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Author: adlismel

Soalan berkaitan Matematik & Matematik Tambahan

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Post time 12-5-2005 10:15 PM | Show all posts
Aku dah kopak monitor nie sampai belakang pon tak jumpa2 lagik tangan ko...?? Ko silap antar alamat kot....?? :lol
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Post time 13-5-2005 09:45 AM | Show all posts
so, ade ape ape lagi soklan yang kite leh bincang?
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Post time 25-5-2005 04:10 PM | Show all posts
terernye beandiesel nih... cikgu ek... ke lect?

aku yg terer math pon tetiba cam lembab...
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Post time 26-5-2005 08:11 AM | Show all posts
aku tengah tercungap cungap abiskan degree ni. tapi bile balik mesia nanti, aku ingat nak gak gi skola rendah aku dulu, jadi cikgu sandaran setaun ke. hmm...
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ancoi This user has been deleted
Post time 28-5-2005 02:11 PM | Show all posts
Soalan Janjang!

Ali memasukkan wangnya ke dalam bank setiap awal tahun sebanyak RM 1000 selama 10 tahun dengan kadar faedah 5% setahun. Berapakah jumlah wang terkumpul dalam bank pada akhir tahun ke 10?
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Post time 29-5-2005 09:22 AM | Show all posts
pergh, susah siot soklan ni. level SPM ke.

meh aku try.

katakan starting duit dalam bank RM0

D=RM1000   <----duit yang di masukkan setiap awal taun
k=5% = 0.05 <----kadar peningkatan setiap akhir taun

J(n)= jumlah terkumpul bagi tahun ke-n

J(1)= 1000+ 1000*0.05 = 1000*(1+k) =1000 * 1.05 = D * a        (where a=1.05)

J(2) = [J(1) + D]*a = Da^2 + Da  = Da^2 + Da     = D *(a^2 + a)

J(3) = [J(2) + D] *a = [D*(a^2+a)]*a= Da^3 + Da^2 + Da = D (a^3 + a^2 + a)

pattern dia, J(n)=D sigma(n, n=1) a^n

or J(n)=RM1000 sigma(n,n=1) 1.05^n

kalo sepuluh taun,  J(10)= RM1000 sigma (10,n=1) 1.05^n

J(10)= RM 13 206. 79

masaelahnya, aku tak sure cammana nak simplify lagi sigma (n, n=1) 1.05^n. aku kira pakai scientific calculator, ok lagi ( tu pon sebab n=10, kalo seratus mati aku). kalo pakai calculator level SPM sakit woo. lagipon ni kira divergent series, so sum dia pon divergent, maknanya membesar tanpa had.

aku takde idea. ade sesapa leh kasi cadangan yang lebih bernas? sila sila.
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ancoi This user has been deleted
Post time 29-5-2005 10:53 AM | Show all posts
Jawapan die dah betul. Pattern die...

Let
D = RM 1000
i = 5%
n = 10

FV = jumlah terkumpul.

Formula:-

FV = D*(1+i)+D*(1+i)^2+...+D*(1+i)^n
     = D*[(1+i)+(1+i)^2+...+(1+i)^n] {siri geometri}
     :
     :
     = D *(1+i)*[(1+i)^n - 1]/i {H/work:- cuba buktikan ungkapan ini}
     :
     :
     :
FV = RM 13,206.79

Selamat berjaya!
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Post time 29-5-2005 06:31 PM | Show all posts
Originally posted by ancoi at 28-5-2005 06:53 PM:
Jawapan die dah betul. Pattern die...

Let
D = RM 1000
i = 5%
n = 10

FV = jumlah terkumpul.

Formula:-

FV = D*(1+i)+D*(1+i)^2+...+D*(1+i)^n
     = D*[(1+i)+(1+i)^2+...+(1+i)^n] {siri  ...



ooh, aku actually cuba nak ungkapkan persamaan tu dalam sebutan yang singkat so that we can just plug in the numbers and get the answer, no matter how big n is. so memang kena wat satu satu ye? it makes sense though, sebab series tu divergent kan?

thanks a lot. that's a good input.:tq:
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ancoi This user has been deleted
Post time 29-5-2005 10:02 PM | Show all posts
Originally posted by BeanDiesel at  06:31 PM:



ooh, aku actually cuba nak ungkapkan persamaan tu dalam sebutan yang singkat so that we can just plug in the numbers and get the answer, no matter how big n is. so memang kena wat satu satu y ...



had          FV = INFINITI
n->infiniti

Series tu memang divergent:ah:

:cak:Aaa... kawan-kawan, kalau nak tau, janjang tu banyak gunanya dalam harian kite... kalau nak buat projection kira hutang rumah kita pun pakai skill janjang tau!
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 Author| Post time 30-5-2005 03:23 PM | Show all posts
Originally posted by ancoi at 5/2005/28 02:11 PM:
Soalan Janjang!

Ali memasukkan wangnya ke dalam bank setiap awal tahun sebanyak RM 1000 selama 10 tahun dengan kadar faedah 5% setahun. Berapakah jumlah wang terkumpul dalam bank pada akhir tahu ...


cuba dolo
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ancoi This user has been deleted
Post time 30-5-2005 08:13 PM | Show all posts
Okay... ade sesape dapat buktikan ungkapan yang aku suruh buat tuh?

FYI,

FV = D *(1+i)*[(1+i)^n - 1]/i              memang satu ungkapan ringkas.
kita just masukkan D = 1000, i = 5% and n = 10 dalam ungkapan tu...

[ Last edited by ancoi on 31-5-2005 at 06:53 AM ]
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buyaogaosuni This user has been deleted
Post time 31-5-2005 11:03 AM | Show all posts
Originally posted by ancoi at 30-5-2005 08:13 PM:
Okay... ade sesape dapat buktikan ungkapan yang aku suruh buat tuh?

FYI,

FV = D *(1+i)*[(1+i)^n - 1]/i              memang satu ungkapan ringkas.
kita just masukkan D = 1000, i = 5% and n =  ...



Guna formula  Sn=a(r^n-1)/(r-1)  where a=D , r=1+i
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ancoi This user has been deleted
Post time 31-5-2005 10:22 PM | Show all posts
Originally posted by buyaogaosuni at  11:03 AM:



Guna formula  Sn=a(r^n-1)/(r-1)  where a=D , r=1+i


Formula tu dah ade... cuma ade sikit modification...
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ancoi This user has been deleted
Post time 1-6-2005 09:21 PM | Show all posts
Orait! Jawapannye bole kata beginilah...

FV = D*(1+i)+D*(1+i)^2+...+D*(1+i)^n
     = D*[(1+i)+(1+i)^2+...+(1+i)^n]
     = D*(1+i)*[1+(1+i)+(1+i)^2+...+(1+i)^(n-1)]
     = D*(1+i)*(r^n-1)/(r-1)
     = D*(1+i)*((1+i)^n-1)/(1+i-1)

where r = 1+i

So,
FV = D*(1+i)*((1+i)^n-1)/i


Adios...

Lepas nie nak tanye soalan berkaitan pinjaman bank lak...
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Post time 4-6-2005 08:33 AM | Show all posts
Originally posted by ancoi at 1-6-2005 05:21 AM:
Orait! Jawapannye bole kata beginilah...

FV = D*(1+i)+D*(1+i)^2+...+D*(1+i)^n
     = D*[(1+i)+(1+i)^2+...+(1+i)^n]
     = D*(1+i)*[1+(1+i)+(1+i)^2+...+(1+i)^(n-1)]
     = D*(1+i)*(r^n-1)/(r-1 ...


thanks a lot. :tq:
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 Author| Post time 5-7-2005 10:31 PM | Show all posts
Cuba buktikan / Tunjukkan

1/log(asas a)abc + 1/log(asas a)abc + 1/log(asas a)abc = 1
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Post time 10-7-2005 04:41 PM | Show all posts

vektor

bleh tak tlg jwbn solan nie....

diberi u=(12)    dan v=(2)  u dan v adlh selari cari nilai p
        ~   (-5)         ~  (p)



tlg jwbn yer...puas cari jwpn tak jumpa ar... guna rumus a=kb bleh ker dpt jwpn?
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Post time 12-7-2005 09:52 AM | Show all posts

Tolong...prof mathematik sekalian...

Nie mungkin bukan soalan add math, mungkin hanya math biasa, tapi tak leh dapat jawapannya... saper bleh tolong??

Nie perbualan antara Mr. Lim ngan Mrs. Chong

Mr. Lim : I have 3 children. The products of my children ages is 72 and the sum of their numbers is same as my number of house. What is my children ages?

Mrs Chong pon keluar tengok no rumah Mr Lim.

Mrs Chong : I still cannot guess it

Mr Lim: I give a small hint. My eldest child like to eat vanilla ice-cream.

Soalannya : Berapakah no rumah Mr. Lim dan berapakah umur anaknya?


Saper bleh tolong... please!!!
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Post time 13-7-2005 06:42 AM | Show all posts
Originally posted by tipahditipu at 10-7-2005 12:41 AM:
bleh tak tlg jwbn solan nie....

diberi u=(12)    dan v=(2)  u dan v adlh selari cari nilai p
        ~   (-5)         ~  (p)



tlg jwbn yer...puas cari jwpn tak jumpa ar... guna rumus a=kb ...


i dont really understand ur notation.

u=12i - 5j   and   v=2i + pj   ke?

kalo betul la,

kalo nak buktikan dia selari, cross product u ngan v mesti 0.

so, u X v = 12p + 10 = 0

p= -10/12 = -5/6
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Post time 13-7-2005 06:44 AM | Show all posts
Originally posted by Eastern at 11-7-2005 05:52 PM:
Nie mungkin bukan soalan add math, mungkin hanya math biasa, tapi tak leh dapat jawapannya... saper bleh tolong??

Nie perbualan antara Mr. Lim ngan Mrs. Chong

Mr. Lim : I have 3 children.  ...


aku ade 4 variable dua equation jek. since aku tak paham hint ni

Mr Lim: I give a small hint. My eldest child like to eat vanilla ice-cream.

tak dapat la nak solve.
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