Answer:[tex]-\frac{4x^2y}{27}[/tex]
Step-by-step explanation:
1.[tex](-3x^{2} y^5)^-3=-(3x^2y^5)^-3[/tex]
2.[tex]\frac{-\left(3x^2y^5\right)^{-3}}{\left(2x^4y^8\right)^{-2}}[/tex]
*Apply The Fraction Rule [tex]\frac{-a}{b} =-\frac{a}{b}[/tex]*
3.[tex](2x^4y^8)^-2=\frac{1}{2^2 x^8 y^16}[/tex]
4.[tex]-\frac{\left(3x^2y^5\right)^{-3}}{\frac{1}{2^2x^8y^{16}}}[/tex]
5.[tex](3x^2y^5)^-3=\frac{1}{3^3 x^6 y^15}[/tex]
6.[tex]=-\frac{\frac{1}{3^3x^6y^{15}}}{\frac{1}{2^2x^8y^{16}}}[/tex]
7.[tex]=-\frac{\frac{1}{3^3x^6y^{15}}}{\frac{1}{2^2x^8y^{16}}}=\frac{2^2 x^8 y^16}{3^3 x^6 y^15}[/tex]
8.[tex]=-\frac{2^2x^8y^{16}}{3^3x^6y^{15}}[/tex]
*Subtract the numbers 8-6=2*
9.[tex]=\frac{2^2x^2y^{16}}{3^3y^{15}}[/tex]
*apply the Exponent Rule [tex]\frac{x^a}{x^b} =x^a-b[/tex]*
[tex]\frac{x^8}{x^6} =x^8^-^6[/tex]
10.[tex]=\frac{2^2y^{16}x^{8-6}}{3^3y^{15}}[/tex]
*Subtract the numbers 8-6=2
11.[tex]=\frac{2^2x^2y^{16}}{3^3y^{15}}[/tex]
apply the Exponent Rule [tex]\frac{x^a}{x^b} =x^a-b[/tex]*
[tex]\frac{y^16}{y^15} =y^16-^15[/tex]
12.[tex]\frac{2^2x^2y^{16-15}}{3^3}[/tex]
*Subtract the numbers 16-15=1
13.[tex]=-\frac{2^2x^2y}{3^3}[/tex]
Final Answer:
14.[tex]=-\frac{4x^2y}{27}[/tex]
I hope this helps :)
Brainlest Please!
