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03 mc ce maths Q. 16 Q18 Q.12
03 mc ce maths Q. 16 Q18 Q.12 解釋比我聽
最佳解答:
12. D 16. D 18. B 12. Total deposits after 4 years =$8000(1+1%)4 =$8324.83 ∴interest earned after 4 years =$8324.83-$8000 =$324.83 =$325(correct to the nearest dollar) 16. θ=360°/8 =45° 3/h=tan45°/2=tan22.5° ∴h=3/ tan22.5°=7.24 Area of the regular 8-sided polygon=(6*7.24/)2*8 = 174cm2 (correct to 3 sig.fig.) 18. Since AB:BC=1:2 and AF:FE=2:1 AB=x and BC=2x, where x is a constant AF=2y and FE=y, where y is a constant Let ∠EAC=θ ∠BCD=180°-θ (int. ∠s,AE//CD) So area of △ABF:area of △BCD =1/2(AB)(AF)sinθ:1/2(CB)(CD)sin(180°-θ) =1/2(x)(2y) sinθ:1/2(2x)(2y+y) sinθ (∵AE=CD) =xy:3xy =1:3
其他解答:63D0B758E2D502CC
03 mc ce maths Q. 16 Q18 Q.12
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發問:03 mc ce maths Q. 16 Q18 Q.12 解釋比我聽
最佳解答:
12. D 16. D 18. B 12. Total deposits after 4 years =$8000(1+1%)4 =$8324.83 ∴interest earned after 4 years =$8324.83-$8000 =$324.83 =$325(correct to the nearest dollar) 16. θ=360°/8 =45° 3/h=tan45°/2=tan22.5° ∴h=3/ tan22.5°=7.24 Area of the regular 8-sided polygon=(6*7.24/)2*8 = 174cm2 (correct to 3 sig.fig.) 18. Since AB:BC=1:2 and AF:FE=2:1 AB=x and BC=2x, where x is a constant AF=2y and FE=y, where y is a constant Let ∠EAC=θ ∠BCD=180°-θ (int. ∠s,AE//CD) So area of △ABF:area of △BCD =1/2(AB)(AF)sinθ:1/2(CB)(CD)sin(180°-θ) =1/2(x)(2y) sinθ:1/2(2x)(2y+y) sinθ (∵AE=CD) =xy:3xy =1:3
其他解答:63D0B758E2D502CC
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