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Question

If y varies direct to square of x and (x,y) = (40,12) then find the value of y when x is 80

Answer 1

y = kx^2

12 =k40^2 = 1600k

k = 12/1600 = 3/400

y = (3/400)x^2

y = (3/400)80^2

y= 3(6400)/400 = 3(16) = 48

y = 48 when x = 800