close
標題:
F.4 variations---
發問:
最佳解答:
(a) let y=ax+bx^2,where a and b is constant. Putting y=33,x=3 . y=-48, x=6. By solving, we have a=-14 b=1 So,y=-14x+x^2 Putting x=16 y=-14(16)+16^2 =32 (b) y=x^2-14x =(x-7)^2-49 So,y is minimum when x=7 y=-49
其他解答:
F.4 variations---
發問:
此文章來自奇摩知識+如有不便請留言告知
y is the sum of two parts. One part varies directly as x, while the other part varies directly as x^2. y=-33 when x=3 and y=-48 when x=6. (a) Express y in terms of x. Hence find the value of y when x=16 (b) Find the minimum possible value of y when x varies.最佳解答:
(a) let y=ax+bx^2,where a and b is constant. Putting y=33,x=3 . y=-48, x=6. By solving, we have a=-14 b=1 So,y=-14x+x^2 Putting x=16 y=-14(16)+16^2 =32 (b) y=x^2-14x =(x-7)^2-49 So,y is minimum when x=7 y=-49
其他解答:
文章標籤
全站熱搜
留言列表
