標題:
hkcee past paper 98 long question 16(a)
發問:
hkcee past paper 98 long question 16(a) 怎樣做啊?
最佳解答:
16a) By similar triangles, the base radius of the melted ice-cream ( rcm ): ( 2x + 3 ) / 8 = r / 4 2x + 3 = 2r r = ( 2x + 3 ) / 2 The volume of the liquid in the cup: ( π/ 3 )r^2 h = ( π/ 3 ){ ( 2x + 3 )^2 / 4 }( 2x + 3 ) = π( 2x + 3 )^3 / 12 Then the total volume of the melted ice-cream: ( 4π/ 3 )( 2 ^3 ) + ( 4π/ 3 )( x^3 ) = ( 4π/ 3 )( 8 + x^3 ) As given, π( 2x + 3 )^3 / 12 = ( 4π/ 3 )( 8 + x^3 ) ( 2x + 3 )^3 / 4 = 4 ( 8 + x^3 ) ( 2x + 3 )^3 = 16 ( 8 + x^3 ) ( 2x )^0( 3 )^3 + 3 ( 2x )( 3 )^2 + 3 ( 2x )^2 ( 3 ) + ( 2x )^3 ( 3^0) = 128 + 16x^3 27 + 54x + 36x^2 + 8x^3 = 128 + 16x^3 Therefore 8x^3 - 36x^2 - 54x + 101 = 0 b) I also do part b for yr reference. From a), 8x^3 - 36x^2 - 54x + 101 = 0 4 ( 2x^3 - 9x^2 ) - 54x + 101 = 0 4y - 54x + 101 = 0 4y = 54x - 101 y = 27x / 2 - 101 / 4 Add this line to the given graph and you will get the value of x.
其他解答:
hkcee past paper 98 long question 16(a)
發問:
hkcee past paper 98 long question 16(a) 怎樣做啊?
最佳解答:
16a) By similar triangles, the base radius of the melted ice-cream ( rcm ): ( 2x + 3 ) / 8 = r / 4 2x + 3 = 2r r = ( 2x + 3 ) / 2 The volume of the liquid in the cup: ( π/ 3 )r^2 h = ( π/ 3 ){ ( 2x + 3 )^2 / 4 }( 2x + 3 ) = π( 2x + 3 )^3 / 12 Then the total volume of the melted ice-cream: ( 4π/ 3 )( 2 ^3 ) + ( 4π/ 3 )( x^3 ) = ( 4π/ 3 )( 8 + x^3 ) As given, π( 2x + 3 )^3 / 12 = ( 4π/ 3 )( 8 + x^3 ) ( 2x + 3 )^3 / 4 = 4 ( 8 + x^3 ) ( 2x + 3 )^3 = 16 ( 8 + x^3 ) ( 2x )^0( 3 )^3 + 3 ( 2x )( 3 )^2 + 3 ( 2x )^2 ( 3 ) + ( 2x )^3 ( 3^0) = 128 + 16x^3 27 + 54x + 36x^2 + 8x^3 = 128 + 16x^3 Therefore 8x^3 - 36x^2 - 54x + 101 = 0 b) I also do part b for yr reference. From a), 8x^3 - 36x^2 - 54x + 101 = 0 4 ( 2x^3 - 9x^2 ) - 54x + 101 = 0 4y - 54x + 101 = 0 4y = 54x - 101 y = 27x / 2 - 101 / 4 Add this line to the given graph and you will get the value of x.
其他解答:
此文章來自奇摩知識+如有不便請留言告知
文章標籤
全站熱搜
留言列表
