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X varies directly with the square of y, where x>0,and y >0. If x = 3 when y = 5, then what is the valueof y when x = 48?

Sagot :

x varies directly with y²

You can express this relationship as:

[tex]x=ky^2\text{ }\forall x>0,y>0[/tex]

Where k is the coefficient of variation.

Using the values of x=3 and y=5 you can calculate the coefficient of variation k:

[tex]\begin{gathered} 3=k(5)^2 \\ 3=k\cdot25 \\ k=\frac{3}{25} \\ k=0.12 \end{gathered}[/tex]

So the equation of the relationship is:

[tex]x=0.12y^2[/tex]

With this you can calculate y when x=48 as:

[tex]\begin{gathered} 48=0.12y^2 \\ \frac{48}{0.12}=\frac{0.12y^2}{0.12} \\ 400=y^2 \\ y=\sqrt[]{400} \\ y=20 \end{gathered}[/tex]

So for this relationship, when x=48, y=20

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