₦airaland Forum

can it serve as fertilizer to plants

Although OP is crazy ....let's be realistic, prostitutes will sha get hooked to someone ni that's sure.. Someone?? Like op yes

Enter madam

dkronicle:
This school go make sense gon. See student carry uniform hang for belle with him shade

lwkmd

Lol

Orunto:
They are bokoharams trying to escape, so shoot them all and the driver.

maggots among us

randwise:
1/x[(x½/2)(1+x)-½ - (x-½/2)(1+x)½]
Pls, simplify.

-1/{2x^2+2x^3}^1/2

agentofchange1:
kinda cheep I guess

I bought 6lamps from the market , positioned them at 6corners of a dark hexagonal-building at about 8:30pm all lamps functioning well, using all/any of the lamps . in how many ways can the building be illuminated ?


answer asap..

56ways?

Mary and Joseph were black?

agentofchange1:
am not with pen/paper now.
Q3 put. u=7+sinx

du=cosxdx
dx= du/sinx

thus => $cosx/u *du/cosx

= $du/u = ln (u) +C

hence we have ln(7+sinx) +C . ...(as expected)
Q4.
A $sinxcosxdx

set u= sinx
dx=du/cosx

=> $u*sinx*du/sinx

= >u^2 /2 + C
= 0.5sin^2 x + C .

B. similarly

let t= 6x
dt= 6dx
dx= dt/6

=>1/6 $sint .dt
= -1/6 cost + C
= -1/6 cos6x + C .

C) guess the question is like this

x^2 e^(3x)

let's use integration by parts (I.B.P)

put u=x^2 ; du =2xdx

dv=e^(3x) dx ; v=e^(3x) /3

by $udv =u.v -$vdu. , we have

x^2 e^(3x) /3 -2/3 $ x e^(3x) dx..............(*)

again let's compute $xe^(3x)dx from. (*) above

set p= x. ; dp = dx

dq=e^(3x)dx ; q=e^(3x) /3

by $pdq = p.q -$qdp ,

we have xe^(3x) /3 - 1/3 $ e^(3x) dx

=>xe^(3x) /3. -e^(3x) /9 ..........(**)

putting (**) in (*) to get the final ans..

=>x^2 e^(3x) /3 -2/3 [ xe^(3x) /3 -e^(3x) /9 ]. + C

simply. further..
rest solution coming later... m not with pen/paper ...will use partial fraction for 1 &2 .

check Q5. are we to integrate or differentiate .?

e^3x/3{x^2-2x/3+2/3}+K

yemstok:
Here

1. Integral 7x-4/2x^2-3x-2 dx

2. Integral x^3/x+1 dx

3. Integral cosx/7+sinx dx = In (7+sinx)+C

4. Integrate with respect to X
A. Sinxcosx
B. Sin6x
C. X^2 e^3

5. Y= sin^-1 (2x-1)
b. Y=tan^-1 (1+x/1-x)

no 1::: 3ln(2x+1)+2ln(x-2)+K

axortedbabe:
Mehn i see am when i go watch world cup i shock....

let see the pictures you took in sau paulo stadium..

axortedbabe:
Mehn i see am when i go watch world cup i shock....

let see the pictures you took in sau paulo stadium..

Ladyboy

agoadiv:
THIS MUST BE THE WORK OF PORNOGRAPHY

watching por.n must definitely be her hobby.. No offense o but u gerrit.. Lol

starlingbawa:
It's not live on Supersport and I'm wondering why

I think it'd be very difficult for Athletico to overturn the 2-0 deficit they suffered at the Camp nou against Barcelona. I don't fancy Barca as a team, but then I see them going through...

2:0??

Juliaann:
The match will be tough but barca is winning this






Barca 2: 1 Alt Madrid

we shall see about that.

tohero:
Hi everyone!
I think we should make this suggestion to seun.

We all write mathematical equations and expressions using our phone's symbols list. Apart from the fact that it is not appropriate, it reduces the readability and the zeal to post and attend to mathematical questions.

There is a solution!
But that depends on seun. There is a mathematical typesetting program called LaTeX which is specifically for typesetting mathematical documents. Not just only available for desktops alone but also available for the web!

This is what is being used by forums like math.stackexchange.com/questions/96151/do-imaginary-and-complex-angles-exist
And the likes of wikipedia when writing anything mathematical.

How do we use it here?
Depending on the platform to which this forum is built, there is a plugin will allows us to write such equations. If the plugin can be integrated to this forum, then we are good to go!

THEN,
We need to learn how to use the language LaTeX. Since we are mathematics student, I believe few if not all of us knew about the use. Unfortunately, since it is not included in our courses at school majority may fail to know about this.

SOLUTION
I can sacrifice my time to teach us all on how to use this beautiful language.

NB: LaTeX is what is being used to write our textbooks, project, thesis and the rest!

whao.!!

jgenius:
bolkay you are wrong,for the solution of if x=sin@,simplify x/(1-x^2)^1/2

lols I know that now.. But pls post the solution.

neighy:
maybe this will help

maybe the answer is just none. Not that sure tho.

(No) got -1/2 as the limit.

neighy:
Bolkay47 and Others...

More

(NB..¥ is Root)

1)Find the root of the equation
2 ¥x =x-3....

2)If F and G are two different functions such that F(x)=x+1 and -1/x=g(x) find gof

3)A train starts from rest and its acceleration t seconds after start is 1/4 (20-t) m/s what is its speed after 20sec...

4)A curve passes through the point (4, -3). Its gradient is 2-x find the equation of the curve...

5) Find the area enclosed between the curves Y=4-x^2 and Y=x^2-2x



To all the maths Gurus here

Do the needFul...

find what in no2?

neighy:
bolkay47...
{1}---note: ¥ means root)
12¥e=x¥7::12*¥12=x*¥7
Divide both sides by ¥7
X=12¥12/¥7 using surd
You will arrive at x=24¥21/7..

My Bad but its actually 12 base e=X base 7... Where e=12 ... Find X


{2}. Y=4x-x^2 at x=1 and x=2
(Note $ means integral)
$4x-x^2dx
$4x^2/2-x^3/3
Therefore$4x-x^2= 2x^2-x^3/3+C
[2(2^2)-2^3/3]-[2-1/3]
(8-8/3)-(2-1/3)
16/3-5/3=11/3
Ans=3.7square units.
Well my book here says the answer is 2/3 only that the Solution wasn't given...

re-write that no 5 question its not clear....but number 3 is weird.
For number (2). Maybe I made a mistake or its a typographical error in your textbook.

neighy:
It might seem simple but guys please do justice to this Questions and give me a mention after every solving... Am Preparing For an Exam Thou... Thanks...


1) If 12 root e = X root 7 Find X where e=12

2) What is the area enclosed by the curve Y=4x-x^2, X=1,X=2, and the X axis

3) the factor of X^99+1

4)If F(x)=x/x-1, what is f(x+1)...

5) Find Lim 4+h-2/h... H}2...

#More To Come...

{1}---note: ¥ means root)
12¥e=x¥7::12*¥12=x*¥7
Divide both sides by ¥7
X=12¥12/¥7 using surd
You will arrive at x=24¥21/7..

{2}. Y=4x-x^2 at x=1 and x=2
(Note $ means integral)
$4x-x^2dx
$4x^2/2-x^3/3
Therefore$4x-x^2= 2x^2-x^3/3+C
[2(2^2)-2^3/3]-[2-1/3]
(8-8/3)-(2-1/3)
16/3-5/3=11/3
Ans=3.7square units.

{3}. NO IDEA

{4}. F(x+1)=(x+1)/(x+1-1)=(x+1)/x

{5}. I DON'T GET THE QUESTION.

Thanks. Mr neighy.

Umartins1:
Hi,,, well done.

Do you know me? This is Umartins on myschool. We do compete together in the classroom. I don't have time for that again cos of poor network.

Anyway, I think you should re-visit the guy's question. The denominator is inside a square root.

I didn't notice that sir..thanks
Long time though..

jgenius:
solve these?if x=sin@,simplify x/(1-x^2).note that @ stands for tita.

sin@/1-sin^2@
But Sin^2@+cos^2@=1
So,cos^2@=1-sin^2@
: sin@/cos^2@
Sin@/cos@*1/cos@
Sin@/cos@=tan@
and 1/cos@=sec@
tan@*sec@
=tan@sec@...

AlphaMaximus:
It wasnt an assumption......"a" represents the initial term, which in this case is 20.

thumbs up .its very clear now...

AlphaMaximus:
NB: I accidentally typed 461,875 instead of 421,875 in my formula justification exercise.......take note

thanks so much sir...but why did you assume "a" to be 20?

AlphaMaximus:
alright.....well in that case, this is how it goes:
We have info that field is depleted by 0.1% with each swipe of the cutlass, thus each swipe leaves the field devoid of 1/1000th its preceding area.
Therefore, mathematically, the first term, a=20, the ratio , r=1/1000, and the number of terms/swipes, n=60........in this case since we are interested in the amount left from whole, we subtract "r" from 1 and we have : 1-1/1000= 999/1000.....lets call 999/1000 "R".....then we use the formula S_remainder= a (R)ⁿ and we have:
S_remainder= 20*(999/1000)⁶⁰=18.83472524
To further corroborate the validity of the formula, lets examine a case where each flow of a pipe depletes a 1,000,000litre tank by 1/4.......the first time, 250,000(one-fourth of 1,000,000) will be removed leaving 750,000,the second time, 187,500 (one-fourth of 750,000), leaving 562,500, the third time,140,625(one-fourth of 562,500), will be removed leaving 461,875
.....instead of this rather cumbersome approach, the previously used formula can be used to attain the 461,875 litres left......a=1,000,000 and n=3, and R=1-r=1-1/4=3/4....therefore;
S_remainder=a(R)ⁿ=1,000,000*(3/4)³=421,875 litres. Thus, the method is justified.

this is just excellent.... Although I don't know the answer yet cos its an assignment but WAO.#respect..

AlphaMaximus:
20-(60*(0.1/100)*20)=18.8 square feet of grass left.....dont really see what was complicated about the question. Open to correction though

the question is on GP....it should somehow form a sequence. That's what I don't understand.

With each roll of a cutlass,a man cut 0.1% of grass in a field. if there were 20 square feet of grass to be cut. how much grass will remain after 60th stroke?

Please help me out...thanks

Allureoftheseas:
Bitccchhhh plz!!!! Dnt you have something meaningful to do rather than come here and create stupid, pointless topics

lol

jgenius:
Solve this:
x^2=2^x, find the value of x

use iteration method::
F(x)=x^2-2^x
F“(x)=2x-2^xln2
use x0=0 as starting point. You will get 2..

I Heard irina and ronaldo don break up...is it true?