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\(\begin{array}{c|c} x 1 4 p \\ \hline y 0.5 1 2.5\end{array}\). The table above ...

\(\begin{array}{c|c} x & 1 & 4 & p \\ \hline y & 0.5 & 1 & 2.5\end{array}\). The table above satisfies the relation y = k\(\sqrt{x}\), where k is a positive constant. Find the value of K.
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  • A 0.5
  • B 1
  • C 1.5
  • D 2
Correct Answer: Option A
Explanation:
y = k\(\sqrt{x}\) when y = 1, x = 4
k = \(\frac{y}{\sqrt{x}}\)
= \(\frac{1}{\sqrt{4}}\)
= \(\frac{1}{2}\)
= 0.5

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