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Maths 2001 ce mc question

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最佳解答:

01#53 x^2+y^2-8x-6y+21=0 (x^2-8x)+(y^2-6y)=-21 (x^2-8x+16)+(y^2-6y+9)=-21+16+9=4 (x-4)^2+(y-3)^2=2^2 So centre(4,3) radius=2 圓心至弦的垂線平分弦 因此 slope of centre to mid-point=(3-2)/(4-5)=-1 slope of chord=-1/-1=1 so the equation (y-2)/(x-5)=1 y-2=x-5 x-y-3=0 ans: E 01#47 (1-x)/(x^2+4x-5)+(x-1)/(x+1) =(1-x)/(x-1)(x+5)+(x-1)/(x+1) =-(x-1)/(x-1)(x+5)+(x-1)/(x+1) =-1/(x+5)+(x-1)/(x+1) =-1(x+1)+(x-1)(x+5)/(x+5)(x+1) =(-x-1+x^2+4x-5)/(x+5)(x+1) =(x^2+3x-6)/(x+5)(x+1) ANS: A 01#41(Mean deviation is out-c in the maths syllabus) Mean deviation=the total of │x-mean│/total Mean=(0+3+4+6+7)/5=4 Mean deviation =[(4-0)+(4-3)+(4-4)+(6-4)+(7-4)]/5 =2 ANS: D 01#26 Join AB,BC, AC Calculate the area of sector ABC three times and then minus two timres of area of triangle ABC Area of sector =3(pi)(1^2)(60/360) =(pi)/2 Area of two triangle ABC =2(1/2)(1)(1)(sin60) =sqrt3/2 The required area =(pi/2)-sqrt3/2 38. a>b -ab a+b>2b So II is not correct a>b a^2>b^2 not accurately correct Because a or b may be a negative number So A

其他解答:

Thx! But can you explain why q.26 is that 'calculate the area of sector ABC three times and then minus two timres of area of triangle ABC'