Page 20 - PRAKTIS HEBAT ADDMATHS TG4

Additional Mathematics Form 4 Answers
12q – 3
(b) arithmetic progression: T = –280, 14. p = ——–––
6
8
geometric progression : T = ——; r 3
125
6
x
not the same for both progressions 15. y = —––––
2 – 4x
janjang aritmetik: T = –280,
1
1 2
6 16. (a) xy = pq — – p
8
janjang geometri: T = ——; x
125
6
10
10
10
tidak sama bagi kedua-dua janjang (b) log y = (log n)x + log m
3
2
(c) 250 17. y = – —x + 9
2
y
8. (a) (i) 149 cm (ii) 2 295 cm 18. — = –x + 9
(b) 25 rectangle/ segi empat tepat ke-25; P x 2
th
9. (a) RM900 (b) RM1 850 19. (a) log y = 2x (b) 10 000
10
(c) RM48 750 20. xy = –2x + 11 1
10. RM6 960 21. (a) 3 (b) ———
1 024
11. (a) 80(0.90) n - 1 , n = 1, 2, 3, 4, …
(b) 31 cm PAPER 2
12. 102 1. (a) Graph 2 because the graph obtained is a straight
line.
HOTS PRACTICES PAK-21 Graf 2 kerana graf yang diperoleh merupakan
9
8
1
1. a = 9, r = —; a = —, r = — satu garis lurus.
9 8 9 (b) Graph 1 because the graph obtained is a curve.
Graf 1 kerana graf yang diperoleh merupakan
satu lengkung.
Pract ice Linear Law 2. (a) For Table 1/ Bagi Jadual 1:
spm 6 Hukum Linear y
30
PAPER 1
1. (a) 2.7 25
(b) 17
3
2. y = – —x + 9 20
2
1
3. (a) y = – —x + 4 15
2
(b) 4
4. (a) 10 10
(b) 1
5
5. p = –4, q = 8
b
6. (a) a = — x
2 –4 –2 0 2 4 6
6 – y
(b) m = 2 ⇒ —––– = 2 ⇒ y = 0
3 – 0 –5
Hence, passing through the origin (0, 0).
Maka, melalui titik asalan (0, 0). For Table 2/ Bagi Jadual 2:
y y q 1 y
1 2
7. (a) — = 4x – 3 (b) — = — x + —
x x p p
(c) log y = (log n)(x + 3) + log m 8
10 10 10
8. h = –4, k = 2 6
m – 1
2
1
2
9. (a) xy = — ––– x + 5
2
(b) m = –5, n = 1 4
10. p = 18 – 3q 2
1
1
11. h = —, k = – — 0 x 2
6
4
2 4 6 8 10 12
12. (a) log y = (log w)x + 3 –2
10
10
1
(b) k = 1, w = ——
100 –4
13. y = –4x + x + 12
2
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