F2 MATHS 急!!!!!!!! 10點

7.a)
(x - 3k)^2 = x^2 - 6kx + 9k^2

b)
let x = a + 2b
(x - 3k)^2
= (a + 2b)^2 - 6k(a + 2b) + 9k^2
= a^2 + 4ab + 4b^2 - 6ka - 12kb + 9k^2


8.
(x^2 - xy + 3x - 3y) / (2x^2 + 2xy + 6x + 6y)
= [x(x - y) + 3(x - y)] / [2x(x + y) + 6(x + y)]
= [(x + 3)(x - y)] / [2(x + 3)(x + y)]
= (x - y) / 2(x + y)


10.a)
AE^2 = AC^2 + AE^2
AE^2 = 288 + 288
AE = 24 cm

OA = 24 / 2
OA = 12 cm

b)
in △OAB
AB^2 = OA^2 + OB^2
169 = 144 + OB^2
OB^2 = 25
OB = 5 cm

Area of △OAB
= (5 x 12) / 2
= 30 cm^2

Area of shaded region
= 30 x 4
= 120 cm^2


11.a)
(3a + 4b)(3a - 4b) = 9a^2 - 16b^2

b)
Area of the rectangular wall
= (3a + 4b)(3a - 4b)
= (9a^2 - 16b^2) m^2

cost of painting
= (9a^2 - 16b^2)(5c - 4)
= 45(a^2)c - 36a^2 - 80b^2c + 64b^2

c)
if a = 2, b = 1 & c = 5
cost of painting
= 45(2^2)(5) - 36(2^2) - 80(1^2)(5) + 64(1^2)
= 900 - 144 - 400 + 64
= $420

* represents square.
7.a.(x-3k)*
=x*-6kx+9k*
b.(a+2b-3k)*
((a+2b)-3k)*
=a*+4b*+9k*-4ab-6ak

8.x*-xy+3x+3y/2x*+2xy+6x+6y
=x*-xy+3x+3y/2(x*+2xy+3x+3y)
=1/2

10a.OA=OE(given)
AG=GE(given)
AG*+GE*=AE*(pyth.thorem)
576=AE*
AE=24
OA+OE=AE
2OA=24
OA=12
b.In tri.AOB
OA*+OB*=AB*(pyth.thorem)
144+OB*=169
OB*=25
OB=5
AOB area:
12‧5
=60cm*
tri.OAB=OAD=OEF=OGH(given)
The area:
60‧4
=240cm*

11a.(3a+4b)(3a-4b)
=9a*-16b*
b.The area:(9a*-16b*)m*
The required cost:(9a-16b)(5c-4)
=45ac+30bc-36a-24b
=15c(3a-2b)+12(3a-2b)
=$(15c+12)(3a-2b)
c.a=2,b=1,c=5
The cost:(15c+12)(3a-2b)
=(15‧5+12)(3‧2-2‧1)
=(75+12)(6-2)
=87‧4
=$348

2007-03-26 21:56:12 補充：
Question11b,c were wrong.Please copy the following. 11bThe required cost:(9a*-16b*)(5c-4)=45a*c+30b*c-36a*-24b*=15c(3a*-2b*)+12(3a*-2b*)=$(15c+12)(3a*-2b*)c. a=2,b=1,c=5The cost:(15c+12)(3a*-2b*) =(15‧5+12)(3‧4-2‧1) =87‧10 =$870