₦airaland Forum

terrywales:
ote author=benji93 post=37276576]DR WIP, I sent an appointnent request to the designated email last friday, but i am yet to receive my appointment day, pls how long does it take to get an appointment date after sending the appointment request I guess by now u should have received your appointment date..

mrdrizzy:
hi thanx for the motivational words

care to explain few things bout the admin u got

-is it for postgraduate or under?

-on METU site it was written boldly ''METU does not provide any scholarships to international students'',how den did u do urz

-does it cover all expense,ie medicals,visa,flight ticket etc?
-it seems its over this year,care to highlight how u did urs exactly.thanx

wip7:
hmmmm I am assuming the person you spoke with is either a newbie, lost or clueless regarding your request. The university in this region dont send emails to embassy. The details sent goes via 2-3 channels right from the university to the embassy. In most cases (93%), it goes via the local authority (ministry or embassy in turkey) and they communicate directly with Turkish embassy Abuja or elsewhere. Or a double fax-system is used especially in case of METU.

One question is have you made your payment? if you have made your payment, the right person in METU will dispatch your details via the right channels to Abuja if you have earlier collected your originals from Nigeria.

wip7:
I have replied you above correctly. If you wish to apply for visa in Turkish embassy in Ghana, you can do that provided you have legal residence permit in Ghana. You will email or call your chosen University to send your details to Turkish embassy in Ghana.

If you want Turkish embassy in Nigeria. Kindly inform them also. Since they are not wizard/witchcraft most of them believe by default that Nigerian holding Nigeria passport lives in Nigeria by DEFAULT. But if you dont live in Nigeria, then re-set them by telling them where you want the details to be sent to.

NOTE: the details I referred to above, means copy of your acceptance letter, payment details if any, submitted docs during application...........all goes to embassy in order to attend to you.

wip7:
You can choose where to send it to provided you have legal residence. If you are living legally (documented) in Ghana and wants to apply in Ghana...you will inform your university to send the details to Turkish embassy in Ghana and it will be done. We do that for our own students that live elsewhere, so is not an issue!

wip7:
For Turkish embassy, the relationship does not really count unlike Cyprus embassy/consulate that you either must be the sponsor or your parents.

Meanwhile your dad's friends should state the extent he will sponsor you, in what capacity and possibly why he will be the one taking the role of your parents (tips; he may talk about your academic goals, educational prowess and need to augment your parents financial status). If he has sponsored you before on any project, he may include or reference that.

Not boring and long epistle, point-based may be good and not very very big English but simple and direct (dont let him show his Queens English too much...so the man wont be scared).

He should state your chosen program and duration, what kind of sponsorship he will provide (full or partial) and possibly his source of income (Chairman OANDO or Primary school teacher or whatever he is)...just in 1-2 lines will be nice.

If he owns company on the letterhead will be perfect or he may borrow letterhead from where he works to support his claims.

These are the points in my template for you.
Wish u Success!!

aman2:
purely for statisticians. if u really guru. prove d simpson1/3 rule n state it axiom

wip7:
Yes you can apply for visa with the current acceptance letter you have. When did you apply for admission and when did you get result? Most international students application have not been evaluated....yours seems strange and too fast! Anyway you can pay towards August and book appointment with embassy. Visit Turkish embassy website for other requirements.

goofyone:
| 3-x | ≥ 1, i.e. 3-x ≥ 1 & -(3-x) ≥ 1
-x ≥ 1 - 3 & -3+x ≥ 1
x ≤ 2 & x ≥ 4




x> x²
x-x²>0
x(1-x)>0
x > 0 & 1 - x > 0
x > 0 & 1 > x or 1 > x > 0. X will be numbers between 1 and 0, e.g. 1/2



Singularities at x = 1 and x = -1 define domain. Pretty much all real numbers except 1 & -1
U ( 2 , infinity ), U ( -infinity , -2 ), 0


f(secθ) - f(tanθ) = sec²θ - 1 - (tan²θ - 1)
= sec²θ - 1 - tan²θ + 1
but sec²θ = 1 + tan²θ
thus, f(secθ) - f(tanθ) = 1



f(2sinx) + f(cosx) = log(2sinx) + log(cosx)
= log(2sinxcosx) = log(sin2x), which at x=π/4, is log(Sin(π/2)) = log1 = 0


cotx = 1/tanx (derivative of same curve)
tan0 = 0
tanπ = 0
Period is π


f(x)= x+1/x
f(1/x)= 1/x+1/(1/x) = x + 1/x
if f(x) = (x+1)/x, then f(1/x) = 1/(1+x)


f(2) = 5/4
f(f(2)) = f(5/4) = (2(5/4)+1)/(3(5/4)-2)
=(⁵/₂+1)/(¹⁵/₄-2) = 4/2=2




lim (1+x)^x as x→∞ = e
x = 1/2h in equation, then
lim (1+h)^2h as h→∞ = e², if approaching x approaching 0 from righted

you are right brother but you misstated, (1 + x)^1/x = e as x approaches 0

(1 + 1/x)^x =e as x approaches infinity.
so as u rightly did we substitute 1/2h for x
(1+2x)^1/x = (1 + 1/h)^2h = ((1 + 1/h)^h)^2 = e^2 as h approaches infinity
but as h approaches infinity x approaches 0, therefore the limit is e^2
but you have really done well by solving all these, God bless this forum
use l'hopital to find solution.
DIfferentiate top and bottom,
(-2sin2x)/(x²(-2sin2x) + 2x(1+cos2x))
differentiate again
(4cos2x)/(x²(-4cos2x)+2x(-2sin2x)+2(1+cos2x)+2x(-2sin2x))
at x=0
three terms in the denominator vanish

lim 1-cos2x/x²(1+cos2x) as x→0 = 4/4 = 1




x = 4y

lim (1+(1/y))^4y as x→∞

= e⁴


lim 2sinx-sin2x/x^3 as x→0

if, lim (2sinx-sin2x)/x³ as x→0
using l'hopital rule...
differentiate num and denom
(2cosx - 2cos2x)/(3x²)
(-2sinx + 4 sin2x)/6x
(-2cosx+8cos2x)/6
with x = 0 (at the origin)
6/6 = 1
hence,
lim (2sinx-sin2x)/x³ as x→0 = 1

if, lim 2sinx-sin2x)/x³ as x→0, then
using l'hopital rule
lim2sinx as x→0 = 0

lim sin2x/x³ as x→0
8 cos2x/6, after differentiating twice
lim sin2x/x³ as x→0 = 8/6 = 4/3



Incomplete question



f(x) = x²/(1+x²)
d/d(x²)(x²/(1+x²))
d/du(u/(1+u)) = 1/(1+u)²
derivative is thus 1/(1+x²)²




dy/dx = 4x³-21x²
d²y/dx² = 12x² - 42x
d³y/dx³ = 24x - 42
at x = 2, d³y/dx³ = 6


2.2sinxcox - 2sin2x
2.sin2x - 2sin2x = 0.

Also cos2x = 1 - 2sin²x
d/dx(2sin²x + 1 - 2sin²x) = 0


y = 2arc tan√x
tan(y/2) = √x
x = tan²(y/2)
dx/dy = 2.tan(y/2). sec²(y/2). 1/2
dy/dx = 1/(dx/dy)
but, sec²u = 1 + tan²u
tan²(y/2) = x
dx/dy = 2√x(1+x).1/2
dx/dy = √x(1+x)
dy/dx = 1/(√x(1+x))




y = cosh^-1(2x)
2x = coshy
2.dx/dy = sinhy
cosh²y - sinh²y = 1
cosh²y - 1 = sinh²y
sinhy =√(cosh²y - 1)

2.dx/dy = √(cosh²y - 1)
2.dx/dy = √(4x² - 1)
dx/dy = √(4x² - 1)/2
dy/dx = 2/√(4x² - 1)
at x = 2,
dy/dx = 2/√15


Even though solvable, question appears wrongly worded


y = x^sinx
log_xy = sinx
Iny/Inx = sinx
Iny = Inxsinx
(1/y)dy/dx = d(Inxsinx)/dx
(1/y)dy/dx = Inx.cosx + (sinx)/x
dy/dx = y.(Inx.cosx + (sinx)/x)
dy/dx = x^sinx(Inx.cosx + (sinx)/x)


sec²x = 1 + tan²x
=∫(sec²x + sec²x - 1)dx
=∫(2sec²x - 1)dx
=2tanx - x + C



=x³/3 - x²/2 + 2√x + C





=arcsin(3x/2)/3 + C
make x = (²/₃)sinu to solve problem

dx/du = (²/₃)cosu

∫ dx/√(4-9x²) = ∫ dx/√(4-4sin²u) = ∫ (2/3)cosudu/2cosu

=∫ (1/3)du
=(1/3)u + C
=(1/3)arcsin(3x/2) + C



∫ dx/(4+x²)
using x = 2sinhu

dx/du = 2coshu
x² = 4 sinh²u

∫ dx/4+x² = (¹/₂)∫ (1/coshu)du = (¹/₂)∫sechudu
using standard tables, maybe
=(¹/₂)arctan(sinhu) + C
=(¹/₂)arctan(^x/₂) + C



∫(^Inx/_In10)²/x dx = (¹/_In10)² ∫ (Inx)²/x dx

u = Inx
du/dx = 1/x
dx = xdu

(¹/_In10)² ∫ ^(Inx)²/_x dx = (¹/_In10)² ∫ ^(u)²/_x .xdu
(¹/_In10)² ∫ ^(u)²/_x .xdu
(¹/_In10)² (u)³/3

(¹/_In10)² ( (Inx)³/3) + C




Interesting solving these questions...

goofyone:
| 3-x | ≥ 1, i.e. 3-x ≥ 1 & -(3-x) ≥ 1
-x ≥ 1 - 3 & -3+x ≥ 1
x ≤ 2 & x ≥ 4




x> x²
x-x²>0
x(1-x)>0
x > 0 & 1 - x > 0
x > 0 & 1 > x or 1 > x > 0. X will be numbers between 1 and 0, e.g. 1/2



Singularities at x = 1 and x = -1 define domain. Pretty much all real numbers except 1 & -1
U ( 2 , infinity ), U ( -infinity , -2 ), 0


f(secθ) - f(tanθ) = sec²θ - 1 - (tan²θ - 1)
= sec²θ - 1 - tan²θ + 1
but sec²θ = 1 + tan²θ
thus, f(secθ) - f(tanθ) = 1



f(2sinx) + f(cosx) = log(2sinx) + log(cosx)
= log(2sinxcosx) = log(sin2x), which at x=π/4, is log(Sin(π/2)) = log1 = 0


cotx = 1/tanx (derivative of same curve)
tan0 = 0
tanπ = 0
Period is π


f(x)= x+1/x
f(1/x)= 1/x+1/(1/x) = x + 1/x
if f(x) = (x+1)/x, then f(1/x) = 1/(1+x)


f(2) = 5/4
f(f(2)) = f(5/4) = (2(5/4)+1)/(3(5/4)-2)
=(⁵/₂+1)/(¹⁵/₄-2) = 4/2=2




lim (1+x)^x as x→∞ = e
x = 1/2h in equation, then
lim (1+h)^2h as h→∞ = e², if approaching x approaching 0 from righted

you are right brother but you misstated, (1 + x)^1/x = e as x approaches 0

(1 + 1/x)^x =e as x approaches infinity.
so as u rightly did we substitute 1/2h for x
(1+2x)^1/x = (1 + 1/h)^2h = ((1 + 1/h)^h)^2 = e^2 as h approaches infinity
but as h approaches infinity x approaches 0, therefore the limit is e^2
use l'hopital rule to find solution.
DIfferentiate top and bottom,
(-2sin2x)/(x²(-2sin2x) + 2x(1+cos2x))
differentiate again
(4cos2x)/(x²(-4cos2x)+2x(-2sin2x)+2(1+cos2x)+2x(-2sin2x))
at x=0
three terms in the denominator vanish

lim 1-cos2x/x²(1+cos2x) as x→0 = 4/4 = 1




x = 4y

lim (1+(1/y))^4y as x→∞

= e⁴


lim 2sinx-sin2x/x^3 as x→0

if, lim (2sinx-sin2x)/x³ as x→0
using l'hopital rule...
differentiate num and denom
(2cosx - 2cos2x)/(3x²)
(-2sinx + 4 sin2x)/6x
(-2cosx+8cos2x)/6
with x = 0 (at the origin)
6/6 = 1
hence,
lim (2sinx-sin2x)/x³ as x→0 = 1

if, lim 2sinx-sin2x)/x³ as x→0, then
using l'hopital rule
lim2sinx as x→0 = 0

lim sin2x/x³ as x→0
8 cos2x/6, after differentiating twice
lim sin2x/x³ as x→0 = 8/6 = 4/3



Incomplete question



f(x) = x²/(1+x²)
d/d(x²)(x²/(1+x²))
d/du(u/(1+u)) = 1/(1+u)²
derivative is thus 1/(1+x²)²




dy/dx = 4x³-21x²
d²y/dx² = 12x² - 42x
d³y/dx³ = 24x - 42
at x = 2, d³y/dx³ = 6


2.2sinxcox - 2sin2x
2.sin2x - 2sin2x = 0.

Also cos2x = 1 - 2sin²x
d/dx(2sin²x + 1 - 2sin²x) = 0


y = 2arc tan√x
tan(y/2) = √x
x = tan²(y/2)
dx/dy = 2.tan(y/2). sec²(y/2). 1/2
dy/dx = 1/(dx/dy)
but, sec²u = 1 + tan²u
tan²(y/2) = x
dx/dy = 2√x(1+x).1/2
dx/dy = √x(1+x)
dy/dx = 1/(√x(1+x))




y = cosh^-1(2x)
2x = coshy
2.dx/dy = sinhy
cosh²y - sinh²y = 1
cosh²y - 1 = sinh²y
sinhy =√(cosh²y - 1)

2.dx/dy = √(cosh²y - 1)
2.dx/dy = √(4x² - 1)
dx/dy = √(4x² - 1)/2
dy/dx = 2/√(4x² - 1)
at x = 2,
dy/dx = 2/√15


Even though solvable, question appears wrongly worded


y = x^sinx
log_xy = sinx
Iny/Inx = sinx
Iny = Inxsinx
(1/y)dy/dx = d(Inxsinx)/dx
(1/y)dy/dx = Inx.cosx + (sinx)/x
dy/dx = y.(Inx.cosx + (sinx)/x)
dy/dx = x^sinx(Inx.cosx + (sinx)/x)


sec²x = 1 + tan²x
=∫(sec²x + sec²x - 1)dx
=∫(2sec²x - 1)dx
=2tanx - x + C



=x³/3 - x²/2 + 2√x + C





=arcsin(3x/2)/3 + C
make x = (²/₃)sinu to solve problem

dx/du = (²/₃)cosu

∫ dx/√(4-9x²) = ∫ dx/√(4-4sin²u) = ∫ (2/3)cosudu/2cosu

=∫ (1/3)du
=(1/3)u + C
=(1/3)arcsin(3x/2) + C



∫ dx/(4+x²)
using x = 2sinhu

dx/du = 2coshu
x² = 4 sinh²u

∫ dx/4+x² = (¹/₂)∫ (1/coshu)du = (¹/₂)∫sechudu
using standard tables, maybe
=(¹/₂)arctan(sinhu) + C
=(¹/₂)arctan(^x/₂) + C



∫(^Inx/_In10)²/x dx = (¹/_In10)² ∫ (Inx)²/x dx

u = Inx
du/dx = 1/x
dx = xdu

(¹/_In10)² ∫ ^(Inx)²/_x dx = (¹/_In10)² ∫ ^(u)²/_x .xdu
(¹/_In10)² ∫ ^(u)²/_x .xdu
(¹/_In10)² (u)³/3

(¹/_In10)² ( (Inx)³/3) + C




Interesting solving these questions...

Mnewton:
tanx alot 6 is the answer right?

RobinHez:
wow! it was actually a Pume question from Delsu...
i just hope other people are going to come up with answers too

RobinHez:
pls help..
A contractor agreeing to finish a work in 150 days, employed 75men
each working 8 hrs daily. after 90 days, only 2/7 of the work was
completed. Increasing the number of men by each working now for
10hrs daily, the work can be completed in time
A. 120 men
B. 100 men
C. 75 men
D. 150 men

akpos4uall:
This problem involves differential equation
We were given an initial value as well as another value after a certain amount of time
Here is the problem once again



The flow rate can be represented as dV/dt
This is proportional to the volume V
i.e. dV/dt = kV
Where k is the constant of proportionality
dV/dt = kV



dV/V = kdt
Integrate to get
InV = kt + c
Where c is constant of integration
Applying the given conditions
When t = 0, V = 2500 &
When t = 10, V = 25
The equation becomes
InV = In2500 - 0.1tIn100
InV = In2500 - In(10)^0.2t
Take exponential of both sides to get
V = 2500*(10)^-0.2t
The above equation is the volume-time relationship.

Using V = 250,
Solve for t to get t = 5
Hence it'll take 5hrs for the water to reduce to 10%

When t = 1, V = 1577.4 which is 63.10%
When t = 9, V = 39.622 which is 1.585%

In a table like form, here is corresponding % left for the first ten 10hrs
Time(hrs). % left
1 63.10
2 39.81
3 25.12
4 15.85
5 10.00
6 6.310
7 3.981
8 2.512
9 1.585
10 1.000

I stand to be corrected