[tex]-\frac{9}{4}[/tex]
Explanation
[tex]-\frac{4x^3}{3y^2}[/tex]
Step 1
to find the answer just replace the values for x and y
for x=3 and y =4
[tex]\begin{gathered} -\frac{4x^3}{3y^2} \\ -\frac{4(3)^3}{3(4)^2}=-\frac{4\cdot(3\cdot3\cdot3)}{3\cdot(4\cdot4)}=-\frac{4\cdot27}{3\cdot16}=-\frac{27}{12}=-\frac{9}{4} \end{gathered}[/tex]
I hope this helps you
