標題:
Maths 2001 ce mc question
發問:
http://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp.jpghttp://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp1.jpghttp://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp2.jpghttp://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp3.jpghttp://i226.photobucke... 顯示更多 http://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp.jpg http://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp1.jpg http://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp2.jpg http://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp3.jpg http://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp4.jpg Could anyone explain why the answer is that? thx~ 更新: 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'
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 -a<-b must be true for both positive and negative numbers So I true a>b 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'
Maths 2001 ce mc question
發問:
http://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp.jpghttp://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp1.jpghttp://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp2.jpghttp://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp3.jpghttp://i226.photobucke... 顯示更多 http://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp.jpg http://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp1.jpg http://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp2.jpg http://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp3.jpg http://i226.photobucket.com/albums/dd260/wingreally/maths2001cepp4.jpg Could anyone explain why the answer is that? thx~ 更新: 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'
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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 -a<-b must be true for both positive and negative numbers So I true a>b 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'
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