# If y is inversely proportional to the square of x and y =1/9 when x =4, find y when x = 12?

When y is inversely proportional to the square of x, it satisfies the equation

y = k/x^2

for some constant k. Using the given information, we can find k to be

k = y*x^2 = (1/9)*4^2 = 16/9

Then when x=12, we have

y = (16/9)/12^2 = 16/(9*16*9) = 1/81

thanked the writer.

When y is inversely proportional to the square of x, it satisfies the equation

y = k/x^2

for some constant k. Using the given information, we can find k to be

k = y*x^2 = (1/9)*4^2 = 16/9

Then when x=12, we have

y = (16/9)/12^2 = 16/(9*16*9) = 1/81

thanked the writer.
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}