|
Kuiz Matematik Mingguan 3 April (Soalan ke-6 Page 41 #1021)
[Copy link]
|
|
* 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 |
|
|
|
|
|
|
|
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..![](static/image/smiley/default/biggrin.gif) |
|
|
|
|
|
|
|
Sudut AOC = 2 x Sudut ADC
= 2 x 75
= 150 deg
2q + 150 = 180
q = (180 - 150)/2 = 15 deg |
|
|
|
|
|
|
|
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 |
|
|
|
|
|
|
|
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 |
|
|
|
|
|
|
|
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 ![](http://eforum6.cari.com.my/images/common/back.gif)
tahniah kepada animaniac.. :pompom:
soalan beginner 1(a) & 1(b) semuanya telah dijawab dgn tepat..![](static/image/smiley/default/biggrin.gif)
soalan ini telah ditutup. |
|
|
|
|
|
|
|
eh..soklan dh abes plak.. |
|
|
|
|
|
|
|
eh..soklan dh abes plak..
budingyun Post at 7-3-2010 21:40 ![](http://eforum6.cari.com.my/images/common/back.gif)
ade lagi soalan 2 dan berikutnya![](static/image/smiley/default/titter.gif) |
|
|
|
|
|
|
|
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 ![](http://eforum6.cari.com.my/images/common/back.gif)
kurang tepat![](static/image/smiley/default/tongue.gif) |
|
|
|
|
|
|
|
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 ![](http://eforum6.cari.com.my/images/common/back.gif)
gunakan petua rantai...![](static/image/smiley/default/smile.gif) |
|
|
|
|
|
|
|
nomot 2:
(i) ![](http://i929.photobucket.com/albums/ad134/rotikek/jwpn1.jpg)
(ii) ![](http://i929.photobucket.com/albums/ad134/rotikek/jwpn2.jpg) |
|
|
|
|
|
|
|
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~~~!!!???? |
|
|
|
|
|
|
|
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 ![](http://eforum6.cari.com.my/images/common/back.gif)
no, tak leh simply plug in the initial value:geram:
D E equation looks like separable differential ![](static/image/smiley/default/smile.gif) |
|
|
|
|
|
|
|
check sign determinant![](static/image/smiley/default/smile.gif)
dauswq Post at 7-3-2010 22:22 ![](http://eforum6.cari.com.my/images/common/back.gif)
bahek~~~ |
|
|
|
|
|
|
|
nomot 2 b
![](http://i929.photobucket.com/albums/ad134/rotikek/jwpn3.jpg) |
|
|
|
|
|
|
|
tamau cek ke........![](static/image/smiley/grapeman/18.gif) |
|
|
|
|
|
|
|
tamau cek ke........![](static/image/smiley/grapeman/18.gif)
kelapamuda Post at 7-3-2010 23:06 ![](http://eforum6.cari.com.my/images/common/back.gif)
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? |
|
|
|
|
|
|
| |
Category: Belia & Informasi
|