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Re: Nairaland Mathematics Clinic by Nobody: 1:04pm On Jul 03, 2014 |
layez: See works ere wey i no fit solve... Chaii,i don suffer ooo.... Ehen abeg make una elp me solve dis palasa quest can do this in multiple ways , lets use a simple one now, considering the function , we re-write as secx .(tanx +secx)/(tanx +sec x ) =( secx.tanx+sec^2 x) /(tanx +secx ) now taking integral = $ [(sec^2 x +tanx.secx) /(tanx +sec x) ]dx by inspection , we see that the denominator (tanx +secx) is the exact derivative of the numerator , since D(tanx + secx) = sec^2 x + tanx.secx => $ du/u = ln |u| + c ..where u = tanx + secx thus we have $secxdx = ln|tanx+secx|+ c hope you get ..? 1 Like |
Re: Nairaland Mathematics Clinic by Nobody: 1:06pm On Jul 03, 2014 |
nice work @ oladistinct . |
Re: Nairaland Mathematics Clinic by layez: 2:11pm On Jul 03, 2014 |
oladistinct: yaga,i gt |
Re: Nairaland Mathematics Clinic by jaryeh(m): 2:12pm On Jul 03, 2014 |
dejt4u:oga mi, I gentle o. How have you been? |
Re: Nairaland Mathematics Clinic by oladistinct(m): 2:58pm On Jul 03, 2014 |
benbuks: nice work @ oladistinct .Yes boss |
Re: Nairaland Mathematics Clinic by oladistinct(m): 4:15pm On Jul 03, 2014 |
benbuks:Integral xlnx/[1+x^2]^2dx Integration by part U=lnx >> du=dx/x dv=x/[1+x^2]^2dx Let p=1+x^2 dp=2xdx >>dx=dp/2x Therefore v=integral [x/p^2]*dp/2x =integral 1/2[p^-2]dp =-1/2p Since p=1+x^2 >>>v=-1/2[1+x^2] Now integral udv=uv- integral vdu >>integral xlnx/[1+x^2]^2dx= -lnx/2[1+x^2]- 1/2integral[-1/1+x^2]*1/xdx =-lnx/2(1+x^2)+1/2integral[1/x(1+x^2)]dx Resolving 1/x(1+x^2) into partial fraction 1/x(1+x^2)=1/x -x/1+x^2 We now have -lnx/2(1+x^2)+1/2integral[1/x -x/1+x^2]dx We can adjust x/1+x^2 to 1/2[2x/x^2+1] >>integral xlnx/[1+x^2]^2=-lnx/2(1+x^2)+1/2lnx-1/4ln[x^2+1]. |
Re: Nairaland Mathematics Clinic by Olarewajub: 4:54pm On Jul 03, 2014 |
I just want to say thank you to all the great mathematicians here, especially @ Benbuks . Wrote my Mth 102, exam today and it was great. Merci Beaucoup |
Re: Nairaland Mathematics Clinic by dejt4u(m): 5:02pm On Jul 03, 2014 |
jaryeh:m very good..tnx..hope u ar enjoyin dis your 'Omole cum madam durosimi' compulsory break |
Re: Nairaland Mathematics Clinic by Nobody: 5:08pm On Jul 03, 2014 |
Olarewajub: I just want to say thank you to all the great mathematicians here, especially @ Benbuks . Wrote my Mth 102, exam today and it was great. Merci Beaucoup wow..that's nice bro. we give all glory to God almighty for wisdom , knowledge & understanding , like one of my bosses will say " A CANDLE LOSSES NOTHING BY LIGHTENING ANOTHER " try also/always to impact / give a helping hand the little you know/have to others in need , don't be selfish !! the more you teach , the more you learn & the more your reward/blessing kudos to you all my mathematical friends/generals , your efforts here is not / shall never be in vain . wish you the best in your exams bruv. trust it shall be glorious . 1luv. |
Re: Nairaland Mathematics Clinic by jaryeh(m): 5:10pm On Jul 03, 2014 |
dejt4u: Ha! I dey manage am o. I have no option. |
Re: Nairaland Mathematics Clinic by dejt4u(m): 5:13pm On Jul 03, 2014 |
jaryeh:abi na.. That is OAU for u ooo my guy..na so 5 yrs go turn to 6 b dat ooo without havin extra...welcome on board |
Re: Nairaland Mathematics Clinic by jaryeh(m): 5:17pm On Jul 03, 2014 |
dejt4u: abi na.. That is OAU for u ooo my guy..na so 5 yrs go turn to 6 b dat ooo without havin extra...welcome on boardSure. We don gba kamu o. I hope you're enjoying life after school sir. |
Re: Nairaland Mathematics Clinic by Nobody: 5:55pm On Jul 03, 2014 |
Four persons are chosen at random from a group containing 3 men , 2women & 3children show that the probability that exactly two of them are children is 10/21 A five digit number is formed by using 0,1,2,3,4&5 without repetition . find the probability that the number is divisible by 6 .. asap.. |
Re: Nairaland Mathematics Clinic by oladistinct(m): 6:41pm On Jul 03, 2014 |
benbuks:Let us consider the expansion of [a+b+c]^2 [a+b+c]^2=a^2+b^2+c^2+2[ab+ac+bc] Now a+b+c=26 and a^2+b^2+c^2=364 >>26^2=364+2[a+b+c] 676-364=2[ab+ac+bc] ab+ac+bc=312/2=156 we now have ab+ac+bc=156.............(i) and a+b+c=26....................(ii) Since the progression is geometric Tn=ar^[n-1] >>b=ar and c=ar^2 since b and c are 2nd and 3rd term respectively Substitute these into equation (i) and (ii) a+ar+ar^2=26 >>>a[1+r+r^2]=26.......(iii) also a^2r+a^2r^2+a^2r^3=156 >>>a^2r[1+r+r^2]=156.........(iv) Divide iv by iii [a^2r(1+r+r^2)]/[a(1+r+r^2)]=156/26 >>>ar=6 therefore r=6/a Now b=ar=6 c=ar^2=[ar]r=6r From a+b+c=26 a+6+6r=26 a+6r=20 Now r=6/a a+6[6/a]=20 a+36/a=20 >>> a^2+36=20a a^2-20a+36=0 Solving the above quadratic equstn a=18 and 2 using a=2 Since c=ar^2 Now r=6/a=6/2=3 >>>c=2*3^2=2*9=18 Therefore a=2, b=6 and c=18. 2 Likes |
Re: Nairaland Mathematics Clinic by Nobody: 9:22pm On Jul 03, 2014 |
oladistinct: great work sir.. |
Re: Nairaland Mathematics Clinic by layez: 9:36pm On Jul 03, 2014 |
oladistinct: i must commend ur effort by typing d soln but u almost made my head go haywire cos of d way u solved... Aiit,i grab cos i did it too.. Firstly, let * = Exp then, Tn=ar*n-1 so you made a mistake dere.... Thumbs up 1 Like |
Re: Nairaland Mathematics Clinic by oladistinct(m): 10:36pm On Jul 03, 2014 |
layez:Thanks bro, it was a typo error |
Re: Nairaland Mathematics Clinic by efficiencie(m): 11:31pm On Jul 03, 2014 |
benbuks: let 's try this welcome back question Given I=∫(arcsin(lnx))dx Let u=arcsinlnx sinu=lnx cosu.du=dx/x dx=e^sinu.cosu du I=∫(u. e^sinu.cosu) du I=u. e^sinu – ∫(e^sinu) du For values of sinu defined as: -1<sinu<1, I can apply the taylor’s rule such that: I=u. e^sinu – ∫(1+sinu+(1/2!)sin^2(u)+(1/3!)sin^3(u)+(1/4!)sin^4(u)+…) du Since for higher powers n of sin^n(u) the integral peters out I stop, albeit arbitrarily at n=4 Hence I≈u. e^sinu – ∫(1+sinu+(1/2!)sin^2(u)+(1/3!)sin^3(u)+(1/4!)sin^4(u)) du I≈u. e^sinu – ∫(1+sinu+(1/2!)( (1/2) – (1/2)cos2u)+(1/3!)( (3/4)sinu – (1/4)sin3u)+(1/4!)( (3/8 ) – (1/2)cos2u + (1/8 )cos4u)) du I≈u. e^sinu – ∫(1+sinu+( (1/4) – (1/4)cos2u)+( (1/12)sinu – (1/24)sin3u)+( (1/64) – (1/48 )cos2u + (1/192)cos4u)) du I≈u. e^sinu – [u – cosu + (u/4) – (1/8 )sin2u – (1/12)cosu + (1/72)cos3u + (u/64) – (1/96)sin2u + (1/768 )sin4u] I≈u. e^sinu – u + cosu – (u/4) + (1/8 )sin2u + (1/12)cosu – (1/72)cos3u – (u/64) + (1/96)sin2u – (1/768 )sin4u I ≈ -81/64 + u. e^sinu + (13/12)cosu + (13/96)sin2u – (1/72)cos3u – (1/768 )sin4u Put the value of u I ≈ -81/64 + x.arcsinlnx + (13/12)√(1-(lnx)^2) + (13/48)lnx√(1-(lnx)^2) – (1/36)(1-(lnx)^2)) √(1-(lnx)^2) – (1/72)√(1-(lnx)^2) – (1/768 )sin[4(arcsinlnx)] I left the last term unexpanded…this question sef! Over to the auditors |
Re: Nairaland Mathematics Clinic by efficiencie(m): 12:40am On Jul 04, 2014 |
benbuks: Four persons are chosen at random from a group containing 3 men , 2women & 3children show that the probability that exactly two of them is 10/21 Four persons are chosen at random from a group containing 3 men , 2women & 3children show that the probability that exactly two of them ARE WHAT is 10/21 incomplete question sir |
Re: Nairaland Mathematics Clinic by efficiencie(m): 1:41am On Jul 04, 2014 |
Sir Chides: Solve d equation::: see pic below sir: 1 Like
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Re: Nairaland Mathematics Clinic by Nobody: 6:56am On Jul 04, 2014 |
efficiencie:"children" modified .. |
Re: Nairaland Mathematics Clinic by Nobody: 11:01am On Jul 04, 2014 |
*Goodluck Jonathan-President of Nigeria (Zoologist) *Olusegun Obasanjo-Ex President of Nigeria (Civil Engineer) *Pastor E.A. Adeboye-General Overseer, RCCG (Mathematician ) *Pastor W.F.Kumuyi-General Overseer, DLBC (Mathematician). *Nkem owoh popularly known as Osofia Nigerian actor and comedian-(Engin eer). *Bishop David O.Oyedepo General Overseer- WINNERS CHAPEL (Architect) *Dr. D.K. Olukoya-General Overseer-MFM (Microbiologist) *Rev Dr Uma Upkai-Evangelist (Electrical Engineer) & Many more. FRIENDS, U need GRACE not human connection. U need GOD'S Favor not much friends. U need GOD'S back up not human standard. In Proverbs, 23 vs 4. He says. Labour not to be rich and cease from your own wisdom. Now, GOD visited Abraham, the farmer and turns him to generational blessings. He visited Moses, the stammerer and turns him generational giant. BEFORE THIS YEAR IS OVER, IF ONLY U HAVE FAITH & BELIEVE, U shall become a living wonders to yourself, your household, your state & your Nation in Jesus mighty name! Ur destiny shall not turn to rags in Jesus name. 1 Like |
Re: Nairaland Mathematics Clinic by jackpot(f): 1:12pm On Jul 04, 2014 |
Dear Math Generals, Lieutenants and Sergeants Evaluate the limit: lim x -> 0 (sin x)sin x 5Star tags: doubleDx, Laplacian, Richiez, benbuks, efficiencie, dejt4u, Amazing Angel, STENON, Mathematical Scientists and Engineers. |
Re: Nairaland Mathematics Clinic by STENON(f): 1:15pm On Jul 04, 2014 |
[quote author=jackpot]Dear Math Generals, Lieutenants and Sergeants Evaluate the limit: lim x -> 0 (sin x)sin x 5Star tags: doubleDx, Laplacian, Richiez, benbuks, efficiencie, dejt4u, Amazing Angel, STENON, Mathematical Scientists and Engineers. [/quote] My Boss!!...I salute you o.....Awwwww.....!!!.............Brb ooo.... Goodpm to you....... |
Re: Nairaland Mathematics Clinic by Nobody: 1:42pm On Jul 04, 2014 |
jackpot: Dear Math Generals, Lieutenants and Sergeants HMM dis my madam don show again oo solution below , my boss don devour d tin i"ll try alternative solution sha.. most limits of indeterminate forms. yields 1 . 1 Like |
Re: Nairaland Mathematics Clinic by efficiencie(m): 2:15pm On Jul 04, 2014 |
jackpot: Dear Math Generals, Lieutenants and Sergeants Lim (sinx)^(sinx) as x→0 Let y=(sinx)^(sinx) lny=sinx.lnsinx lny=lnsinx/cosecx Which results in -∞/∞ as x→0 Limlny=lim(lnsinx/cosecx) Limlny=lim((dlnsinx/dx)/(dcosecx/dx)) Limlny=lim(cotx/-cosecxcotx) Limlny=-lim(1/cosecx) Limlny=-lim(sinx) lny=0 as x→0 y=1 Hence Lim (sinx)^(sinx)=1 as x→0 1 Like |
Re: Nairaland Mathematics Clinic by Nobody: 3:06pm On Jul 04, 2014 |
efficiencie: just on point bruv. your great man.. |
Re: Nairaland Mathematics Clinic by Nobody: 3:30pm On Jul 04, 2014 |
dp |
Re: Nairaland Mathematics Clinic by Nobody: 3:31pm On Jul 04, 2014 |
jackpot: Dear Math Generals, Lieutenants and Sergeants Yes, it's = 1 I think general efficiencie solved it correctly!...I will check out later. 1luv I'm a little busy now! |
Re: Nairaland Mathematics Clinic by efficiencie(m): 5:28pm On Jul 04, 2014 |
Someone posted the below question, I quess Sir Chides and I think the doctors in the clinic need to diagnose and profer 'curative' measures for the problem: Evaluate: ∫dx/(cosk + cosx) |
Re: Nairaland Mathematics Clinic by Hayormeah(f): 6:03pm On Jul 04, 2014 |
how much glucose syrup with 20 % concentration has to be mixed with 100kg glucose syrup with 40 % concentration so that the mixture will have 36 % glucose? |
Re: Nairaland Mathematics Clinic by jackpot(f): 6:22pm On Jul 04, 2014 |
efficiencie:I am having problems with the bolded. How can you equate "limit" both sides and cancel out? Next, that LimIny is as y tends to what? |
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