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If $y=kx^{\frac{1}{4}}$ and $y=3\sqrt{2}$ at $x=81$, what is the value of $y$ at $x=4$?

 Jun 19, 2023
 #1
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We can find k by substituting the known values of x and y into the equation.

3\sqrt{2} = k(81)^(1/4)

3\sqrt{2} = k(3)

k = 3\sqrt{2}/3 = \sqrt{2}

Now we can find y at x=4 by substituting this value of k into the equation.

y = \sqrt{2}(4)^(1/4) = \sqrt{2}(2) = 2\sqrt{2}

Therefore, the value of y at x=4 is 2\sqrt{2}.

 Jun 19, 2023
 #2
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None of your LaTex is displaying properly for me.

Melody  Jun 19, 2023

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