cigie:

Summation of (4*10^(n-1) + 4)
Abi?

.......not so correct......
from your equation, the sum of the first two terms is 44.....first three terms,404
but 4+44=48 and 4+44+444=492
so u see it's quite wrong
in fact,it is not even equal to the equation of the nth term.....
....the first term is 4 and making n=1 in your above equation gives 8........

........so my question again is what the sum of the nth term of the series below
........4+44+444+4444+44444+......+nth

........note that the formula for the nth term in a series is different from the formula for the sum of the nth term of a series.....

Math gurus pls solve dis 4 me.
Resolve to partial fractions. 2x^3-2x^2-2x-7/x^2-x-2.
Couldnt attempt it cos the numerator is more than the denominator. Shed more lite pls. Bless u.

topmostg: Math gurus pls solve dis 4 me.
Resolve to partial fractions. 2x^3-2x^2-2x-7/x^2-x-2.
Couldnt attempt it cos the numerator is more than the denominator. Shed more lite pls. Bless u.

........since the numerator is more than the denominator, u will have to go through the long division first.......
........To be shown sooooooooooooon

.......here it is.....

Krucifax: Okay here's one for the board. Solve below using equations of absolute value


a) | x | = 5

b) | x - 2 | = 7

c) | 89 x + 4 | = -2

d) | -11 x - 9 | = 0

e) 2 | x + 5 | = 10

f) 3 | -9 x - 7 | - 2 = 13

g) -5 | -x + 2 | + 8 = -12

h) | -3 x - 2 | / 3 = 7

i) | (x - 5) / 4 | = 6

j) | x + 1 | = |-4 x + 3|

k) | x + 3 | = - |-x - 3 |

Note: |x|=a then -a=x=a

A. |x|=5
-5=x or x=5

B. |x-2|=7
-7=x-2=7
-7=x-2 or x-2=7
-5=x or x=9

C. |89x+4|=-2
2=89x+4=-2
2=89x+4 or 89x+4=-2
X=-2/89 or x=-6/89
Absolute value not -ve
No solution

D. |-11x-9|=0
0=-11x-9=0
X=-9/11

E. 2|x+5|=10
Divide both sides by 2
|x+5|=5
-5=x+5=5
-5=x+5 or x+5=5
X=-10 or x=0

F. 3|-9x-7|-2=13
3|-9x-7|=15
Divide bothsides by 3
|-9x-7|=5
-5=-9x-7=5
-5=-9x-7 or -9x-7=5
X=-2/9 0r x=-12/9=-4/3

G. -5|-x+2|+8=-12
-5|-x+2|=-20
Divide both sides by -5
|-x+2|=4
-4=-x+2=4
-4=-x+2 or -x+2=4
X=6 or x=-2

H. |-3x-2|/3=7
Cross multiply
|-3x-2|=21
-21=-3x-2=21
-21=-3x-2 or -3x-2=21
X=19/3 or x= -23/3


I. |x-5|/4=6
Cross multiply
|x-5|=24
-24=x-5=24
-24=x-5 or x-5=24
X=-19 or x=29


J. |x+1|=|-4x+3|
-(-4x+3)=x+1=-4x+3
4x-3=x+1 or x+1=-4x+3
X=4/3 or x=2/5


K. |x+3|=-|-x-3|
-{-(-x-3)}=x+3=-(-x-3)
-x-3=x+3 or x+3=x+3
X=-3.

ositadima1:

No fall hand ooo, like moi. You have the greatest study aid this days, yeah, the INTERNET. Brother everything is there. Even if u cant afford all required school books, with the internet in ur hands ur are king.

Bros you 're right, I apologise, I was merely joking

@coolexy,maths is applied in all spheres of lyf ranging 4rm its industrial application to day to day.e.g in d construction of aircraft lots of equations on projectiles would av bn solved

At cryptic,It is finitely generated.A group is said to be finitely generated when there exist a^n=e. a is d generator and n is d order of d group.for a polynomial ring generated by a group dere will b a point where a^n will be equal to the identity element.

@solomon,As 4 me stat is simpler than maths.I do mts,csc and sts courses and stats is d simplest of d three.If u do groups and rings in maths and u start dealing with letters,u will knw hw complex maths is

its getting jangkles here
Thereshd be a central source to which we submit questions too
A set of questions are pasted and solved

Then another set
The way it is, not organised

awesome_benji: @OP especially, simultaneous eqn in 3unknowns.
Given that:
a1x + b1y + c1z + d1 = 0
a2x + b2y + c2z + d2 = 0
a3x + b3y + c3z + d3 = 0
find x, y & z and leave d answer in matrix form.
Please, please & please, help me 2 solve dis as d lecturer has promised 2 bring it in d exam. Thanks.

a1x + b1y + c1z + d1 = 0
a2x + b2y + c2z + d2 =0
a3x + b3y + c3z + d3 = 0
SOLUTION
Note that the terms a, b, and c are coefficients while the ‘d’ terms are all constants
First we shall find the determinant of the coefficients arranged in matrix form lDl
a1 b1 c1
a2 b2 c2
a3 b3 c3
lDl= a1[ b2c3 – b3c2] – b1[ a2c3 – a3c2] + c1[a2b3 – a3b2]
secondly we shall replace column of a, b, and c with d then obtain the corresponding determinants.
d1 b1 c1
d2 b2 c2
d3 b3 c3
lAl = d1[ b2c3 – b3c2] – b1[ d2c3 – d3c2] + c1[ d2b3 – d3b2]
a1 d1 c1
a2 d2 c2
a3 d3 c3
lBl = a1[ d2c3 – d3c2] – d1[ a2c3 – a3c2] + c1[ a2d3 – a3d2]
a1 b1 d1
a2 b2 d2
a3 b3 d3
lCl = a1[ b2d3 – b3d2] – b1[ a2d3 – a3d2] + d1[ a2b3 – a3b2]
now the values of x, y and z can be gotten thus,
X = lAl/ lDl
Y = lBl/ lDl
Z = lCl/ lDl
You can make a check by inserting values for a, b, c and d in the question

^^^ cool

eddiegracia: Prove that 7/2 = 5... In whatever way u can or even use any formular u can...

7/2 = 3.5

Ok am not a mathematicians or infact am on d opposite side...ART not english though.but I must say this is one of the most matured trend and discussion I hav seen on nairaland in a while..'Udos

Great thread here. I'm a topologist. Mathematics is the sluggish, little brother of topology. I'm offering myself to the house. If time permits, I'd be here to answer some of the problems.

Keep up the good work guys

Richiez:

a1x + b1y + c1z + d1 = 0
a2x + b2y + c2z + d2 =0
a3x + b3y + c3z + d3 = 0
SOLUTION
Note that the terms a, b, and c are coefficients while the ‘d’ terms are all constants
First we shall find the determinant of the coefficients arranged in matrix form lDl
a1 b1 c1
a2 b2 c2
a3 b3 c3
lDl= a1[ b2c3 – b3c2] – b1[ a2c3 – a3c2] + c1[a2b3 – a3b2]
secondly we shall replace column of a, b, and c with d then obtain the corresponding determinants.
d1 b1 c1
d2 b2 c2
d3 b3 c3
lAl = d1[ b2c3 – b3c2] – b1[ d2c3 – d3c2] + c1[ d2b3 – d3b2]
a1 d1 c1
a2 d2 c2
a3 d3 c3
lBl = a1[ d2c3 – d3c2] – d1[ a2c3 – a3c2] + c1[ a2d3 – a3d2]
a1 b1 d1
a2 b2 d2
a3 b3 d3
lCl = a1[ b2d3 – b3d2] – b1[ a2d3 – a3d2] + d1[ a2b3 – a3b2]
now the values of x, y and z can be gotten thus,
X = lAl/ lDl
Y = lBl/ lDl
Z = lCl/ lDl
You can make a check by inserting values for a, b, c and d in the question

Your used the determinant method but I think he want the Guassian elimination method which Ositadima1 tried to solve earlier by replacing the coefficient of x,y and z i.e(a,b and c) with numbers.

Someone directed me to this thread.

https://www.nairaland.com/1151301/experienced-mathematics-teachers-good-command

Of course, it's not for teh big boys, as I'm seeing my O&G bigs boys in this thread already.

But hey, somebody else in need may be reading.

solomon111: I dread statistics.
That is one area of mathematics,i have not mastered.
As far as am concerned,statisticians are the real mathematicians.

R U B B I S H ! ! ! o

2good:

Your used the determinant method but I think he want the Guassian elimination method which Ositadima1 tried to solve earlier by replacing the coefficient of x,y and z i.e(a,b and c) with numbers.

I don't think he'll ask you to use Gaussian elimination method since we aren't working with numbers here, there is no way you can multiply a row by a letter and subtract it from another row, it might be too ambiguous.

jackpot: R U B B I S H ! ! ! o

U come here to fight, hmmm?
***scratching his 2-days old stubble*** I need to shave.

Richiez:

I don't think he'll ask you to use Gaussian elimination method since we aren't working with numbers here, there is no way you can multiply a row by a letter and subtract it from another row, it might be too ambiguous.

you make me remember a formulae I use for guassian rule.

olawalebabs: you make me remember a formulae I use for guassian rule.

please can you share it with us

Richiez:

I don't think he'll ask you to use Gaussian elimination method since we aren't working with numbers here, there is no way you can multiply a row by a letter and subtract it from another row, it might be too ambiguous.

You can actually solve it in terms of a, b and c using Guassian elimination method. Its as straight forward as you'll do when you use number. To eliminate a2 from row 2 for example, multiply row 1 by a2 and multiply row 2 by a1 and subtract row 1 from row 2, to get a new row 2. In that way you can make element(2,1) of the matrix zero. Doing similar manipulation with other rows will help you successfully eliminate {(3,1) and (3,2)}

pseudonomer:

Is correct, all he did is take log of both sides.

If you take the log of both sides, the laws of log has to be
Considered; thus the addition "to" multiplication.
But, I guess u guys are over that now.

solomon111: I dread statistics.
That is one area of mathematics,i have not mastered.
As far as am concerned,statisticians are the real mathematicians.

$
I understand why u say so, but its wrong. As engineers, we had so many
Maths courses, but non relating to statistics. Staticians have their
Own brand of maths, it makes them unique but it doesn't mean they
Are the 'real mathematicians'

eddiegracia: Prove that 7/2 = 5... In whatever way u can or even use any formular u can...

hehehe...
Well, to solve this the knowledge of roman numerals is required. Converting the numbers to roman numerals we have
7 = vii(roman numerals)
2= ii; and 5 = v.
7/2 = 5; vii/ii
ii will cancel out ii.
Hence as we are left with v which means 5, therefore 7/2 = vii/ii = v.

coolexy2: ](1)what is the real application of mathematics in our live?

(2) prove that:
a. Sin^3 x = 1/4(3sinx - sin3x)
b. Find 2nd derivative of:
y = cos^3 x

I'll take question 2 (a) for now.

Since: sin (A + B) = sin A cos B + cos A sin B

sin 2x = sin x cos x + cos x sin x
sin 2x= 2 sin x cos x

Likewise, cos (A+B) = cos A cos B - sin A sin B
cos 2x = cos^2 x - sin^2 x
Also since, sin^2 x + cos^2 x = 1, cos^2 x = 1 - sin ^2 x
Substituting in the above,
cos 2x becomes = 2 cos^2 x - 1
= 1 - 2 sin^2 x

Having established that:

Sin 2x = 2sin x cos x and
cos 2x = 1-2sin^2 x

sin 3x can now be expressed as follows :
= sin (2x + x)
= sin 2x. cos x + cos 2x. sin x
= 2 sin x cos x. cos x + (1 - 2 sin^2 x) . sin x
= 2 sin x cos x. cos x + sin x - 2 sin^3 x
= 2 sin x cos^2 x + sin x - 2 sin^3 x
= 2 sin x (1 - sin^2 x) + sin x - 2 sin^3 x
= 2 sin x - 2 sin^3 x + sin x - 2 sin^3 x
Hence, sin 3x = 3 sin x - 4 sin^3 x

Back to your question:

Prove that sin^3 x = 1/4( 3 sin x - sin 3x )

Taking from the right hand side and substituting the expression for sin 3x above,

We have 1/4 [ 3sin x - ( 3 sin x - 4 sin^3 x ) ]
= 1/4 [3sin x - 3sin x + 4sin^3 x ]
= sin^3 x

Thus,
sin^3 x = 1/4 ( 3sin x - sin 3x )

olasesi:

.......not so correct......
from your equation, the sum of the first two terms is 44.....first three terms,404
but 4+44=48 and 4+44+444=492
so u see it's quite wrong
in fact,it is not even equal to the equation of the nth term.....
....the first term is 4 and making n=1 in your above equation gives 8........

........so my question again is what the sum of the nth term of the series below
........4+44+444+4444+44444+......+nth

........note that the formula for the nth term in a series is different from the formula for the sum of the nth term of a series.....

Guy u made me sweat, but finally I got it

First, nth term

Tn=4(1-10^n)/(1-10)

Second, sum of the nth terms

Sn=4[10(1-10^n)/(1-10)-n]/(1-10)

U can simplify this, i didn't bother cos it will still look scary.

Do u want to c my workings?
NB: I hope u like bodmas, I am saying follow the rules of precedence, bracket before division, division before minus blah blah

I can't remember d last time I sat down, to go through a 10-paged thread, reading every dam.n post; I just did dat nw.
It is Official----> Richiez and Ositadima are d Bomb on here! They deserve credits. Thumbs up brahs!