Respuesta :
We have that a varies directly as the square of b.
It means that when a increases, also b will increase.
Then, we can write the next equation:
[tex]a=kb^2[/tex]Where k is the constant of variation.
If a=5 when b=3
Then:
[tex]\begin{gathered} 5=k(3)^2 \\ 5=k9 \\ \text{Solve for k:} \\ k=\frac{5}{9} \end{gathered}[/tex]Now, we need to find the a value when b=2.
Hence:
[tex]\begin{gathered} a=kb \\ a=k(2)^2 \\ \text{Where the contant k=(}\frac{5}{9}) \end{gathered}[/tex]Replacing
[tex]\begin{gathered} a=(\frac{5}{9})2^2 \\ a=(\frac{5}{9})4 \\ a=\frac{20}{9} \end{gathered}[/tex]Therefore, when b=2, a value will be equal to 20/9.
