₦airaland Forum

masperano: Stennon how are you? and Jackpot why did u steal my profile pic . lol. I love clever splakss(chicks). where is Doubledx(a great guy). and what of efficiencie(wanted to ask what you studied. mathematics or engineering or physics) you got the age question i threw in. you indeed have a rigorous mathematical mind. To all the generals masperano sends his greetings. between am one hell of a lazy dude else i would have been typing in solutions . may be when am less busy will start contributing. You guyz are really doing a great job. Kudos!!!

AlphaMaximus: The circumcentre of a triangle is the point in the triangle equidistant from all three vertices/the point where any two perpendicular bisectors of any two of the three sides of a triangle intersect. Therefore to find the circumcentre of a triangle, we have to obtain the midpoints of two sides, their gradients and find the equations of each of their perpendicular bisectors and equate them to solve simultaneously for the x and y coordinates of the circumcenter.
To find midpoint of side AB:
(X1+X2/2 , Y1+Y2/2)<====midpoint of AB
The gradient ====>(y2-y1/x2-x1)
Therefore the equation of the parallel bisector of AB is found using:
Y-y1=m(x-x1)
Y-y1=[y2-y1/x2-x1](x-x1)........eq1
Now repeating the same procedure for line BC:
Midpoint====>(x2+x3/2,y2+y3/2)
Gradient=====>(y3-y2/x3-x2)
Equation of perpendicular bisector===>
Y-y2=m(x-x2)
Y-y2=[y3-y2/x3-x2](x-x2).......eq2
After both equations have been obtained , you simply solve simultaneously and obtain the values of x and y which are the x and y coordinates of the circumcentre.

benbuks: Draw lines from one vertex to a mid point. You'll do
this three times. Then get equations for each line and
solve the three simultaneous equations.

Agybabe: Please explain. I don't get it

STENON: The Boss of all Orthogonals!!.....I hail you oo...

MizMyColi:

AW: No matter how hard I try, I barely get the porridge texture when cooking porridge yam

AS: Ah can cook delicious jollof rice

benbuks: see my ideal girl ...*wetin carry u come here my darling .*?

hmm guess same reason am here too ; think av answered d question ryt...?

Missmossy: Aw - Easily emotional

As - Highly innovative

Tags - Jackpot,Darlynlara

STENON: Honest and Optimistic.
Funny in Words and in action, Someone who smiles always.

Lastly....Someone that is very good and skillful in solving Mathematics.

SmellOfMonkey: WHEN EVER I HUG MEN, MY NIPPPLE STANDS AND I GET WET

lawa18: The whole advice boils down to the fact that I should let them be. God knws am jst concerned about his future happiness. I promise to always remember them in prayers n let them be

ElGuapo: @Op, Bar the genotype incompatibility, is there anyother concrete reason you have for your bruh not to get married to his fiancee?

If your answer to the above question is a capital NO then I'd advice you to let 'em be.

Listen, I once met a lady who had every feature I craved for in a woman. To me, she puts the P in perfection but dang! She was a sickler.

Friends expected me to back off at this point but man I didnt, for pete's sake she's got what I need in a lady, do I have to lose gold just b'cos of a lil scratch? She could be the last of her kind, so I couldn't allow people's views get to me - theories were propagated anyway and one was that El Guapo is serving some serious Ogoja spell

Your bro could be in such situation and I dont see your opinion or that of your parents changing anything.

I'd advice that you lot stray off his union. Instead of the incessant interferance, support him and his decision, they could become the perfect couple earth has been searching for aeons.

You dont have to push the emergency button when there aint any hazard.

Missmossy: Thats the spirit,ow did ur day go?

RedClay: Permit me to say night. I'm in bed now with one eye closed, the other struggling to stay awake.
Lately I've been hearing rumors of ghost roaming around in neighborhood so I'm trying not to fall asleep.
There is a graveyard close nearby.
OMG! As I'm typing this, the security lights in my backyard are playing with themselves, and I just saw a dark shadow creeping near my window.
Now I'm hearing strange noises of kids playing and giggling near my door!
OO my God! My door just flung open!
An Old Woman is staring at me!!!
Gotcha!
Was Only kidding. Goodnight.
****
Herald9, Goodnight also.
# I love writing.

herald9: Lolz...
It could only scare kids...

Am immuned against fear...
So that sounds like some moonlight tales



Abeg stop joor... my ba3 z dwn nd am sleeping all alone...
its not funny

Missmossy: Great, i see you lost all the quiz competition stuff. Ndo,we just have to forge ahead.

herald9: Je ne sais pas pour rétablir cette chose o ... Mec est jaloux!

Missmossy: Good to see you ow was ur nite?

herald9: Je suis désolé ... justi4jesu spammeur m'ont interdit!

Bonjour, J'espère que vous avez eu une bonne nuit ...

Amusez-vous! Je vais être effacée demain.

Missmossy: Softsparkky

jackpot ""

efficiencie: ∫(√tanx/(sinx.cosx) dx
=∫{√sinx/[√cos (√sinx)^2).(√cosx)^2)] dx
=∫{1/[√cos (√sinx)).(√cosx)^2)] dx
=∫{1/[(sinx)^(1/2)].[(cosx)^(3/2)]} dx
=∫{[cosx/((sinx)^(1/2))] + (sinx)^(3/2)/(cosx)^(3/2)]} dx
=∫{[cosx/((sinx)^(1/2))]dx + ∫(sinx)^(3/2)/(cosx)^(3/2)]} dx
=2√sinx + ∫(tan)^(3/2) dx
=2√sinx + ∫(tan^2(x).√tanx) dx
=2√sinx + ∫(sec^2(x).√tanx) dx - ∫√tanxdx
=2√sinx+(2/3).tanx.√tanx)-(√2/4)ln[(1-√(2tanx)+tanx)/(1+√(2tanx)+tanx)]- (√2/2)arctan[√(2tanx)/(1-tanx)]


proof of that:
∫√tanxdx=(√2/4)ln[(1-√(2tanx)+tanx)/(1+√(2tanx)+tanx)]+ (√2/2)arctan[√(2tanx)/(1-tanx)]

Given:
∫√tanxdx
Put u^2=tanx
2udu=sec^2(x)dx
(2u/1+u^4)du=dx
2∫(u^2/(1+u^4))du
2∫(u^2/[(1+u^2)^2-2u^2)]du
2∫(u^2.du/[(1-√2.u +u^2)(1+√2.u +u^2)]
2∫(u^2.du/[(1-√2.u +u^2)(1+√2.u +u^2)]
u^2/[(1-√2.u +u^2)(1+√2.u +u^2)=(a+bx)/(1-√2.u +u^2) + (c+dx)/(1+√2.u +u^2)
u^2=(a+bu)(1+√2.u +u^2) + (c+du)(1-√2.u +u^2)
u^2=(a+√2.ua +au^2) + (bu+b√2.u^2 +bu^3) + (c-c√2.u +cu^2)+ (du-d√2.u^2 +du^3)
u^2=a + c + (√2a + b + d -c√2)u + (b√2 + a + c - d√2)u^2 + (b + d)u^3
a+c=0
√2a + b + d -c√2=0
b√2 + a + c - d√2=1
b+d=0
from these we have:
a - c=0
b - d=√2/2
hence we have that:
a=0, c=0, b=√2/4 and d=-√2/4
(√2/2)∫[u/(1-√2.u +u^2)du - (√2/2)∫u/(1+√2.u +u^2)]du
(√2/4)∫[(2u-√2+√2)/(1-√2.u +u^2)du - (√2/4)∫(2u+√2-√2)/(1+√2.u +u^2)]du
(√2/4)∫[(2u-√2)/(1-√2.u +u^2)]du + (√2/4)∫√2/(1-√2.u +u^2)du - (√2/4)∫(2u+√2) /(1+√2.u + u^2)]du +(√2/4)∫√2/(1+√2.u +u^2)]du
(√2/4)ln(1-√2.u +u^2) + (1/2)∫1/(1-√2.u +u^2)du - (√2/4)ln(1+√2.u+u^2)] + (1/2)∫1/(1+√2.u +u^2)]du
(√2/4)ln(1-√2.u +u^2) + (1/2)∫du/[(u-√2/2)^2+1/2] - (√2/4)ln(1+√2.u+u^2)] + (1/2)∫du/[(u+√2/2)^2+1/2]
Hence we have that:
(√2/4)ln(1-√2.u +u^2) - (√2/4)ln(1+√2.u+u^2)] + (√2/2)arctan(√2.(u-√2/2)) + (√2/2)arctan(√2(u+√2/2))
(√2/4)ln[(1-√2.u+u^2)/(1+√2.u+u^2)]+(√2/2)arctan[(√2.(u-√2/2)+√2(u+√2/2))/(1-√2(u+√2/2) √2(u-√2/2)]
put u=√tanx
(√2/4)ln[(1-√(2tanx)+tanx)/(1+√(2tanx)+tanx)]+(√2/2)arctan[√(2tanx)/(1-tanx)]

over to the auditors general!

efficiencie: Alternatively!
∫(√tanx/(sinx.cosx) dx
= ∫(tanx/(sinx.cosx. √tanx) dx
= ∫(1/(cos^2(x). √tanx) dx
= ∫(sec^2(x)/√tanx) dx
= 2√tanx + c
Sometimes the way the brain functions could be funny yet scary . I didn’t envisage this when I tackled the problem initially!

benbuks: i need complete solution dear .

benbuks: Find d period if f{x}=cot[x+pie/2]



@ doubleDx, efficiencie, ,jackpot e.t.c.

try....asap.