₦airaland Forum

Jossy4luv1: u can be offered admission into two institutions,you will hav to pick one of your choice and reject the other by getting a from 4rm jamb office at Abuja,in which you will feel and take to the school in which you dont want...I HOPE THIS IS CLEAR

tnx a lot bros....i really appreciate!!!

d citizen:
@emmyeuler1 ur solution to the problem 4^x = 8x show dat u re no a math general but a math major. Look log2(2x -3) = (2x-3)log 2

Or log2 (2x -3) =log2^(2x-3) stop dat ur wuru to the answer. Your solution to the problem is wrong.

ok boss......we are all learning.....no one knows it all

@amb fitsaint,,,,do u know if its feasible to have double admission...i.e getting admission into ui after ife has given 1 admision initially?

d citizen:
Solve 4^x = 8x

4^x=8x
2^(2x)=(2^3)x
take logarithm of both sides
log2^(2x) =log2^3 +logx
2xlog2=3log2 +logx
2xlog2-3log2=logx
log2(2x-3)=logx
2(2x-3)=x
4x-6=x
3x=6
x=2

d citizen:
simplify sqr (6 +sqr (6 + sqr(6+ to infinite)))

Note sqr mean square root.

LET x=\/6+\/(6+\/(6+....=\/6+\/(6+\/(6+\/(6+...
then,x^2=6+\/(6+\/(6+\/6+...
x^2=6+x
x^2-x-6=0
x^2-3x+2x-6=0
x(x-3)+2(x-3)=0
(x-3)(x+2)=0
therefore,x=3 and the other root is a complex root is not considered

general double dx i hail ooo...its bin a long time

join us here https://www.nairaland.com/1147658/nairaland-mathematics-clinic/70#18549921 nairaland maths clinic......

Calculusf(x):
For the benbuks question...£ is my integral sign.y=∫√1+√x.dx.....let √x be a²...then x=a^4...dx=4a^3da...substitute to give ∫4a^3√1+a².da....let u be 1+a²...du/2a=da...then ∫4a²√u...4∫a²√u but u-1=a²...then,4∫(u-1)√u...4∫(u^3/2)-(u^1/2)...4{[2u^(5/2)]/5-[2u^(3/2)]/3}...8/5.u^(5/2)-8/3.u^(3/2)

guy u said u got admission to read mechanical engr......but how come u know all this.....did u do any A-LEVEL program?

d citizen:
Am still expecting solution to 3^x + 4^x = 5^x

i have two solutions but i'll start wit d easier one. Let # denote "congruency" then since 3^x+4^x=5^x, 4^x#5^x(mod3) or by Fermat's Little theorem (4^x)^2#(5^x)^2#1(mod3), i.e 4^2x#5^2x#1(mod3) or 16^x#1(mod3) and 25^x#1(mod3) so by FLT, x must divide @(3)=3-1=2 Euler's phi function, i.e x|2 hence either x=1 or 2 but since 1 does not satisfy the eqn, x=2,

Amb Fitsaint: UI no dey respect person at all. It seems most peeps dat re already admitted in2 d dept of dia choice tuk d exam with levity Cant saywheda i even gave it a 4 hr prep.
Pharm mi,ife lo sure ju bayi oo

bros nasoo

Attah Ojoma :
Farmer cee nd biokemi respectively

supposing fallback is still editable,i would have advised u to pick physiotherapy and nursing as fallbacks...cos pharm itself is highly competitive..and those nursing ppl fail alot

Attah Ojoma :
Sumbdy help o!


Am runnin short of Adrenalin

what were ur first and second fallback courses?

Oluwatola5: My co-Prof, its been a while now oo, how are you doing?

prof Tola i am fine oooo............how is everything?

Samtolly4JESUS: Evening Pharm!

barrister werin dey 4 ur side na.........?hope u are kul??

evening felaz.........

Ifyx: boss of lyf,ah haıl oo..hw tınz?

i just dey where ui hang me......

afobam: euler dy bobo u ni, dnt mind d guy wassup emmy? U no fit bobo me sha

my guy afobam na so i c am ooooo.......i com dey wonder wetin i do ui ?

Ifyx: ...as we awaıt d lıst nd cutoffz 2mao...#probably

xup dude?.............hope u're kul?

benatex: to those dat made d oau pd cutoffs jus like I did,pli say congratulations oo. #electelect. I greet ma ogas and ogaresses at d top,Chiamaka,Oluwatola,Miracy,Emmyeuler,papadinho,fynestboi,T.ice,dokdave,Felcocom and ma friend in pd Kejilee,we ate,played,even went to d same church,we were nightcrashers,Keji I miss u jare.. all those pd wahala. Does this thread have a whatsapp group,if yes,na my number b dis 07032327097,if not,everyone here shud DROP HIS/HER NUMBER OO,pls oo

congratz bro....................

shakur07: pharm Emmyeuler abeg werin UI jam yhu nao *****

38%

omo na wa oooo

afobam: Greetings to evry1 esp ifyx, euler, namesake *gets a barley for evry1 to cool d tension * bless!B-)

afobam how ur side na?.....hope u're doing gr8?

Chicksy: if d result doesn,t come out today,den it shud b wednesday cos 2mao is a public holiday

u bin argue drink/eat ice cream 4 gate on saturday abi?

d citizen:
Solve the exponential equation: 3^x + 4^x = 5^x

guy....this is not a maths thread....go to maths thread!!

Messilo: Please, someoe should give me two(2) questions inchemistry-------meweatook small!

(1)calc the minimum mass of elemental nitrogen dat wil combine with 7moles of barium.(2)rxn betw ketone and hydroxylamine produces......

Oysdam: me sef leave am ni,i dnt think i eventually pick a guess

my guy hope u bin krest d exam?

Oysdam: me sef no do that resonance freq sha o

i knw say na 1/2¥ rootLC....but i cant imagine myself cramin square root table

shakur07: yhu dey do me aro ni ..yhu cant compare d thaid oau"x ownh of oau was harder

dat na true sha,oau own hard pass....though calculations plenty 4 d physics and chem nd dem no allow calculator...hw man 4 do resonance freq?

mhen dis u.i putme hard wella....omo me dey fear ooo.......

Mismak: viewing this topic: Mismak(f)
Na wa o...una dey slip or dey wak book?...
01:46am.

how vigil?

johnpaul1101: gud morning everyone, plz i need ur help here.
A small oil drop falls with a terminal velocity of 3*10^-4m/s through air, calculate the radius of the drop, and what is the new terminal velocity for an oil drop of half it's radius.
(viscocity of air=1.8*10^-5N/SIN^2, density of oil=900kgm^-3, g=10m/s^2, ignore density of air)
plz, "WORKINGS/SOLUTION" will be gr8ly appreciated.

4/3#(r^3)(rho)g =6#rUV
WHERE rho = density of oil,#=pie,u=viscosity of air,v=terminal velocity
making r subject of relation
r=sqrt[(9*1.8*10^-5 *10^-4)/(2*900*10)]
r=2.12 * 10^-6 m
PART B
if the radius halved r=2.12*10^-6/2 =1.06 *10^-8m
v=[4/3*r^2 *rho*g/(6u)] =1.25 * 10^-4 m/s

Mikebis: Pls ma oga at d top help me wt dis little questn..(2y+3)^-5.its a binomial questn

(a+b)^-n = a^-n + (-5)(b)(a)^-n-1 +[ (-n)(-n-1)/2](b^2) (a)^-n-2 + [(-n)(-n-1)/6](b^3)(a^-n-3)......endless
where a=2y and b=3
therefor,(2y + 3)^-5 =-32(y^-5) -5/64(y^-6) + 135/128(y^-7)-945/256(y^-.........endless