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

dauswq

  
  
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Masa menjawab:
4.30 p.m. 6 Mac - 6 a.m. 11 Mac
SILA TUNJUKKAN JALAN KIRA
rujuk SINI untuk panduan

jie_kumiko

jap ek daus...jie nk try jwp...ehehehe...

dauswq

jap ek daus...jie nk try jwp...ehehehe...
jie_kumiko Post at 11-3-2010 12:29

silakan

jie_kumiko

silakan
dauswq Post at 11-3-2010 12:38

jwp yg no 2 b taw...xsure betul ke dak..huhu...

|v|=√(3^2)+(-2^2)+(5^2) = 

(uxv) = |i j k|
            |3 -2 5|
            |-7 2 0|
         = i (0-10) -j(0-5(-7)) + k(6-14)
         = -1oi -35j - 8k

(uxv) . |v| = √38 . (-10, -35, -8)
                = (-61.6 i, -215.75j, -49.4k)


ntah btul ke tdak...

jie_kumiko

yg no3 2 ngah try....huhu....

jie_kumiko

eh2 clap...yg no 2 b 2 otw....yg no 3 xingt lerrr

yg jwpn kat ataih 2 no 2a i)

jie_kumiko

yg ni pon xsure la..
jie main try je....hihi..

dauswq

jwp yg no 2 b taw...xsure betul ke dak..huhu...

|v|=√(3^2)+(-2^2)+(5^2) =

(uxv) = |i j k|
            |3 -2 5|
            ...
jie_kumiko Post at 11-3-2010 13:23

jawapan sket lagi betul...sign problem utk salah satu component

dauswq

yg ni pon xsure la..
jie main try je....hihi..


jie_kumiko Post at 11-3-2010 14:26

sign problem ...
check integration ln y

patu find the intial value

jie_kumiko

erkkkk..alamak...silap lagik ek????
cer daus tgk final answer jie yg soalan no 2 b tu....
td screen cptur half je....
ni baru edit...

jie_kumiko

sign problem ...
check integration ln y

patu find the intial value
dauswq Post at 11-3-2010 14:33

integrate ln y mukan ke dpt y ln y - y + c???
cer tgk final jwpn jie 2.....
ade initial value ka???
jap2....nk tgk soalan..takut salin separuh je td 

jie_kumiko

line cam sput vavi...
xleh nk loading soalan  smpai abis....
sakit ati

dauswq

integrate ln y mukan ke dpt y ln y - y + c???
cer tgk final jwpn jie 2.....
ade initial value ka???
jap2....nk tgk soalan..takut salin separuh je td
jie_kumiko Post at 11-3-2010 14:55

yup..ade intial value

sori integration tu betul..mata tak berapa nk nmpk td

dauswq

line cam sput vavi...
xleh nk loading soalan  smpai abis....
sakit ati
jie_kumiko Post at 11-3-2010 15:17

sabar yo jie

jie_kumiko

yup..ade intial value

sori integration tu betul..mata tak berapa nk nmpk td
dauswq Post at 11-3-2010 15:18

leh bg initial value dia x????
line giler pny sengal....mmg smpai skunk xleh loading full nye soalan 2..huhuhu..


lorhhh..wt saspen je..
kalo mata dah xleh nk celik...sila p tdoq.....kah3.....

jie_kumiko

sabar yo jie
dauswq Post at 11-3-2010 15:19

ngah sabo la ni daus oi....
if not....da lama da royter stimix ni masuk longkang dpn uma

dauswq

leh bg initial value dia x????
line giler pny sengal....mmg smpai skunk xleh loading full nye soalan 2..huhuhu..


lorhhh..wt saspen je..
kalo mata dah xleh nk celik...sila p tdoq.....kah3... ...
jie_kumiko Post at 11-3-2010 15:29

y[0]=1

jie_kumiko

y ln y - y= 1/2 e^X(sin x + cos x) + C

given y(0)=1
thus,
1 ln 1 - 1=1/2e^0(sin 0 + cos 0) + C
           -1= 1/2(1) + C

c= -3/2

hence, y ln y - y = 1/2e^x (sin x+cos X) - 3/2

dauswq

y ln y - y= 1/2 e^X(sin x + cos x) + C

given y(0)=1
thus,
1 ln 1 - 1=1/2e^0(sin 0 + cos 0) + C
           -1= 1/2(1) + C

c= -3/2

hence, y ln y - y = 1/2e^x (sin x+cos X) - 3/2
{ ...
jie_kumiko Post at 11-3-2010 15:49

yup jawapan betul

dauswq

jie..betulkan sign jawapan 2 (b) tu