.

The aptitude test form is out for full time ND program. So rush and get form at the rate of 1500 naira per one. The date for the test will be communicated soon. Beware OF Scammer.
07030799567

The aptitude test form is out for full time ND program. So rush and get form at the rate of 1500 naira per one. The date for the test will be communicated soon. Beware OF Scammer.
07030799567

The aptitude test form is out for full time ND program. So rush and get form at the rate of 1500 naira per one. The date for the test will be communicated soon. Beware OF Scammer.
07030799567

Aptitude test form for ND full time is out. 1500 naira only , at the fidelity bank nekede branch which is inside the school. The date for the test will be communicated soon.
BEWARE OF SCAMMERs instead. Inquiries: ehumphery@gmail.com

If you didn't choose nekede as a choice of institution during your JAMB registration, then you 've two options: you may buy evening program form, or you can buy the full time aptitude test form then write the test, Jst contact me, Ehumphery@gmail.com

If you didn't choose nekede as a choice of institution during your JAMB registration, then you 've two options: you may buy evening program form, or you can buy the full time aptitude test form then write the test, Jst contact me, Ehumphery@gmail.com

BEWARE OFScammers. Instead Don't call any one,

The forms for full time ND Is not out yet,

It's okay that 's good attempts

Secadmin Bleep u.

Actually the forms for hnd part time and full time program is out.
ND Part time form is out but for full time is on the way.
BEWARE OF FRAUD.
No date have been fixed for any screening exam. We're even under strike tension.but not yet on strike.

.

Na wah for youooo, na so u go take right the aptitude test?

Na wah for youoooo, na so u go take write the aptitude test?

Please beware of who u cal on admission matters.
FEDPONEK has commenced selling there HND full time and parttime program while part time ND form is out, full time is still on the way.
PLEASE COME AND BUY THE FORMS UR SELF. Don't give or send money to any one, It 've happened to me

Can someone tell us how ACID RAIN is formed.

[color=#550000][/color]
(a) Note that some constant will be given:c=speed of light(3x10^m/s,h=placks const,h,(6.6x10^-34)Js, 1electron volt,P=(1.6x10^-19)joules. Caesium is a metal of workfntn=1.9xP,let threshold wavelenght=L=c/f,W=hf, L=ch/W, Substitue correctly to have (6.5x10^-7)m.
(b)max energy of liberated electrons=hf-W=(hc/L)-W, on correct substitution (1.4x10^-19)J is the max energy.
(c) the stopping pd(V) needs a diagramatic analogy 4u to understand. Post ur mail address here so can send u a material that will help u.

You can go on with the calculations, jst try to understand the question. [color=#990000][/color]

Gartos u missolved the whole thing

Treat the arctan as a trig function let see,

This is it:
photoelectricity talks about ejection of electrons from a metal surface when they got illuminated.
The process of that is called photoelectric effect, ie the emission of electrons out from a metal surface when light beam(photons) strikes on them. But the beam of light must posses a minimum amount of energy for photoelectric effect to occur, this energy is called the work function,w. This w will be sufficient enough to EJECT the electrons out and aswell give them KINETIC energy(k).the w depends on the type of metal, i.e w varies for d/f metals.therefore w=h¥, h is placks constant,¥ is the threshold frequency which is minimum freq the photon must posses for the process to occur.
The topic is a very good one, jst read and understand, and God must surely see u through,

This is it:
photoelectricity talks about ejection of electrons from a metal surface when they got illuminated.
The process of that is called photoelectric effect, ie the emission of electrons out from a metal surface when light beam(photons) strikes on them. But the beam of light must posses a minimum amount of energy for photoelectric effect to occur, this energy is called the work function,w. This w will be sufficient enough to EJECT the electrons out and aswell give them KINETIC energy(k).the w depends on the type of metal, i.e w varies for d/f metals.therefore w=h¥, h is placks constant,¥ is the threshold frequency which is minimum freq the photon must posses for the process to occur.
The topic is a very good one, jst read and understand, and God must surely see u through,

Am a student of fedponek, mechanical engineering department. Follow this tread https://www.nairaland.com/nigeria/topic-472885.0.html to know all about our admission and to avoid thieves.

Am a student of fedponek, mechanical engineering department. Follow this link to know all about our admission and to avoid thieves.

The possibility of gettin admi. Is 98% with that ur JAMButme score, then depending on what u gotto score on the oncoming screening text.
Never think of going for evening program, I 'll tell u when the form is out for full time ND students(those that took JAMButme and scored 160 and above.
Never mind the so called SNAKEADMIN he is a fool. We don't even 've anything like management science. BEWARE OF Fools and NEVER call or SEND MONEY TO ANY ONE,

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if u =arctan(1+x)     du/dx can be precisely gotten,because if   for example:
y=arctanu, therefore dy/du =1/ (1+u^2) from first principle.

[color=#990000][/color]
please still send the settings for opera 4.2
ehumphery@gmail.com

[color=#990000][/color]
please still send the settings for opera 4.2
ehumphery@gmail.com

Always be very free to post ur questions in any subject for the house to help u out,

arctan(1+x) means the tan inverse of (1+x).
Let f(x) =arctan(1+x)/(1-x) also let 'u'=arctan(1-x) and 'v'=(1-x)
> f(x)=u/v applying quotient rule of differentiation:
> f'(x)={vdu/dx - udv/dx}/v^2
>du/dx =? dfrenting 'u' wrt x, du/dx = 1/{1+(1+x)^2}
vdu/dx =(1-x)(du/dx)
>dfrenting 'v' wrt x, dv/dx =-1, udv/dx = -arctan(1+x),
v^2 =(1-x)^2,
so substituting the values obtained in the quotient rule above and taking 'n'=(1-x), 'm'=(1+x) to enhance simplification gives:
>{(n+(1+(m)^2)(arctanm)}/((1+m^2)n^2} as the answer.
This solution can be understood at least by an average Alevel mathematician.

arctan(1+x) means the tan inverse of (1+x).
Let f(x) =arctan(1+x)/(1-x) also let 'u'=arctan(1-x) and 'v'=(1-x)
> f(x)=u/v applying quotient rule of differentiation:
> f'(x)={vdu/dx - udv/dx}/v^2
>du/dx =? dfrenting 'u' wrt x, du/dx = 1/{1+(1+x)^2}
vdu/dx =(1-x)(du/dx)
>dfrenting 'v' wrt x, dv/dx =-1, udv/dx = -arctan(1+x),
v^2 =(1-x)^2,
so substituting the values obtained in the quotient rule above and taking 'n'=(1-x), 'm'=(1+x) to enhance simplification gives:
>{(n+(1+(m)^2)(arctanm)}/((1+m^2)n^2} as the answer.
This solution can be understood at least by an average Alevel mathematician.

This ur differential question is very much good and i luv it, i 'll solve that as soon as possible and paste the solution to u.