SOLUTION: A compact disc (CD) is made such that the shortest distance between the edge of the centre hole and the edge of the disc is 53.0 mm. Find the radius of the centre-hole if 1.36% of

Algebra ->  Length-and-distance -> SOLUTION: A compact disc (CD) is made such that the shortest distance between the edge of the centre hole and the edge of the disc is 53.0 mm. Find the radius of the centre-hole if 1.36% of       Log On


   



Question 771243: A compact disc (CD) is made such that the shortest distance between the edge of the centre hole and the edge of the disc is 53.0 mm. Find the radius of the centre-hole if 1.36% of the disc is removed in making the hole.
Answer by oscargut(2103) About Me  (Show Source):
You can put this solution on YOUR website!
Let r = radius uf the center hole

r^2(pi) = 0.0136(r+53)^2 (pi)
r^2 = 0.0136(r+53)^2
(r/(r+53))^2 = 0.0136
r/(r+53) = 0.116619
r = (r+53)0.116619
0.883381r = 6.180809
r = 6.996765
(Rounding it r = 7)
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