₦airaland Forum

Wow, i'm glad to be back again, feeling the ingenuity of all the mathematicians and maths lovers in the house, I revere oo!

^^^

The dept. of Eco. ni oo. . They announced it in the multipurpose hall and also told us to be on the look-out for the online display. Your dept. was scheduled na, I think!

Hmmm, it's unbelievable that UNILAG can encamp me in Akoka for almost two weeks now, terminating my appointments, spoiling my flenjors (good times). . . All in the name of screening bla bla. The worse part of it is that another schedule has been created but my department which is having a very large number of students wasn't scheduled. We're told to come back by next week's Monday probably everyday, we are typically like the Israelites when they'd no land, o ma shey oo seeing that you guys have gone far far! Abeg make una help us beg them oo.

Prove the couple's formula: 1 + 1 = 1. . . . . . . . PS: your proof shouldn't be less than a page, Lolzzzzzzz

smurfy: As punishment for missing most of Nairaland Maths Quiz Competition, General Richiez asked General Q to list all possible four-letter words that can be formed from the word MEDITERRANEAN such that the first letter is E and the last letter is R.

How many of such four-letter words are possible?

They're listed below but here is how to calculate how many there are
E _ _ R We want to fill those two blanks.
Case 1. The two letters in the blanks are the same.
The only such possibilities are these 3: EAAR, EEER, ENNR.
Case 2. The two letters in the blanks are different. Then
we must choose them from these 8: M,E,D,I,T,R,A,N. We can
choose the letter for the first blank in 8 ways, and for
each of those ways we can choose the letter for the second
blank 7 ways. That's 8×7 or 56 ways.
So 3 ways from case 1 plus 56 ways from case 2, gives us
59 ways total.
Here are all 59 ways listed:
1. EAAR
2. EADR
3. EAER
4. EAIR
5. EAMR
6. EANR
7. EARR
8. EATR
9. EDAR
10. EDER
11. EDIR
12. EDMR
13. EDNR
14. EDRR
15. EDTR
16. EEAR
17. EEDR
18. EEER
19. EEIR
20. EEMR
21. EENR
22. EERR
23. EETR
24. EIAR
25. EIDR
26. EIER
27. EIMR
28. EINR
29. EIRR
30. EITR
31. EMAR
32. EMDR
33. EMER
34. EMIR
35. EMNR
36. EMRR
37. EMTR
38. ENAR
39. ENDR
40. ENER
41. ENIR
42. ENMR
43. ENNR
44. ENRR
45. ENTR
46. ERAR
47. ERDR
48. ERER
49. ERIR
50. ERMR
51. ERNR
52. ERTR
53. ETAR
54. ETDR
55. ETER
56. ETIR
57. ETMR
58. ETNR
59. ETRR

@factorial, there were nothing wrong with my answers. The difference was that I expressed the change in units. E.g. 8(pi)cm2 as my first answer is the same answer as 8(3.142)cm2 = 25.136

oladoya: pls pls pls and pls, the following people should pls drop their number or better still flash me. Dr Asiko, oga labod, mr yemmy etc, pls we need ur help we nevereven do anytin plssssss dis my number 08171862611

Hmmmmmm, Oga mi, chill, na pocco a pocco. Friday on my Holy Mind.

Richiez: We need Generals who will help us with some very short yet very technical questions which we are going to use for the final round of the nairaland math quiz....oya jackpot, ortarico, calculusfx, rhydex, ositadima and the rest pm me if you're ready with the question. I need the on or before 4:00pm 2mao
Thanks Generals!

I've been too busy these days, but I'll see what I can do before then, even if 5 Sir.

smurfy: How many even numbers greater than 4000 can be formed using the digits 0, 1, 2, 3, 4, 5?

Hmmm, the answer will be an infinite number. You can put 1 million of the 4's in line, eg, 444444444444.... 44444 and it would be even. Or a billion. there is no limit.

factorial1: Play with the questions below:
1. If the radius of a sphere is increased from 10cm to 10.1cm, what is the approximate increase in the surface area?

2. The height of a cylinder is 10cm and its radius is 4cm. Find the approximate increase in volume when the radius increases to 4.02cm

3. An error of 3% is made in measuring the radius of a sphere. Find the percentage error in volume?

4. The side of a square is 5cm. How much will the area of the square increase when the side expands by 0.01cm? And lastly

5. If x=tany; find dy/dx?

Hmmmmm. . . . Let's go:

1. Let A = 4r2 be the surface area of the sphere in cm2 . Then dA = 8r. The small dr change in A when the radius changes from 10cm to 10.1cm is then given by small change in A dA (small change in r) dr r=10 = 8 10 0.1 = 8. Thus the increase in the surface area is approximately 8 cm2


2. The volume of the cylinder (in cm2 ) is given by V = r2 h where h is the height and r is the radius (in cm). Thus dV = 2rh. Hence the change in V when the radius changes dr from 4cm to 4.02cm and the height is 10cm is given by change in V dV (change in r) dr r=4 = 2 4 h 0.02 = 2 4 10 0.02 40 = . 25 Thus the volume increases by 40 25 cm3

3. For a sphere, the volume V = (4/3) pi r^3
Take log, log V = 3 log r + log[(4/3) pi]
Differentiate both sides
dV/V = 3 dr/r
If dr/r = 0.03 = 3%, the dV/V is 0.09 or 9%.
You may also find it arithmetically.
Error in V,
eV = abs(V-V0)/V0 = abs( r^3/r0^3 – 1) = abs[(1+- 0.03)^3 -1 ]
Take 1+0.03, then eV = 1.03^3 – 1 = 0.093 = 9.3%
Take 1-0.03, then eV = 1 – 0.97^3 = 0.087 = 8.7%

4. x = tan(y)
y = arctan(x)
dy/dx = 1/(x² + 1)

prodam: Alright she is a smart girl. Am even going with her to reduce the stress.

Pls don't, we were instructed against it. Don't let her forfeit her admission automatically. As long as you've testified that she's smart, then she can take care of herself.

Yeh, so many infos, I've really lagged behind! @damoskiy, pls kindly forward your essay to oketundericharda@yahoo.com, thanks master.

aysuccess99: This is the solution:
6x^2+7xy-5y^2=0
6x^2+7xy-30y^2=0
6x^2+10xy-3xy-5y^2=0
you will now factorisebt pls first arrange accordn 2 lyk terms)
6x^2-3xy+10xy-5y^2
3x(2x-y)+5y(2x-y)
(3x+5)(2x-y)
i hope it is correct.

Correct!

oladoya: anyone that will be going on 23rd and 29th, oga ortarico said you should add me with 08171862611: so that we can seee. Bye for now

Hehe, whatsapp finz. . . Make the rat for town tell the rat for village sey clearance bin wan start ooo

Alas, to God be the glory! Sir Labodz, thank you oo

benbuks: the sum of the first two terms of a g.p is x and the sum of last two terms is y .If there are n terms.
Find common ratio in terms of x , y and n

Another approach:

Sn = a(1 - r^n)/1 - r
Sum of the first 2 terms = S2
S2 = a(1 - r^2)/1 - r or a(1 - r^2) = x(1 - r)
Sum of the last 2 terms from n terms = Sn - Sn-2
Thus, a(1-r^n)/(1-r) - a(1-r^n-2)/(1-r) = y
Multiplying through by LCM (1-r)
a(1-r^n) - a(1-r^n) = y(1-r)
a[1-r^n-1+r^n-2] = y(1-r)
a(r^n-2 - r^n) = y(1-r)
a(r^n . r^-2 - r^n) = y(1-r)
ar^n(r^-2 - 1) = y(1-r)
ar^n[1/r^2 - 1] = y(1-r)
a^n(1-r^2)/r^2 = y(1-r)
r^n/r^2 = y(1-r)/a(1-r^2) = y . 1/x = y/x
r^n-2 = y/x
r = (y/x)^1/n-2

rhydex 247: @ laplacian here is my solution.
sin2x+sinx=1.
Subtract 1 4rm bth sides. We av -1+sinx+sin2x=0.
Substitute y=tan(x/2). Then sinx=2y/y^2+1 and cosx=1-y^2/y^2+1.
-1+4y/(y^2+1)^2-4y^3/(y^2+1)^2+2y/y^2+1 =0
takin d lcm nd simplifyin we av
-y^4-2y^3-2y^2+6y-1=0.
-(y-1)(y^3+3y^2+5y-1)=0.
multiply thru by -1.
y-1=0 or y^3+3y^2+5y-1=0.
y=1. Recal dat y=tan(x/2).
tan(x/2)=1. x/2=tan^-1.
x/2=45. Finally x=90. For the y^3+3y^2+5y-1=0. I will leave dat to my fellow maths general. All is well. I'm off to MARINA.

Seen!

Alpha Maximus: ...Not yet, I'm saving the question for harder rounds!! No one even attempted it!! They all ran to Richiez's question!!

Deal, thank you!

2nioshine: pls recheck ur soln sire...
2sinxcosx +sinx -1=0
does not reduce to
sinx(2cosx+1)=0..on factorization because the -1 is ignored by that approach.
Also chk when x=0
=>sin0+sin0=1
which is nt true.....rechk bro while i wuk on mine
but for i think of a graphical approach where
sin2x=1 +sinx
and a table for sin2x and 1+sinx are obtaind 4rm a given range f values and plotd...just finking

Oops, Thanks a lot bro, that part was truly an error. I have modified my answer. I'll upload the new solution when chanced. Thanks!

Laplacian: nice try bro....but ur solution is flawed

Hmmmm. . . How bro?

Alpha Maximus: ah!! You chance me for this solution abi! solved like a pro! I heard you wanted to become a participabt instead of a quiz master in the Math Competition. There's yet another screening test if you are interested. 1 more page for the thread to be a CENTENOPAGERIAN!!

I feel like performing the two ni ooo but it really pains me gan that I won't be doing either. Na work cause am cause na field work. But Sir Richiez has e-mailed me again to be his assistant score keeper and will give out my contributions whenever i'm available. Shey make I help you do the correction for that your question on mixtures?

Laplacian: ...sin2x+sinx=1...find x...

A graphical solution is the most accepted approach, I'll upload it when chanced. Thanks!

phayvorr: Abeg help me tell dis fresher d truth oo.....e b like say him neva know as Unilag system dey.....oboy u beta go find work dey do make u dey use wait till march...

I don hear na, thanks bro.

tshabalala: I made a mistake the answer is 45degree

Lol

Alpha Maximus: Alright here are the names of the qualified participants:
JARYEH, DENNIECE1, IDENYIJOSHUA and PAPINDINHO.

only and only four, will it do for the quiz?

echibuzor: It would not be fair if you are allowed to participate, its like enrolling a University graduate for an O level exam...

Haba, I was only joking ooo my oga, that question was so cheap but only 5 answered it correctly.

The time is up, I think.

Alpha Maximus: @Ortarico, the Math Quiz has already started....were you aware of this? If not check Richiez topics and port

unaware ooo, thanks. Hurriedly going there.

Alpha Maximus: ...of course: facebook, whatsapp and bbm...I don't really use twitter, it seems bogus to me....whatsapp: 07017413450

Thanks bro.

haibe: DLI students will use 1 month+ after we finish our semester, so be expecting march for ur resumption

Hmmmmmm. . .

shol: I'm an economics major student , i'll whatsapp u.

Nice bro, I'll add you through him then, thanks.

oladoya: Prof. Ortarico so this is your hide-out? Well well well. You are wanted on the thread asap. Pls ooo any economic student in the house or if you know any of them that can help me and my oga, just flash this number(08171862611) i will call you back cos we really need help. May God help you tooo

Oga mi, i'm not in a hide-out ooo. Thanks for this, we really need to hook-up with a staylite for proper orientations. I greet you powerfully sir.

Alpha Maximus: Your secret is safe with me ! Hahahaha!! You no want make those Clinic guyz know your class abi? As far we no dey operate on O Level brain again it still dey good at least! We done prove our worth for there!! That withstanding, no yarn anyone for clinic o!! I think 'statusnet' in an Akokite fresher as well.

Deal! But some masters already know sey I be fresher ooo. Statusnet, im sef be one. Are you on any social networking chats?