MATHS FOR THREE QUESTIONS

1)8(a^2 -a)^3 +42(a^2 -a)^2 -36a^2 +36a
= 8(a^6-a^3)+42(a^4-a^2)-36a^2+36a
= 8a^6 - 8a^3 + 42a^4 - 42a^2 - 26a^2 + 36a
= 8a^6 + 42a^4 - 8a^3 -68a^2 + 36a


2) 3x-4y=2...............(1)
kx+12y=6..........(2)

3X=2+4Y
X=2/3 + 4/3Y..........(3)


Put (3) into (2),


kx+12y=6
k (2/3 + 4/3Y) + 12y = 6
2/3k + 4/3ky + 12y = 6
2/3k + 4/3ky = 6-12y
(2/3k + 4/3ky) / k = (6-12y) / k
2/3 + 4/3y = (6-12y) / k
0 = (6-12y-2/3-4/3y) / k
k = 6-12y-2/3-4/3y
k = 5/1/3 - 13/1/3y

3) Cut the the rectangle to a triangle.

LET X BE THE LENGTH OF THE TRINAGLE AND Y BE THE

2X + 2Y =90...(1)
XY = 450...(2)

(2),
X = 450/y...(3)

(3)→(1),
2(450/Y) + 2Y = 90
900/Y + 2Y = 90
900 + 2Y^2 = 90Y
2Y^2 - 90Y + 900 = 0
(Y-15)(Y-30) = 0
X=15 OR 30 Y = 15 OR 30

SO THE LENGTH OF THE TRIANGLE IS 30 AND THE WIDTH IS 15

用畢氏定理~
C^2 = x^2 + y^2
C^2 = 30^2 + 15^2
c^2 = 900+225
c^2 = 1125
c = 33.541019




although it is long= = hope it help you

2009-06-20 23:28:05 補充：
sor= =
3)

c^2 = 1125
= 15√5



sorry= =
頭一二條...抱歉...但the last one 我用左15 min 計 ga~ 保證一定對的^^

1)Factorize 8(a^2 -a)^3 +42(a^2 -a)^2 -36a^2 +36a

8(a^2 - a)^3 + 42(a^2 - a)^2 - 36a^2 + 36a
= 8(a^2 - a)^3 + 42(a^2 - a)^2 - 36(a^2 - a)
= (a^2 - a) [8(a^2 - a)^2 + 42(a^2 - a) - 36]
= 2(a^2 - a) [4(a^2 - a)^2 + 21(a^2 - a) - 18]
= 2a(a - 1) [(a^2 - a) + 6] [4(a^2 - a) - 3]
= 2a(a - 1) [ a^2 - a + 6] [4a^2 - 4a -3]
= 2a(a - 1) [ a^2 - a + 6] (2a + 1) (2a - 3)

2)If the two lines 3x-4y=2 and kx+12y=6 do not intersect,then k=?

If two lines do not intercept to each other
It implies they are both parallel,
i.e. their gradient are the same
So
k / 12 = 3 / 4
k = 3 * 12 / 4
k = 9


3)A rectangle has perimeter 90 cm and area 450 cm^2 .Find the length of the diagonal of the rectangle.

Let the Length be X and Y respectively

X + Y = 90 / 2 = 45
XY = 450

The length of diagonal = √(X^2 + Y^2)
√(X^2 + Y^2)
= √[(X + Y)^2 - 2XY]
= √[45^2 - 2(450)]
= √(2025 - 900)
= √1125
= 15√5

1)2a(a^2-1)(4a^2+21a-18)

2)9

3)6cm

2009-06-20 23:22:38 補充：
好多,唔知點寫=.=