It is given that y^2 varies directly as (x-b), where y>0 and b is a constant. y=2 when x=3; y=4 when x=9. Find the value of b
做呢D數有咩技巧?
Variations
2012-10-28 10:07 pm
回答 (2)
2012-10-28 10:56 pm
✔ 最佳答案
y² ∝ x – b{Set a variable k}y² = k(x – b) {Substitute x, y}2² = k(3 – b) ⇒ k = 4/(3 – b) … (i)4² = k(9 – b) ⇒ k = 16/(9 – b) … (ii){Combine (i) & (ii)} 4/(3 – b) = 16/(9 – b) 4(9 – b) = 16(3 – b) 9 – b = 12 – 4b b = 1 I HOPE THIS CAN HELP YOU!~ ^_^2012-10-30 21:09:56 補充：
Yeah, k should be constant.
參考: My Maths World
2012-10-29 6:53 pm
Let y^2 = k(x-b), where k is a variation constant.
[note: k should not be a variable. In another word, must be a constant]
when x=3, y=2
2^2 = k(3-b)
k(3-b) = 4 ... (1)
when x=9, y=4
4^2 = k(9-b)
k(9-b) = 16 ... (2)
(2)/(1):
(9-b)/(3-b) = 16/4
After solving,
b = 1
(k=2)
[note: k should not be a variable. In another word, must be a constant]
when x=3, y=2
2^2 = k(3-b)
k(3-b) = 4 ... (1)
when x=9, y=4
4^2 = k(9-b)
k(9-b) = 16 ... (2)
(2)/(1):
(9-b)/(3-b) = 16/4
After solving,
b = 1
(k=2)
參考: maeducation.edu.hk
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