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AllQuestion and Answers: Page 1965

Question Number 5838    Answers: 0   Comments: 3

∫(1/(1+cot x))dx

11+cotxdx

Question Number 5835    Answers: 0   Comments: 2

Evaluate the integral. ∫[(x−(x^3 /2)+(x^5 /(2.4))−(x^7 /(2.4.6))+...)(1−(x^2 /2^2 )+(x^4 /(2^2 .4^2 ))−(x^6 /(2^2 .4^2 .6^2 ))+....)]dx for 0<x<∞ Please help

Evaluatetheintegral.[(xx32+x52.4x72.4.6+...)(1x222+x422.42x622.42.62+....)]dxfor0<x<Pleasehelp

Question Number 5834    Answers: 0   Comments: 2

Find all positive integers n for which there exist non−negative integer . a_(1 ) a_2 a_3 ....... a_n . Such that (1/2^a_1 ) + (1/2^a_2 ) + (1/2^a_3 ) + .... + (1/2^a_n ) = (1/3^a_1 ) + (2/3^a_2 ) + .... + (n/3^a_n ) = 1 Please help.

Findallpositiveintegersnforwhichthereexistnonnegativeinteger.a1a2a3.......an.Suchthat12a1+12a2+12a3+....+12an=13a1+23a2+....+n3an=1Pleasehelp.

Question Number 5825    Answers: 1   Comments: 1

Question Number 6081    Answers: 0   Comments: 1

∫e^(−st) (t^n /(n!))dt=?

esttnn!dt=?

Question Number 5822    Answers: 0   Comments: 0

Prove that among all triangles, which have same circum-radius, the equilateral triangle has maximum area.

Provethatamongalltriangles,whichhavesamecircumradius,theequilateraltrianglehasmaximumarea.

Question Number 5821    Answers: 1   Comments: 0

L=lim_(x→0) x^x

L=limx0xx

Question Number 5809    Answers: 1   Comments: 1

Question Number 5807    Answers: 0   Comments: 0

Town A is 36 kilometers from town B. Two persons start to travel at the same time, one from A to B and other from B to A. After meeting in a way the first person covers the remaining distance in 2 hours and the second in 8 hours. Determine their speed. (From a book)

TownAis36kilometersfromtownB.Twopersonsstarttotravelatthesametime,onefromAtoBandotherfromBtoA.Aftermeetinginawaythefirstpersoncoverstheremainingdistancein2hoursandthesecondin8hours.Determinetheirspeed.(Fromabook)

Question Number 5802    Answers: 0   Comments: 0

Solve simultaneously 2x + y − z = 8 ........... (i) x^2 − y^2 +2z^2 = 14 .......... (ii) 3x^3 + 4y^3 + z^3 = 195 ......... (iii) please help me.

Solvesimultaneously2x+yz=8...........(i)x2y2+2z2=14..........(ii)3x3+4y3+z3=195.........(iii)pleasehelpme.

Question Number 5800    Answers: 1   Comments: 1

∫((cos2x)/(sin^2 x∙cos^2 x))dx

cos2xsin2xcos2xdx

Question Number 5816    Answers: 0   Comments: 2

Prove that among all cyclic n-gons, which have same radius, regular n-gon has maximum area.

Provethatamongallcyclicngons,whichhavesameradius,regularngonhasmaximumarea.

Question Number 5814    Answers: 0   Comments: 1

Please help me with that simultaneous equation

Pleasehelpmewiththatsimultaneousequation

Question Number 5796    Answers: 0   Comments: 6

How many ways can you express 30,030 as the product of 4 positive numbers? (excluding 1)

Howmanywayscanyouexpress30,030astheproductof4positivenumbers?(excluding1)

Question Number 5792    Answers: 0   Comments: 0

Solve simultaneously 2x + y − z = 8 ...... equation(i) x^2 − y^2 + 2z^2 = 14 .... equation(ii) 3x^3 + 4x^3 + z^3 = 195 ..... equation(iii) please help thanks.

Solvesimultaneously2x+yz=8......equation(i)x2y2+2z2=14....equation(ii)3x3+4x3+z3=195.....equation(iii)pleasehelpthanks.

Question Number 5785    Answers: 0   Comments: 0

Solve simultaneously 2x + y − z = 8 ........... (i) x^2 − y^2 + 2z^2 = 14 .......... (ii) 3x^3 + 4y^3 + z^3 = 195 ........... (iii) Please help. though equation. Thanks for your help.

Solvesimultaneously2x+yz=8...........(i)x2y2+2z2=14..........(ii)3x3+4y3+z3=195...........(iii)Pleasehelp.thoughequation.Thanksforyourhelp.

Question Number 5784    Answers: 1   Comments: 0

Determine: 1+((a+r)/(ar))+((a+2r)/(ar^2 ))+...+((a+(n−1)r)/(ar^(n−1) ))

Determine:1+a+rar+a+2rar2+...+a+(n1)rarn1

Question Number 5783    Answers: 0   Comments: 0

Determine: a^a +(ar)^(a+r) +(ar^2 )^(a+2r) +...+(ar^(n−1) )^(a+(n−1)r)

Determine:aa+(ar)a+r+(ar2)a+2r+...+(arn1)a+(n1)r

Question Number 5782    Answers: 2   Comments: 0

Why x+x=2x ? Explain by properties/laws.

Whyx+x=2x?Explainbyproperties/laws.

Question Number 5780    Answers: 0   Comments: 0

at developers can we please have capital greek symbols? It would make things so much better when writing functions <3

atdeveloperscanwepleasehavecapitalgreeksymbols?Itwouldmakethingssomuchbetterwhenwritingfunctions<3

Question Number 5777    Answers: 0   Comments: 2

∃x∈Z^+ :x=p_1 p_2 ...p_n ∧∀p≠x, p∈Z^+ x=(factors of p_1 )(fact. p_2 )...(fact. p_n ) How could you formally write the number of ways you can write x in terms of the product of n integers?

xZ+:x=p1p2...pnpx,pZ+x=(factorsofp1)(fact.p2)...(fact.pn)Howcouldyouformallywritethenumberofwaysyoucanwritexintermsoftheproductofnintegers?

Question Number 5824    Answers: 0   Comments: 5

The exponent of 12 in 100! is

Theexponentof12in100!is

Question Number 5772    Answers: 0   Comments: 1

q+

q+

Question Number 5771    Answers: 0   Comments: 1

Find to the nearest hundredth the positive cube-root of 29 .

Findtothenearesthundredththepositivecuberootof29.

Question Number 5765    Answers: 0   Comments: 1

(√((x−1)/(3x+2))) + 2 ((√((3x+2)/(x−1))) ) = 3 Find the value of x

x13x+2+2(3x+2x1)=3Findthevalueofx

Question Number 5764    Answers: 0   Comments: 1

2^x = 4x Solution 2^x = 4x This can be re write as (1+1)^x = 4x Using combination to epand from the identity. (1+x)^n = 1+nx+((n(n−1))/(2!))x^2 +((n(n−1)(n−2))/(3!))x^3 +....+nCrx^(r ) Therefore. (1+1)^x = 4x 1+x+((x(x−1))/(2!))(1^2 )+((x(x−1)(x−2))/(3!))(1^3 )+......+1 = 4x ignore the continuity (what rule is ..... ignore the +...+) is it linear approximation 1+x+((x^2 −x)/(2×1))+(((x^2 −x)(x−2))/(3×2×1))+1 = 4x 1+x+((x^2 −x)/2)+((x^3 −2x^2 −x^2 +2x)/6)+1 = 4x 1+x+((x^2 −x)/2)+((x^3 −3x^2 +2x)/6)+1 = 4x Multiply through by 6 6+6x+3(x^2 −x)+x^3 −3x^2 +2x+6 = 24x 12+6x+3x^2 −3x+x^3 −3x^2 +2x+6 = 24x 12+5x+x^3 = 24x 12+5x+x^3 −24x = 0 x^3 −19x+12 = 0 Factorize x^3 −4x^2 +4x^2 −16x−3x+12 = 0 (x^3 −4x^2 )+(4x^2 −16x)−(3x+12) = 0 x^2 (x−4)+4x(x−4)−3(x−4) = 0 Factor out (x−4) (x−4)(x^2 +4x−3) = 0 x−4 = 0 or x^( 2) +4x−3 = 0 x = 4 or x = 0.6458 or x = −4.6458 The only real solution is x = 4 Therefore x = 4 DONE! Please confirm the solution. is it correct or please corect it or show me alternative. This is my trial. Thanks.

2x=4xSolution2x=4xThiscanberewriteas(1+1)x=4xUsingcombinationtoepandfromtheidentity.(1+x)n=1+nx+n(n1)2!x2+n(n1)(n2)3!x3+....+nCrxrTherefore.(1+1)x=4x1+x+x(x1)2!(12)+x(x1)(x2)3!(13)+......+1=4xignorethecontinuity(whatruleis.....ignorethe+...+)isitlinearapproximation1+x+x2x2×1+(x2x)(x2)3×2×1+1=4x1+x+x2x2+x32x2x2+2x6+1=4x1+x+x2x2+x33x2+2x6+1=4xMultiplythroughby66+6x+3(x2x)+x33x2+2x+6=24x12+6x+3x23x+x33x2+2x+6=24x12+5x+x3=24x12+5x+x324x=0x319x+12=0Factorizex34x2+4x216x3x+12=0(x34x2)+(4x216x)(3x+12)=0x2(x4)+4x(x4)3(x4)=0Factorout(x4)(x4)(x2+4x3)=0x4=0orx2+4x3=0x=4orx=0.6458orx=4.6458Theonlyrealsolutionisx=4Thereforex=4DONE!Pleaseconfirmthesolution.isitcorrectorpleasecorectitorshowmealternative.Thisismytrial.Thanks.

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