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This blog has been surviving strong since o3o2o8 ;D



16 Girls + 24 Boys =
Bananas of o4 unite,
WHOOSH :D
Chairman; Guo Dong
Vice-chairman: Li Qing

♀
01; Ang Man Ling; Band; 26o494'
02; Cao Lei Lei; CO; o11092'
03; Chay Pui Shan; Band; 10o594'
04; Cheryl Lim; CO; o3o894'
05; Cheryl Yeo; NPCC; 15o294'
06; Cindy Ong; Band; o3o294'
07; Eong Rui Xuan; SJAB; 19o594'
08; Lee Yue Yu; DMC; 31o894'
09; Leow Ke Yun; CO; o2o694'
10; Lew Hui Ling; Netball; 18o994'
11; Lim Jia Min; NPCC; 171194'
12; Madeline Koh; SJAB; 18o994'
13; Na Xiao Shuang; Netball; 10o194'
14; Ng Li Qing; Netball; 22o594'
15; Niketa S.K; Science Team; (?)
16; Prisillia Lee; SJAB; 101294'

♂
17; Ang Xiu Zhi; NPCC; 25o394'
18; Benhardi Susanto; DMC; (?)
19; Cheong Chong You; SJAB; o6o194'
20; Chua Ming Lun; Dragon Boat; o8o294'
21; Javier Er; Science Team; 18o794'
22; Ewe Zi Yi; NPCC; Battle of Cowshed'
23; Goh Zheng Lun; Volleyball; (?)
24; Kester Teo; NPCC; (?)
25; Lee Jin Wee; NPCC; (?)
26; Benjamen Lim; Science Team; (?)
27; Lim Wei Hao; Science Team; o5o794'
28; Marcus Liau; Volleyball; (?)
29; Ong Yu Bryan; Dragon Boat; 241294'
30; Qian QiKai; Volleyball; o71293'
31; Sim Sze Kai; Dragon Boat; (?)
32; Soh Kin Hiong; NPCC; 24o194'
33; Tan Guo Dong; NPCC; o9o294'
34; Tan Qing Xi; CO; o8o294'
35; Tan Ri Zheng; DMC; (?)
36; Thon Bok Woon; NPCC; 23o994'
37; Victor Chang; Volleyball; (?)
38; Wan Ming Kang; Band; 30o494'
39; Yim Chia Hui; SJAB; (?)
40; Tommy Yuen; Volleyball; (?)





Form & Maths: Ms Koo
English & Literature: Mrs Lim
HMT & MT: 张老师 & 李老师
Physics: Mr Hisham & Ms Ting
Chemistry: Mrs Ang & Ms Tay
History: Ms Lim
CME: Mr Senan
Art: Ms Ow
Home Econs: Mdm Yati & Mrs Chiang






Benjamen
Bokwoon
Bryan
CherylL
Chongyou
Huiling
Javier
Liqing
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Manling
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Puishan
Qi Kai
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Ruixuan
Yueyu






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Q1: triangle A is the right-angle triangle. As the square of the hypotenuse of a right-angle triangle is equal to the sum of the squares of the other two sides that is called pythagoras' theorem.In this case, triangle A matches the Pythagoras’ theorem.
triangle A:6²+8²=10²

Q2:

x²=4²+3²
=16+9
=25
x=5cm

Q3:
3)a
x² +(2x - 1)²=(2x + 1)²
x²+4x²-4x+1=4x²+4x+1
x²-8x=0
x(x-8)=0
x=0 (reject) or x-8=0
x=8
3)b
area=1/2 × 8 × [2( 8 ) -1]
=4×15
=60cm²



Q4:
AE: 13²-5²=144
√144=12
EB:6²-5²=11
√11=3.3
AB:12+3.3=15.3cm
AREA: 15.3×5=76.5cm²

Q5:
a)
(3x-1)²+(2x)²=(x+7)²
9x²-6x+1+4x²=x²+14x+49
9x²-6x+1+4x²-x²-14x-49=0
12x²-20x-48=0
3x²-5x-12=0

b)
(3x-1)²+(2x)²=(x+7)²
9x²-6x+1+4x²=x²+14x+49
9x²-6x+1+4x²-x²-14x-49=0
12x²-20x-48=0
3x²-5x-12=0
(3x+4)(x-3)=0
(3x+4)=0 or (x-3)=0
x=4/3 (reject) x=3
BC= 9-1
=8cm

Q6:
a)10²+24²=26²

b) 4²+6²≠7²

thus, a is the right-angled triangle.

Q7:
a)
BC: 13²-5²=12²
√12²=12cm
b)
AD: 15²-12²=9²
√9²=9cm

Q8:
17²-8²=15²
√15²=15
10²-8²=36
√36=6
15-6=9cm

Q9:

x: 3²+4²=25
√25=5cm

Q10:
Pythagoras Theorem :
a²+b²=c²
in this case:
8²+7²=x²

12:59 PM