*Simplify first 3*
1. (-2)^4 X^3 Y^6
----------------------
(-2)^2 X^-2 Y^-3
2. Pretend that the little paranthesis that i'm going to insert into this problem are one BIG one that cover both of the number. You know..like the gigantic ( and ).
(3X^2)
---------
(5Y^3) and this entire fraction is squared by -2
3. (4x^2 y^3)(5x^-3 y^-2 Z^0)
1. (-2)^4 X^3 Y^6
----------------------
(-2)^2 X^-2 Y^-3
2. Pretend that the little paranthesis that i'm going to insert into this problem are one BIG one that cover both of the number. You know..like the gigantic ( and ).
(3X^2)
---------
(5Y^3) and this entire fraction is squared by -2
3. (4x^2 y^3)(5x^-3 y^-2 Z^0)
-
1.
16x^3y^6
---------------
4x^-2y^-3
you can bring the x and y to the top if you change the negative to a positive
16x^3 y^6 x^2 y^3
------------------------
4
=4x^5 y^9
2. since x^-2 = 1/x^2, you can flip the fraction and take it to the power of +2 instead
(5y^3)^2
------------
(3x^3)^2
25y^6
---------
9x^6
3. anything to the power of 0 = 1, so you can ignore that Z term
then just multiply the corresponding constant, x and y
4*5 = 20
x^2 * x^-3 = x^-1 = 1/x
y^3 * y^-2 = y^1 = y
therefore the final answer is:
= 20y/x
16x^3y^6
---------------
4x^-2y^-3
you can bring the x and y to the top if you change the negative to a positive
16x^3 y^6 x^2 y^3
------------------------
4
=4x^5 y^9
2. since x^-2 = 1/x^2, you can flip the fraction and take it to the power of +2 instead
(5y^3)^2
------------
(3x^3)^2
25y^6
---------
9x^6
3. anything to the power of 0 = 1, so you can ignore that Z term
then just multiply the corresponding constant, x and y
4*5 = 20
x^2 * x^-3 = x^-1 = 1/x
y^3 * y^-2 = y^1 = y
therefore the final answer is:
= 20y/x