Doing a great job here,I must confess.I like maths though i'm not too good at applying calculus (lagrange,laplace,etc to solving real life problems but I want to learn o.And I hope this thread would help,so I'm *following*.
@jackport:I raise my hat.Welldone ma'am.
Thanks to everybody too.

Alpha Maximus: ......Ehen, you dodge am abi!!!Sharp guy!! Alright , I'll search for more misleading questions.....might even make up a few myself,that have a reasonable solution of course!!! Thanks for the attempt

Welcome bro,solve this. x+y=5-----------(i)
x^x+y^x=13---(ii)

jackpot:
*feels so happy*
Alright dear.

I dey expect your solution ooo

**rolls my mat on the floor of the thread and lies down on it, and with phone in my hand, refreshing the page**

Cool, seems no one is tryna' do something about your question....I will post my own solution today anyway! My eyes dey see competition here

jackpot:

Given that x,y,z depends on three variables, viz,
x=x(u,v,w), y=y(u,v,w), z=z(u,v,w),
we know that the three transformed equations
u=u(x,y,z), v=v(x,y,z), w=w(x,y,z) is non-degenerate if the Jacobian of the transformation doesn't vanish.

Now, consider the equation
x=x(u,v,w).
Differentiating both sides partially w.r.t. x, we have
1=Xu Ux+ Xv Vx+ Xw Wx

so, the relationship is

Xu*Ux=1-(Xv Vx + Xw Wx)

So, ma'am your question is a crazy one....My cousin & I had to share ideas to bring this up! I'm uploading in .PDF format => It's hand written & 5 pages any way . I'll take out my time to type it out for the benefit of other viewers. I was rolling in circles in the past days, but after sharing ideas with my Cousin, I finally got it! Here you go=>


Opss the size! I gotta look for a way to reduce them....


Loading....

@ lanrexlan questn.
x+y=5... (1) nd x^x+y^x=13...(2)
Solutn.
frm eqn 1. x=5-y. Put x=5-y in eqn 2. We av
(5-y)^(5-y)+y^(5-y)=13... eqn * Nw we av reduced d problem to what value of y that must be put in eqn * to obtain 13. Hmmm. A thorough look at eqn 1 nd eqn 2 show dat d values of x nd y must b positive integers within the range 0<x<5 nd 0<y<5.
clearly, y=3. i.e
(5-3)^(5-3)+3^(5-3)=13. Nw dat we av y=3. It is more convenient to get x.
recal dat x=5-y. x=5-3. x=2.
hence d soln is x=2 nd y=3. All is well. @ richiez u knw wat i mean.

rhydex 247: @ lanrexlan questn.
x+y=5... (1) nd x^x+y^x=13...(2)
Solutn.
frm eqn 1. x=5-y. Put x=5-y in eqn 2. We av
(5-y)^(5-y)+y^(5-y)=13... eqn * Nw we av reduced d problem to what value of y that must be put in eqn * to obtain 13. Hmmm. A thorough look at eqn 1 nd eqn 2 show dat d values of x nd y must b positive integers within the range 0<x<5 nd 0<y<5.
clearly, y=3. i.e
(5-3)^(5-3)+3^(5-3)=13. Nw dat we av y=3. It is more convenient to get x.
recal dat x=5-y. x=5-3. x=2.
hence d soln is x=2 nd y=3. All is well. @ richiez u knw wat i mean.

Fabulous,you are very correct.Many Mathematics Gurus here o,more coming though.Thanks.

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So, @ jackpot, I hope you get it now. I'll type it out with explanation if the uploaded is not clear enough. 1luv

yeah. I got it clearly. Your solution is excellent, compact and you used a totally different approach from mine. Even though your solution looks short, you lifted some things like the formula for the scale factors(h1, h2, h3) and that of the unit vectors Tu, Tv, Tphi. I bet if you included the derivations of those formulas, your solution would have clocked more pages

welldone, DoubleDx. Your solution is cool.

@DoubleDx, check Page 81. I posted my solution on that question. I used a totally different elementary approach. You will see the upload on my signature. It is a pdf file.

Download it. I need your comment on the solution.

@ laplacian, alpha, d citizen, double dx, jackpot nd other maths general i'm still waiting for the solution to my first question.

rhydex 247: @ laplacian, alpha, d citizen, double dx, jackpot nd other maths general i'm still waiting for the solution to my first question.

.....the one containing something like V=IR^2?

rhydex 247: @ laplacian, alpha, d citizen, double dx, jackpot nd other maths general i'm still waiting for the solution to my first question.

.....the one containing something like V=IR^2?

yea @ alpha miximus

rhydex 247: yea @ alpha miximus

I believe its been solved already by Laplacian

he only solved the no 2 question

rhydex 247: he only solved the no 2 question

on which page is the question?...just new here...@laplacian
and co, u guys re great!

here is the question again
Show that V=IR^2 is not a vector space over IR with respect to operation (a,b)+(c,d)=(a+c, b+d) and k(a,b)=(ka,kb)

jackpot: yeah. I got it clearly. Your solution is excellent, compact and you used a totally different approach from mine. Even though your solution looks short, you lifted some things like the formula for the scale factors(h1, h2, h3) and that of the unit vectors Tu, Tv, Tphi. I bet if you included the derivations of those formulas, your solution would have clocked more pages

welldone, DoubleDx. Your solution is cool.

It would have been more than 10 pages . I went with the Cartesian method all the way. Thanks

I love this trend

rhydex 247: here is the question again
Show that V=IR^2 is not a vector space over IR with respect to operation (a,b)+(c,d)=(a+c, b+d) and k(a,b)=(ka,kb)

If k is a real number, then the plane IR² is a vector space over the scalar field IR with respect to addition and scalar multiplication. IR² is then said to be a real vector space since the underlying scalar field is IR, the set of real numbers.

This question doesn't make sense. It is like asking you to prove that black is equal to white. There's nothing to prove really. Tear that question or use eraser wipe the "not" in the question.

Alpha Maximus: ....No offence, but I have to say the answer was kind of obvious! Though the 'trial and error' method worked, it becomes impractical when a similar question has much higher values of 'x' and 'y'.....thus, it would be better to stick to strictly analytical , non-graphical and algebraic processes AKA the HARD NONSENSE and more directquasi-direct way....
This is my own solution:
X+y=5, x^x+y^X=13
From(I), x=5-y(3), input (3) into (2)
(5-y)^(5-y) +y^(5-y)=13
Now converting the rather tedious equation into function form, we have:
F(y)=(5-y)^(5-y) + y^(5-y)-13=0
Using the remainder theorem, we have that y=3
Since the use of the long division method would still leave the index/power of (5-y) unknown, I will substitute y=3 strictly into the index and convert the equation back to algebraic form giving:
(5-y)^(5-3)+y^(5-3)-13=0
(5-y)^2+y^2-13=0
Expanding and collecting like terms, we have a quadratic expession of:
2y^2-10y+12=0
Factorising we have:
(2y-4)(y-3)=0
Thus,y=3(already established) and,
2y=4, y=4/2=2(this is also the value of 'x')
Therefore, x=2, y=3
Though this is more tedious than your method, it is more practical when larger values are involved

The bolded is just cock-and-bull.

First of all, what does the remainder theorem say?

Once a polynomial f(x) is divided by x-a, the remainder is another polynomial of 1 degree less than that of f(x) together with a remainder R still to be divided by x-a.

i.e., f(x)/(x-a)=g(x)+R/(x-a)
therefore f(x)=(x-a)g(x)+R
substituting x=a yields that
the remainder R=f(a)

back to the question. What is a polynomial? I would leave that for you to find out.

The equation x^x+y^y=13 is not a polynomial.

Why then are you applying the remainder theorem which specifically talks about polynomials?

And the funny thing is that you said the application yields y=3. SMH

Rubbish. Bros, no just dey form anyhow, you are still a BIG learner.

Converts his answer sheet to toilet paper

Find x and y
x^1/3 + y^1/3 = 6..........(i)
X + y = 72...................(ii)

jackpot: Here comes the full solution in PDF format.


@Readers, pls download it. I need your reviews

Perfecto ma'am...nice one!

jackpot: The bolded is just cock-and-bull.

First of all, what does the remainder theorem say?

Once a polynomial f(x) is divided by x-a, the remainder is another polynomial of 1 degree less than that of f(x) together with a remainder R still to be divided by x-a.

i.e., f(x)/(x-a)=g(x)+R/(x-a)
therefore f(x)=(x-a)g(x)+R
substituting x=a yields that
the remainder R=f(a)

back to the question. What is a polynomial? I would leave that for you to find out.

The equation x^x+y^y=13 is not a polynomial.

Why then are you applying the remainder theorem which specifically talks about polynomials?

And the funny thing is that you said the application yields y=3. SMH

Rubbish. Bros, no just dey form anyhow, you are still a BIG learner.

Converts his answer sheet to toilet paper

...@Jackpot...fear no allow me talk ooo...

@jackpot, u use lagrange multipler to solve the problem posted by laplacian

Let me use simple differentiation to solve it

Let p = xy

Sum of the digit,

X+ y = 12
Y= 12-x

P= x(12-x)
P=12x -x^2

dp/dx=12 -2x
d^2p/dx^2 = -2

It implies dat the function is maximum

for thr number to be maximum, dp/dx=0
2x=12
X=6

thus, y=6

The number=10x+y
=10(6)+6= 66