Q1 : Given that y varies directly as x , where y>x, is it ture that (x+2) varies directly as (x+2)? Explain briefly
Q2 : It is given that y varies directly as x^2 where x>0. y=16 when x=t ; y=4 when x=t-1 . find the two possible value of t
find solution_Variation
2011-03-04 7:52 am
回答 (1)
2011-03-04 8:07 am
✔ 最佳答案
Q1 y varies directly as x. Let y = kxy + 2 = kx + 2
Since (kx + 2)/(x + 2) is not a constant, we conclude that (y + 2) do not vary directly as (x+2).
Q2 Let y = kx^2
kt^2 = 16
k(t - 1)^2 = 4
So, kt^2 - 2kt + k = 4
16 - 2kt + k = 4
2kt = 12 + k
2t = 12/k + 1
2t = (3/4)t^2 + 1
3t^2 - 8t + 4 = 0
(3t - 2)(t - 2) = 0
t = 2/3 or t = 2
收錄日期: 2021-04-26 14:06:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110303000051KK01494