\(\begin{array}{c|c} x & 1 & 4 & p \\ \hline ...
x14py0.512.5. The table above satisfies the relation y = k√x, where k is a positive constant. Find the value of K.
0.5
1
1.5
2
Correct answer is A
y = k√x when y = 1, x = 4
k = y√x
= 1√4
= 12
= 0.5