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Author: dauswq

Kuiz Matematik Mingguan 3 April (Soalan ke-6 Page 41 #1021)

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Post time 7-3-2010 08:43 AM | Show all posts
* aku jawab apa yg boleh dulu la ye?

(a.i )  PT = h – 6

(a.ii)  LUAS PTUV , A = (h-6)(h+8) = h^2 + 8h – 6h – 48 = h^2 + 2h – 48

b. LUAS PTUV = 207 cm2

207 = h^2 + 2h – 48
0 = h^2 + 2h – 255

h =[-b +/- (b^2 – 4ac)]/2a = 15 &  -17

PV = 15 + 8 = 23 cm
PT = 15 - 6 = 9 cm

b.i

75 + p = 180
p = 180 – 75 = 105

b.ii.

115 + 2q = 180
q = (180-115) / 2 = 32.5 deg

b.iii

Sudut ADE = 180 – 75 = 105 deg.
25 + Sudut ADE + Sudut DAE = 180
25 + 105 + Sudut DAE = 180
Sudut DAE = 180 – 105 – 25 = 50 deg
Sudut r = 90 – 50 = 40 deg
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Post time 7-3-2010 10:02 AM | Show all posts
tahniah.... :pompom:

soalan 1 (a)
telah dijawab dgn tepat oleh animaniac.. :pompom:

soalan 1(b)
(a) dan (c) telah dijawab dgn tepat jugak oleh animaniac.. :pompom:


soalan 1(b)(b) adalah kurang tepat..
soalan ini masih dibuka untuk dijawab oleh forumer..
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 Author| Post time 7-3-2010 11:17 AM | Show all posts
Post Last Edit by dauswq at 7-3-2010 12:05





  
  
SOALAN KEENAM
:
Participation:
+5 kredit


Bonus:

Jawab tp jwpn tak betul: (depend on jln kira, mod bg berapa byk kurang dr point sbnr)
+N/a


1) Beginner:
(a) +15 credit - tahniah animaniac:pompom:
(b) +10 credit - tahniah animaniac:pompom:
    (b)(ii) Terbuka
(c) + 20 credit


2) Pre-Intermediate:
(a) (i) +20 kredit
     (ii) +20 kredit
(b)
+40 kredit

3) Intermediate
(a) (i) +25 kredit
     (ii) +25 kredit

(b) (i) +50 kredit
     (ii) +25 kredit

4) Advance:

+100 kredit



Masa menjawab:
4.30 p.m. 6 Mac - 6 a.m. 11 Mac
SILA TUNJUKKAN JALAN KIRA
rujuk
SINI untuk panduan
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Post time 7-3-2010 05:02 PM | Show all posts
Sudut AOC = 2 x Sudut ADC
                    = 2 x  75
                    = 150 deg

2q + 150 = 180
q = (180 - 150)/2 = 15 deg
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Post time 7-3-2010 05:27 PM | Show all posts
1(c)

r  = 3 cm
h = 5 cm

kadar kenaikan = 7.9 cm3 / s
kadar penyusutan jejari = 0.03 cm/s
kadar penyusutan tinggi + ?

Isipadu cone = (1/3) * 3.142 * r^2 * h
                     = (1/3) * 3.142 * 9 * 5
                     = 47.13 cm3

Dalam tempoh 1 saat, r = 3 - 0.03 = 2.97 cm
Dalam tempoh 1 saat , Isipadu cone yg berkurang = 47.13 cm3 - 7.9 cm3 = 39.23cm3

39.23  = (1/3) * 3.142 * 2.97^2 * h
           = (1/3) * 3.142 * 8.82 * h
         
h selepas 1 saat = (39.23 * 3 ) / (3.142*8.82)
                           = 117.69 / 27.71 = 4.247 cm


kadar penyusutan tinggi = 4.247 cm / s
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Post time 7-3-2010 05:58 PM | Show all posts
2.a.i

v = 3i - 2j + 5k & u = -7i + 2j

cos (tetha) = v.u / |v||u|

v.u = -21 - 4+ 0 = -25

|v|= SQRT (9 + 4 + 25)  = SQRT (38)

|u| = SQRT (49 + 4) = SQRT (53)


tetha = INV_cos [(-25) / (SQRT (53) * SQRT (38))]
      = INV_cos [(-25) / (7.28*6.16)]
      = INV_cos [(-25) / 44.88] = INV_cos (-0.56)
      = 248.63 deg
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Post time 7-3-2010 06:52 PM | Show all posts
Sudut AOC = 2 x Sudut ADC
                    = 2 x  75
                    = 150 deg

2q + 150 = 180
q = (180 - 150)/2 = 15 deg
animaniac Post at 7-3-2010 17:02


tahniah kepada animaniac.. :pompom:

soalan beginner 1(a) & 1(b) semuanya telah dijawab dgn tepat..

soalan ini telah ditutup.
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Post time 7-3-2010 09:40 PM | Show all posts
eh..soklan dh abes plak..
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 Author| Post time 7-3-2010 10:00 PM | Show all posts
eh..soklan dh abes plak..
budingyun Post at 7-3-2010 21:40


ade lagi soalan 2 dan berikutnya
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 Author| Post time 7-3-2010 10:03 PM | Show all posts
2.a.i

v = 3i - 2j + 5k & u = -7i + 2j

cos (tetha) = v.u / |v||u|

v.u = -21 - 4+ 0 = -25

|v|= SQRT (9 + 4 + 25)  = SQRT (38)

|u| = SQRT (49 + 4) = SQRT (53)


tetha = INV_cos [(-25) / ...
animaniac Post at 7-3-2010 17:58


kurang tepat
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 Author| Post time 7-3-2010 10:03 PM | Show all posts
1(c)

r  = 3 cm
h = 5 cm

kadar kenaikan = 7.9 cm3 / s
kadar penyusutan jejari = 0.03 cm/s
kadar penyusutan tinggi + ?

Isipadu cone = (1/3) * 3.142 * r^2 * h
                     = (1/3) *  ...
animaniac Post at 7-3-2010 17:27


gunakan petua rantai...
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Post time 7-3-2010 10:14 PM | Show all posts
nomot 2:
(i)

(ii)
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 Author| Post time 7-3-2010 10:22 PM | Show all posts
nomot 2:
(i)

(ii)
kelapamuda Post at 7-3-2010 22:14


check sign determinant
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Post time 7-3-2010 10:25 PM | Show all posts
tak de orang nak check ke??

mmomod daus..
nak tnye nomot 2(b) tu..
betul tak pemahaman mud..

when x=0, y=1..
but the purpose of solving this equation and the y(0)=1, we will plug in the value kah~~~!!!????
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 Author| Post time 7-3-2010 10:36 PM | Show all posts
tak de orang nak check ke??

mmomod daus..
nak tnye nomot 2(b) tu..
betul tak pemahaman mud..

when x=0, y=1..
but the purpose of solving this equation and the y(0)=1, we will plug in the value ...
kelapamuda Post at 7-3-2010 22:25


no, tak leh simply plug in the initial value:geram:
D E equation looks like  separable differential
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Post time 7-3-2010 10:48 PM | Show all posts
check sign determinant
dauswq Post at 7-3-2010 22:22


bahek~~~
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Post time 7-3-2010 10:48 PM | Show all posts
nomot 2 b
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Post time 7-3-2010 11:06 PM | Show all posts
tamau cek ke........
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 Author| Post time 8-3-2010 05:01 AM | Show all posts
tamau cek ke........
kelapamuda Post at 7-3-2010 23:06


integrate ln y mane leh jadi 1/y  ...guna integration by part

lagi satu: integrate e^x cos x guna by part mmg betull...tp jwpn tak betul lagi, perlu letak satu constant depan e^x (cos x +sin x), apa dia?
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Post time 8-3-2010 06:31 AM | Show all posts
integrate ln y mane leh jadi 1/y  ...guna integration by part

lagi satu: integrate e^x cos x guna by part mmg betull...tp jwpn tak betul lagi, perlu letak satu constant depan e^x (cos x +sin x) ...
dauswq Post at 8-3-2010 05:01


adeh...ni la akibat tak tgok notes...hahaha~~~

nnt mud betulkan ea...
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