maths 再次..... thank you!希望有心人士幫忙!

2.
a. (100-20)(1-4%)^t
= 80(0.96)^t
b. the temp. difference after 15mins:
= 80(0.96)^15
= 43.37
the temp. of hot water after 15 mins:
= 43.37 + 20
= 63 degree C (correct to the nearest degree C)
c. when the hot water drops to 30degree C, the temp. defference = 30 - 20 = 10 degree C
let t be the time required,
80(0.96)^t = 10
0.96^t = 10/80
0.96^t= 0.125
tlog0.96 = log0.125
t = log0.125/log0.96
t = 50.9
t = 51 mins (correct to the nearest min)


2008-08-12 18:03:23 補充：
1.a.
i. 10mins = 1/6 hr.
the dist. in the first part = speed x time
= xkm/hr (1/6)hr
= x/6 km

2008-08-12 18:18:11 補充：
ii. speed in second part = (x-18)km/hr
time for the second part = (1/6+x/60)hr
the distance in second part :
= (x - 18)(1/6 + x/60)
= (x-18)[(10+x)/60]
= (x^2 - 8x -180)/60
= x^2/60 - 8x/60 - 180/60
= x^2 - 2x/15 - 3

2008-08-12 18:25:59 補充：
iii.
the total dist. is 59
x/6 + x^2 - 2x/15 - 3 = 59
10x/60 + (x^2 - 8x -180)/60 = 59
(x^2 +2x -180)/60 = 59
x^2 + 2x - 180 = 3540
x^2 + 2x - 3720 = 0
(x-60)(x+62) = 0
x = 60 or x = -62(rejected)
x = 60
so the drivng speed in first part is 60km/hr