₦airaland Forum

benbuks: your friend is ryt bro


medicine cobo. is eng , phy , bio & chem ,

unless you wanna study engr. den combo is eng. math , phy , & chem.

that's it .


don't you have a brochure ?

Nahhhh..........thanks anyway.

Good afternoon all gurus present........plz i would like to confirm something important from you guys; i'm preparn for nxt year jamb but i dnt know whose subjects to focus on as i would like to study medicine. i know i have to write test for english, biology, chemistry, and one other. i would like to write maths but a frnd of mine said i would rather write physics. plz wat should i do

What the hell...........am i paying for "school fees" or for the "tickets to heaven"

Hon patrick messege to ASSU
Okay, its either my dictionary is
incomplete or this man has his
own special dictionary. Anyways
can anyone simplify this for us. This ASUU strike is a miasma of a
deprecable apothesis of an
hemorrhaging plutocracy,
cascadingly oozing into a
malodorous excrescence of
mobocracy. With all termagant ossifying
proclivities of a kakistocracy, our
knowledgia centura is enveloped
in a paraphlegic crinkum
crankum. Therefore ASUU, cest in
dejavu, dejavu peret ologomabia

goodevening every1, plz i need ur help here:
find the value of k given that logk - log(k-2) = log5
(WASSCE JUNE 1995 GENERAL MATHEMATICS THEORY, QUESTION 6b)

lets keep solving problem!!!

goodday everyone, plz someone should help me solve this:
a piece of metal of mass 50g heated to 100degree celcius is dropped into 200g of water @ 14degree celcius in copper calorimeter weighing 80g and raises the temperature to 70degrees celcius. What is the heat capacity of the piece of metal?(specific heat capacity=0.4J/gK)

goodday everyone. Plz i saw this question on this thread but couldnt find a response.
A lift of weight 30kN moves upward through a height of 20m, if 25% of the workdone on the lift is used in overcoming friction, calculate the workdone.

Ortarico: Past question, here's the solution.

thanks very much bro, i really appreciate. Yea i know it's a past question but i didnt see any solution, dat's why i had 2 post it again. Have a nice day and remember 2 keep my salah meat

goodday and happy salah 2 everyone.
Plz i could use some here:
The numerator of a fraction is 5 less than it's denominator. If 6 is added to the numerator and 4 to the denominator. The fraction is doubled. What is the fraction?
Plz solution/workings will be greatly appreciated.

emmyeuler1: 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

thanks a lot.....plz can i get ur number, or u can reach me @ 08137456323

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.

What is the total volume of the residual gas formed when 20CM^3 of hydrogen is sparked in 15CM^3 of oxygen at a temperature of 150 degrees celcius according to the following equations: 2 moles of hydrogen gas + 1 mole of oxygen gas giving a product of 2 moles of water.
(a) 10CM^3
(b) 15CM^3
(c) 25CM^3
(d) 30CM^3
plz help me out

goodday everyone. Plz how do you find the third side of a triangle when two sides and the area of the triangle is given?

johnpaul1101: Greetings all generals. Plz i would like us to play with this practical question:
if 8 people handshake each other, what is the total number of handshakes that will be made?

kudos everyone, actually u re all ryt. 28 handshakes will be made in totality!

Greetings all generals. Plz i would like us to play with this practical question:
if 8 people handshake each other, what is the total number of handshakes that will be made?

[color=#006600][/color]
ewwwwwww. she look awful on that hair

Salute to everyone.
Plz help me make t the subject from:
s=ut+1/2*at^2

Gudevening every1, plz i would appreciate your help here:
the diagram associated with this shows a 6V battery of internal resistance 0.5ohm, connected to three resistors of values 3ohm, 4ohm, and 5ohm as show. Calculate the current in each resistor, the p.d across the 3ohm resistor, and across the parallel resistor
(question 13, exercise 4, book 3, page 397 of new school physics for senior secondary schools by anyakoha)

emmyeuler1: E=V/d=4.2/(15/100)=28N/C

thanks emmy

Calculate the magnitude of the electric field intensity between two plates, 15cm apart, if the P.D between the plates is 4.2V.

Richiez: One very useful application of the binomial theorem is in the finding of the squares, cubes e.t.c of numbers especially those with decimals. For example;
(1.5)³

This can be rewritten as
(1+0.5)³
Here, a=1 b=0.5 and n=3

Thus expanding will yield;
1³ + 3(1)²0.5 + 3*2/2!(1)(0.5)² + 0.5³

=1 + 1.5 + 0.75 + 0.125
=3.375

thanks a lot, i'm getting it little by little now, expecting more examples

Salute to everyone, plz how can i convert voltmeter to ammeter and vice versa

Thanks

echibuzor: .
.
Take your time and welcome back to the thread..

doubleDx: Right on point bruv! LMAO @ the bolded part





^ Binomial expansion: If √(z) = √(a(x + 1/a) = √(a).[(x + 1/a)]^(1/2). Expanding [(x + 1/a)]^(1/2) using binomial theorem and multiplying the result by √(a) gives you the answer. It's quite easy if 'z' can be expressed in terms of 'a and x' such that a = perfect square!


Eg===>>


√10 can be rewritten as =>

= √[9(1 + 1/9)]
= 3√(1 + 1/9)
= 3 (1 + 1/9)^(1/2)
Expanding the bolded part yields =>
= 3[1 + (1/2)(1/9) + (1/2)(-1/2)(1/9)^2/2! + (1/2)(-1/2)(-3/2)(1/9)^3/3! + ....]
Simplifying further yields =>
= 3[ 1.0541 approx]
= 3.162 approx

thanks a lot richiez, but i'm not familiar with this binomial expansion, plz i would appreciate more examples. Thanks!!!

doubleDx: Right on point bruv! LMAO @ the bolded part





^ Binomial expansion: If √(z) = √(a(x + 1/a) = √(a).[(x + 1/a)]^(1/2). Expanding [(x + 1/a)]^(1/2) using binomial theorem and multiplying the result by √(a) gives you the answer. It's quite easy if 'z' can be expressed in terms of 'a and x' such that a = perfect square!


Eg===>>


√10 can be rewritten as =>

= √[9(1 + 1/9)]
= 3√(1 + 1/9)
= 3 (1 + 1/9)^(1/2)
Expanding the bolded part yields =>
= 3[1 + (1/2)(1/9) + (1/2)(-1/2)(1/9)^2/2! + (1/2)(-1/2)(-3/2)(1/9)^3/3! + ....]
Simplifying further yields =>
= 3[ 1.0541 approx]
= 3.162 approx

thanks richiez, i appreciate, but i'm not familiar with this binomial theorem. I would appreciate more examples, thanks!!!

echibuzor: That's Cos 15 in the fourth quadrant.... Solution Loading


cos 2x = 2cos^2 15 - 1
2cos^2x = cos2x + 1 ... x = 15

2cos^2 (15) = cos 30 + 1 .... Cos 30 = (√3)/2

2cos^2 (15) = [(√3)/2] + 1

cos^2 (15) = ([(√3)/2] + 1) /2

2cos^2 15 = [(√3) + 2] /2

cos^2 15 = [(√3) + 2] /4

Square root both sides...
cos15 = ±√([(√3) + 2] /4)

cos15 = (√[(√3) + 2])/2

and that, my friend, is cos 345 = cos 15 = (√[(√3) + 2])/2...

Did u notice that i removed the ± in the final answer? That's because Cos is always positive in the fourth quadrant....

At first I was solving for sin15 before i looked at the question again... But, Hey, U might need it too someday right?

So, Here goes, Its quite lengthy and questions about further explanations are welcomed... Lets go there....
.
.
.
.
.
.


Sin (2y) = 2sinycosy ------ (by putting y=15)
Sin30 = 2sin15cos15 ------(ii) (cos^2 (y) = 1 – sin^2 (y))
Hence, cosy = √(1 – sin^ (y))
By putting this into eq (ii) we have;
Sin30 =2sin15 * √1-sin^2 (15)
Square both sides…
Sin^2 (30) = 4sin^2 (15) (1 – Sin^2 (15))
Open Bracket to give…
Sin^2 (30) = 4sin^2 (15) – 4sin^4 (15)
Remember that Sin ^2 (30) = Sin30 * Sin30 and Sin 30 = ½ (abi?)
So: ½ * ½ = 4Sin^2 (15) - 4Sin^4 (15)
¼ = 4Sin^2 (15) - 4Sin^4 (15)
Cross Multiply…..
1 = 16Sin^2 (15) – 16Sin^4 (15) =>>> 16Sin^4 (15) -16Sin^2 (15) +1 = 0
What we have here is a quadratic equation of the Sin^2 form
Let Sin^2 (15) = x
Our new equation becomes…..
16x2 – 16x + 1 = 0
Using Quadratic Formulae (Popularly known as Almighty Fomular)
x = [-(-16) ± √162 - 64 ] /32
==>> [16±√256 – 64] /32
==>> [16 ± √192] /32 ==>> [16 ± √64 * √3] / 32
==>> [16 ± 8√3] /32 …… By Simplifying with a common factor (
==>> [2 ± √3] /4 ]
Sin^2 (15) = [2 ± √3] /4]
Sqrt both sides….
Sin15 = ± (√[2± √3] /4 )
This is where the logic comes in… Remember that I said Sin345 is Sin 15 in the fourth Quadrant (Fourth Quadrant is angles between 2710 and 3600) and any Sine in the Fourth Quadrant gives a negative answer… This point eliminates + √[2± √3] /4 from being a valid answer, which gives us with -√[2± √3] /4… Also (√2+√3) / 4 gives a value too extreme to be relevant… Hence, we are left with
-√[2 - √3] /4 as our answer for sin15
To have a more simplified answer… -√[2 - √3] /4 is also the same as [√(2 - √3)] / √4 which gives -[√(2 - √3)] /2 …….. That is your answer… Confirm with a calculator… Questionand request on further explanations are welcome…

Chibuzor..

thanks @chibuzor, although quite lengthy and confusing, but i'm still on it

Thanks

salute to you all
plz i need a hand here
(1)simplify sin90 + cos¤
(2)using the laws of indices and logarithm, find x if 2^x=x^2

salute to you all
plz i need a hand here
(1)simplify sin90 + cos¤
(2)using the laws of indices and logarithm, find x if 2^x=x^2
thanks ;Dsalute to you all
plz i need a hand here
(1)simplify sin90 + cos¤
(2)using the laws of indices and logarithm, find x if 2^x=x^2
thanks

Edis4christ: Dats true, d sun doesnt move, nd shadow is an image cast by an object

Thanks a lot, but d argument here is that sun moves around it's axis and that shadow is an object

Mr Calculus: x=antilog of bar 2.3675
y=antilog of 0.9750..
anytin u get add it together

thanks very much