y varies partly as the square of x and partly as the inverse of the square root of x.Write down the expression for y if y = 2 when x = 1 and y = 6 when x = 4
a
y = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\)
b
y = x2 + \(\frac{1}{\sqrt{x}}\)
c
y = x2 + \(\frac{1}{x}\)
d
y = \(\frac{x^2}{31} + \frac{1}{31\sqrt{x}}\)
Explanation
Correct Option
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