F4 maths ~~~~~~

2009-08-29 12:01 am

1. Energy E (J) transmitted from a source through a medium is given by the following equation:
E = a * 6^bt
where a , b are constants and t (s) is the time of transmission.
(a) Initially, the energy transmitted from the source is 500 J. Find the value of a.
(b) After 2 s, the energy transmitted is dropped by 200 J, find the value of b.
(c) Find the time taken when the energy transmitted is dropped to half of the initial value.
(d) An instrument is used to detect the energy level. However, the energy level below 1 * 10^-9 J could not be detected. How long will it take the reading of the instrument to drop to zero?
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2. The cost of organizing a party ($P) is partly constant and partly varies as the number of participants (N). If 20 people join the party, the cost is $1200; if 50 people join the party, the cost is $2700.
(a) Find the equation connecting P and N.
(b) The admission fee for each participant is $60. To ensure that it will not be a loss, find the minimum number of participants.
(c) It is estimated that 70 people will join the party. To ensure that it
will not be a loss, find the minimum admission fee ( correct to the nearest dollar).

回答 (1)

用戶9112

2009-08-29 8:00 am

✔ 最佳答案

1. (a) At t = 0, E = a60 = a = 500
(b) At t = 2, E = a62b = 500*62b = 200
62b = 0.4
2b(log6) = log0.4
b = log0.4 / (2log6) = -0.2557
(c) Half the value = 250J
250 = 500*6-0.2557t
0.5 = 6-0.2557t
log(0.5) = (-0.2557t)(log(6)
-0.2557t = log(0.5)/log(6) = -0.3869
t = 1.513s
(d) 1 x 10-9 = 500*6-0.2557t
2 x 10-12 = 6-0.2557t
log(2 x 10-12) = (-0.2557t)log(6)
-0.2557t = log(2 x 10-12)/log(6) = -15.034
t = 58.8s
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2. (a) P = C + KN where C and K are constants
1200 = C + 20K … (1)
2700 = C + 50k … (2)
(2) – (1) => 1500 = 30K
K = 50
Sub into (1), 1200 = C + (20)(50)
C = 200
Therefore P = 200 + 50N
(b) Total admission fee = 60N
To ensure it will not be a loss, 60N >= P
60N >= P
60N >= 200 + 50N
10N >= 200
N >= 20
Minimum number of participants is 20
(c) Total cost P = 200 + (70)(50) = 3700
To balance the cost the admission fee should be 3700/70 = $53

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