英文數學工課,今晚急要答案,數學高手幫幫忙

1.
(i)
C = kr²
k is constant.

When r = 3, C = 150 :
150 = k(3)²
k = (50/3)
Hence, C = (50/3)r²

(ii)
When r = 4 :
C = (50/3)(4)²
Cost, C = $266.67

(iii)
When C = 600 :
600 = (50/3)r²
Radius, r = 6 cm

(iv)
When r = 1 :
C = (50/3) x 1²
C = 50/3

When r = 2
C = (50/3) x 2²
C = 200/3

% change of C
= {[(200/3) - (50/3)]/ (50/3)} x 100%
= +300%


=====
2.
(i)
T = k√L
k is constant.

When L = 64, T = 1.6 :
1.6 = k√64
k = 0.2Hence, T = 0.2√L

(ii)
When L = 121 :
T = 0.2√121
T = 2.2

(iii)
When T = 2.8
2.8 = 0.2√L
L = 196

(iv)
When L = 1 unit:
T = 0.2√1
T = 0.2

When L = 1 x (1 + 300%) = 4 :
T = 0.2√4
T = 0.4

% change in T
= [(0.4 - 0.2)/0.2] x 100%
= +100%


=====
3.
(i)
P = kT/V
k is variation.

When T = 280 and V = 0.035, P = 100000 Pa :
100000 = k(280)/0.035
Variation constant, k = 25/2

(ii)
When T = 400 and V = 0.032 :
P = (25/2)(400)/0.032
P = 156250 Pa


=====
4.
V = khr²

When h = 1 and r = 1 :
V = k

When h = 1 x (1 - 40%) = 0.6 and r = 1 x (1 + 20%) = 1.2 :
V = k(0.6)(1.2)² =0.864k

% change in volume
= [(0.864 - 1)/1] x 100%
= -13.6%


=====
5.
e : expenses
n : no. of guests

e = k1 + k2n
where k1 and k2 are variation constant.

when n = 50, e = 1850 :
1850 = k1 + 50k2 ...... [1]

When n = 150, e = 4350
4350 = k1 + 150k2 ...... [2]

[2] - [1] :
100k2 = 2500
k2 = 25

Put into [1] :
1850 = k1 + (50)(25)
k1 = 600

e = 600 + 25n

(ii)
When n = 200 :
e = 600 + 25(200)
cost, e = $5600

(iii)
When e ≤ 7000 :
600 + 25n ≤ 7000
n ≤ 256
The maximum no. of guests = 256


=====
6.
t = k1 + (k2/n)

When n = 5, t = 39 :
39 = k1 + (k2/5) ..... [1]

When n = 12, t = 25 :
25 = k1 + k2/12 ...... [2]

[1] - [2] :
7k2 / 60 = 14
k2 = 120

Put into [1] :
39 = k1 + (120/5)
k1 = 15

t = 15 + (120/n)

(ii)
Wen n = 15 :
t = 15 + (120/15)
Time taken, t = 23 hours

(iii)
When T = 20 :
20 = 15 + (120/n)
No. of workers, n = 24

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1.
(i)
C = kr²
k is constant.

When r = 3, C = 150 :
150 = k(3)²
k = (50/3)
Hence, C = (50/3)r²

(ii)
When r = 4 :
C = (50/3)(4)²
Cost, C = $266.67

(iii)
When C = 600 :
600 = (50/3)r²
Radius, r = 6 cm

(iv)
When r = 1 :
C = (50/3) x 1²
C = 50/3

When r = 2
C = (50/3) x 2²
C = 200/3

% change of C
= {[(200/3) - (50/3)]/ (50/3)} x 100%
= +300%


=====
2.
(i)
T = k√L
k is constant.

When L = 64, T = 1.6 :
1.6 = k√64
k = 0.2
Hence, T = 0.2√L

(ii)
When L = 121 :
T = 0.2√121
T = 2.2

(iii)
When T = 2.8
2.8 = 0.2√L
L = 196

(iv)
When L = 1 unit:
T = 0.2√1
T = 0.2

When L = 1 x (1 + 300%) = 4 :
T = 0.2√4
T = 0.4

% change in T
= [(0.4 - 0.2)/0.2] x 100%
= +100%


=====
3.
(i)
P = kT/V
k is variation.

When T = 280 and V = 0.035, P = 100000 Pa :
100000 = k(280)/0.035
Variation constant, k = 25/2

(ii)
When T = 400 and V = 0.032 :
P = (25/2)(400)/0.032
P = 156250 Pa


=====
4.
V = khr²

When h = 1 and r = 1 :
V = k

When h = 1 x (1 - 40%) = 0.6 and r = 1 x (1 + 20%) = 1.2 :
V = k(0.6)(1.2)² =0.864k

% change in volume
= [(0.864 - 1)/1] x 100%
= -13.6%


=====
5.
e : expenses
n : no. of guests

e = k1 + k2n
where k1 and k2 are variation constant.

when n = 50, e = 1850 :
1850 = k1 + 50k2 ...... [1]

When n = 150, e = 4350
4350 = k1 + 150k2 ...... [2]

[2] - [1] :
100k2 = 2500
k2 = 25

Put into [1] :
1850 = k1 + (50)(25)
k1 = 600

e = 600 + 25n

(ii)
When n = 200 :
e = 600 + 25(200)
cost, e = $5600

(iii)
When e ≤ 7000 :
600 + 25n ≤ 7000
n ≤ 256
The maximum no. of guests = 256


=====
6.
t = k1 + (k2/n)

When n = 5, t = 39 :
39 = k1 + (k2/5) ..... [1]

When n = 12, t = 25 :
25 = k1 + k2/12 ...... [2]

[1] - [2] :
7k2 / 60 = 14
k2 = 120

Put into [1] :
39 = k1 + (120/5)
k1 = 15

t = 15 + (120/n)

(ii)
Wen n = 15 :
t = 15 + (120/15)
Time taken, t = 23 hours

(iii)
When T = 20 :
20 = 15 + (120/n)
No. of workers, n = 24

C = kr^2
150 = k(3)^2
k = 50/3

C = (50/3)(4^2) = 800/3

600 = (50/3)r^2 => r = 6

C' = k(r')^2 = k(2r)^2 = 4kr^2 = 4C
% change = (4C - C)/C x 100% = 300%

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