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Re: Nairaland Mathematics Clinic by Fanccy: 4:05am On May 26, 2013
2nioshine, thanks for starting off this equation. I have been trying it out but still hanging on the way. Am waiting for you to complete it.
doubledx and Ortarico. Am waiting for you guys to make a contribution.
Re: Nairaland Mathematics Clinic by Richiez(m): 1:59pm On May 26, 2013
@fanccy, wot equation? can u retype?
Re: Nairaland Mathematics Clinic by 2nioshine(m): 2:27pm On May 26, 2013
Fanccy: 2nioshine, thanks for starting off this equation. I have been trying it out but still hanging on the way. Am waiting for you to complete it.
doubledx and Ortarico. Am waiting for you guys to make a contribution.
..4 d engine values check d previous post
if A.X=¥X
it implies dat multiplyn d matrix A wt d colum matrix X=d engine values multipyn X
oR
Using d [A-¥]X=0
when ¥=4 we have
[-2 2 -2][X1] [0]
[1 -1 1][X2]=[0]
[1 2 -2][X3] [0]
Nb:this values r obtaind by substutn ¥=4 in d xteristic eqn. and X(engine vector) will av 3values since itz a 3*3 matrix
but since when ¥=4 is required we have

-2X1+2X2-2X3=0.......1

X1-X2+X3=0....2

X1+2X2 -2X3=0......3
4Rm eqn 2
x1=x2-x3......*
substitutn n eqn 3
gvs
3X2-3X2=0
hence X2=X3.....**
subt n eqn 1
gvs 2X1+2X3-2X3=0
=>2X1=0
X1=0
Since x2=x3
their ratio is 1:1
hence x2=x3=1 and x1=0
:- the engn vector for ¥=4
=>[0]
[1]
[1]g
Re: Nairaland Mathematics Clinic by 2nioshine(m): 2:31pm On May 26, 2013
Fanccy: 2nioshine, thanks for starting off this equation. I have been trying it out but still hanging on the way. Am waiting for you to complete it.
doubledx and Ortarico. Am waiting for you guys to make a contribution.
hope it helps...my loyalty 2 all em xo impresd
Re: Nairaland Mathematics Clinic by fazbuk(m): 4:20pm On May 26, 2013
Help me wit dis Fourier series question,:f(x)=(-2A/L(1+x) for -L<x<-L/2,,,,....2Ax/L for -L/2<x<L/2,........2A/L(1-x) for L/2<x<L pls z very important n urgent.tanx in advance.
Re: Nairaland Mathematics Clinic by Fanccy: 5:52pm On May 26, 2013
2nioshine: ..4 d engine values check d previous post
if A.X=¥X
it implies dat multiplyn d matrix A wt d colum matrix X=d engine values multipyn X
oR
Using d [A-¥]X=0
when ¥=4 we have
[-2 2 -2][X1] [0]
[1 -1 1][X2]=[0]
[1 2 -2][X3] [0]
Nb:this values r obtaind by substutn ¥=4 in d xteristic eqn. and X(engine vector) will av 3values since itz a 3*3 matrix
but since when ¥=4 is required we have

-2X1+2X2-2X3=0.......1

X1-X2+X3=0....2

X1+2X2 -2X3=0......3
4Rm eqn 2
x1=x2-x3......*
substitutn n eqn 3
gvs
3X2-3X2=0
hence X2=X3.....**
subt n eqn 1
gvs 2X1+2X3-2X3=0
=>2X1=0
X1=0
Since x2=x3
their ratio is 1:1
hence x2=x3=1 and x1=0
:- the engn vector for ¥=4
=>[0]
[1]
[1]g
from the above, you did not use identity matrix I. I used it and got different answer from the one you have. What I mean is using the characteristic equation [A -¥I] = 0.
Re: Nairaland Mathematics Clinic by OgeneDom(m): 6:18pm On May 26, 2013
Oh guys, am most grateful.
Re: Nairaland Mathematics Clinic by 2nioshine(m): 8:02pm On May 26, 2013
Fanccy:
from the above, you did not use identity matrix I. I used it and got different answer from the one you have. What I mean is using the characteristic equation [A -¥I] = 0.
tnks 4 d observation but d identity of any matrix only gives value at d principal diagonal
¥[1 0 0]
[0 1 0]
[0 0 1]
and this xplains y its values r only subtratd 4rm its principal diagonals....so by doing xo it az already bn used

1 Like

Re: Nairaland Mathematics Clinic by Fanccy: 9:18pm On May 26, 2013
^^ thanks man. I got it correctly.

1 Like

Re: Nairaland Mathematics Clinic by Ortarico(m): 11:31pm On May 29, 2013
Fanccy: 2nioshine, thanks for starting off this equation. I have been trying it out but still hanging on the way. Am waiting for you to complete it.
doubledx and Ortarico. Am waiting for you guys to make a contribution.

I'm so sorry bruv, the best of my knowledge on matrices is scalars's. Like I said in one of my posts that I'm an aspirant and green on some further topics. I really hate it when summoned on things like this but can't be of help. Lookin' forward to attaining a Guru's rank anyway. Thanks to one of my Ogas at the top for helpin' you at last!
Re: Nairaland Mathematics Clinic by sniperwolf(m): 7:28am On May 30, 2013
A manufacturing company is independently working on two
separate jobs. If the probability that either of the jobs will be
finished on time is 0.02, the probability that just one of the jobs
will be finished on time will be ....
[ICAN May 2013. Costing and Quantitative Techniques]
Re: Nairaland Mathematics Clinic by echibuzor: 1:28pm On May 30, 2013
sniperwolf: A manufacturing company is independently working on two
separate jobs. If the probability that either of the jobs will be
finished on time is 0.02, the probability that just one of the jobs
will be finished on time will be ....
[ICAN May 2013. Costing and Quantitative Techniques]

Is this Question complete? before i even attempt...
Re: Nairaland Mathematics Clinic by biolabee(m): 1:49pm On May 30, 2013
sniperwolf: A manufacturing company is independently working on two
separate jobs. If the probability that either of the jobs will be
finished on time is 0.02, the probability that just one of the jobs
will be finished on time will be ....
[ICAN May 2013. Costing and Quantitative Techniques]

This is the 1 - P (2 jobs) or P (no jobs)

P(2 jobs) = 0.2 *0.2 = 0.04
P(no jobs) = 0.8 * 0.8 = 0.64

1 - 0.64-0.04 = 1- 0.68 = 0.32

You can also do P(one job, no job = 0.8 * 0.2 = 0.16
Multiply by 2 = 0.32

Possiblity

P (JJ), P(JN), P(NJ)and P(NN)

1 Like

Re: Nairaland Mathematics Clinic by echibuzor: 3:21pm On May 30, 2013
biolabee:

This is the 1 - P (2 jobs) or P (no jobs)

P(2 jobs) = 0.2 *0.2 = 0.04
P(no jobs) = 0.8 * 0.8 = 0.64

1 - 0.64-0.04 = 1- 0.68 = 0.32

You can also do P(one job, no job = 0.8 * 0.2 = 0.16
Multiply by 2 = 0.32

Possiblity

P (JJ), P(JN), P(NJ)and P(NN)



Good man, I was reading the question as dependent probability... I need to see my optician ASAP.. anyway, good job..
Re: Nairaland Mathematics Clinic by biolabee(m): 3:59pm On May 30, 2013
echibuzor:


Good man, I was reading the question as dependent probability... I need to see my optician ASAP.. anyway, good job..

i be small boy o,,, na una be the main men for hia....
Re: Nairaland Mathematics Clinic by Boladearo(m): 3:01pm On May 31, 2013
How many numbers greater that 3000 can be made from the digits, 1,2,3,4,5 without repeting any of them
Re: Nairaland Mathematics Clinic by Boladearo(m): 3:02pm On May 31, 2013
How many numbers greater that 3000 can be made from the digits, 1,2,3,4,5 without repeting any of them
pls, sum 1 should help me to solve it, have tried my best, but i still couldn't solve it
Re: Nairaland Mathematics Clinic by Raylight2(m): 3:25pm On May 31, 2013
Help Solve This Pls X^4 - 3x^2 + 1 = 0. I need the solution to pass a course I'll be writing very soon.
Re: Nairaland Mathematics Clinic by Ortarico(m): 5:09pm On May 31, 2013
Boladearo: How many numbers greater that 3000 can be made from the digits, 1,2,3,4,5 without repeting any of them
pls, sum 1 should help me to solve it, have tried my best, but i still couldn't solve it

First of all, we can have a five number digit greater than 3,000 i.e 5!

:. number greater than 3,000 can either be taken from 3, 4 or 5 which is 3 ways > 3000, and the remaining 3 digits of the 4 digit numbers. . .

=> 5! + 3(4*3*2)
=> 120 + 72
=> 192 ways
Re: Nairaland Mathematics Clinic by Ortarico(m): 5:13pm On May 31, 2013
Raylight_1: Help Solve This Pls X^4 - 3x^2 + 1 = 0. I need the solution to pass a course I'll be writing very soon.

I can attempt this but don't want my answer to contradict with my Ogas @ d top own. . . Plz wait for them
Re: Nairaland Mathematics Clinic by Boladearo(m): 7:10pm On May 31, 2013
Ortarico, and other maths gurus, pls drop ur 2go user name so that we can add u guys. You guys are mathematician of the highest order.
Re: Nairaland Mathematics Clinic by sniperwolf(m): 9:42pm On May 31, 2013
biolabee:

This is the 1 - P (2 jobs) or P (no jobs)

P(2 jobs) = 0.2 *0.2 = 0.04
P(no jobs) = 0.8 * 0.8 = 0.64

1 - 0.64-0.04 = 1- 0.68 = 0.32

You can also do P(one job, no job = 0.8 * 0.2 = 0.16
Multiply by 2 = 0.32

Possiblity

P (JJ), P(JN), P(NJ)and P(NN)


Guy you are good. Option B is 0.32. I never even understood the question talk less of attempting it

1 Like

Re: Nairaland Mathematics Clinic by Mbahchiboy(m): 9:50pm On May 31, 2013
Ortarico:

First of all, we can have a five number digit greater than 3,000 i.e 5!

:. number greater than 3,000 can either be taken from 3, 4 or 5 which is 3 ways > 3000, and the remaining 3 digits of the 4 digit numbers. . .

=> 5! + 3(4*3*2)
=> 120 + 72
=> 192 ways
Kudoz man
Re: Nairaland Mathematics Clinic by Mbahchiboy(m): 9:55pm On May 31, 2013
Raylight_1: Help Solve This Pls X^4 - 3x^2 + 1 = 0. I need the solution to pass a course I'll be writing very soon.
Re: Nairaland Mathematics Clinic by biolabee(m): 11:29pm On May 31, 2013
Nice one bro

sniperwolf:

Guy you are good. Option B is 0.32. I never even understood the question talk less of attempting it
Re: Nairaland Mathematics Clinic by Raylight2(m): 7:27am On Jun 01, 2013
Raylight_1: Help solve this Pls X^4 - 3x^2 + 1 = 0. I need the solution to pass a course I'll be writing very soon.
Re: Nairaland Mathematics Clinic by Edis4christ(m): 9:13am On Jun 01, 2013
doubleDx: ^
Yes, whatever works Chief!

Thanks.

y = cos^2t .....(1)
then dy/dt = -2costsint
x = sin^2t ........(2)
then, dx/dt = 2costsint
dy/dx = dy/dt/dx/dt
= -2costsint/2costsint
dy/dx = -1
:. d^2y/dx^2 = 0
9ice wrk general doubledx........respect sir
Re: Nairaland Mathematics Clinic by Nobody: 10:57am On Jun 01, 2013
9ice work @2nioshine, Ortarico & biolabee! I hail ona!

Edis4christ:
9ice wrk general doubledx........respect sir

Thanks Chief; respek!

Raylight_1: Raylight_1: Help solve this Pls X4 - 3x2 + 1 = 0. I need the solution to pass a course I'll be writing very soon.


Solution

x4 - 3x2 + 1 = 0

Let’s simply put
s = x2

so that the equation can be rewritten as follows =>>>

s2 - 3s + 1 = 0
Applying quadratic formula to the above equation yields=>

a = 1, b = -3 and C = 1
s = {-b ±√(b2 - 4ac)}/2a
s = { -(-3)±√((-3)2 - 4(1)(1))}/2(1)
s = {3 ±√(9 - 4)}/2
s = 1/2{3 ±√(9 - 4)}
s = 1/2{3 ±√5}

:. s = 1/2{3 + √5) or 1/2({3 - √5)
s = 1/2 ( 3 + 2.236) or 1/2 ( 3 - 2.236)
s = 2.618 or 0.618 approx.

Since s = x2
x = ±√s

Hence,
x = ±√2.618 or ±√0.618
x = ±1.618 or ±0.786
:. The values of x are = 1.618, -1.618, 0.786, -0.786 approx.

4 Likes

Re: Nairaland Mathematics Clinic by Richiez(m): 11:22am On Jun 01, 2013
kudos to the reliable Generals in the house, Doubledx, Biolabee and ortarico, keep up the good work!

2 Likes

Re: Nairaland Mathematics Clinic by Ortarico(m): 11:40am On Jun 01, 2013
Boladearo: Ortarico, and other maths gurus, pls drop ur 2go user name so that we can add u guys. You guys are mathematician of the highest order.
oketunde2
Re: Nairaland Mathematics Clinic by Ortarico(m): 11:41am On Jun 01, 2013
Mbahchiboy: Kudoz man
thanks boss
Re: Nairaland Mathematics Clinic by Ortarico(m): 11:46am On Jun 01, 2013
doubleDx: 9ice work @2nioshine, Ortarico & biolabee! I hail ona!



Thanks Chief; respek!




Solution

x^4 - 3x^2 + 1 = 0

Let’s simply put
s = x^2

so that the equation can be rewritten as follows =>>>

s^2 - 3s + 1 = 0
Applying quadratic formula to the above equation yields=>

a = 1, b = -3 and C = 1
s = {-b ±√(b^2 - 4ac)}/2a
s = { -(-3)±√((-3)^2 - 4(1)(1))}/2(1)
s = {3 ±√(9 - 4)}/2
s = 1/2{3 ±√(9 - 4)}
s = 1/2{3 ±√5}

:. s = 1/2{3 + √5) or 1/2({3 - √5)
s = 1/2 ( 3 + 2.236) or 1/2 ( 3 - 2.236)
s = 2.618 or 0.618 approx.

Since s = x^2
x = ±√s

Hence,
x = ±√2.618 or ±√0.618
x = ±1.618 or ±0.786
:. The values of x are = 1.618, -1.618, 0.786, -0.786 approx.

I said it, exponential eqn! It rhymes!! Thanks for the eulogy @general doubleDx and Richiez as well as all, one love!!!

1 Like

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