X varies inversely with square of Y and when X is 3, Y is 4. What is the value of X when Y is 16?
We can set up the equation for the inverse variation as follows:
X = k/Y^2
We are given that when X is 3, Y is 4. Plugging these values into the equation, we can solve for the constant k:
3 = k/4^2
3 = k/16
To solve for k, multiply both sides of the equation by 16:
48 = k
Now that we know the value of k, we can find the value of X when Y is 16:
X = 48/16^2
X = 48/256
X = 3/16
Therefore, when Y is 16, X is 3/16.