The value V of a certain kind of circular gold medal varies jointly as its thickness T and the square of its diameter D.If the ratio of the thickness of two such medals is 3:4 and the ratio of their values is 25:48,find the ratio of their diameter.
Answer : 5:6
f.4 Variations
2007-02-16 6:25 am
回答 (1)
2007-02-16 7:08 am
✔ 最佳答案
V=kTD^2 where k is a non-zero positive constantLet the quantity of first medal be q1 and the second be q2.
i.e. T1:T2=3:4
Let T1=3m,T2=4m
V1:V2=25:48
Let V1=25n,V2=48n
25n=k(3m)(D1)^2
48n=k(4m)(D2)^2
(D1)^2=(25n)/(3mk)
(D2)^2=(48n)/(4mk)=(12n)/(mk)
Therefore (D1)^2:(D2)^2=(25/3):12=25:36
D1:D2=5:6
Therefore the ratio of their diameter us 5:6
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