標題:
發問:
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
最佳解答:
V=kTD^2 where k is a non-zero positive constant Let 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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f.4 Variations發問:
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
最佳解答:
V=kTD^2 where k is a non-zero positive constant Let 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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