Answer:
[tex]y=5[/tex]
Step-by-step explanation:
[tex]\textsf{If }y \textsf{ is \underline{inversely proportional} to the square root of }x, \textsf{ then}:[/tex]
[tex]y \propto\dfrac{1}{\sqrt{x}} \implies y=\dfrac{k}{\sqrt{x}}\quad \textsf{(where k is some constant)}[/tex]
[tex]\textsf{When }x=25, y = 3:[/tex]
[tex]\implies 3=\dfrac{k}{\sqrt{25}}[/tex]
[tex]\implies 3=\dfrac{k}{5}[/tex]
[tex]\implies k=15[/tex]
Inputting the found value of k into the equation:
[tex]\implies y=\dfrac{15}{\sqrt{x}}[/tex]
To find the value of y when x is 9, substitute x = 9 into the found equation:
[tex]\implies y=\dfrac{15}{\sqrt{9}}[/tex]
[tex]\implies y=\dfrac{15}{3}[/tex]
[tex]\implies y=5[/tex]
