Step-by-step explanation:
Since, Y varies inversely as the square root of x.
[tex]\therefore \:y \: \alpha \: \frac{1}{ \sqrt{x} } \\ \therefore \: y \: = \: \frac{k}{ \sqrt{x} } ...(1)\: \\ ( \: where \: k = constant) \\ now, \: at \: x = 25 \: \: and \: \: y = 2 \\ 2 = \frac{k}{ \sqrt{25} } \\ \therefore \: 2 = \frac{k}{ \sqrt{25} } \\ \therefore \: 2 = \frac{k}{ 5 } \\ \therefore \: \: k = 2 \times 5 \\ \huge \red{ \boxed{ \therefore \: \: k = 10}} \\ \\ substituting \: k = 10 \: in \: eq. \: (1) \:we \: find : \\ y \: = \: \frac{10}{ \sqrt{x} } \\ \\ next, \: at \: x = 4 \\ y \: = \: \frac{10}{ \sqrt{4} } \\ \therefore \: y \: = \: \frac{10}{ 2 } \\ \huge \purple{ \boxed{\therefore \: y = 5}}[/tex]
