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a =
3+𝑏
5+2𝑏
a (5+2b) = 3+b
5a+2ab = 3+b
3+b-2ab = 5a
b(1-2a) = 5a-3
b =
5𝑎−3
1−2𝑎
log10 3 = 0.4771 → 100.4771 = 3
log10 𝑥 = 3.4771 → 103.4771
= 𝑥
𝑥 = 103
× 100.4771
= 1000 × 3
= 3000
𝑎2 + 𝑏2 = 𝑐2
∆𝐶𝐷𝐹: 𝐷𝐹2
= 52
+ (5 3)2
𝐷𝐹 = 10
DE = 10
tan 𝜃 =
𝐶𝐹
𝐶𝐷
=
5
5 3
𝑎𝑛𝑔𝑙𝑒 𝐶𝐷𝐹 = 𝑎𝑛𝑔𝑙𝑒 𝐶𝐷𝐹 = 𝜃
𝜃 = 30 °
𝜃 = 30°
∆𝐷𝐸𝐹
𝑎𝑛𝑔𝑙𝑒 𝐶𝐷𝐹 =
180 ° − 30 °
2
= 75 °
mean =
3 4 +6 10 +9 4 +12(2)
20
4,5,6
V = 144 𝜋
V hemisphere =
2
3
𝜋𝑟3
144 𝜋 =
2
3
𝜋𝑟3
r = 6
𝑥 = original price
𝑦 = selling price
𝑦 − 𝑥
𝑥
× 100 % = 20 %
𝑦 - 𝑥 = 0.2 𝑥
𝑦 = 1.2 𝑥−−−−①
0.9 𝑦 - 𝑥 = 24 −−②
?
?
CD = 𝐵𝐷 = 15 cm
angle ABC = 75°
angle ABE = 30°
angle EBD = 30°
angle CBD = 15°
∆𝐴𝐵𝐷
sin 30° =
𝐴𝐵
𝐵𝐷
1
2
=
𝐴𝐵
15
𝐴𝐵 = 7.5 cm
∆𝐴𝐵𝐸
sin 60° =
𝐴𝐵
𝐵𝐸
3
2
=
7.5
𝐵𝐸
𝐵𝐸 =
15
3
=
15 3
3
= 5 3 cm
= 9 − 1 2 + 2 − 8 2
= 8 2 + −6 2
AB = 10 AC = 𝐵𝐶 = 50
= 5 2
𝑎2 + 𝑏2 = 𝑐2
1
4
=
𝐴𝐷𝐵
2𝜋𝑟

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JUEC 初中 (2013 Past Year)

  • 1. a = 3+𝑏 5+2𝑏 a (5+2b) = 3+b 5a+2ab = 3+b 3+b-2ab = 5a b(1-2a) = 5a-3 b = 5𝑎−3 1−2𝑎
  • 2. log10 3 = 0.4771 → 100.4771 = 3 log10 𝑥 = 3.4771 → 103.4771 = 𝑥 𝑥 = 103 × 100.4771 = 1000 × 3 = 3000
  • 3. 𝑎2 + 𝑏2 = 𝑐2 ∆𝐶𝐷𝐹: 𝐷𝐹2 = 52 + (5 3)2 𝐷𝐹 = 10 DE = 10 tan 𝜃 = 𝐶𝐹 𝐶𝐷 = 5 5 3 𝑎𝑛𝑔𝑙𝑒 𝐶𝐷𝐹 = 𝑎𝑛𝑔𝑙𝑒 𝐶𝐷𝐹 = 𝜃 𝜃 = 30 ° 𝜃 = 30° ∆𝐷𝐸𝐹 𝑎𝑛𝑔𝑙𝑒 𝐶𝐷𝐹 = 180 ° − 30 ° 2 = 75 °
  • 4. mean = 3 4 +6 10 +9 4 +12(2) 20
  • 6. V = 144 𝜋 V hemisphere = 2 3 𝜋𝑟3 144 𝜋 = 2 3 𝜋𝑟3 r = 6
  • 7. 𝑥 = original price 𝑦 = selling price 𝑦 − 𝑥 𝑥 × 100 % = 20 % 𝑦 - 𝑥 = 0.2 𝑥 𝑦 = 1.2 𝑥−−−−① 0.9 𝑦 - 𝑥 = 24 −−②
  • 8. ? ? CD = 𝐵𝐷 = 15 cm angle ABC = 75° angle ABE = 30° angle EBD = 30° angle CBD = 15° ∆𝐴𝐵𝐷 sin 30° = 𝐴𝐵 𝐵𝐷 1 2 = 𝐴𝐵 15 𝐴𝐵 = 7.5 cm ∆𝐴𝐵𝐸 sin 60° = 𝐴𝐵 𝐵𝐸 3 2 = 7.5 𝐵𝐸 𝐵𝐸 = 15 3 = 15 3 3 = 5 3 cm
  • 9. = 9 − 1 2 + 2 − 8 2 = 8 2 + −6 2 AB = 10 AC = 𝐵𝐶 = 50 = 5 2 𝑎2 + 𝑏2 = 𝑐2 1 4 = 𝐴𝐷𝐵 2𝜋𝑟