Good work guys. Inspiring, motivating & educative thread. Keep up the good work. I will post some questions later!

1 a. Solve the simultaneous equation: x + 2y = 6 and y√x - y = 1

b. And find the value(s) of (x - 2y)

1 a. Solve the simultaneous equation: x + 2y = 6 and y√x - y = 1

b. And find the value(s) of (x - 2y)

y(√x-1) = 1

First of all GO DOWN LOW

set p = √x
p2 = x

y(p2 - 1 ) = 1
y = 1/(p2-1)

2y = 6-x = 6-p2

thus substituting

2y = 6 - p2 = 2/(p2-1)

6 - p2 = 2/(p2-1)

expanding
6p-6-p3+p2 = 2

Arranging
p3 -p2-6p+8= 0

Using remainder theorem and polynomial division it yields

(p-2)(p2+p-4) =0
p =2 the second equation comprises surds -> useless

x = 2^2 = 4 hence y =1

(x- 2y) = 4-2 =2

For compact 2-dimensional surfaces without boundary, if every loop can be continuously tightened to a point, then the surface is topologically homeomorphic to a 2-sphere (usually just called a sphere). Is thesame true for 3 dimensionals surface? Pls any guru here?ans d question

Ogbeozioma: For compact 2-dimensional surfaces without boundary, if every loop can be continuously tightened to a point, then the surface is topologically homeomorphic to a 2-sphere (usually just called a sphere). Is thesame true for 3 dimensionals surface? Pls any guru here?ans d question

And how is poincare conjecture supposed to help folks here? What are you trying to prove eh? #smh#

http://en.wikipedia.org/wiki/Poincaré_conjecture

Solution: http://en.wikipedia.org/wiki/Grigori_Perelman

^^^^ thank u o we said maths not quantum theory

mschewwwww

biolabee:


First of all GO DOWN LOW

set p = √x
p2 = x

y(p2 - 1 ) = 1
y = 1/(p2-1)

2y = 6-x = 6-p2

thus substituting

2y = 6 - p2 = 2/(p2-1)

6 - p2 = 2/(p2-1)

expanding
6p-6-p3+p2 = 2

Arranging
p3 -p2-6p+8= 0

Using remainder theorem and polynomial division it yields

(p-2)(p2+p-4) =0
p =2 the second equation comprises surds -> useless

x = 2^2 = 4 hence y =1

(x- 2y) = 4-2 =2

Ok. Thanks bro

Ghettoguru: 1 a. Solve the simultaneous equation: x + 2y = 6 and y√x - y = 1

b. And find the value(s) of (x - 2y)

@biolabee, nice work! Lemme add more to what you did.

@ Ghettoguru,

Question 1(a)
x + 2y = 6 ....equation 1
y√x - y = 1.....equation 2

From equation 1

x = 6 - 2y ....equation 3

From equation 2,

y√x = y + 1
squaring both sides to get rid of the square root,

(y√x)^2 = (y + 1)^2
xy^2 = y^2 + 2y + 1 .....equation 4
Subtituting equation 3 in 4,

(6 - 2y)y^2 = y^2 + 2y + 1
6y^2 - 2y^3 = y^2 + 2y + 1
2y^3 + y^2 - 6y^2 + 2y + 1 = 0
2y^3 - 5y^2 + 2y + 1 = 0 ... equation 5

If we assume that y = 1 and substitute in equation 5, the result will be 0, implying that y - 1 is a factor of equation 5.

Using polynomial remainder theorem and division on equation 5 with its factor (y -1)

= [2y^3 - 5y^2 + 2y + 1] / [y - 1]
= (y - 1)(2y^2 - 3y - 1) = 0
:. y - 1 = 0 or 2y^2 - 3y - 1 = 0
y1= 1, solving 2y^2 - 3y - 1 = 0 for y2 and y3,

2y^2 - 3y - 1 = 0
using quadratic formula, a = 2, b = -3 and c = -1
y = - b ±√(b^2 - 4ac)/2a
y = - ( -3) ±√ (-3)^2 - 4.(2)(-1)/2(2)
y = 3 ±√17/4

:. y2= 3+ √17/4 and y3 = 3 - √17/4

Hence the values of y are : y1 =1, y2= 3+ √17/4 and y3 = 3 - √17/4

Substituting each value of y in equation 3,

x = 6 - 2y ..... equation 3
x1 = 6 - 2(1)
x1 = 4

x2 = 6 - 2 (3+ √17)/4
x2 = (12 -3 - √17)/2
x2 = (9 - √17)/2

x3 = 6 - 2( 3 - √17)/4
x3 = (12 - 3 + √17)/2
x3 = (9 + √17)/2

Hence the values of x are: x1 = 4, x2 = (9 - √17)/2 and x3 = (9 + √17)/2

Question 1(b)

Having calcaluted the values of x and y. We can now substitute each value of x and y to evaluate (x - 2y).

Substituting x1 = 4 and y1 = 1 in (x1 - 2y1)
= 4 - 2(1)
= 2

Substituting x2 = and y2 = in (x2 - 2y2)
= (9 - √17)/2 - 2(3+ √17)/4
= (9 - √17 - 3 - √17)/2
= 6/2
= 3

Substituting x3 = and y3 = in (x3 - 2y3)
= (9 + √17)/2 - 2(3 - √17)/4
= (9 + √17 - 3 + √17)/2
= (6 + 2√17)/2
= 2(3 + √17)/2
= 3 + √17

Hence the values of (x - 2y) are 2, 3 and 3 + √17

biolabee:


First of all GO DOWN LOW

Yeah, GO DOWN LOW, hahaha

@doubledx, thanks.

Yes Sir

^^^ do u do excel

Nice one doubledx
That surd tin just vex me so I no bother
Nice one again any more questions I feel pumped up

^Yes, I figured that bruv! The surd thing is difficult to type esp on a mobile.

Whao ! Dis a d kind of trend i lik... U guyz r rellin doin gr8 job.. Pls i nid solution on dis question : 3^n+1 + 6.3^n-2 - 3^n = 3 find n.

ositadima1: ^^^ do u do excel

No, I just use it minimally. I have templates for everything I need to do

math-diva:

No, I just use it minimally. I have templates for everything I need to do

U designed this templates?

ositadima1:


U designed this templates?

No I don't

lanre074: Whao ! Dis a d kind of trend i lik... U guyz r rellin doin gr8 job.. Pls i nid solution on dis question : 3^n+1 + 6.3^n-2 - 3^n = 3 find n.

3^n +1 + 6.3^n - 2 - 3^n = 3 . There are no brackets to show that: eg. 3 is to the power of (n+1). But I believe your question is the same as the one below? :

3^(n +1) + 6.3^(n - 2) - 3^n = 3

If it is then we can proceed by applying the law of indices:

3^n.3^1 + 6.3^n/3^2 - 3^n = 3
Lets put 3^n = x, substituting yields:

(x).3 + 2.(x)/3 - x = 3
[3x.3 + 2x - (x).3]/3 = 3
9x + 2x - 3x = 3.(3)
8x = 9
x = 9/8

Since 3^n = x
Then 3^n = 9/8
Taking log of both sides gives :

Log3^n = Log 9/8
nLog3 = Log9 - Log8
n = (Log3^2 - Log8 )/log3
n = 2Log3/Log3 - (Log8 )/Log3
n = 2 - Log8 ₃
n = 2 - Log2^3 ₃
n = 2 - 3Log2 ₃

Nice one
i used p = 3^n

arrived at 8/3p = 3

p = 9/8 (advanced log) I no get that kain time

I stopped there
This guys r just trying to fry people brains

^ lol yeah, you are right!

doubleDx:


3^n +1 + 6.3^n - 2 - 3^n = 3 . There are no brackets to show that: eg. 3 is to the power of (n+1). But I believe your question is the same as the one below? :

3^(n +1) + 6.3^(n - 2) - 3^n = 3

If it is then we can proceed by applying the law of indices:

3^n.3^1 + 6.3^n/3^2 - 3^n = 3
Lets put 3^n = x, substituting yields:

(x).3 + 2.(x)/3 - x = 3
[3x.3 + 2x - (x).3]/3 = 3
9x + 2x - 3x = 3.(3)
8x = 9
x = 9/8

Since 3^n = x
Then 3^n = 9/8
Taking log of both sides gives :

Log3^n = Log 9/8
nLog3 = Log9 - Log8
n = (Log3^2 - Log8 )/log3
n = 2Log3/Log3 - (Log8 )/Log3
n = 2 - Log8 ₃
n = 2 - Log2^3 ₃
n = 2 - 3Log2 ₃

dre s no any bracket in it n d options r a. -2 b. -1 c. 1 d. 3/8 pls i nid ur elp cos dis s a kind of question I'll b facing soon in d exam hall n av tried alot buh i couldn't get it n it's just 45sec per each question...! Futminna bad !

CHOI.....
See as pple dey destroy maths like say na Beanz ..
I hail d badt guyz in here ...
BRAVO!

1.Solve d simultaneous equationz usin the system of Matrix Inversion ....
10x1+3x2+6x3=76
4x1 + 5x3y=41
5x1+2x2 + 2x3 =5


2.
a renewable natural resources R will allow an estimated maximum consumtion rate of 200 million units per annum. Current annual usage is 6.5 million units. If d annual level of usage grows continually at an annual rate of 7.5%,will there be sufficient R to satisfy annual demand after

a)5 years b)10 years c)15 years and d)20 years ?

.

lanre074: dre s no any bracket in it n d options r a. -2 b. -1 c. 1 d. 3/8 pls i nid ur elp cos dis s a kind of question I'll b facing soon in d exam hall n av tried alot buh i couldn't get it n it's just 45sec per each question...! Futminna bad !

If the options are 1, -2, -1, 3/8 then I'm afraid something must be wrong with the question. It's either you did not copy it correctly or there may be some kind of tpyo-errors in the question (If it's a past question). I tried checking by substituting 'n' with each of the options given but none equals 3.

Try copying the question properly or write with a pen, snap or scan and upload!

Goodboy2012: 1.Solve d simultaneous equationz usin the system of Matrix Inversion ....
10x1+3x2+6x3=76
4x1 + 5x3y=41
5x1+2x2 + 2x3 =5


2.
a renewable natural resources R will allow an estimated maximum consumtion rate of 200 million units per annum. Current annual usage is 6.5 million units. If d annual level of usage grows continually at an annual rate of 7.5%,will there be sufficient R to satisfy annual demand after

a)5 years b)10 years c)15 years and d)20 years ?

solve dis maths gurus

doubleDx:

If the options are 1, -2, -1, 3/8 then I'm afraid something must be wrong with the question. It's either you did not copy it correctly or there may be some kind of tpyo-errors in the question (If it's a past question). I tried checking by substituting 'n' with each of the option given but none of the result equals 3.

Try copying the question properly or write with a pen, snap or scan and upload!

yeah non of d options s right... Mayb dre s a mistake wit d options.

Goodboy2012:

solve dis maths gurus

...plz check d equation 2 of d first question and ensure it is correct...mainwhile 4 question 2, since rate increase annualy it can be computed compoundly using p(1+R)^n i.e..6.5(1+7.5/100)^n...where n=no of years(5,...etc)and d respective ans are use 2 compare d maximum(200)..if values are greater than it implies R won't b suficient 4 dat no of yrs...just a tip sha

OMG i tot i know maths, mayb its because 'm stl in year1. 'm jst speechles dey way my felow black man is demistifyin maths. Wil jst sit back n add more complex maths knowledge 2 my archives in brain.

odun joy:

A,B.C will use 5mins

.........how did Ɣ☺u get it