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| 27201. |
Can you suggestme how can i get the 2019 2020 10th syllabus |
| Answer» www. cbse. nic. in | |
| 27202. |
Find the zeros, the sum of zeros and the productof the zeros of the quadratic polynomial p(x)=3x-x-4 |
| Answer» Sorry, I think the question is wrong | |
| 27203. |
If a and b are two integers, show that root2 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}. | |
| 27204. |
149 ×789 |
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Answer» 117561 117561 |
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| 27205. |
IFind the angle of x if 2sinx=sin2x |
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| 27206. |
k constant is multiplied or deleted from the polynomial before dividing it |
| Answer» The number which is been divided with the polynomial is the value of k and been cut off as they both are same .....EXAMPLE:- 2( x+y/2)2 will be cancelled out | |
| 27207. |
Express the HCF of 468 and 222 as 468x + 222y where x,y are integersin two different waxs. |
| Answer» Given integers are 468 and 222 where 468 > 222.By applying Euclid’s division lemma, we get 468 = 222 × 2 + 24 …(i)Since remainder ≠ 0, apply division lemma on division 222 and remainder 24 222 = 24 × 9 + 6 …(ii)Since remainder ≠ 0, apply division lemma on division 24 and remainder 6 24 = 6 × 4 + 0 …(iii)We observe that the remainder = 0, so the last divisor 6 is the HCF of the 468 and 222From (ii) we have6 = 222 – 24 × 9⇒ 6 = 222 – [468 – 222 × 2] × 9 [Substituting 24 = 468 – 222 × 2 from (i)]⇒ 6 = 222 – 468 × 9 – 222 × 18⇒ 6 = 222y + 468x, where x = −9 and y = 19⇒ 6 = 222 × 19 – 468 × 9 | |
| 27208. |
Show that only one out of the n n plus two and n + 4 is divisible divisible by 3 |
| Answer» Let the number be (3q + r){tex}n = 3 q + r \\quad 0 \\leq r < 3{/tex}{tex}\\text { or } 3 q , 3 q + 1,3 q + 2{/tex}{tex}\\text { If } n = 3 q \\text { then, numbers are } 3 q , ( 3 q + 1 ) , ( 3 q + 2 ){/tex}{tex}3 q \\text { is divisible by } 3{/tex}.{tex}\\text { If } n = 3 q + 1 \\text { then, numbers are } ( 3 q + 1 ) , ( 3 q + 3 ) , ( 3 q + 4 ){/tex}{tex}( 3 q + 3 ) \\text { is divisible by } 3{/tex}.{tex}\\text { If } n = 3 q + 2 \\text { then, numbers are } ( 3 q + 2 ) , ( 3 q + 4 ) , ( 3 q + 6 ){/tex}{tex}( 3 q + 6 ) \\text { is divisible by } 3{/tex}.{tex}\\therefore \\text { out of } n , ( n + 2 ) \\text { and } ( n + 4 ) \\text { only one is divisible by } 3{/tex}. | |
| 27209. |
What is coincident |
| Answer» If a1 upon a2 = b1 upon b2= c1 upon c2 then these lines are concident (lines lies on one another) | |
| 27210. |
Ak |
| Answer» | |
| 27211. |
Why is 0!=1 ??? |
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Answer» Aastha ??? ?? 0=(1/2^2-1/2^2)/(1/2-1/2) =(1/2+1/2)(1/2-1/2)/(1/2-1/2) =1/2+1/2=1 Because....of real number. |
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| 27212. |
What is the formula of (a+b)² |
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Answer» Thanks dear... a²+b²+2ab a^2 + b^2 +2ab a sq + b sq +2ab |
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| 27213. |
What do you meant by the pair of linear equation is consistent |
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Answer» It means that both equation will have a unique solution Hi |
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| 27214. |
Pair of linear equations exercise 2 |
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| 27215. |
Chapter 3whole |
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| 27216. |
The mean of 20 is 17 if 3is added to each other then find the mean |
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| 27217. |
what is a fundamental theorem of arithmatic |
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Answer» Every composite number can be expressed (factorised ) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.This theorem also says that the prime factorisation of a natural number is unique, except for the order of its factors.For example 20 can be expressed as 2×2×5Using this theorem the LCM and HCF of the given pair of positive integers can be calculated.LCM = Product of the greatest power of each prime factor, involved in the numbers.HCF = Product of the smallest power of each common prime factor in the numbers. Every composite number is expressed as a product of prime and this factorisation is unique apart from the order at which the prime factors occur... |
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| 27218. |
Show that reciprocal of 3+2√2 is an irrational number |
| Answer» First of all, rationalise the denominator of the reciprocal of 3 + 2√2.{tex}\\sf 3 + 2 \\sqrt{2} \\\\ \\\\ \\sf \\frac{1}{3 + 2 \\sqrt{2} } \\times \\frac{3 - 2 \\sqrt{2} }{3 - 2 \\sqrt{2} } \\\\ \\\\ \\sf \\frac{3 - 2 \\sqrt{2} }{(3 ){}^{2} - (2 \\sqrt{2} ) {}^{2} } \\\\ \\\\ \\sf \\frac{3 - 2 \\sqrt{2} }{9 - 8} \\\\ \\\\ \\bf 3 - 2 \\sqrt{2}{/tex}After rationalising its denominator, we get ( 3 - 2√2 ) as a result.Now, let us assume that ( 3 - 2√2 ) is an irrational number. So, taking a rational number i.e., 3 and subtracting from it.We have ;[ 3 - 2√2 - 3 ]⇒ - 2√2As a result, we get ( - 2√2 ) which is an irrational number.Hence, the reciprocal of ( 3 + 2√2) is an irrational number.\xa0 | |
| 27219. |
Solve the equation by subtuition method3x-5y-4=09x=2y+7 |
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Answer» Thnkew 3x - 5y - 4 = 0 .....(1)9x = 2y + 7 .......(2)So multiply equation 1 by 3 to get9x - 15y - 12 = 0 ......... (3)Put (2) in (3)2y + 7 - 15y - 12 = 0- 13y - 5 = 0- 13y = 5y = 5/-13Put value of y in (2)9x = 2(5/-13) + 79x = - 10/13 + 79x = (-10+ 91)/139x = 81/13x = 9/13 |
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| 27220. |
Solve the equation by subtuition method.x+y=5x-3y=4 |
| Answer» x\xa0+\xa0y\xa0= 5 ...\xa0(i)we getx\xa0= 5 -\xa0y Putting the value of\xa0x\xa0in equation\xa0(ii)\xa0we get5 –\xa0y – 3y\xa0= 4-4y\xa0= - 1y\xa0= 1/4\xa0Putting the value of\xa0y\xa0in equation\xa0(i)\xa0we getx\xa0= 5 – 1/4x\xa0= 19/4Hence,\xa0x\xa0= 19/4 and\xa0y\xa0= 1/4 | |
| 27221. |
Ex 14.1qno6 |
| Answer» Actually there are 2 methods One is direct method of sigma f x /sigma fThen the answer is 8.1Second is step deviation method of A+sigma f d/sigma d× hThen the answer is also 8.1 | |
| 27222. |
If cossec 5/3 ,prove that tan /1+tan×tan=sin/sec |
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| 27223. |
If 3cot=2 ,find the value of 5sin-3cos/2sin+6cos |
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| 27224. |
If tan=24/7,find sin +cos |
| Answer» (24/25)+(7/25)=31/25 | |
| 27225. |
If cot=3/4 ,find 4 tan-5cos/sec-4cot |
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| 27226. |
How to prove √2 is irrational number? |
| Answer» Suppose\xa0{tex}\\sqrt2{/tex}\xa0is a rational number. That is ,\xa0{tex}\\sqrt2{/tex}\xa0=\xa0{tex}\\frac{p}{q}{/tex}\xa0for some p{tex}\\in{/tex}Z and q {tex}\\in{/tex}Z.\xa0We can assume the fraction is in lowest fraction, That is p and q shares no common factors.Then {tex}\\sqrt2q=p{/tex}\xa0Squaring both side we get,\xa0{tex}2q^2=p^2{/tex}So\xa0{tex}p^2{/tex}\xa0is a multiple of 2,let\'s assume\xa0{tex}p=2m{/tex}\xa0Then,\xa0{tex}2q^2=\\left(2m\\right)^2{/tex}\xa0{tex}2q^2=4m^2{/tex}Or {tex}q^2=2m^2{/tex}So {tex}q^2{/tex}\xa0is a multiple of 2,{tex}\\therefore{/tex} q is multiple of 2Thus p and q shares a common factor.This is contradiction.{tex}\\Rightarrow {/tex}{tex}\\sqrt { 2 }{/tex}\xa0is an irrational number. | |
| 27227. |
All solutions of unit 3 |
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| 27228. |
17+4[19-41(3-2)] |
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| 27229. |
Find the zeros of polynomial x square -m(m+3) |
| Answer» Wrong question | |
| 27230. |
2 years ago salim was thrise three time age and 6 year later he will 4 year older than twice her age |
| Answer» Let Salim\'s present age be x years and his daughter\'s age be y years.Two years ago,Salim\'s age = (x - 2) yearsDaughter\'s age = (y - 2) yearsAs per given conditionTwo years ago, Salim was thrice as old as his daughter .\xa0x - 2 = 3(y - 2)x - 2 = 3y - 6{tex}\\Rightarrow{/tex} x - 3y = -4 .........(i)Six years hence,Salim\'s age = (x + 6) yearsDaughter\'s age = (y + 6) yearsAs per second conditionSix years later, he will be four years older than twice her age.x + 6 = 2(y + 6) + 4x + 6 = 2y + 12 + 4{tex}\\Rightarrow{/tex} x - 2y = 10 ...........(ii)Subtracting (i) from (ii), we havex - 2y - x + 3y =10 - ( -4)y = 14Put y = 14 in (i){tex}\\Rightarrow{/tex} x - 3(14) = -4{tex}\\Rightarrow{/tex} x - 42 = - 4{tex}\\Rightarrow{/tex} x = 38Therefore, the present age of Salim is 38 years and that of his daughter is 14 years. | |
| 27231. |
Find the hcf of 1999 and 2105 |
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| 27232. |
How do Euclid division lemma |
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| 27233. |
Sum of two no is 1000 and difference between their Square is 256000 find the no |
| Answer» Let us suppose that the numbers are x and y.According to question it is given thatThe sum of the two numbers is 1000.Thus, we have x + y = 1000The difference between the squares of the two numbers is 256000.Therefore, we have x2 - y2 = 256000{tex} \\Rightarrow{/tex}\xa0(x + y)(x - y) = 256000{tex}\\Rightarrow{/tex}\xa01000(x - y) = 256000{tex}\\Rightarrow x - y = \\frac{{256000}}{{1000}}{/tex}{tex}\\Rightarrow{/tex}\xa0x - y = 256Therefore, we have two equationsx + y = 1000 ......(1)x - y = 256 .....(2)Here x and y are unknowns.We have to solve the above equations for x and y.Adding equation (1) and (2), we get(x + y) + (x - y) = 1000 + 256{tex}\\Rightarrow{/tex}\xa0x + y + x - y = 1256{tex}\\Rightarrow{/tex}\xa02x = 1256{tex}\\Rightarrow x = \\frac{{1256}}{2}{/tex}x = 628Substituting the value of x in the equation (1) we get,\xa0628 + y = 1000{tex}\\Rightarrow{/tex}\xa0y = 1000 - 628{tex}\\Rightarrow{/tex}\xa0y = 372Therefore the numbers are 628 and 372. | |
| 27234. |
Give rational approximation of root5 to two places |
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| 27235. |
How to get full mark in math\'s |
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Answer» Solve sample paper Everyday solve at least 21 question By solving questions in question paper???? By solving questions ?? |
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| 27236. |
How many two digit natural number is divisible by 7? |
| Answer» Two digit numbers which are divisible by 7 are 14,21,28,.....98.It forms an A.P.{tex}a = 14, d = 7, a_n = 98{/tex}{tex}a_n = a + (n - 1)d{/tex}{tex}98 = 14 + (n - 1)7{/tex}{tex}98 -14 = 7n - 7{/tex}{tex}84 + 7 = 7n{/tex}or, {tex}7n = 91{/tex}or, {tex}n = 13{/tex} | |
| 27237. |
X 2y_+_ =-1and x-y _ =3 solve it by elimination method 2 3 3 |
| Answer» {tex}\\frac{x}{2} + \\frac{{2y}}{3} = - 1{/tex} ....(1){tex}x - \\frac{y}{3} = 3{/tex} ...(2)\tElimination method: Multiplying equation (2) by 2, we get (3)\t{tex}2x - \\frac{2}{3}y = 6{/tex} ....(3)\t{tex}\\frac{x}{2} + \\frac{{2y}}{3} = - 1{/tex} ....(1)\tAdding (3) and (1), we get\t{tex}\\frac{5}{2}x = 5{/tex}\t\xa0⇒ x =2\tPutting value of x in (2), we get\t2− {tex}\\frac{y}{3}{/tex}= 3\t⇒ y =−3\tTherefore, x =2 and y =−3\tSubstitution method:{tex}\\frac{x}{2} + \\frac{{2y}}{3} = - 1{/tex} ....(1)\t{tex}x - \\frac{y}{3} = 3{/tex} ....(2)\tFrom equation (2), we can say that {tex}x = 3 + \\frac{y}{3} = \\frac{{9 + y}}{3}{/tex}\tPutting this in equation (1), we get\t{tex}\\frac{{9 + y}}{6} + \\frac{2}{3}y = - 1{/tex}\t{tex} \\Rightarrow \\;\\frac{{9 + y + 4y}}{6} = - 1{/tex}\t⇒ 5y +9=−6\t⇒ 5y =−15\t⇒ y =−3\tPutting value of y in (1), we get\t{tex}\\frac{x}{2} + \\frac{2}{3}( - 3) = - 1{/tex}\t⇒ x = 2\tTherefore, x =2 and y =−3. | |
| 27238. |
Show that the square of any positive integer connot be form of 5q+2, 5q+3 |
| Answer» Let n be any positive integer. Applying Euclids division lemma with divisor = 5, we get{tex}\\style{font-family:Arial}{\\begin{array}{l}n=5q+1,5q+2,5q+3\\;and\\;5q+4\\;\\\\\\end{array}}{/tex}Now (5q)2 = 25q2 = 5m, where m = 5q2, which is an integer;{tex}\\style{font-family:Arial}{\\begin{array}{l}(5q\\;+\\;1)^{\\;2}\\;=\\;25q^2\\;+\\;10q\\;+\\;1\\;=\\;5(5q^2\\;+\\;2q)\\;+\\;1\\;=\\;5m\\;+\\;1\\\\where\\;m\\;=\\;5q^2\\;+\\;2q,\\;which\\;is\\;an\\;integer;\\\\\\;(5q\\;+\\;2)^2\\;=\\;25q^2\\;+\\;20q\\;+\\;4\\;=\\;5(5q^2\\;+\\;4q)\\;+\\;4\\;=\\;5m\\;+\\;4,\\\\\\;where\\;m\\;=\\;5q^2\\;+\\;4q,\\;which\\;is\\;an\\;integer;\\\\\\;(5q\\;+\\;3)^{\\;2}\\;=\\;25q^2\\;+\\;30q\\;+\\;9\\;=\\;5(5q^2\\;+\\;6q+\\;1)\\;+\\;4\\;=\\;5m\\;+\\;4,\\\\\\;where\\;m\\;=\\;5q^2\\;+\\;6q\\;+\\;1,\\;which\\;is\\;an\\;integer;\\\\\\;(5q\\;+\\;4)^2\\;=\\;25q^2\\;+\\;40q\\;+\\;16\\;=\\;5(5q^2\\;+\\;8q\\;+\\;3)\\;+\\;1\\;=\\;5m\\;+\\;1,\\;\\\\where\\;m\\;=\\;5q^2\\;+\\;8q\\;+\\;3,\\;which\\;is\\;an\\;integer\\\\\\end{array}}{/tex}Thus, the square of any positive integer is of the form 5m, 5m + 1 or 5m + 4 for some integer m.It follows that the square of any positive integer cannot be of the form 5m + 2 or 5m + 3 for some integer m. | |
| 27239. |
Find the value of a and b so that x^4 +x^3+ 8x^2 +ax+b is divible by x^2 +1. |
| Answer» If {tex}x^4+x^3+8x^2+ax +b{/tex}\xa0is exactly divisible by x2 + 1, the remainder after division should be zero.Now let us perform long divisionWe get, remainder = x (a - 1) + (b - 7)\xa0x (a - 1) + (b - 7 ) = 0{tex}\\Rightarrow{/tex}\xa0x (a - 1) + (b - 7) = 0x + 0{tex}\\Rightarrow{/tex}\xa0a - 1 = 0 and b - 7 = 0\xa0[On equating the coefficients of like powers of x]{tex}\\Rightarrow{/tex}a = 1 and b = 7 | |
| 27240. |
Find H. C. F at 592 and 252. Express it in linear combination of 592 and 252 |
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Answer» 592=252×2+88252=88×2+7688=76×1+1276=12×6+412=4×3+0Here,r=0So,HCF OF 592 & 252 is 4 We need to find the H.C.F. of 592 and 252 and express it as a linear combination of 592 and 252.By applying Euclid’s division lemma592 = 252 x 2 + 88Since remainder ≠ 0, apply division lemma on divisor 252 and remainder 88252 = 88 x 2 + 76Since remainder ≠ 0, apply division lemma on divisor 88 and remainder 7688 = 76 x 1 + 12Since remainder ≠ 0, apply division lemma on divisor 76 and remainder 1276 = 12 x 6 + 4Since remainder ≠ 0, apply division lemma on divisor 12 and remainder 412 = 4 x 3 + 0.Therefore, H.C.F. = 4.Now, 4 = 76 – 12 x 6= 76 – 88 – 76 x 1 x 6= 76 – 88 x 6 + 76 x 6= 76 x 7 – 88 x 6= 252 – 88 x 2 x 7 – 88 x 6= 252 x 7- 88 x 14- 88 x 6= 252 x 7- 88 x 20= 252 x 7 – 592 – 252 x 2 x 20= 252 x 7 – 592 x 20 + 252 x 40= 252 x 47 – 592 x 20= 252 x 47 + 592 x (-20)Hence obtained. |
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| 27241. |
Dive p(x) =x^2+1-x by g(x)=x^2+1-x find qx and rx |
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| 27242. |
If√3tanthita=1 then find the value ofsin2thita-cos2thita |
| Answer» Root under 3tanthitha=1Tanthitha=1/root under 3》p=1 & b=rootunder 3》h=root under 4= 2Sin^2thitha-cos^2thitha》(p/h)^2 - (b/h)^2》1/4 - 3/4》- 2/4 = - 1/2 | |
| 27243. |
Write 3 integers between -5/4 and 11/3 |
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| 27244. |
Define euclid\'s division algorithem |
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Answer» a=bq+r For a pair of given positive integers ‘a’ and ‘b’, there exist unique integers ‘q’ and ‘r’ such thata=bq+r, where 0≤r |
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| 27245. |
(A+b)^2 identity |
| Answer» A^2+b^2+2Ab | |
| 27246. |
Rs Aggarwal exercise 3b question number 26 |
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Answer» Abhi book niii h Ques. Likh do please Book nhi he? |
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| 27247. |
(1) 5x - 4y + 8=0 7x + 6y - 9 =0 |
| Answer» What should we have to do in this question to find solution or anything else? | |
| 27248. |
Solve the follwing |
| Answer» But what we have to solve.....?????? | |
| 27249. |
1,1 |
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| 27250. |
Mathematics class10 ch 3ex.3.1 ka pahla question |
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