Welcome, Guest: Register On Nairaland / LOGIN! / Trending / Recent / New
Stats: 3,155,868 members, 7,828,091 topics. Date: Wednesday, 15 May 2024 at 12:08 AM

Nairaland Mathematics Clinic - Education (101) - Nairaland

Nairaland Forum / Nairaland / General / Education / Nairaland Mathematics Clinic (481790 Views)

Mathematics Clinic / 2015 Cowbell Mathematics Champion, Akinkuowo Honoured By School. / Lead City University Clinic Welcomes First Ever Baby Since 10 Years Of Opening (2) (3) (4)

(1) (2) (3) ... (98) (99) (100) (101) (102) (103) (104) ... (284) (Reply) (Go Down)

Re: Nairaland Mathematics Clinic by Richiez(m): 11:09am On Nov 14, 2013
QUIZ MASTERS IN THE QUIZ THREAD PLS CHECK YOUR MAILS
Re: Nairaland Mathematics Clinic by Mozarinia(m): 11:48am On Nov 14, 2013
benbuks: ^ 9c Try guys, bt they gat 2b anoda method....mayb 'series'
9/10 plus sum to infinity of gp with first term 7/100 and ratio of 1/10. Dat gives u 9/10 plus 7/90 wich gives 44/45
Re: Nairaland Mathematics Clinic by AlphaMaximus(m): 12:34pm On Nov 14, 2013

6 Likes

Re: Nairaland Mathematics Clinic by Aliciakeys(f): 12:36pm On Nov 14, 2013
Re: Nairaland Mathematics Clinic by Calculusfx: 12:51pm On Nov 14, 2013
Laplacian: ...sin2x+sinx=1...find x...
.hmmm,here is my own approach to it...sin2x+sinx=1...2sinxcosx+sinx=1...2sinx.sqrt(1-sin^2x)+sinx=1...{let sinx be a...(i)}...2a.sqrt(1-a^2)+a=1...2a.sqrt(1-a^2)=1-a...square both sides...4a^2(1-a^2)=1-2a+a^2...4a^2-4a^4=1-2a+a^2...4a^4-3a^2-2a+1=0,it's obvious that a=1(one of the four roots) then we divide 4a^4-3a^2-2a+1 by a-1 to give 4a^3+4a^2+a-1=0...then transform the equation to reduced cubic equation to get by representing a as y-1/3...(ii) to give y^3-y/12-7/27...you can use any method for solving cubic polynomial to do it but for me,i used tartaglia's method for finding real root of a cubic equation and got y as 0.68114(approximated),don't forget that a=y-1/3 and by substituting that a=0.34781(the first root of the cubic polynomial)...to get the other two roots...from relationship between the co-efficients and roots of a polynomial...a1+a2+a3=-1 a1.a2.a3=0.25(from the polynomial 4a^3+4a^2+a-1=0)...let a1 be 0.34781,by solving a2+a3=-1.34781 and a2.a3=0.7188 and by solving the simultaneous equation a={-1.34781±j1.0287}/2...with all we have done so far,the two real roots got were a=1 and a=0.34781,don't forget that sinx=a...let a be 1,then sinx=1 and x=90 and if a=0.34781,sinx=0.34781 and x=20.35...i rest my case

1 Like

Re: Nairaland Mathematics Clinic by Laplacian(m): 1:06pm On Nov 14, 2013
Calculusf(x):
.hmmm,here is my own approach to it...sin2x+sinx=1...2sinxcosx+sinx=1...2sinx.sqrt(1-sin^2x)+sinx=1...{let sinx be a...(i)}...2a.sqrt(1-a^2)+a=1...2a.sqrt(1-a^2)=1-a...square both sides...4a^2(1-a^2)=1-2a+a^2...4a^2-4a^4=1-2a+a^2...4a^4-3a^2-2a+1=0,it's obvious that a=1(one of the four roots) then we divide 4a^4-3a^2-2a+1 by a-1 to give 4a^3+4a^2+a-1=0...then transform the equation to reduced cubic equation to get by representing a as y-1/3...(ii) to give y^3-y/12-7/27...you can use any method for solving cubic polynomial to do it but for me,i used tartaglia's method for finding real root of a cubic equation and got y as 0.68114(approximated),don't forget that a=y-1/3 and by substituting that a=0.34781(the first root of the cubic polynomial)...to get the other two roots...from relationship between the co-efficients and roots of a polynomial...a1+a2+a3=-1 a1.a2.a3=0.25(from the polynomial 4a^3+4a^2+a-1=0)...let a1 be 0.34781,by solving a2+a3=-1.34781 and a2.a3=0.7188 and by solving the simultaneous equation a={-1.34781±j1.0287}/2...with all we have done so far,the two real roots got were a=1 and a=0.34781,don't forget that sinx=a...let a be 1,then sinx=1 and x=90 and if a=0.34781,sinx=0.34781 and x=20.35...i rest my case
...the indomitable f(x)...nice wrk...
Re: Nairaland Mathematics Clinic by pickabeau1: 1:08pm On Nov 14, 2013
well done Gurus grin
Re: Nairaland Mathematics Clinic by Richiez(m): 1:11pm On Nov 14, 2013
Re: Nairaland Mathematics Clinic by Calculusfx: 1:13pm On Nov 14, 2013
Laplacian:
...the indomitable f(x)...nice wrk...
.i respect you master,though only the deep can come into the deep.hmmm,i pray i find myself in maths world like you o...thanx for the compliment sir
Re: Nairaland Mathematics Clinic by Laplacian(m): 1:23pm On Nov 14, 2013
...an alternate solution to sin2x+sinx=1...(eqn1) let
cos2x+cosx=p...(eqn2) then square each of d eqns and add to get
(sin2x)^2+2sin2x*sinx+sin^2x=1..........(3)
(cos2x)^2+2cos2x*cosx+cos^2x=p^2...(4)
addin,
1+2(cos2xcos+sin2xsinx)+1=1+p^2
or from d addition formular for cosine;
2cos(2x-x)=p^2-1,

cosx=(p^2-1)/2....(6)
substitut into (2), knowin dat cos2x=2cos^2x-1...one can then arrive @ d polynomial of Rhydex and F(x)...much tanks @Rhydex&F(x)
Re: Nairaland Mathematics Clinic by jackpot(f): 4:00pm On Nov 14, 2013
Hi all, please solve this.

If three fair dice are rolled simultaneously ONCE, what is the probability of getting:
(i) a sum of 11?
(ii) a sum of 12?



Show FULL working cheesy
Re: Nairaland Mathematics Clinic by 2nioshine(m): 4:01pm On Nov 14, 2013
Calculusf(x):
.hmmm,here is my own approach to it...sin2x+sinx=1...2sinxcosx+sinx=1...2sinx.sqrt(1-sin^2x)+sinx=1...{let sinx be a...(i)}...2a.sqrt(1-a^2)+a=1...2a.sqrt(1-a^2)=1-a...square both sides...4a^2(1-a^2)=1-2a+a^2...4a^2-4a^4=1-2a+a^2...4a^4-3a^2-2a+1=0,it's obvious that a=1(one of the four roots) then we divide 4a^4-3a^2-2a+1 by a-1 to give 4a^3+4a^2+a-1=0...then transform the equation to reduced cubic equation to get by representing a as y-1/3...(ii) to give y^3-y/12-7/27...you can use any method for solving cubic polynomial to do it but for me,i used tartaglia's method for finding real root of a cubic equation and got y as 0.68114(approximated),don't forget that a=y-1/3 and by substituting that a=0.34781(the first root of the cubic polynomial)...to get the other two roots...from relationship between the co-efficients and roots of a polynomial...a1+a2+a3=-1 a1.a2.a3=0.25(from the polynomial 4a^3+4a^2+a-1=0)...let a1 be 0.34781,by solving a2+a3=-1.34781 and a2.a3=0.7188 and by solving the simultaneous equation a={-1.34781±j1.0287}/2...with all we have done so far,the two real roots got were a=1 and a=0.34781,don't forget that sinx=a...let a be 1,then sinx=1 and x=90 and if a=0.34781,sinx=0.34781 and x=20.35...i rest my case
Nice approach bro,rhydex also did a gr8 job..
Re: Nairaland Mathematics Clinic by emekakelvin(m): 4:39pm On Nov 14, 2013
jackpot: Hi all, please solve this.

If three fair dice are rolled simultaneously ONCE, what is the probability of getting:
(i) a sum of 11?
(ii) a sum of 12?



Show FULL working cheesy
My working does nt look coherent and hypothesis is involve.
1) p(sum of 11)= 1/72
2) p(sum of 12)= 1/108
Waiting for correction!
Re: Nairaland Mathematics Clinic by rhydex247(m): 5:08pm On Nov 14, 2013
@ NAIRALAND MATHS GENERAL.
Happy CENTENOPAGERIAN me too dey Pop Mathematical bottle of ALOMO ND VODKA FOR HERE. My EYES DON DEY TINY. Lolllz.
Re: Nairaland Mathematics Clinic by Nobody: 5:39pm On Nov 14, 2013
benbuks: ^ 9c Try guys, bt they gat 2b anoda method....mayb 'series'

Okay.

Here goes...

0.977777... =
0.9 + (0.07 + 0.007 + ...) = 9/10 + (0.07 + 0.007 + ...)
Using sum to infinity for expression in the bracket: a = 0.07, r = 1/10.
Sum to infinity = a/(1-r)
= 0.07/(1-1/10)
= 7/100 * 10/9
= 7/90
So we have 9/10 + 7/90
= 880/900
= 44/45
Re: Nairaland Mathematics Clinic by Nobody: 6:09pm On Nov 14, 2013
jackpot: Hi all, please solve this.

If three fair dice are rolled simultaneously ONCE, what is the probability of getting:
(i) a sum of 11?
(ii) a sum of 12?



Show FULL working cheesy

Here goes...

(a) Sample space = 6^3 = 216

Set of combinations which gives a sum of 11 include:
164, 146, 461, 416, 641, 614, 155, 515, 551, 245, 254, 425, 452, 524, 542, 236, 263, 326, 362, 623, 632, 344, 434, 443, 335, 353, 533.

That's 27 ways of getting the sum 11. So answer to question (a) is 27/216 = 3/24

(b) Set of combination which gives the sum 12 include:
156, 165, 516, 561, 615, 651, 246, 264, 426, 462, 624, 642, 255, 525, 552, 336, 363, 633, 345, 354, 435, 453, 534, 543, 444.

That's 25 ways of getting the sum of 12. That gives 25/216.
Re: Nairaland Mathematics Clinic by Laplacian(m): 6:24pm On Nov 14, 2013
Laplacian: ...a two digit number is divisible by 4, if d unit digit is also divisible by 4, show that d tens digit MUST be even...
Re: Nairaland Mathematics Clinic by Laplacian(m): 6:25pm On Nov 14, 2013
Laplacian: ...a two digit number is divisble by 4 prove that d unit digit MUST be even...
Re: Nairaland Mathematics Clinic by Onyeoma3: 6:29pm On Nov 14, 2013
Nice job RICHIEZ...keep the good work going
Re: Nairaland Mathematics Clinic by Nobody: 6:50pm On Nov 14, 2013
smurfy:

Okay.

Here goes...

0.977777... =
0.9 + (0.07 + 0.007 + ...) = 9/10 + (0.07 + 0.007 + ...)
Using sum to infinity for expression in the bracket: a = 0.07, r = 1/10.
Sum to infinity = a/(1-r)
= 0.07/(1-1/10)
= 7/100 * 10/9
= 7/90
So we have 9/10 + 7/90
= 880/900
= 44/45
....on point..
Re: Nairaland Mathematics Clinic by Nobody: 7:19pm On Nov 14, 2013
Alpha Maximus: Finally!!! The clinic is a CENTENOPAGERIAN!!! *pops mathematical bottle of champagne* I wanna give a shoutout to Sivasubramaniam, the author of SS Further Math textbooks and Co-author Tuttuh Adegun,Leonhard Euler(who discovered the Euler's constant, is the Father of Mathematics) , Euclid(the Prince Of Math) , Pythagoras Of Samos(triangles would have been a pain in the calculator if not for this guy), Rene Descartes for his beautiful work which gave birth to Cartesian Planes, Blaise Pascal( a true prodigy, made his own formulae at 20 or younger), Isaac Newton( laid the foundation for Albert Einstein's works and published the single most important document in Mathematics- Principia de Mathematica), Pierre de Fermat(constructed Fermat's last conjecture:stating that no 3 positive integers a, b and c can satisfy a^n+b^n=c^n for positive integer values of n which are equal to or greater than 3, which was disproven by Andrew Wiles who worked on this for 7 years!!), Andrew Wiles, British mathematician who solved Fermat's notorious Conjecture and rejected the 1 million dollar prize money, Chike Obi of Nigeria who showed that whites were not the only capable problem-solvers, Grace Alele, pioneer of Nigerian female mathematicians, Albert Einstein( E=mc^2 says it all), the author's of new general mathematics for preparing the Nigerian youth for WAEC, KA Stroud for his insightful Advanced Engineering Math textbook and all other great mathematician whom I have failed to mention. We can't do without math in life! Even space is made up of numbers! Mathematics is the very back-bone of all sciences, the blood that pumps through the veins of engineers! Indeed what would we be without math? Merry solving and cheers to the Nairaland Mathematics Clinic on its 100 page anniversary grin *moonwalks on stage*


Nice one Comrade Alpha!!! grin grin Happy CENTENOPAGERIAN!!!! grin grin grin grin
Re: Nairaland Mathematics Clinic by Nobody: 7:22pm On Nov 14, 2013
Wow wow wow....first of introduction..., introduction, x2 ....
Congratulations to this geat thread "MATHEMATICS CLINIC" a.k.a nairaland math gurus..on your 10th decade anniversary (by pages) ..to all my tireless mathematical doctors who always receives patients to meticulously and indomitalely treat maladies involved.., though some generals and doctors run away from some enigmatic maladies (e.g benbuks) ..all the same you guys are great oo, i remove my cap, shirt, shoes, trouser..for una o...if care is not taken i will remove the last one o..lolz.,.dont tempt me to do that...hahaha,

really glad to be among this great family..
@master richiez my big bros ,i respect o...may God reward you for all u,ve been doin.


ASUU don make sm pple turn to math freaks oo,..i will really miss this thread when school resume....we go dey meet sha.....

.
All glory be to almighty God the giver of life and knowledge to him alone be all praise and adoration
Amen.
1love ... Al z wel

cool..# Benbuk#...

1 Like

Re: Nairaland Mathematics Clinic by jackpot(f): 7:50pm On Nov 14, 2013
smurfy:

Here goes...

(a) Sample space = 6^3 = 216

Set of combinations which gives a sum of 11 include:
164, 146, 461, 416, 641, 614, 155, 515, 551, 245, 254, 425, 452, 524, 542, 236, 263, 326, 362, 623, 632, 344, 434, 443, 335, 353, 533.

That's 27 ways of getting the sum 11. So answer to question (a) is 27/216 = 3/24

(b) Set of combination which gives the sum 12 include:
156, 165, 516, 561, 615, 651, 246, 264, 426, 462, 624, 642, 255, 525, 552, 336, 363, 633, 345, 354, 435, 453, 534, 543, 444.

That's 25 ways of getting the sum of 12. That gives 25/216.
Nice!!!
Re: Nairaland Mathematics Clinic by Nobody: 8:40pm On Nov 14, 2013
jackpot: Nice!!!

Really? I thought my method was watery and was expecting some rebukes from thee followed by a grandiose way of solving it.

Is there another method?
Re: Nairaland Mathematics Clinic by Cashio(m): 9:11pm On Nov 14, 2013
And this thread just made a century of pages!!!
Pls if x is a positive integer divisible by 3 but less than 100,what are the possible square roots of x
Re: Nairaland Mathematics Clinic by Nobody: 9:45pm On Nov 14, 2013
Cashio: And this thread just made a century of pages!!!
Pls if x is a positive integer divisible by 3 but less than 100,what are the possible square roots of x

Here goes...

x includes all multiples of 3, i.e. 3, 6, 9, 12,..., 93, 96, 99.
Answer is the square root of each of the numbers above.

Note that you never said x is a perfect square. However, if x is a perfect square divisible by 3, i.e. 9, 36 or 81, then the required square roots are 3, 6 or 9.
Re: Nairaland Mathematics Clinic by jackpot(f): 10:23pm On Nov 14, 2013
smurfy:

Really? I thought my method was watery and was expecting some rebukes from thee followed by a grandiose way of solving it.

Is there another method?
well, you could have used permutations anyway.

Like arrangement of 6,5, 1 is 3P3=6
then, do the same for all and add.
Re: Nairaland Mathematics Clinic by winbyforce: 10:54pm On Nov 14, 2013
Alpha Maximus: Finally!!! The clinic is a CENTENOPAGERIAN!!! *pops mathematical bottle of champagne* I wanna give a shoutout to Sivasubramaniam, the author of SS Further Math textbooks and Co-author Tuttuh Adegun,Leonhard Euler(who discovered the Euler's constant, is the Father of Mathematics) , Euclid(the Prince Of Math) , Pythagoras Of Samos(triangles would have been a pain in the calculator if not for this guy), Rene Descartes for his beautiful work which gave birth to Cartesian Planes, Blaise Pascal( a true prodigy, made his own formulae at 20 or younger), Isaac Newton( laid the foundation for Albert Einstein's works and published the single most important document in Mathematics- Principia de Mathematica), Pierre de Fermat(constructed Fermat's last conjecture:stating that no 3 positive integers a, b and c can satisfy a^n+b^n=c^n for positive integer values of n which are equal to or greater than 3, which was disproven by Andrew Wiles who worked on this for 7 years!!), Andrew Wiles, British mathematician who solved Fermat's notorious Conjecture and rejected the 1 million dollar prize money, Chike Obi of Nigeria who showed that whites were not the only capable problem-solvers, Grace Alele, pioneer of Nigerian female mathematicians, Albert Einstein( E=mc^2 says it all), the author's of new general mathematics for preparing the Nigerian youth for WAEC, KA Stroud for his insightful Advanced Engineering Math textbook and all other great mathematician whom I have failed to mention. We can't do without math in life! Even space is made up of numbers! Mathematics is the very back-bone of all sciences, the blood that pumps through the veins of engineers! Indeed what would we be without math? Merry solving and cheers to the Nairaland Mathematics Clinic on its 100 page anniversary grin *moonwalks on stage*
Add Grigory Perelman to the list. The russian recluse of jewish heritage who solved the poincare conjecture and rejected both the 2006 fields medals(highest honor in mathematics) and a 1million dollar clay institute's millenium prize. Retired from maths because of the 'politics of academia'
Re: Nairaland Mathematics Clinic by LogoDWhiz(m): 11:09pm On Nov 14, 2013
How can 0.9999...=1?
without approximation.
Re: Nairaland Mathematics Clinic by brainiacjp(m): 11:36pm On Nov 14, 2013
Gosh...i fink I belong here.....here is brainiac 4rm unilag....i really commend u guys'$ work....i wonder how u managed 2 type such lengthy mathematical manipulations.....u r gr8....
......one love.....
brainiac noni....
Re: Nairaland Mathematics Clinic by brainiacjp(m): 11:40pm On Nov 14, 2013
@ logoDWhiz.....he no get apprxmtn sign 4 im system ni....~
. -
Re: Nairaland Mathematics Clinic by Nobody: 7:05am On Nov 15, 2013
jackpot: well, you could have used permutations anyway.

Like arrangement of 6,5, 1 is 3P3=6
then, do the same for all and add.

Noted.

(1) (2) (3) ... (98) (99) (100) (101) (102) (103) (104) ... (284) (Reply)

DIRECT ENTRY Admission. / Mastercard Foundation Scholarship, Enter Here / 2016/2017 University of Ibadan Admission Thread Guide.

(Go Up)

Sections: politics (1) business autos (1) jobs (1) career education (1) romance computers phones travel sports fashion health
religion celebs tv-movies music-radio literature webmasters programming techmarket

Links: (1) (2) (3) (4) (5) (6) (7) (8) (9) (10)

Nairaland - Copyright © 2005 - 2024 Oluwaseun Osewa. All rights reserved. See How To Advertise. 63
Disclaimer: Every Nairaland member is solely responsible for anything that he/she posts or uploads on Nairaland.