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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.
| 27401. |
How to represent quadratic polynomial on a graph? |
| Answer» By finfing out the solutions only as we used to do for making the graphs of a linear polynomial. Here, the shape of the graph will be U shaped which is called parabola... | |
| 27402. |
For what least value of \'n\' a natural number (24)power n is divisible by 8 solution |
| Answer» 1 because 1 is a natural number and 24x1 is divisible by 8.so n=1 | |
| 27403. |
7×11×13×15=15 is what and it solution |
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| 27404. |
the speed of a car is 171/2 kilometre per hour how much distance will it cover in 22 upon 3 hours |
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| 27405. |
Exercise 3.3 by substituting the value |
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| 27406. |
Plz give me 14.1 full ex with assumed method |
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| 27407. |
Use Euclid\'s algorithm to find HCF of 1651and 2032.Express the HCF in the form of 1651m+2032n |
| Answer» \xa02032 = 1651 {tex} \\times{/tex} 1 + 381 .1651 = 381 {tex} \\times{/tex} 4 + 127\xa0381 = 127 {tex} \\times{/tex} 3 + 0.\xa0Since the remainder becomes 0 here, so HCF of 1651 and 2032 is 127.{tex} \\therefore{/tex}\xa0HCF (1651, 2032) = 127.Now,{tex} 1651 = 381 \\times 4 + 127{/tex}{tex} \\Rightarrow \\quad 127 = 1651 - 381 \\times 4{/tex}{tex} \\Rightarrow \\quad 127 = 1651 - ( 2032 - 1651 \\times 1 ) \\times 4{/tex}\xa0[from 2032 = 1651 {tex} \\times{/tex} 1 + 381]{tex} \\Rightarrow \\quad 127 = 1651 - 2032 \\times 4 + 1651 \\times 4{/tex}{tex} \\Rightarrow \\quad 127 = 1651 \\times 5 + 2032 \\times ( - 4 ){/tex}Hence, m = 5, n = -4. | |
| 27408. |
Maths real numbers |
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| 27409. |
1$ =how much ponds |
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Answer» yes....0.77 0.77 pounds |
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| 27410. |
May please give sample papers of last 5 years class 10th |
| Answer» You may find it in cbse guide as well as in google too | |
| 27411. |
2× 2=? |
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Answer» 4 It is always 4 4 Hii Anushka Obviously 4 The answer is 4 |
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| 27412. |
What is the difference between root 2 and root 3 |
| Answer» i think nothing any difference between them | |
| 27413. |
Show that every positive integer is either even or odd. |
| Answer» According to the question, we have to show that every positive integer is either even or odd.Let us assume that there exists a smallest positive integer that is neither odd nor even, say n. Since n is the least positive integer which is neither even nor odd, n - 1 must be either odd or even.Case 1: If n - 1 is even, n - 1 = 2k for some k.But this implies n = 2k + 1This implies n is odd.Case 2: If n - 1 is odd, n - 1 = 2k + 1 for some k.But this implies n = 2k + 2 = 2(k + 1)This implies n is even.Therefore,In both cases , we arrive at a contradiction.Thus, every positive integer is either even or odd | |
| 27414. |
Prove that n (n+1) is always even for any positive integer value n |
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| 27415. |
Obtain all the zeros of f(x)=x^3+13x^2+3x+20is one of its zeroes is -2 |
| Answer» Here\xa0f(x) = x3 + 13x2 + 32x + 20-2 is a zero of f(x), so x+2 will be factor of f(x)On long division of f(x) by x+2 we getSo, f(x) = x3 + 13x2 + 32x + 20= (x2 + 11x + 10)(x + 2)= (x2 + 10x + x + 10)(x + 2)= (x + 10)(x + 1)(x + 2)Hence f(x)=0 if x+10 =0 or x+1=0 or x+2=0Therefore, the zeroes of the polynomial are -1, -10, -2. | |
| 27416. |
Prove that Tan1 Tan2 Tan3 ..... Tan89 is equal to 1 |
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| 27417. |
Who made maths and why |
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| 27418. |
Cos square 47 +sin square 47 is equal to |
| Answer» 1 | |
| 27419. |
What is the value of sin60 |
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Answer» √3/2 Root 3 by 2 Root 3 by 2 |
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| 27420. |
Find six rational numbers between -2/3and4/3 |
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Answer» Given,to find six rationals b/w -2/3 and 4/3 so answer is -2/3×7/7=-14/21 4/3×7/7=28/21Therefore the six rationals from above are 1/21,2/21,3/21,4/21,5/21,6/21 2/3×7/7=14/214/3×7/7=28/21 |
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| 27421. |
What perfect definition of lemma |
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Answer» A lemma can be termed as a proved statement which is used to prove other statement or question. Its proved statment use to prove another statement |
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| 27422. |
Solve the following linear equation by the substitution method0.2x+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 | |
| 27423. |
Whatr natural numbers |
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Answer» Thanks A natural number is an integer greater than zero natural number began at 1 and increment to infinity:1,2,3,4,5,6etc natural number is also called counting numbers The positive integers (whole numbers) and zero as well are called natural numbers |
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| 27424. |
64^1÷2 |
| Answer» 32 | |
| 27425. |
Show that any positive integer is of the form 6q+1,6q+3,6q+5where q is some integer |
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Answer» We Consider An positive integer as a.On dividing a by b .Here , let q is the quotient and r is the remainder.Now, a = bq + r , 0 ≤ r < b ..... ( 1 )[ by using Euclid\'s division lemma]Here we putting b = 6 in eq ( 1 )Here we find , a = 6q + r , 0 ≤ r < b ..... ( 2 )so here possible values of r = 1 , 2, 3, 4, 5.If r = 0, then find Equation (2) , a = 6q.Here, 6q is is divisible by 2 , so 6q is here Even .If r = 1 , then find Equation (2) , a = 6q + 1.Here, 6q + 1 is not divisible by 2 , so 6q + 1 is here odd.If r = 2 , then find Equation (2) , a = 6q + 2.Here, 6q + 2 is not divisible by 2 , so 6q + 2 is here even.If r = 3 , then find Equation (2) , a = 6q + 3.Here, 6q + 3 is not divisible by 2 , so 6q + 3 is here odd.If r = 4 , then find Equation (2) , a = 6q + 4.Here, 6q + 4 is not divisible by 2 , so 6q + 4 is here even.If r = 5 , then find Equation (2) , a = 6q + 5.Here, 6q + 5 is not divisible by 2 , so 6q + 5 is here odd.so , a is odd , so a cannot be 6q,6q+2, 6q+4.Therefore any positive odd integer is of the form 6q+1 , 6q +3, 6q + 5. Aese kyun bol rahe ho, deepak to sirf question puche hein, ye kaise answer hota hai |
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| 27426. |
Prove that sinA+cosA the whole square+cosA+cosecA the whole square =1+secAcosecA the whole square |
| Answer» (sin A + sec A)² + (Cos A + Cosec A)²\xa0= Sin² A + sec² A + 2 Sin A Sec A + Cos² A + Cosec² A + 2 Cos A Cosec A= Sin²A + COs²A + Sec²A + Cosec²A + 2 Sin A SecA + 2 Cos A Cosec A= 1 + [ 1/Cos²A + 1/ Sin²A ] + [ 2 Sin A / Cos A + 2 Cos A / SIn A ]= 1 + (Sin²A + COs²A)/ [Cos²A Sin²A ] + 2 [ SIn² A + Cos²A ] / [ SinA CosA= 1 + 1/Cos²A 1/Sin²A + 2 1/SinA 1/CosA= 1 + Sec²A Cosec²A + 2 COsecA Sec A= (1 + SecA CosecA )²====================alternately,(sin A + sec A)² + (Cos A + Cosec A)²\xa0= (Sin A + 1/Cos A)² + (COs A + 1/ SinA)²= (Sin A Cos A + 1)² / Cos² A + (SinA COsA + 1)² / Sin² A= [ SIn A Cos A + 1]² [ 1/Cos² A + 1/Sin² A ]\xa0= [ SIn A Cos A + 1]² [ Cos² A + Sin² A ] / [Sin²A Cos² A ]= [ Sin A Cos A + 1 ]² / [Sin²A Cos² A ]= [ (Sin A Cos A + 1) / (Sin A Cos A )]²= (1 + 1/Sin A 1/Cos A)²\xa0= (1 + Sec A Cosec A)²\xa0 | |
| 27427. |
Prove that2903^N-803^N-464^N+261^N is divisible by 1897 |
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| 27428. |
55555 H.C.F |
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Answer» ??... question puch kar pagl bolte ho... Esa agar kisi or ke sath kroge na...to answer na dene ke sath sath pit jaoge.. Kya mst question hai.... mila kahan se... ya khud ki imagination hai batao???? Paglo |
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| 27429. |
Check whether 4n can end with the digit 0 for any natural number n. |
| Answer» Nope it cannot end with 0 because 4 has none of its factors as 5... | |
| 27430. |
X square + 2root 2x +6 |
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| 27431. |
Find conditions that zeroes of a quadratic polynomial are reciprocal of each other |
| Answer» The condition that zeroes of a polynomial are reciprocal of each other are a=c | |
| 27432. |
Show that root 7 is irrational number |
| Answer» let us assume that √7 be rational.then it must in the form of p / q [q ≠ 0] [p and q are co-prime]√7 = p / q=> √7 x q = psquaring on both sides=> 7q2= p2 ------ (1)p2 is divisible by 7p is divisible by 7p = 7c [c is a positive integer] [squaring on both sides ]p2 = 49 c2 --------- (2)Subsitute p2 in equ (1) we get7q2 = 49 c2q2 = 7c2=> q is divisible by 7thus q and p have a common factor 7.there is a contradictionas our assumsion p & q are co prime but it has a common factor.So that √7 is an irrational. | |
| 27433. |
Find the H.C.F of 867 and 225 |
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Answer» 867 is grater then 225867=225×3+192225=192×1+33192=33×5+2733=27×1+627=6×4+36=3×2+0Therefore HCF of 867 and 225 = 3 3.... H.C.F. of 867 and 225We know , a =bq+r 876=255×3+102 255=102×2+51102=51×2+0As r = 0Nd b = 51So , H.C.F. of 867 nd 255 is 51 |
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| 27434. |
Anyone on ????? |
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Answer» Yeah frnd Hmm... Yup ... |
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| 27435. |
What is the regioning |
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| 27436. |
Find the HCF and LCM 96 and 21 |
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Answer» 96 =\xa02\xa0×\xa02\xa0×\xa02\xa0×\xa02\xa0×\xa02\xa0×\xa0321 = 7 ×\xa03HCF = 3LCM = 2\xa0×\xa02\xa0×\xa02\xa0×\xa02\xa0×\xa02\xa0×\xa03× 7 = 672 LCM is 672 and it\'s HCF is 3 ....??????????? |
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| 27437. |
2x+3y=4 and 3x-y=-5 represent on the graph |
| Answer» Ram ram | |
| 27438. |
Find a quadratic polynomial whose zeros are -3 and 4 |
| Answer» Let x= 3 | |
| 27439. |
Tan45 |
| Answer» Value is 1 | |
| 27440. |
Use Euclid\'s division algorithm to find the HCF of |
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Answer» Complete the question then ask Complete yoir question first and then ask. Complete yur question.... |
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| 27441. |
What is the remainder theorm |
| Answer» Let p(x) be any polynomial of degree greater than or equal to 1 and let α be any real number. If p(x) is divided by the polynomial (x - α), then the remainder is p(α).In other words:If the polynomial f(x) is divided by x - α then the remainder R is given by f(x) = (x - α) q(x) + R, where q(x) is the quotient and R is a constant (because the degree of the remainder is less than the degree of the divisor x - α).Putting x = α, f(α) = (α - α)q(α) + R or f(α) = RWhen the polynomial f(x) is divided by x - α, the remainder R = f(α) = value of f(x) when x is α. | |
| 27442. |
Artamatic progrations lesson is there or not |
| Answer» | |
| 27443. |
How to substitute 13x+26y-13=0and 8x+4y-5=0 |
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Answer» X= 1/2 X=1/2.... Y=1/4 X=1/2 and y=1/4 |
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| 27444. |
Is mathematics easy for class 10? |
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Answer» Its easy..But it is also depends on you...If you like maths then it is easy to you and if not than... Bhgvaan hi maalik hai...??? Easy Its easy.... Its easy....? Easy* Very east |
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| 27445. |
The sum of first 10 terms of an ap is -150 and the sum of its next 10 terms is -550 . Find the ap. |
| Answer» Minus 700 | |
| 27446. |
class 9 ka 1.1 ka exaple |
| Answer» | |
| 27447. |
2x-3y=1solve by all methods done in chapter 3 |
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| 27448. |
The class 9 with example |
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| 27449. |
9 class ka example |
| Answer» Solve my question. | |
| 27450. |
If sinA=cosA then find the value of tanA÷cotA |
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Answer» It is 1 1 |
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