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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.
| 37201. |
From the top of a 75 m high |
| Answer» | |
| 37202. |
3.7 Ka 7 no koi bhej dega bhai |
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Answer» Okk Ap iss app ke solutions me check kar lijiye.. |
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| 37203. |
Determine the value of k so that linear equation have no solution (3k+1)x+3y-2=0,(k2 +1)x+(k-2)y-5 |
| Answer» 3+1× | |
| 37204. |
Find the value of k ifK(Sn-1)=Sn+Sn-2-d |
| Answer» | |
| 37205. |
Find the digit at units place of 8n if n is 9 |
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Answer» Here, question is 8^n So 8^9= 134217728 So unit place is 8 n=9 8n8×9=72Unit place digit =2 No answer is 8 but how to solve it 9 |
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| 37206. |
In an AP if d=4,n=7,an=4,then a is |
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Answer» an = a+(n-1)d4 = a+(7-1)×44 = a+ 6×44 = a+ 124-12= a-8 = aSo,a= -8 I think -20 |
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| 37207. |
Find the value of \'a\' so that the point (3,9) lies on the line represented by 2x-3y=5 |
| Answer» Since the point (3, a) lies on the line 2x - 3y = 5, we have2\xa0{tex}\\times{/tex}\xa03 - 3\xa0{tex}\\times{/tex}\xa0a = 5{tex}\\Rightarrow{/tex}\xa06 - 3a = 5{tex}\\Rightarrow{/tex}\xa03a = 1{tex}\\Rightarrow{/tex}\xa0{tex}a = \\frac { 1 } { 3 }{/tex} | |
| 37208. |
If 1-tan theta/1+tan theta = root 3-1/root3+1 show that sin theta/cos^2theta=1 |
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| 37209. |
Find the cubic polynomial whose zeroes are 5,-1and 3 |
| Answer» x^3-7x^2+7x+15 | |
| 37210. |
Write the zeroes of the polynomial x²-x-6 |
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Answer» -2 and 3 (x - 3) (x + 2) = 0x = 3 and x = -2 x2 - x - 6 = 0x2 - 3x + 2x - 6 = 0x(x - 3) + 2 (x - 3) = 0(x - 3) (x + 2) = 0 -2 and 3 are the zeroes of the given polynomial. The zeroes are 3 and -2 |
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| 37211. |
Express the 0.12 and bar on 2 as a fraction in simplest form |
| Answer» Let we have ,{tex}x= 0.1 \\overline { 2 }{/tex}, then{tex}x{/tex}\xa0= 0.12222....{tex}\\implies{/tex}10x\xa0= 1.2222.... ...(i){tex}\\implies{/tex}100x= 12.2222... ...(ii)Subtracting (i) from (ii), we get{tex}\\implies{/tex}100x-10x=11{tex}\\implies90 x = 11 \\quad \\therefore x = \\frac { 11 } { 90 }{/tex}Thus the value of x ={tex}0.1\\overline {2}{/tex} is equal to\xa0{tex}\\frac {11}{90}{/tex} | |
| 37212. |
Find the value of x for which (8x+4),(6x-2),(2x+7) are in A.P. |
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Answer» Use the formula 2b = a+c...then solve it..you will get yur answer 7 |
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| 37213. |
156+200=156*(-73) verify and name property |
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| 37214. |
What is the unit digit place of number38 power20 |
| Answer» For this, you need to understand the repetition or cyclicity of unit digits coming in the powers of number 8Powers of 8 = 8, 64, 512, 4096, 32768, 262144, 2097152, 16777216 and so on...Unit digits of powers of 8 = 8, 4, 2, 6, 8, 4, 2, 6, ..........So the repetition cycle is = {tex}\\large8\\to4\\to2\\to6{/tex}\xa0So the cyclicity quantity is\xa04.So,{tex}for \\space38^{20}\\\\We \\space have\\space to\\space do =\\frac{20}4 ,where \\space remainder=0\\\\The \\space cyclicity\\space unit\\space digit\\space number\\space for\\space this\\space is\\space 6.{/tex} | |
| 37215. |
I want all 1,2,3,4,5.marks questions. and exampler questions |
| Answer» 8rthmL | |
| 37216. |
In triangle DEllBC, AD=1cm and BD |
| Answer» It is given that AD = 1 cm,BD = 2 cm and\xa0{tex}D E \\| B C{/tex}In {tex}\\triangle{/tex}ADE and\xa0{tex}\\triangle {/tex}ABC{tex}\\angle A D E = \\angle A B C{/tex}\xa0(Corresponding angles){tex}\\angle A = \\angle A{/tex}\xa0[Common]Therefore, by A.A. similar condition{tex}\\triangle \\mathrm { ADE } \\sim \\triangle \\mathrm { ABC }{/tex}Ratio of areas of similar triangles is equal to the square of the ratio of the corresponding sides.{tex}\\therefore \\quad \\frac { \\operatorname { ar } ( \\triangle \\mathrm { ABC } ) } { \\operatorname { ar } ( \\triangle \\mathrm { ADE } ) } = \\frac { \\mathrm { AB } ^ { 2 } } { \\mathrm { AD } ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad \\frac { \\operatorname { ar } ( \\triangle A B C ) } { \\operatorname { ar } ( \\triangle A D E ) } = \\left( \\frac { 3 } { 1 } \\right) ^ { 2 }{/tex}{tex}\\Rightarrow \\quad \\frac { \\operatorname { arc } \\triangle A B C ) } { \\operatorname { ar } ( \\triangle A D E ) }{/tex}{tex}= \\frac { 9 } { 1 }{/tex} | |
| 37217. |
Mathematics is a father of all subjects |
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Answer» Yes Kisne kaha? |
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| 37218. |
How many two digit no can be divisible by 3 ???? |
| Answer» Two digit number divisible by 3 are12,15.......................99This is an Ap and a=12,d=3 ,l=99,n=?l=a+(n-)d99=12+(n-1)*3(n-1)*3=99-12=87n-1=87/3=29n=30Total no are+30 | |
| 37219. |
If tan A=cot B, prove that A+B=90° |
| Answer» tan A = cot B• tan A = tan(90-B)• A = 90 - B• A + B = 90....Hence proved | |
| 37220. |
If tan theta+1/tan theta=2;show that:tan^theta+¹/tan^theta=2 |
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Answer» We have,{tex} \\tan \\theta + \\frac { 1 } { \\tan \\theta } = 2{/tex}Squaring both sides, we get{tex}\\Rightarrow \\left( \\tan \\theta + \\frac { 1 } { \\tan \\theta } \\right) ^ { 2 } = 2 ^ { 2 }{/tex}{tex}\\Rightarrow\\quad\\tan^2\\theta+\\frac1{\\tan^2\\theta}+2\\times\\tan\\theta\\times\\frac1{\\tan\\theta}=4{/tex}{tex}\\Rightarrow \\quad \\tan ^ { 2 } \\theta + \\frac { 1 } { \\tan ^ { 2 } \\theta } + 2 = 4{/tex}{tex}\\Rightarrow \\quad \\tan ^ { 2 } \\theta + \\frac { 1 } { \\tan ^ { 2 } \\theta } = 2{/tex}Alternate method, We have{tex}\\tan \\theta + \\frac { 1 } { \\tan \\theta } = 2{/tex}{tex}\\Rightarrow \\quad \\tan ^ { 2 } \\theta + 1 = 2 \\tan \\theta{/tex}{tex}\\Rightarrow \\quad \\tan ^ { 2 } \\theta - 2 \\tan \\theta + 1 = 0{/tex}{tex}\\Rightarrow \\quad ( \\tan \\theta - 1 ) ^ { 2 } = 0{/tex}{tex}\\Rightarrow \\quad \\tan \\theta = 1{/tex}{tex}\\therefore \\quad \\tan ^ { 2 } \\theta + \\frac { 1 } { \\tan ^ { 2 } \\theta } = 1 + 1 = 2{/tex} Thank? |
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| 37221. |
How to get previous year \'squestion papers in pdf |
| Answer» Search in Google.." previous years sample paper class 10 pdf" Simple!!! | |
| 37222. |
Define Mathematics |
| Answer» It is a group of various formulas and equations by which we can solve various day to day life problems like calculating, bill area.......... | |
| 37223. |
Yadi tanA=cotB proof A+B=90 |
| Answer» tanA=cotBtanA=1/tanBtanAtanB=1tan(A+B)=tanA+tanB/1-tanATab=tanA+tanB/1-1=tanA+tanB/0tan(A+B)={tex}\\infty{/tex}or tan(A+B)=tan90hence A+B=90 | |
| 37224. |
Cos 30°+sin60°÷1+sin 30°+cos 60° |
| Answer» {tex}\\frac { \\cos 30 ^ { \\circ } + \\sin 60 ^ { \\circ } } { 1 + \\sin 30 ^ { \\circ } + \\cos 60 ^ { \\circ } }{/tex}{tex}= \\frac { \\frac { \\sqrt { 3 } } { 2 } + \\frac { \\sqrt { 3 } } { 2 } } { 1 + \\frac { 1 } { 2 } + \\frac { 1 } { 2 } } = \\frac { \\sqrt { 3 } } { 2 }{/tex}\xa0{tex}\\Rightarrow\\cos 30 ^ { \\circ } = \\frac { \\sqrt { 3 } } { 2 }{/tex}Hence,\xa0{tex}\\frac { \\cos 30 ^ { \\circ } + \\sin 60 ^ { \\circ } } { 1 + \\sin 30 ^ { \\circ } + \\cos 60 ^ { \\circ } } = \\cos 30 ^ { \\circ }{/tex} | |
| 37225. |
Explain why 7×11×13 &7×6×5×4×3×2×1+5 are composite numbers |
| Answer» Because they have factors more than 2 | |
| 37226. |
In a right angle triangle at b find cos a add b |
| Answer» Since ABC is right angled and angle C is 90°therefore,A+B=180°-CA+B=180°-90°A+B= 90°Therefore,cos (A+B)= cos90° =0 | |
| 37227. |
Solve the equation 10x^-3x=1 by method of completing the square |
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| 37228. |
ex 8.2 Qno1(2) |
| Answer» [email\xa0protected]&#:#@;@,@,@*,288#,#,- | |
| 37229. |
For any integer a and b there exists unique integers q and r such that a=3q+r.find the value of r |
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Answer» 0,1,2 A=bq+r0< or =r |
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| 37230. |
Find zero of the polynomial 4u2 8u |
| Answer» 4u2\xa0- 8u = 04u(u - 2) = 0Therefore, u = 0 and u = 2 | |
| 37231. |
If sum of squares of zeros of polynomial x^2 + 8x+k is-40 then find the value of k |
| Answer» Let {tex}\\alpha,\\beta{/tex}\xa0be the zeros of the polynomial {tex}f(x)=x^2-8x+k{/tex}.Sum of zeroes =\xa0{tex} \\alpha + \\beta = - \\left( \\frac { - 8 } { 1 } \\right) = 8{/tex}\xa0and, Product of zeroes =\xa0{tex} \\alpha \\beta = \\frac { k } { 1 } = k{/tex}Now,\xa0{tex} \\alpha ^ { 2 } + \\beta ^ { 2 } = 40{/tex}{tex} \\Rightarrow \\alpha ^ { 2 } + \\beta ^ { 2 }+2 \\alpha\\beta-2 \\alpha\\beta= 40{/tex}{tex} \\Rightarrow \\quad ( \\alpha + \\beta ) ^ { 2 } - 2 \\alpha \\beta = 40{/tex}{tex} \\Rightarrow \\quad 8 ^ { 2 } - 2 k = 40{/tex}{tex} \\Rightarrow \\quad 2 k = 64 - 40 {/tex}{tex}\\Rightarrow 2 k = 24 {/tex}{tex}\\Rightarrow k = 12{/tex} | |
| 37232. |
Find the hcf of 225 and 575. Express it as linear eq 225m + 575n find the value of m and n |
| Answer» First find the HCF of 575 and 225 by Using Euclid\'s division algorithm,575 = 225{tex}\\times{/tex}\xa02\xa0+ 125225 = 125{tex}\\times{/tex} 1 + 100125\xa0= 100{tex}\\times{/tex}\xa01\xa0+ 25100 = 25\xa0{tex}\\times{/tex}4 + 0So, HCF of 575\xa0and 225 = 25 | |
| 37233. |
why 3+2 is not equal to 6 |
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Answer» NO , because when we add any odd and even number it is not possible that answer will even number so, 3+2 is not 6 3+2=5 is not equal to 6 Because 3+2=5 Because if we add 3 and 2 it will give us answer as 5 it\'s common if we take the product of these two number so it will give us answer 6 Because3+2=5 Because 2+2=4 |
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| 37234. |
Solve for \'x\'1/2a+b+2x = 1/2a+1/b+1/2x ,\'x\' is not equal to zero.. |
| Answer» {tex}\\frac{1}{2a + b + 2x}{/tex}\xa0=\xa0{tex}\\frac{1}{2a}{/tex}\xa0+\xa0{tex}\\frac{1}{b}{/tex}\xa0+\xa0{tex}\\frac{1}{2x}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{1}{2a + b + 2x}{/tex}\xa0-\xa0{tex}\\frac{1}{2x}{/tex}\xa0=\xa0{tex}\\frac{1}{2a}{/tex}\xa0+\xa0{tex}\\frac{1}{b}{/tex}\xa0{tex}\\Rightarrow{/tex}{tex}\\frac { 2 x - 2 a - b - 2 x } { ( 2 a + b + 2 x ) ( 2 x ) }{/tex}\xa0=\xa0{tex}\\frac{b + 2a}{2a \\times b}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { - ( 2 a + b ) } { ( 2 a + b + 2 x ) 2 x }{/tex}\xa0=\xa0{tex}\\frac{b + 2a}{2ab}{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac { - 1 } { 4 a x + 2 b x + 4 x ^ { 2 } }{/tex}\xa0=\xa0{tex}\\frac{1}{2ab}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}4x^2 + 2bx + 4ax = -2ab{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}4x^2 + 2bx + 4ax + 2ab = 0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}2x(2x + b) + 2a(2x + b) = 0{/tex}{tex}\\Rightarrow{/tex}\xa0(2x + b)(2x + 2a) = 0{tex}\\Rightarrow{/tex}\xa0x = -{tex}\\frac{b}{2}{/tex} or x = -a | |
| 37235. |
Show that the points (1,7),(4,2),(-1,-1) and (-4,4) are the vertices of a square. |
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Answer» Given points are (1,7),(4,2),(-1,-1),(-4,4)Let the points are A,B,C,D.Distance formula = √(x₂-x₁)²+(y₂-y₁)²\xa0AB = √(4-1)²+(2-7)² = √9+25\xa0= √34\xa0BC = √(-1-4)²+(-1-2)²\xa0= √25+9\xa0= √34\xa0CD = √(-4-(-1))²+(4-(-1))²\xa0= √9+25 = √34\xa0DA = √(1-(-4))²+(7-4)² = √25+9\xa0= √34We know that the all sides of the square are equal.Here by the distance formula, we got the for sides equal .So,from this we can say that these points are the vertices of a square.\xa0 What is polynomial |
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| 37236. |
Solve for x and y : 6x + 3y = 7xy, 3x + 9y = 11xy. |
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Answer» The given equations are as follows: 6x + 3y = 7xy ………….(i)3x + 9y = 11xy …………(ii)For equation (i), we have: 6x+3y/ xy = 7 On dividing each of the given equations by xy, we get{tex}\\frac { 6 } { y } + \\frac { 3 } { x } = 7{/tex}.......(i){tex}\\frac { 3 } { y } + \\frac { 9 } { x } = 11{/tex}.....(ii)Putting {tex}\\frac 1x{/tex}\xa0= u and {tex}\\frac 1y{/tex}\xa0= v\xa0in (i) and (ii), we get{tex}6v + 3u\xa0= 7{/tex}..... (iii){tex}3v\xa0+ 9u\xa0= 11{/tex}.. ..(iv)Multiplying (iii) by 3 and subtracting (iv) from the result, we get{tex}18v - 3v = 21 - 11{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}15v = 10{/tex}{tex}\\Rightarrow \\quad v = \\frac { 10 } { 15 } = \\frac { 2 } { 3 }{/tex}Putting v\xa0=\xa0{tex}\\frac 23{/tex}\xa0in (iii), we get{tex}\\left( 6 \\times \\frac { 2 } { 3 } \\right) + 3 u = 7{/tex}{tex}\\Rightarrow 4 + 3 u = 7 \\Rightarrow 3 u = 3 \\Rightarrow u = 1{/tex}{tex}u = 1{/tex}\xa0{tex}\\Rightarrow \\frac { 1 } { x } = 1 \\Rightarrow x = 1{/tex}{tex}v = \\frac { 2 } { 3 } \\Rightarrow \\frac { 1 } { y } = \\frac { 2 } { 3 }{/tex}{tex}\\Rightarrow 2 y = 3 \\Rightarrow y = \\frac { 3 } { 2 }{/tex}{tex}\\therefore{/tex}{tex}x = 1\\ and\\ y =\xa0{/tex}{tex}\\frac 32{/tex} |
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| 37237. |
In tossing two coins find the probability of getting 2 heads |
| Answer» 2 | |
| 37238. |
Find the value of K will the equation have infinity many solution. |
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Answer» Where is eq^n Where is the equation?? |
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| 37239. |
If tanA =tanP ,than prove that angleA =angle P |
| Answer» Consider two right triangles XAY and WBZ such that tan A = tan BWe have,tan A =\xa0{tex}\\frac { X Y } { A Y }{/tex}\xa0and tan B =\xa0{tex}\\frac { WZ } { B Z }{/tex}Since, tan A = tan B{tex}\\Rightarrow \\frac { X Y } { A Y } = \\frac { W Z } { B Z }{/tex}{tex}\\Rightarrow \\frac { X Y } { W Z } = \\frac { A Y } { B Z } = k{/tex}(say)......(i){tex}\\Rightarrow X Y = k \\times W Z \\text { and } A Y = k \\times B Z{/tex}.......(ii)Using pythagoras\xa0theorem in triangles XAY and WBZ, we haveXA2\xa0= XY2\xa0+ AY2\xa0and WB2\xa0= WZ2\xa0+ BZ2{tex}\\Rightarrow{/tex}\xa0XA2\xa0= k2WZ2\xa0+ k2BZ2\xa0and WB2\xa0= WZ2\xa0+ BZ2{tex}\\Rightarrow{/tex}\xa0XA2\xa0= k2(WZ2\xa0+ BZ2) and WB2\xa0= WZ2\xa0+ BZ2{tex}\\Rightarrow\\frac { X A ^ { 2 } } { W B ^ { 2 } } = \\frac { k ^ { 2 } \\left( W Z ^ { 2 } + B Z ^ { 2 } \\right) } { \\left( W Z ^ { 2 } + B Z ^ { 2 } \\right) } = k ^ { 2 }{/tex}{tex}\\Rightarrow \\frac { X A } { W B } = k{/tex}...........(iii)From (i), (ii) and (iii), we get{tex}\\frac { X Y } { W Z } = \\frac { A Y } { B Z } = \\frac { X A } { W B }{/tex}{tex}\\Rightarrow \\Delta A Y X \\sim \\Delta B Z W{/tex}{tex}\\Rightarrow \\angle A = \\angle B{/tex} | |
| 37240. |
How many three digit natural numbers are divisible by 7 |
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Answer» 128 three digit numbers 128 4 105,112,119 |
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| 37241. |
Is polynomial x4+4x2+5 have zeroes or not |
| Answer» X ki power 4 h ka ki x4 h | |
| 37242. |
X/x-1 + x-1/x =17/4 |
| Answer» Given,{tex} \\frac { x + 3 } { x - 2 } - \\frac { 1 - x } { x } = \\frac { 17 } { 4 }{/tex}Taking LCM, we get{tex} \\Rightarrow \\quad \\frac { x ( x + 3 ) - ( 1 - x ) ( x - 2 ) } { x ( x - 2 ) } = \\frac { 17 } { 4 }{/tex}{tex} \\Rightarrow \\quad \\frac { x ^ { 2 } + 3 x - \\left( x - 2 - x ^ { 2 } + 2 x \\right) } { x ^ { 2 } - 2 x } = \\frac { 17 } { 4 }{/tex}{tex} \\Rightarrow \\quad \\frac { 2 x ^ { 2 } + 2 } { x ^ { 2 } - 2 x } = \\frac { 17 } { 4 }{/tex}After cross multiplication,we get{tex} \\Rightarrow{/tex}\xa08x2 + 8 = 17x2 - 34x{tex}\\Rightarrow{/tex}\xa09x2 - 34x - 8 = 0{tex}\\Rightarrow{/tex}\xa09x2 - 36x + 2x - 8 = 0{tex}\\Rightarrow{/tex}\xa0{tex}9 x ( x - 4 ) + 2 ( x - 4 ) = 0{/tex}{tex} \\Rightarrow{/tex}\xa0(x - 4)(9x + 2) = 0\xa0{tex} \\Rightarrow{/tex}\xa0x - 4 = 0 or, 9x + 2 = 0\xa0{tex} \\Rightarrow \\quad x = 4 \\text { or, } x = - \\frac { 2 } { 9 }{/tex} | |
| 37243. |
Use Euclid division algorithm to find if the following pair of number is coprime 121,576 |
| Answer» Here we have to find out HCF of 576 and 121\xa0by Using Euclid’s division Lemma, we get576 = 121\xa0{tex}\\times {/tex}4 + 92121 = 92\xa0{tex}\\times {/tex}1 + 2992 = 29\xa0{tex}\\times {/tex}\xa03 + 529 = 5\xa0{tex}\\times {/tex}\xa05 + 45 = 4\xa0{tex}\\times {/tex}1 + 14 = 1{tex}\\times {/tex}4 + 0{tex}\\therefore{/tex}HCF = 1.Hence the numbers are co-prime. | |
| 37244. |
Smallest whole number |
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Answer» 0 Hi Zero is the smallest whole number. O is smallest whole no. 0 0 The numbers are 0,1,2,… These numbers are called whole nos.The number 0 is the first and the smallest whole nos. 0 is smallest whole no 2 |
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| 37245. |
Alfa square + bita square axsquare + bx + c |
| Answer» Sorry but karna kya h | |
| 37246. |
if x=a^3 + ab^2/a^2 + b^2. find the value of x. |
| Answer» x=a | |
| 37247. |
Which term of the AP : 3,8,13,18,..., is 78? |
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Answer» D=5 , a=3 , let an = 78 put values an = a + ( n-1)d , and answer is here Common difference is 5 |
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| 37248. |
Prove that product of two consecutive positive integers is divisible by 2. |
| Answer» Try to generalize and imagine the numerical. | |
| 37249. |
If 1 and -2 are two zeros of the polynomial(x^3-4x^2-7x+10), finds its third zero |
| Answer» Given zeroes are- 1, -2Or, (x-1) and (x+2).(x-1)(x+2)=x^2+x-2If we divide x^3-4x^2-7x+10 by x^2+x-2 we get x-5, i.e., required answer | |
| 37250. |
Show that any number of form 4^n, n€N can never end with digit 0. |
| Answer» {tex}{4^n} = {\\left[ {{2^2}} \\right]^n} = {2^{2n}}{/tex}It does not contains \'5\'. So\xa0{tex}{4^n},n \\in N{/tex} can never end with the digit 0 | |