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2014-03-28 10:39 am
✔ 最佳答案
1.
(i) a = kbc²

(ii) p = kq³/r²

(iii) t = ku³/v²

(iv) z = k(√x)/y²


2.
(i)
a = kbc where k is a constant.

When b = 2 and c = 6, a = 24 :
24 = k(-3)(4)
k = -2
a = -2bc

(ii)
When b = -3 and c = 4 :
a = -2(-3)(4)
a = 24


3.
(i)
z = kx√y

When x = 3 and y = 4, z = 12
12 = k(3)√4
k = 2
z = 2x√y

(ii)
When x = 24 and y = 1/9
z = 2(24)√(1/9)
z = 16


4.
(i) p = k1q + k2r²

(ii) z = k1x³ + (k2/y²)

(iii) a = k1 + k2b + (k3/c)

(iv) v = k1 + k2tu


5.
r = k(√p)/q

Originally, when p = po and q = qo, r = ro
ro = k(√po)/qo

When p = 2po and q = 2qo :
r = k(√2po)/2qo
r = [(√2)/2]k(√po)/qo
r = [(√2)/2]ro

% change of r
= {[(√2)/2] - 1} * 100%
= [50(√2- 2)]%
≈ -29.29%(4 sig. fig.)
(decrease 29.29%)


6.
(i)
b = k1a + k2a²

When a = 1, b = 4 :
4 = k1(1) + k2(1)²
k1 + k2 = 4 ..... [1]

When a = 2, b = 22 :
22 = k1(2) + k2(2)²
k1 + 2k2 = 11 ..... [2]

[2] - [1] :
k2 = 7

Put into [1] :
k1 + 7 = 4
k1 = -3

Hence, b =7a² - 3a

(ii)
When a = 4
b = 7(4)² - 3(4)
b =100


7.
z = k(x + 3)² / (y - 1)

When x = -1 and y = 7, z = 1 :
1 = k(-1 + 3)² / (7 - 1)
k = 3 / 2

Hence, z = 3(x+ 3)² / 2(y - 1)


8.
S = k1t + (k2/t)

When t = 1, S = 8 :
8 = k1(1) + (k2/1)
k1 + k2 = 8 ..... [1]

When t = 6, S = 13 :
13 = k1(6) + (k2/6)
36k1 + k2 = 78 ..... [2]

[2] - [1] :
35k1 = 70
k1 = 2

Put into [2] :
2 + k2 = 8
k2 = 6

S = 2t +(6/t)
參考: fooks


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