Page 36 - Hybrid PBD 2021 Form 1 - Matematik

Matematik Tingkatan 1 Bab 1

1.4 Perpuluhan Positif dan Perpuluhan Negatif Buku Teks
PBD
PBD Positive and Negative Decimals ms. 19 – 23
PBD
20. Bulatkan nilai P, Q dan R yang betul. SP 1.4.1 TP1
Circle the correct values of P, Q and R.

(a) (b)

–6.4 P Q 1.28 R –9.54 P –5.24 Q 1.21 R

P –3.84 –5.12 P –7.39 –5.24
Q –2.56 0 Q 1.21 –0.94
R 2.56 3.84 R 3.36 3.72

21. Banding dan susun perpuluhan berikut mengikut tertib yang dinyatakan. SP 1.4.2 TP1
Compare and arrange the following decimals in the stated order.

(a) 10.6, –15.9, 31.8, 21.2, –21.2 (Tertib menurun / Descending order)

–21.2 –15.9 10.6 21.2 31.8 Tip Penting
Perpuluhan positif ialah
Tertib menurun / Descending order : perpuluhan yang lebih besar
31.8, 21.2, 10.6, –15.9, –21.2 daripada sifar. Perpuluhan
negatif ialah perpuluhan yang
(b) –0.231, –0.105, –0.147, –0.021, –0.168 (Tertib menaik / Ascending order) kurang daripada sifar.
Positive decimal is decimal that is
bigger than zero. Negative decimal
is decimal that is less than zero.
–0.231 –0.168 –0.147 –0.105 –0.021
Tertib menaik / Ascending order :
–0.231, –0.168, –0.147, –0.105, –0.021

22. Rajah di bawah menunjukkan satu garis nombor. SP 1.4.3 Modul HEBAT M21 TP2
The diagram below shows a number line.

p q –0.8 0 r 1.2 1.6 s
–2.0 –1.6 –1.2 –0.4 0.4 0.8 2.0 2.4

Cari nilai bagi setiap yang berikut berdasarkan garis nombor itu.
Find the value of each of the following based on the number line.

(a) p + q × r – s (b) p × s
p × q – r = –2.0 + (–1.2) × 0.8 – 2.4 q –5
= –2.0 × (–1.2) – 0.8 = –2.0 – 0.96 – 2.4 = –2.0 × 2.4
= –2.0 – 1.2 – 0.8 = –5.36 –1.2 –3
5
= –4.0 = × 2.4 0.8
3 1
= 4
(c) (q – r + s) ÷ 0.4 (d) 2p ÷ 100 (e) q + r + s
= (–1.2 – 0.8 + 2.4) ÷ 0.4 3q 81 5
= 0.4 ÷ 0.4 = 2(–2.0) × 81 = –1.2 + 0.8 + 2.4
= 1.0 3(–1.2) 100 2 5
–4 –1 81 9 =
= × 5
–3.6 –0.4 100 25 = 0.4
= 9
10
= 0.9

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01 Hybrid PBD Mate Tg1.indd 9 03/09/2021 3:24 PM