SOLUTION: The intensity of illumination on a surface varies inversely as the square of the distance of the light from the surface. If the illumination from a source is 64 foot candles when t

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: The intensity of illumination on a surface varies inversely as the square of the distance of the light from the surface. If the illumination from a source is 64 foot candles when t      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 202901: The intensity of illumination on a surface varies inversely as the square of the distance of the light from the surface. If the illumination from a source is 64 foot candles when the distance is 9 feet, then find the illumination when the distance is 12 feet. Would it be 6, 8, 32 or 36 foot candles. Could someone please help me with this been trying but need some assistance, thanks.
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Intensity changes inversely with distance squared
I%5B%28d%29%5D+=+K%2Fd%5E2
I%5B%289%29%5D+=+K%2F9%5E2
K+=+81%2AI%5B%289%29%5D
I%5B%2812%29%5D+=+K%2F12%5E2
I%5B%2812%29%5D+=+%2881%2AI%5B%289%29%5D%29%2F144
You are given I%5B%289%29%5D+=+64. Thus
I%5B%2812%29%5D+=+81%2A64%2F144
I%5B%2812%29%5D+=+36