1.Write an irrational number whose square smaller than itself.State your reasoning.
2.Here is a metal sphere of radius R.You are required to recast it into smaller spheres to increase the total surface area as much as possible.The number of final products is unlimited.Suggest what you would do.
3.For the inequality 5×10^‐X>0.0000488
4.A worker wants to make a frustum from the cone such that the volume of the frustum is 87.5% of the volume of the original .Suggest a way .
5.Suggest a way to trisect the line segment joining A(-3,-1) and b(6,5) in a rectangular coordinates plane.
SOME MATHS Q(Good at English and Maths come here)
2008-08-01 7:24 am
回答 (2)
2008-08-05 8:31 am
✔ 最佳答案
Please refer to the solution below:圖片參考：http://hk.geocities.com/stevieg_1023/FRU1.gif
圖片參考：http://hk.geocities.com/stevieg_1023/FRU2.gif
2008-08-05 00:31:55 補充：
I am not able to finish no.4 , sor...>
2008-08-10 8:44 am
4. Aftermaking a frustum from it, the worker can get a smaller cone. It is similar to the large cone. Let’s the ratio of the heights and theradiuses of the cones is 1:x.
(a)(ii) Volume of large cone = πR^2H/3
Volume of small cone = π(R/x)^2(H/x)/3 = πR^2H/3x^3
Ratio = (πR^2H/3) / ( πR^2H/3x^3) = 8
(πR^2H/3) / ( πR^2H/3x^3) = 8
(πR^2H)/(3) x (3x^3)/ ( πR^2H) = 8
(3x^3)/ (3) = 8
x^3= 8
x =2
The ratio of the heights and the radiusesof the cones is 1:2.
He can make a frustum which the height ishalf of the original cone.
(a)(ii) Volume of large cone = πR^2H/3
Volume of small cone = π(R/x)^2(H/x)/3 = πR^2H/3x^3
Ratio = (πR^2H/3) / ( πR^2H/3x^3) = 8
(πR^2H/3) / ( πR^2H/3x^3) = 8
(πR^2H)/(3) x (3x^3)/ ( πR^2H) = 8
(3x^3)/ (3) = 8
x^3= 8
x =2
The ratio of the heights and the radiusesof the cones is 1:2.
He can make a frustum which the height ishalf of the original cone.
參考: myself
收錄日期: 2021-04-25 17:28:01
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