₦airaland Forum

aman2:
no. u gat it wrong. u never told us abt d Newton Gregory that came abt d Simpson. u never told us abt the integration stuff in Simpson. blablabla

I was trying to be specific ,

your question was on trapezoidal vs Simpsons ..

that's it.

I know all formulas , just didn't see it necessary to start typing them.

abi any problem with that ?.

Wow!!!.good memory.....

4me
#Banana gum sweet pass ...sha...

smurfy:
Awesome! Missed you o.

yea. same here , can I have your digit .?

smurfy:
Awesome! Missed you o.

yea. same here , can I have your digit .?

timonski:
All the mathemagicians, help me integrate x^x dx

guy, no analytical solution such yet.

browse about sophomore dream function . guess could help.

smurfy:
Are you benbuks?

sure man , its me.

aman2:
just prove n let learn cos we here to learn. nd again. d axioms n d application. y d diff BTW simpson1/3 n trapezodia rule

the proofs are quite cheep sha , but kinda long (though I could summarize.)

Simpsons rule gives a better approximation over trapezoidal

we use S-1/3 ...for ....Even n

we use. S-3/8 . for n being multiple of 3

( n= data point )

we use trapezoidal rule for close step size

but they're cases where n is neither even nor multiple of 3 ( I.e prime n ) , for such case , we combine S -1/3 & S-3/8

(step size , s= (x-x0)/h

here

S-1/3. => h/3[ £wf*f(x)]

S-3/8 => 3h/8[£wt*f(x)]

T.R= h/2 [£wt.*f(x)]

£= sigma notation for summation

where h= interval ; wt.= wight factor ( for S-1/3 => 1 4 2 4 2 4 2 . . .1 ,

for S-3/8 => 1 3 3 2 3 3 2. . .1,

for T.R..=> 1 2 2 2 2 . . .1 )

f(x)= given function .


guess that's it.

[quote author=smurfy post=35492061][/quote]hey dude , the prof. himself . its me "benbuks" .
you wanted to contribute something. ..its been a while howdy .?

toobby:
Please help the first term of arithmetic progression is 3 and d fifth term is 9 calculate d number of term if d sum is 8

given a= 3 , a+4d=9

=> 3+4d=9

=>4d=6
d=6/4=3/2


now the series goes thus

first term. , a=3
second term, a+d=> 3 +3/2 =(6+3)/2 = 9/2
third term , a+2d , => 3 +2(3/2) = 3+3=6
fourth term , a+3d , =>3+3(3/2) =(6+9)/2 = 15/2
firth term , a+4d , => 3+4(3/2) =3+6 =9

Now.

though my boss above have done Justice so far.


but re-check the question from " if the sum is 8"

are you sure its 8?

jozeph1:
Rehdeokrafee

radiography .


m41hrim4r1c35...

hey guys. ...try out

compute the Laplace transform
given f(t)= tan(kt )

tobillionaire:
tanx boss

Uwc man.

dejt4u:
m coming bk to solve some of ur Questions..

Pending the time.. There is a mistake in ur Question one.. It shld be like this
Simplify (x² - 1)/(x³ -2x² - x + 2)

yea man you're right ....


here ------$01u10n from down

*)
Given |2x-3|=13

which => +/- (2x-3)=13

i.e 2x-3=13 , => 2x=16. => x=8

or -(2x-3)=13 , => -2x+3=13 , => -2x=10 , => x=-5

hence we have (-5 , 8 ) as solution set .

*)
y= 3x^2 -x^2 +x

we compute integral for 1<x<3

=> $(3x^2 -2x+1)dx = x^3 -x^2 +x for 1<x<3
now for F(x)=x^2 -x^2 +x

we compute F(b)-F(a) (where a=1 , b=3)

so that

=> [(3^3 -3^2 +3)-(1^3 -1^2 +1)] = 21-1= 20

*)
dy/dx = 4x-3 , computing 'y' at (2,5) , we take integral on both sides

=> y=$(4x-3)dx = 4x^2 -3x +K

y=2x^2 -3x +k , now at (2,5) .

for x=2. , y=5

=> 5=2(2)^2 -3(2) +k
=> 5=2+k , thus k= 3

hence we have. y= 2x^2 -3x +3

*)
For two perpendicular lines ,

m1*m2=-1
equation of PQ = x-2y+4=0

or y= x/2 +2 , so that m1 =1/2

=> m2/2=-1. or m2=-2

now , by definition, m=(y-y1)/(x-x2)

or y-y1=m(x-x1)

Given PQ's coordinate as (1 ,-1) ..I.e x1=1 , y1=-1

thus we have

y-(-1)= -2(x-1)

or y+1=-2x+2

or 2x+y-1=0

*)

Given two angles 44° & 112°

known that sum of angles in a triangle gives 180°

thus let's call the third angle @

=> @+44+112=180
=> @= 180-156=24°

since its a bisector , => 24/2=12 °

*)
by testing a.b=a+b-ab. . is commutative .

*) option D

(wait for solution from my boss.)

ikoyila:


oooohk! nd am a gurl not a bro.
so ur answer is a single fraction, i am truely learning, like criusli..

oh..sorry , its OK my dear ...

happy learning/solving. . .


#1luv.

ikoyila:
clearify all dis ur calculations cos its not d same wit my final answer " (x-y)^2/3" OR "√(x-y)^3". i believe u trying to compare my answer wit numerals. d expression u usin 2 compare is not my answer nd thus wrong.


ur answer: x^2/3 +(xy)^1/3 +y^2/3 i dnt tink it answers d question wch says express as a "single fraction". cheeerss..

hmm..haba my bro, this isn't something we should be arguing on.

every mathematical expression must be testable & hold for all basic mathematical axioms .


A fraction is the ratio of at least two real numbers (say a/b. , for b=/=0 )

if your answer is correct , irrespective of the number we input to test , it'll still hold .

Now, my answer is still a single fraction , (since x =x/1 )

check: let's still use x=27 & y= 8 ( I used them , because they're perfect cubes , you could use others )

I got x^2/3 + (xy)^1/3 +y^2/3 which we can write as

[ (x^1/3 )^2 + x^1/3 * y^1/3 +(y^1/3) ^2 ]/1

for x=27 , y=8

=> (27^1/3)^2 +27^1/3 * 8^1/3 + (8^1/3)^2

=3^2 +(3*2 ) +2^2

= 9+6+4=19

now the original expression was (x-y)/(x^1/3. -y^1/3 )

thus =( 27 - 8. ) / (27^1/3. - 8^1/3 )

= 19/(3-2)=19 /1 =19 .


hence , my result holds for all x,y € Z

Q.E D .


We are all learning bro.

ikoyila:
(x-y) / (x 1/3 - y 1/3 ) = (x-y)÷(x-y)^1/3
= (x-y)^1-1/3
= (x-y)^2/3 or √(x-y)^3

Chairman,,!!


27^(1/3) - 8^(1/3) = 3-2=1

& (27- 8.)^1/3 = (19)^1/3

so why am I not getting the same thing ?....


sorry bro , your solution is wrong !!

(x^1/3 - y^1/3 ) =/=
(x-y)^1/3


happy learning ...

#greetings to you all.....happy Sunday...

Umartins1:
If I know it, I won't call on u.

^^ modified
check above

Umartins1:
Pls, someone should help me solve this.




Thank you.

Cc: agentofchange1

apply difference of two cubes ..then simplify

by defn,

f^3 -g^3 = (f-g)(f^2 +fg +g^2)

now

put x=(x^1/3)^3 =f & y=(y^1/3)^3 = g

thus we get x^2/3 +(xy)^1/3 +y^2/3


guess that's it.

Try this. Find dy /dx of. (√x)^-(√y)

Mightymanna:
pic below

3! +11+13
(3!=1*2*3=6)

Nwiboazubuike:
Integrate cos^4x dx

^^^^^^^^SolutioN^^^^^^^^^^
let's rewrite as
$(cos² x )²dx
known from trig. ( Pythagorean) identities ,
cos² =0.5(cos2x+1)
thus ,=> $[(0.5(cos2x+1)]^2 dx
=> 1/4$((cos^2 (2x) +2cos2x +1))dx we can also
have cos^2(2x) = 0.5(cos4x +1)
thus , get have 1/4$[0.5(cos4x +1) +2cos2x +1] dx
now on integrating we get
=> 1/4[ 1/2*1/4*sin4x + x/2 +sin2x + x ] +k
=> 1/32* sin4x +1/4*sin2x+ 3x/8 + k

naturalwaves:
Sup Benbuks?

Am cool.

The gurus are resurfacing ...

nice work guys..., wish I have enough time...will try sha..


BTW: Benkuks=agentofchange1

@ sir richiez I sight thee boss.

The gurus are resurfacing ...

nice work guys..., wish I have enough time...will try sha..


BTW: Benkuks=agentofchange1

@ sir richiez I sight thee boss.

bolkay47:
y~n=5/2+C1{10/3}^n+C2{-7/10}^n

Is that your final answer .?

cut their dicks off , Period !!!

WoW, beautiful!!!!, & sexy....

WoW, beautiful!!!!,

omo ..this one is strong!!..




#just:passing :by



++shalom++

lots of questions to solve....dono...we're.2 start from.....

.

some1 help out with this difference equation

solve:

y_n+2 -5y_n+1 +6y_n=2n+ (-1)ⁿ

@jackpot, Doubledx , laplacian, e.t.c..


the particular solution is the issue..


thanks..

hmmmmm..

greetings.