Could you solve these and explain?
(3x^5 y^0)^ -1
(2x^2)(2x^-4)^2
(-2p^3)(3p^5)
=1/(3x^5 y^0)
=1/(3x^5 )
=1/3 (x^-5)
=(2x^2)(4x^-8)
=8x^-6
= -6 p^8
1).
(3x^5 y^0)^-1
= 1/3x^5
2).
= 8/x^6
3).
= -6p^8
1) 1/(3x^5)
2) (2x^2)(4x^-8)
8/(x^6)
3) -6p^8
I hope this information was very helpful.
Comments
(3x^5 y^0)^ -1
=1/(3x^5 y^0)
=1/(3x^5 )
=1/3 (x^-5)
(2x^2)(2x^-4)^2
=(2x^2)(4x^-8)
=8x^-6
(-2p^3)(3p^5)
= -6 p^8
1).
(3x^5 y^0)^-1
= 1/3x^5
2).
(2x^2)(2x^-4)^2
= 8/x^6
3).
(-2p^3)(3p^5)
= -6p^8
1) 1/(3x^5)
2) (2x^2)(4x^-8)
8/(x^6)
3) -6p^8
I hope this information was very helpful.