I greet everyone in this noble group.
Please, I really need your individual help in raising my obligatory fees( school fees) for this academic session.
I'm in my 300L in one of the western Universities in Nigeria, and countlessly, things have been drastically so hard and rough for me on campus ever since I lost my precious mother to the cool hands of death. No sponsors anymore, not even an iota of assistance of any kind; I'm just left to my fate of "taking the bull by the horn".

Please, I need your helping hands at a time like this. I don't want to drop out of school after so many years of struggle on campus.
In any capacity, lend me your helping hands and God in Heaven will supply all your needs in times of your adversity.

Bank: FIDELITY
NO: 6321174106

God bless you richly!

UNILAG, my school, my pride.
By my projection, one day we shall top UI in world ranking

This sounds vague. This means the school doesn't trust in the credibility of her system.

If they deem this fit, they should also give the test to students who graduated with 2.1, and if they merit it, they should be awarded First Class. Rubbish school

I wish I could be considered.
The God of the orphans would richly bless you if you find me worthy of this favour

I'm not always online to follow up post.
You could reach me @ 09077123695, if God touches your heart to render me the help

In my school, UNILAG, H. K DASS(Advanced Engineering Maths) is more preferable for engineering students

Solution.
y=3x² -7x-4
Change in y=change in x:
(y+Δy)=3(x+Δx)² -7(x+Δx)-4
Subtract y from both sides:
y+Δy-y=3(x² +2xΔx+Δx²)-7(x+Δx)-4-y
Where y=3x² -3x-4
y+Δy-y=3x²+6xΔx+3Δx²-7x-7Δx-4-(3x² -7x-4)
Δy=3x² +6xΔx+3Δx² -7x-7Δx-4-3x²+7x+4
Δy=6xΔx-7Δx+3Δx²
Divide both sides by Δx:
Δy/Δx=6xΔx/Δx -7Δx/Δx +3Δx²/Δx
Δy/Δx=6x -7 +3Δx
Lim(Δy/Δx) =6x-7 +3(0) Δx--> 0
Lim(Δy/Δx) =6x -7 .....Q. E. D Δx-->0

UNILAG Petroleum and Gas Engineering

pulsatingpen:
University of Lagos, English

What level are you? I'm also an AKOKITE

Ojams1:
Please am try do help my younger sister with her home work
Question: The electrical resistance of a wire varies inversely as the square of the radius if the resistance is 0.24πwhen the radius is 0.5mm and the radius when the resistance is 0.32π

Solution

Let R=electrical resistance

Let r=radius

R~1/r²

R=K/r²

K=Rr²

When R=0.24 Π, r=0.5mm, K=?

K=(0.24Π)(0.5mm)²

K=0.06Πmm²

Here, we are required to find the radius when R=0.32Π

From R=Kr²

r=√(R/K)

r=√(0.32/0.06)

r=√(5.33333)

r=2.31mm ///

The polity is gradually going in contravention of Tinubu's expectations

The story gets interesting per second per second

What the Bleep is wrong with this PTF?

JAMB keeps demeaning the standard of education every year during and after the policy meeting