Variations

1.
(i) z = k(√x)y²
(ii) c = kb³/a²
(iii) t = ku²/v³
(iv) z = k(√x)/y³


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2.
(i)
a ∝ bc
Then a = kbc, where k is a constant.

a = 12 when b = 9 and c = 3.
12 = k(9)(3)
k = 4/9

a = (4/9)bc

(ii)
When b = -9 and c = 5 :
a = (4/9)(-9)(5)
a = -20


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3.
(i)
z ∝ x√y
z = kx√y, where k is a constant.

z = 64 when x = 4 and y = 16 :
64 = k(4)√16
k = 4

z = 16x√y

(ii)
When x = 10 and y = 9/4 :

z = 16(10)√(9/4)
z = 240


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4.
(i) p = k1q²+ k2r²
(ii) m = (k1/n³)+ k2q²
(iii) z = k1 + k2xy
(iv) a = k1 + (k2/b³)+ k3c


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5.
r ∝ p/√q
r = kp/√q where k is a constant.

ro = kpo/√qo

When p = po/2 and q = qo/2
r = k(po/2)/√(qo/2)
r = (1/√2)kpo/√qo
r = (√2)/2ro

% change in r
= [(√2)/2 - 1] × 100%
= {[100(√2) - 200]/2}%
≈ -29.289%


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6.
(i)
u = k1t + k2t³
where k1 and k2 are constants.

When t = 1, u = 3 :
k1 + k2 = 3 ... [1]

When t = 3, u = 21 :
k1(3) + k2(3)³ = 21
k1 + 9k2 = 7 ... [2]

[2] - [1] :
8k2 = 4
k2 = 1/2

Put into [1] L
k1 + (1/2) = 3
k1 = 5/2

u = (5t/2) + (t³/2)
What are a and b ?

(ii)
What are a and b ?


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7.
z ∝ (3x - 2)²/(2y + 1)
z = k(3x - 2)²/(2y + 1), where k isa constant.

When x = -1 and y = 2, z = 1 :
1 = k[3(-1) - 2]²/[2(2)+ 1]
25k/5 = 1
k = 1/5

z = (3x - 2)²/[5(2y+ 1)]

8.
S = k1t²+ (k2/t)
where k1 and k2 are constants.

S = 5 when t = 1 :
5 = k1(1)²+ [k2/(1)]
k1 + k2 = 5 ... [1]

S = 13 when t = 2 :
13 = k1(2)²+ [k2/(2)]
4k1 +(k2/2) = 13
8k1 + k2 = 26 ... [2]

[2] - [1] :
7k1 = 21
k1 = 3

Put into [1] :
3 + k2 = 5
k2 = 2

S = 3t²+ (2/t)

多謝提點 如果當初有努力聽書 今日也不會與你相見

其實我每次都很想答你，但一看到每個帖都有那麼多題就覺得沒有精神和時間去確定自己可以準確地幫你那麼多題。

如果你可以考慮用
　一個帖當中用 20 點問 4 題
改成
　四個帖每個用 5 點問 1 題
那麼回答你的網友會有更多，而且也會增加別人回答你的機會。
（對你無分別，但對回答者有很大的分別）
（未必人人有心機一次做那麼多，但如果題目簡短、回答容易，大家都會嘗試幫你。）

當然也希望你不要只靠別人做作業。

2015-04-19 18:27:22 補充：
你又幾有文采～
祝你學業進步～