The value of y varies inversely as sqrt(x) and when x=2, y=4. What is x when y=16?
+9519 y varies inversely as sqrt(x) means that y = k / sqrt(x) for some non-zero constant k.
Substitute x = 2 and y = 4 into y = k / sqrt(x),
\(4 = \dfrac k{\sqrt 2}\\ k = 4 \sqrt 2\)
So k is 4 * sqrt(2). When y = 16 and k = 4 * sqrt(2), we have
\(16 = \dfrac{4\sqrt 2}x\\ \)
Can you take it from here?
