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Re: Nairaland Mathematics Clinic by MaxGraviton: 12:57pm On Jul 31, 2015
Antoinne:


He has to be right. In maths, you don't think it half of the way. You must think it through. if there were no women in the committee, the number of men will still be greater than the number of women, i.e. 5>0. And that's a valid mathematical statement, except of course the question introduced the condition that "there must be at least 1 woman".

Pls sir note that before a comparison can be made, the quantities must have something in common,

For example,
•3G is better than EDGE, what is common? They're both internet speed connections.
• Mark Henry is Stronger than big show, what is common? They're both wrestlers.
• A Lamborghini is better than a Volvo 504, what is common? They're both cars.


in this case their ACTUAL PRESENCE in the committee is what they have in common, so comparison is valid.

• 3Men and 2Women. Men>Women. What is common? They're both present.

When you say 0 women.That means NO WOMAN is present and so comparison btwn the men and women is and I say again sir. . . invalid.


• 5Men and 0 Women. And you go like " the no. Of Men are more than the no. Of women". . . a reasonable person would ask you a question. " Excuse Me sir, Where are the Women?" .

I hope I was able to paint a picture in your head. Lools
Re: Nairaland Mathematics Clinic by Antoinne: 1:33pm On Jul 31, 2015
MaxGraviton:


Pls sir note that before a comparison can be made, the quantities must have something in common,

For example,
•3G is better than EDGE, what is common? They're both internet speed connections.
• Mark Henry is Stronger than big show, what is common? They're both wrestlers.
• A Lamborghini is better than a Volvo 504, what is common? They're both cars.


in this case their ACTUAL PRESENCE in the committee is what they have in common, so comparison is valid.

• 3Men and 2Women. Men>Women. What is common? They're both present.

When you say 0 women.That means NO WOMAN is present and so comparison btwn the men and women is and I say again sir. . . invalid.


• 5Men and 0 Women. And you go like " the no. Of Men are more than the no. Of women". . . a reasonable person would ask you a question. " Excuse Me sir, Where are the Women?" .

I hope I was able to paint a picture in your head. Lools

You are not wrong and i didn't say that you were. Arguing from a common sense perspective, you are correct. But maths doesn't derive from these kinds of argument. Maths is based on numbers, that's why a number called zero was created.
The probability of an event can be 0. The mere fact that such event can't happen is not enough reason not to consider it or its probability. This is the reason you have to consider 5>0 in order to give a complete solution.

If you had to solve this question in order to explain something in nature, the incompleteness of your solution will definitely show if you didn't consider that last possibility.


• 5Men and 0 Women. And you go like " the no. Of Men are more than the no. Of women". . . a reasonable person would ask you a question. " Excuse Me sir, Where are the Women?" .
Maybe. Maybe a reasonable person will ask where the women are. But a mathematician will not. Especially as you had started from 3,2; 4,1. 5,0 is a natural in that sequence. Especially as the question doesn't limit you. Clearly 5men is greater than 0 women. 0 is a number, don't forget. Not a word.
Re: Nairaland Mathematics Clinic by Nobody: 6:38pm On Jul 31, 2015
Please guys, click on the link below to help solve the question there.
www.nairaland.com/2492974/solve-maths-me
If you get the answer, PLEASE solve it on the other thread and then quote me if you want to solve it here
Re: Nairaland Mathematics Clinic by MaxGraviton: 6:41pm On Jul 31, 2015
Antoinne:


You are not wrong and i didn't say that you were. Arguing from a common sense perspective, you are correct. But maths doesn't derive from these kinds of argument. Maths is based on numbers, that's why a number called zero was created.
The probability of an event can be 0. The mere fact that such event can't happen is not enough reason not to consider it or its probability. This is the reason why you have to consider 5>0 in order to give a complete solution.

If you had to solve this question in order to explain something in nature, the incompleteness of your solution will definitely show if you didn't consider that last possibility.


Maybe. Maybe a reasonable person will ask where the women are. But a mathematician will not. Especially as you had started from 3,2; 4,1. 5,0 is a natural in that sequence. Especially as the question doesn't limit you. Clearly 5men is greater than 0 women. 0 is a number, don't forget. Not a word.


Lools. Yea 0 is a number, but a valueless number. When the probability of something is 0, that is no longer probability. . . that is certainty. . . probability is btwn 0 and 1.(btwn 0.1 & 0.9)

Anyways,
Comparingng 0 to a number with an actual value, in this aspect of maths(combination & Permutation), is Invalid. This aspect of maths has a real life application, so if you're using mathematical theories to approach this, you should think about it in its real life application.

You might say I'm correct in the realm of common sense, but if you look at it, maths is a deeper concept of common sense.

I'm still open to your counter response.

2 Likes

Re: Nairaland Mathematics Clinic by Antoinne: 8:37pm On Jul 31, 2015
MaxGraviton:



Lools. Yea 0 is a number, but a valueless number. When the probability of something is 0, that is no longer probability. . . that is certainty. . . probability is btwn 0 and 1.(btwn 0.1 & 0.9)

And that it's certainty means you'll never consider it? Probability is between 0 and 1. Some redefine it, except you want to recreate your own maths. Use wikipedia or wolfram and you'll see probability is defined as between 0-1. You shouldn't even be arguing that. 0 is not a valueless number. If it was, it wouldn't have been created.


Anyways,
Comparingng 0 to a number with an actual value, in this aspect of maths(combination & Permutation), is Invalid. This aspect of maths has a real life application, so if you're using mathematical theories to approach this, you should think about it in its real life application.

You might say I'm correct in the realm of common sense, but if you look at it, maths is a deeper concept of common sense.

I'm still open to your counter response.
Maths itself explains things in nature, so you can't separate it from real world applications. Combination and permutation have real world applications; it's in this sense that you have to consider the 5,0. Don't try to play a fast one.

Anyways, see how others have solved similar problems. Your reasoning in this regard can't supported.

https://answers.yahoo.com/question/index?qid=20080708101257AACtrGL

http://math.stackexchange.com/questions/1183067/forming-a-committee-but-there-must-be-more-or-equal-amount-of-female-to-males
Re: Nairaland Mathematics Clinic by nonsso: 12:22pm On Aug 01, 2015
please i need answers to these questions ASAP

1: How many different three digit numbers can be formed using the integers occurs twice in a number?

2: A cyclist rode for 30 min at xkm/h and due to a breakdown he had to push the bike for 2hrs at X -5km/hr. if the total distance covered is less than 60km, what is the range of values of X
a) x< 14
b) x< 20
c) x< 29
d) x< 28
Re: Nairaland Mathematics Clinic by Geofavor(m): 3:54pm On Aug 01, 2015
Hehe. Interesting arguement.

Maxgravition, nice points you got. But antoinne is more on point.

I've solved that kind of question few times before.

@ max,
They would state that at least 1 woman be included if they dont want you to consider the last way - which in this case is 5C0.

You said 0 is valueless. Have you forgotten that 0! = 1

1 Like

Re: Nairaland Mathematics Clinic by ayodijex(m): 8:31pm On Aug 01, 2015
nonsso:
please i need answers to these questions ASAP

1: How many different three digit numbers can be formed using the integers occurs twice in a number?
not clear ??

[/quote]2: A cyclist rode for 30 min at xkm/h and due to a breakdown he had to push the bike for 2hrs at X -5km/hr. if the total distance covered is less than 60km, what is the range of values of X
a) x< 14
b) x< 20
c) x< 29
d) x< 28[/quote]
first change time T=30mins to hrs
=> 30/60 = 0.5hrs
distance = velocity * time
first distance = x*0.5=0.5x
2nd distance=(x-5)*2=2x-10
since total distance<60
0.5x + (2x-10) <60
2.5x<70
x<28.

C'est nes pas??.

1 Like

Re: Nairaland Mathematics Clinic by jackpot(f): 8:39pm On Aug 01, 2015
Find the inverse laplace transform of 1/(s^5+a^5)
tags: Laplacian, agentofchange1, doubleDx, dejt4u, Alphamaximus, jaryeh
Re: Nairaland Mathematics Clinic by Nobody: 9:15pm On Aug 01, 2015
jackpot:
Find the inverse laplace transform of 1/(s^5+a^5)

Re: Nairaland Mathematics Clinic by Laplacian(m): 10:39pm On Aug 01, 2015
Am having trouble with an example in; Advanced Engineering Mathematics, K.A stroud: page 759, example 2; i understand their method but on applying a different method (which looks logical), am getting a different answer (dS=5dxdy). So, using any other method apart from the one provided, find the element of Area dS on the triangle. See fig. Below
tags; jackpot, agentofchange1, doubleDx, dejt4u, Alphamaximus, jaryeh

1 Like

Re: Nairaland Mathematics Clinic by MaxGraviton: 2:49am On Aug 02, 2015
Antoinne:


And that it's certainty means you'll never consider it? Probability is between 0 and 1. Some redefine it, except you want to recreate your own maths. Use wikipedia or wolfram and you'll see probability is defined as between 0-1. You shouldn't even be arguing that. 0 is not a valueless number. If it was, it wouldn't have been created.



Maths itself explains things in nature, so you can't separate it from real world applications. Combination and permutation have real world applications; it's in this sense that you have to consider the 5,0. Don't try to play a fast one.

Anyways, see how others have solved similar problems. Your reasoning in this regard can't supported.

https://answers.yahoo.com/question/index?qid=20080708101257AACtrGL

http://math.stackexchange.com/questions/1183067/forming-a-committee-but-there-must-be-more-or-equal-amount-of-female-to-males


On this ground, I concur. grin cheesy
Thanks Man.
Na small thing dey make person fail exam.
I studied those links well, and the spirit ministered to me. grin Thanks once again

1 Like

Re: Nairaland Mathematics Clinic by MaxGraviton: 2:57am On Aug 02, 2015
Geofavor:
Hehe. Interesting arguement.

Maxgravition, nice points you got. But antoinne is more on point.

I've solved that kind of question few times before.

@ max,
They would state that at least 1 woman be included if they dont want you to consider the last way - which in this case is 5C0.

You said 0 is valueless. Have you forgotten that 0! = 1



Lools, Yea, but the factorial of 0 is not a way of proving the value of zero.
if it proves anything at all, it should be that 0 is a real number and not that 0 is not valueless.
My brother in Combination and Permutation, 0 indicates no availability of the considered parameters, invariably saying, no value.

1 Like

Re: Nairaland Mathematics Clinic by jackpot(f): 6:38am On Aug 02, 2015
Karmanaut:
The details in the pix isn't complete. Please, resnap or cite.
Is that Advanced Engine Maths by H.K. Dass?
Re: Nairaland Mathematics Clinic by jackpot(f): 6:38am On Aug 02, 2015
Karmanaut:
The details in the pix isn't complete. Please, resnap or cite.
Is that Advanced Engine Maths by H.K. Dass?
Re: Nairaland Mathematics Clinic by jackpot(f): 6:52am On Aug 02, 2015
Laplacian:
Am having trouble with an example in; Advanced Engineering Mathematics, K.A stroud: page 759, example 2; i understand their method but on applying a different method (which looks logical), am getting a different answer (dS=5dxdy). So, using any other method apart from the one provided, find the element of Area dS on the triangle. See fig. Below
tags; jackpot, agentofchange1, doubleDx, dejt4u, Alphamaximus, jaryeh
Take f(x,y,z)=2x+2y+z-2=0.

dS= sqrt[(fx)2+(fy)2+1]dxdy=sqrt[4+4+1]dxdy=sqrt[9]dxdy=3dxdy.



Formula reference: www.math.lsa.umich.edu/~glarose/classes/calcIII/web/16_6/
Re: Nairaland Mathematics Clinic by Nobody: 9:56am On Aug 02, 2015
jackpot:
The details in the pix isn't complete. Please, resnap or cite.

Is that Advanced Engine Maths by H.K. Dass?
Better- it's WolframAlpha.
Source: http://www.wolframalpha.com/input/?i=Find+the+inverse+laplace+transform+of+1%2F%28s%5E5%2Ba%5E5%29&x=7&y=11
Re: Nairaland Mathematics Clinic by Geofavor(m): 11:42am On Aug 02, 2015
MaxGraviton:




Lools, Yea, but the factorial of 0 is not a way of proving the value of zero.
if it proves anything at all, it should be that 0 is a real number and not that 0 is not valueless.
My brother in Combination and Permutation, 0 indicates no availability of the considered parameters, invariably saying, no value.
I was being precise with the topic/question being argued, not maths in general. Not numbers.

You said earlier that, 5C0 is invalid in this aspect of maths(permutation n combination) because of the comparison of an actual number with zero. You said they shouldn't be compared because 0 is valueless.

Which was why i reminded you that in this case, it isn't. Why? Because factorial is involved!

5C0 = 5!/(5 - 0)! 0!
Re: Nairaland Mathematics Clinic by possibilita(m): 1:13pm On Aug 02, 2015
Pls help me on these

Re: Nairaland Mathematics Clinic by Antoinne: 1:36pm On Aug 02, 2015
Laplacian:
Am having trouble with an example in; Advanced Engineering Mathematics, K.A stroud: page 759, example 2; i understand their method but on applying a different method (which looks logical), am getting a different answer (dS=5dxdy). So, using any other method apart from the one provided, find the element of Area dS on the triangle. See fig. Below
tags; jackpot, agentofchange1, doubleDx, dejt4u, Alphamaximus, jaryeh
I suppose the way you were doing it was to resolve dx on one hand, and dy on the other to the lines at y=0 & x=0 respectively, so that you obtained dSx = sqrt(5)dx & dSy = sqrt(5)dy

And you then did, dS = dSx.dSy = 5dxdy. if this is how you did it, then you are wrong.

you should understand that even though, you had rightly resolved the elemental distances, you can't multiply them on that surface, because the surface is also bounded by y = 1-x (at z=0) which means in that plane, x is still dependent on y. If you had the snapshot below (Fig 1) where the elemental distance dy will not be dependent on x anymore on that surface, then it would have made sense.

So, it's easier to use vectors. Cos what you want to do is simultaneously apply the relationships given by the curves while still taking their dependences into consideration. In this way, you'd almost be resolving the surface itself dS and not the elemental distances.

Look at Fig2, where the same surface has been resolved onto the xy plane. On that xy plane, the last coordinate that is the projection has its last coordinate at (1,1,0). So, if you had an elemental area on that surface dS1, you can then obtain the angle between that area and the surface with its top coordinate (1,1,2). Note that that is a mirror image of the surface in question, but in a different quadrant.

So, the angle between the two surfaces will be
(-0.5i-0.5j).(-0.5i-0.5j-2k) = sqrt(0.5).sqrt(4.5).cos(phi)
cos(phi) = 1/3.

dS.cos(phi) = dS1
dS = dS1/cos(phi) = 3dS1

dS1 = dxdy

So, dS = 3dxdy.

This elemental area will be the same as the elemental area on the main surface (they are simply mirror images).

Note: you could have resolved to the other side (270-360deg), and you'd have the same answer

Edited.

Re: Nairaland Mathematics Clinic by naturalwaves: 3:00pm On Aug 02, 2015
Antoinne:

I suppose the way you were doing it was to resolve dx on one hand, and dy on the other to the lines at y=0 & x=0 respectively, so that you obtained dSx = sqrt(5)dx & dSy = sqrt(5)dy

And you then did, dS = dSx.dSy = 5dxdy. if this is how you did it, then you are wrong.

you should understand that even though, you had rightly resolved the elemental distances, you can't multiply then on that surface, because the surface is also bounded by y = 1-x (at z=0) which means in that plane, x is still dependent on y. If you had the snapshot below (Fig 1) where the elemental distance dy will not be dependent on x anymore on that surface, then it would have made sense.

So, it's easier to use vectors. Cos what you want to do is simultaneously apply the relationships given by the curves while still taking their dependences into consideration. In this way, you'd almost be resolving the surface itself dS and not the elemental distances.

Look at Fig2, where the same surface has been resolved onto the xy plane. On that xy plane, the last coordinate that will make the area the same is (2,2,0). So, if you had an elemental area on that surface dS1, you can then obtain the angle between that area and the surface with the coordinate (1,1,2). Note that that is a mirror image of the surface in question, but in a different quadrant.

So, the angle between the two surfaces will be
(-1.5i-1.5j).(-0.5i-0.5j+2k) = sqrt(4.5).sqrt(4.5).cos(phi)
cos(phi) = 1/3.

dS.cos(phi) = dS1
dS = dS1/cos(phi) = 3dS1

dS1 = dxdy

So, dS = 3dxdy.

This elemental area will be the same as the elemental area on the main surface (they are simply mirror images).

Note: you could have resolved to the other side (270-360deg), and you'd have the same answer
Nice analysis. Bravo!
Re: Nairaland Mathematics Clinic by Antoinne: 3:17pm On Aug 02, 2015
naturalwaves:

Nice analysis. Bravo!
Thanks

1 Like

Re: Nairaland Mathematics Clinic by agentofchange1(m): 3:18pm On Aug 02, 2015
My eyes are seeing things, Good job guys....

Happy sunday
Re: Nairaland Mathematics Clinic by agentofchange1(m): 4:58pm On Aug 02, 2015
possibilita:
Pls help me on these

Q 1a

(i) given y^2 =Ax^2 + Bx +c

We need three derivatives (since, we Gat 3unknowns)

=>2yy'=2Ax+B

Again

=>2(y')^2 +2yy" = 2A

Or (y') ^2 +yy"=A

Again,

=>2y"*y' +y"*y' +y*y"'

Or 3y' y"+yy"'=0

ii) y=Ax+A^2..... (*)

y'=A

Thus we have

Y= xy'+(y') ^2

Or y=xdy/dx +(dy /dx)^2


b (i)

pix below

1c) m I don't think its a DR except it's written like this

(2xy + x^2 )y' = 3y^2 + 2xy is that ?

1 Share

Re: Nairaland Mathematics Clinic by Laplacian(m): 6:43pm On Aug 02, 2015
jackpot:

Take f(x,y,z)=2x+2y+z-2=0.

dS= sqrt[(fx)2+(fy)2+1]dxdy=sqrt[4+4+1]dxdy=sqrt[9]dxdy=3dxdy.



Formula reference: www.math.lsa.umich.edu/~glarose/classes/calcIII/web/16_6/
ur method luks pretty, but d link didnt giv me d result...does it have name so i can google?
Re: Nairaland Mathematics Clinic by agentofchange1(m): 7:10pm On Aug 02, 2015
Q2 a i

****Solution****
First standardise by dividing through by '(x+1)' to be of
the form
Y' +py=q ..(where p & q are functions of x)
=> y' -y/(x+1) = e^x/(x+1) ......(*)
Integrating factor IF
= e^($pdx)
Evaluating $pdx
=> $-dx /(x+1) =-ln(x+1) =ln[1/(x+1)]
Thus IF= e^ln[1/(x+1)]
By definition, e^lna =a
Thus, IF=1/(x+1)
Now on multiplying IF by (*) , we get
D(y/(x+1)= e^x
Integrating both sides yields
Y/(x+1)= e^x+ C
Hence. Y=(x+1)(e^x + C)
Re: Nairaland Mathematics Clinic by possibilita(m): 7:42pm On Aug 02, 2015
agentofchange1:

Q 1a
(i) given y^2 =Ax^2 + Bx +c
We need three derivatives (since, we Gat 3unknowns)
=>2yy'=2Ax+B
Again
=>2(y')^2 +2yy" = 2A
Or (y') ^2 +yy"=A
Again,
=>2y"*y' +y"*y' +y*y"'
Or 3y' y"+yy"'=0
ii) y=Ax+A^2..... (*)
y'=A
Thus we have
Y= xy'+(y') ^2
Or y=xdy/dx +(dy /dx)^2

b (i)
pix below
1c) m I don't think its a DR except it's written like this
(2xy + x^2 )y' = 3y^2 + 2xy is that ?
yes that's hw 1c is written
Re: Nairaland Mathematics Clinic by Laplacian(m): 8:57pm On Aug 02, 2015
Antoinne:

I suppose the way you were doing it was to resolve dx on one hand, and dy on the other to the lines at y=0 & x=0 respectively, so that you obtained dSx = sqrt(5)dx & dSy = sqrt(5)dy

And you then did, dS = dSx.dSy = 5dxdy. if this is how you did it, then you are wrong.

you should understand that even though, you had rightly resolved the elemental distances, you can't multiply them on that surface, because the surface is also bounded by y = 1-x (at z=0) which means in that plane, x is still dependent on y. If you had the snapshot below (Fig 1) where the elemental distance dy will not be dependent on x anymore on that surface, then it would have made sense.

So, it's easier to use vectors. Cos what you want to do is simultaneously apply the relationships given by the curves while still taking their dependences into consideration. In this way, you'd almost be resolving the surface itself dS and not the elemental distances.

Look at Fig2, where the same surface has been resolved onto the xy plane. On that xy plane, the last coordinate that will make the area the same is (2,2,0). So, if you had an elemental area on that surface dS1, you can then obtain the angle between that area and the surface with the coordinate (1,1,2). Note that that is a mirror image of the surface in question, but in a different quadrant.

So, the angle between the two surfaces will be
(-1.5i-1.5j).(-0.5i-0.5j+2k) = sqrt(4.5).sqrt(4.5).cos(phi)
cos(phi) = 1/3.

dS.cos(phi) = dS1
dS = dS1/cos(phi) = 3dS1

dS1 = dxdy

So, dS = 3dxdy.

This elemental area will be the same as the elemental area on the main surface (they are simply mirror images).

Note: you could have resolved to the other side (270-360deg), and you'd have the same answer
i honestly appreciate ur effort sir! u explained exactly what i did, i know u have explained why it is not proper to resolve distances and obtain elemental area on the surface but i cant understand ur explanation on account of ur diagrams, i cant fit them together, pls snap a neat copy......i felt even if y depend on x at z=0, it should stil not affect d result because altering d original elementary area dS will also alter its projection on the surface...
Re: Nairaland Mathematics Clinic by jackpot(f): 9:43pm On Aug 02, 2015
Karmanaut:

Better- it's WolframAlpha.
Source: http://www.wolframalpha.com/input/?i=Find+the+inverse+laplace+transform+of+1%2F%28s%5E5%2Ba%5E5%29&x=7&y=11
do you trust that long answer? Lol. Even imaginary number entered into the result.
Re: Nairaland Mathematics Clinic by jackpot(f): 9:46pm On Aug 02, 2015
Laplacian:

ur method luks pretty, but d link didnt giv me d result...does it have name so i can google?
Try copy and paste everything in the link into the browser. Note that some words in the link weren't hyperlinked.
Re: Nairaland Mathematics Clinic by Antoinne: 9:47pm On Aug 02, 2015
Laplacian:

i honestly appreciate ur effort sir! u explained exactly what i did, i know u have explained why it is not proper to resolve distances and obtain elemental area on the surface but i cant understand ur explanation on account of ur diagrams, i cant fit them together, pls snap a neat copy......i felt even if y depend on x at z=0, it should stil not affect d result because altering d original elementary area dS will also alter its projection on the surface...
I took it you'd understand my little explanations.
Let's get the first part sorted. Look at the attached pic.

After resolving the elemental distances, dx & dy and multiplying. Follow the extension down to y = 1-x at z=0, you'll see that the distance dx on the other side is not exactly parallel to the distance you just resolved. I marked it with "A". There's still some kind of triangle there, showing that you elemental distance is not representative of the surface.

If that's too difficult, look at the basic 2D triangle below it. If you took a small portion on the x-axis called dx. Do you think it'll be equal to the length marked "A" on that same triangle merely by tracing it up?

It's the same thing happening on that surface above. You get it?

Re: Nairaland Mathematics Clinic by Nobody: 11:27pm On Aug 02, 2015
jackpot:
do you trust that long answer? Lol. Even imaginary number entered into the result.
Well two supercomputers can't be wrong, right?

Two supercomputers, just about 10,000 processor cores, hundreds of terabytes of disks, a heck of a lot of bandwidth, and what seems like enough air conditioning for the Sahara to host a ski resort.
One of our launch partners, R Systems, created the world’s 44th largest supercomputer (per the June 2008 TOP500 list). They call it the R Smarr. We call it the most powerful computer that will be running Wolfram|Alpha on launch day!
http://blog.wolframalpha.com/2009/05/12/the-computers-powering-computable-knowledge/

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