k } ^ { 2 } \\left( \\mathrm { a } ^ { 2 } + \\mathrm { b } ^ { 2 } \\right) ^ { 2 } = \\mathrm { AB } ^ { 2 } + \\mathrm { k } ^ { 2 } \\left( \\mathrm { a } ^ { 2 } - \\mathrm { b } ^ { 2 } \\right) ^ { 2 }{/tex}{tex}\\Rightarrow A B ^ { 2 } = k ^ { 2 } \\left\\{ \\left( a ^ { 2 } + b ^ { 2 } \\right) ^ { 2 } - \\left( a ^ { 2 } - b ^ { 2 } \\right) ^ { 2 } \\right\\}{/tex}{tex}\\Rightarrow A B ^ { 2 } = k ^ { 2 } \\left( 4 a ^ { 2 } b ^ { 2 } \\right) \\Rightarrow A B = 2 a b k{/tex}Therefore,{tex}\\cos \\theta = \\frac { A B } { A C } = \\frac { 2 a b k } { k \\left( a ^ { 2 } + b ^ { 2 } \\right) } = \\frac { 2 a b } { a ^ { 2 } + b ^ { 2 } }{/tex}{tex}\\tan \\theta = \\frac { B C } { A B } = \\frac { k \\left( a ^ { 2 } - b ^ { 2 } \\right) } { 2 a b k } = \\frac { a ^ { 2 } - b ^ { 2 } } { 2 a b }{/tex}{tex}cosec \\theta = \\frac { A C } { B C } = \\frac { k \\left( a ^ { 2 } + b ^ { 2 } \\right) } { k \\left( a ^ { 2 } - b ^ { 2 } \\right) } = \\frac { a ^ { 2 } + b ^ { 2 } } { a ^ { 2 } - b ^ { 2 } }{/tex}{tex}\\sec \\theta = \\frac { A C } { A B } = \\frac { k \\left( a ^ { 2 } + b ^ { 2 } \\right) } { 2 a b k } = \\frac { a ^ { 2 } + b ^ { 2 } } { 2 a b }{/tex}{tex}\\cot \\theta = \\frac { A B } { B C } = \\frac { 2 a b k } { k \\left( a ^ { 2 } - b ^ { 2 } \\right) } = \\frac { 2 a b } { a ^ { 2 } - b ^ { 2 } }{/tex} | |
| 38842. |
What is the greatest number |
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Answer» Well friends 99999is also a number Infinite Brother Rajiv , 9 is greatest digit not number You dont ask greatest 5 digit no you only ask number then 9 is the right answer Tell the Digits 9 99999 |
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| 38843. |
2x-3/y=93x+7/y=2 |
| Answer» Let 1/y be a(2x-3a=9 )3(3x+7a=2. )2By elimination method6x-9a=27. 6x+14a=4- - -__________0x-23a=23-23a=23a=23/-23a=-12x-3(-1)=92x+3=92x=9-3x=6/2x=31/y=a1/y=-1y=-1(x,y)=(3,-1) | |
| 38844. |
Which is heavier 1 kg iron or 1 kg cotton ? Give explanation/solution for your answer. |
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Answer» JIN KAZAMA you are right but in that following question there is no specific condition then on what behalf you have given this type of explanation because your explanation consists of terms ,cases and conditions .but what ever it is i would like to become your homework friend and one of the compatitor thanks Jin Kazama Well if you go for a bit more accuracy and practical experience, it is not true!1kg of cotton will weigh slightly less than 1kg Iron. . Here is the explanation.1. Density of cotton is much less than iron.2. Hence the total volume of 1 kg cotton will be much higher than 1 kg iron.3. So 1 Kg cotton will displace more air than 1 kg iron.4. A larger buoyant force will act on cotton in relative to iron. - Buoyant force being equal to the weight of the fluid being displaced.5. Result will be slight decrease in weight.In vacuum both will weigh the same may be you can able to understand my explanation in this case both have same mass but different weight due to there different surface areas the force of drag(frictional force applied by air ) will be less and also relative density of cotton is more and surface area is also more so the force is diverted to a larger area (p=m/a) Both weight are same |
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| 38845. |
Prove that √6+√8+√9 is a irrational number |
| Answer» let it a ration no. and then prove it according to the ex of NCERT book | |
| 38846. |
please tell me how to find LCM of any two numbers |
| Answer» It\'s very simple Firstly take 2 numbers Let we take 5 and 10 Now make the factors of 5 and 10 5= 5×1 10= 2×5 now we take 5 common and multiply the left values by 5 LCM = 2×1×5 = 10 So by this way u can find the lcm of 2 numbers | |
| 38847. |
What is potential |
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Answer» work doen to carry a unit positive charge from infinity to a point in electric field is called electric potential at that point I mean stored Something which is storef |
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| 38848. |
Show that 3√2 is irrational. |
| Answer» Let us assume to the contrary that 3√2 is a rational This is we can find coprime a and b (b is not equal to 0) such that 3√2=a/bRearranging we get √2=a/3bSince 3,a and b are. Integer a/3b is rational and so √2 is rationalBut this contradicts the fact that √2 is irrationalSo we conclude that 3√2 is irrational | |
| 38849. |
I mean rd sharmaś level examples |
| Answer» | |
| 38850. |
I can\'t improve my math what can i do |
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Answer» Don\'t worry...u can do it...ALL THE BEST I was also having the same situation till class 8. But after that i practised math every day and now I\'m getting very good marks in maths. Practice makes a man perfectU practice maths as much as u can. Then see.... Try doing all sums from your text book and then go your additional reference books You should not opt it |
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