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Sagot :
Given:
[tex]y\propto \dfrac{1}{\sqrt{x}}[/tex]
y = 4 when x = 64
To find:
The value of y when x=16.
Solution:
We have,
[tex]y\propto \dfrac{1}{\sqrt{x}}[/tex]
[tex]y=\dfrac{k}{\sqrt{x}}[/tex] ...(i)
Where, k is the constant of proportionality.
We have, y = 4 when x = 64.
[tex]4=\dfrac{k}{\sqrt{64}}[/tex]
[tex]4=\dfrac{k}{8}[/tex]
[tex]4\times 8=k[/tex]
[tex]32=k[/tex]
Putting k=32 in (i), we get
[tex]y=\dfrac{32}{\sqrt{x}}[/tex]
Putting x=16, we get
[tex]y=\dfrac{32}{\sqrt{16}}[/tex]
[tex]y=\dfrac{32}{4}[/tex]
[tex]y=8[/tex]
Therefore, the value of y is 8 when x=16.
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