Answer:
[tex]\frac{3}{y^{22}}[/tex]
Step-by-step explanation:
To simplify the expression we need to understand one rule of exponents.
That is:
[tex](ab)^x=a^xb^x[/tex]
Using this property, we can write:
[tex](3y^5)^{-2}*(2y^{-4})^{3}\\=(3)^{-2}(y^5)^{-2}*(3)^3 (y^{-4})^3\\[/tex]
Now, we use 2 other properties to simplify this further:
1. [tex]a^{-x}=\frac{1}{a^x}[/tex]
2. [tex](a^b)^c=a^{b*c}[/tex]
So, now we simplify further:
[tex](3)^{-2}(y^5)^{-2}*(3)^3 (y^{-4})^3\\=\frac{1}{3^2}y^{-10}*(27)y^{-12}\\=\frac{1}{9}y^{-10}(27)y^{-12}\\=(\frac{1}{9}*27)y^{-10}y^{-12}\\=3y^{-10-12}\\=3y^{-22}[/tex]
Or in positive exponents, that is:
[tex]3y^{-22}\\=\frac{3}{y^{22}}[/tex]
