20 marks - maths

2010-02-16 10:24 pm

1. t is proportional to the square root of d. (t=12 when d=4)
a) Find a formula for t in terms of d. b) Calculate the value of t when d=9
2. The positions of two ships, A and B and a lighthouse L. Ship A is 5 km from L on a bearing of 070°from L. ship B is 3km from L on a bearing of 210° from L. Calculate the distance between ship A and ship B. Hive your answer cor to 3 sig fig.
3.In a race, Paula runs 25 laps of a track. Each lap of the track is 400m, cor to the nearest metre. Paula's average speed is 5.0 m/s, cor to 1 dec place. CAlculate the upper bound for the time that Paula takes to run the race. Give your answer in mins and secs, cor to the nearest sec.
4. f(x) = x^2, g(x) = x-3
a) Find g^-1(x) b) Solve the equation gf(x)= g^-1 (x)
5a) (√a)^7 =k√a, where k+a^n. Find the value of n.
b) Express 1/(2√2) as a power of 2.

回答 (1)

老爺子

2010-02-16 11:20 pm

✔ 最佳答案

1.
a)
t is proportional to the square root of d:
t = k√d (where k is proportional constant)

When t = 12, d = 4:
12= k√4
k = 6
Hence, t = 6√d

b)
When d = 9:
t = 6√9
t = 18

2.
圖：
http://i320.photobucket.com/albums/nn347/old-master/100216-2.jpg?t=1266304710

圖片參考：http://i320.photobucket.com/albums/nn347/old-master/100216-2.jpg?t=1266304710

θ = 210° - 70° = 140°

By cosine law:
AB
= √(LA^2 + LB^2 - 2LA•LB•cosθ) km
= √(5^2 + 3^2 - 2 x 5 x 3 xcos140°) km
= 7.55 km (3 sig. fig.)

Distance between ship A and ship B = 7.55 km

3.
400 m: from 399.5 m to 400.5 m
5.0 m/s: from 4.95 m/s to 5.05 m

Upper bound for the time
= 25 x400.5 / 4.95 s
= 2023 sec (to the nearest sec)
= 33 min 43 sec

4.
a)
Let y = g(x)

y = x - 3
x = y + 3
g^-1(x) = x + 3

b)
g[f(x)] = g^-1(x)
g(x^2) = x + 3
x^2 - 3 = x + 3
x^2 - x - 6 = 0
(x + 2)(x - 3) = 0
x = -2 or x = 3

5
a)
(√a)^7 =k√a

Since k = a^n
(√a)^7 =a^n√a
(a^1/2)^7 = a^n(a^1/2)
a^7/2 = a^n + 1/2
n + 1/2 = 7/2
n = 3

b)
1/(2√2)
= (2√2)^-1
= (2^1•2^1/2)^-1
= (2^1+1/2)^-1
= (2^3/2)^-1
= 2^-3/2

2010-02-16 15:21:01 補充：
Reference: 老爺子

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