omablogs:
i want to apply for direct entry, and i did not apply through Jamb, do i still have hope?

Learn1:
Hello. I helped my nephew register for the post time at a cyber cafe at main gate his waec details were not required except date of exam and exam number the grades were not invited we've submitted though. Hope they is no problem they said those grades were for clearance time

Joeysage:
Hey guys I tried applying for the Uniben PostUtme but my application record is not found what do i do as the registration ends on 22nd Sunday.

SwayG:


guy I dey o, I've just bin busy with some urgent work..
afa na

Phayteprecious1:

Is it post utme online screening or direct entry online screening?

Joeysage:
Post utme online screening. I applied for a change of institution on the 16th of this month though. Any solutions please..

BiafranDel:
Dra.ei please do you have any idea the returning school fee for current 100lvl students. That is their 200level fees

Draei:

According to the new development brought about by the school, their returning fee will be around 30+ i.e, thirty something thousand.

BiafranDel:
okay thanks. But has it been truly confirmed and stamped or just speculations.

Draei:

The former.

miracood:
pls am have problem solving questions usin this formular


1/f =1/u+1/v

pls putme me true

BiafranDel:
u= object distance. v= image distance. f= focal length. Two of the the variables will be given, one is unknown. You will be asked to find the unknown variable.

Sometimes, with just the formula above, you can still be asked to find magnification too. So when you have gotten the answers for the three variables i.e F, U, V.

Use the formula m= image distance/object distance. i.e m = v/u. That gives you the magnification.

miracood:
kindly solve this with explanation


find F when v=3, u=5

BiafranDel:
Use 1/f= 1/u + 1/v. Substitute the numbers for the variables. U=5, V=3, F=? i.e 1/f= 1/5 + 1/3. Find the l.c.m of 5 and 3 which is 15, in other words, solve for the r.h.s then cross multiply with the l.h.s, you will arrive at the answer.

miracood:
OK pls solve this post ume que


A converging lens of focl length of 27cm. an object is placed 9cm from the lens

what is the image distnce

miracood:
pls am have problem solving questions usin this formular


1/f =1/u+1/v

pls putme me true

Learn1:
Hello when is it advisable for the aspirant to be in school. My nephew is writing the exam and when will the exam start and when will dates be given. Thanks

Draei:
Holla, registration closes today for post-utme, take note. Also for those that have registered, log in your kofa page and make sure your application is showing SUBMITTED. (((DEADLINE DAY))))

wi5dom:



he should keep checking his portal for his exam date as from Monday. He can be in school after seeing his exam date & venue.

BiafranDel:
1/f = 1/u + 1/v i.e 1/27= 1/9 + 1/v.


1/v = 1/27 - 1/9. L.c.m of 27 and 9 which is 27. Solve the r.h.s then cross multiply with the l.h.s

1/v = (1 - 3)/27

1/v = -2/27


-2v = 27, v = 27/-2

V = -13.5cm. (the image is virtual)

Mimmie2301:
A friend of mine v bin tryng 2 reg 4 post ume since on tuesdae,he didnt do c.o.c or anioda tin nd its showng record nt found....what cn he do as d deadline 4 reg is 2dae?

Learn1:

Thank you. And when is the exam starting proper

Mimmie2301:
A friend of mine v bin tryng 2 reg 4 post ume since on tuesdae,he didnt do c.o.c or anioda tin nd its showng record nt found....what cn he do as d deadline 4 reg is 2dae?