Can you help?

Question

1. (3x^-2y)^-1/(2xy^-2)^0

a) 1/3x^2y
b) 3xy
c) 3y/x^2
d) x^2/3y



2) [2pq^-1/4q^2]^-1

a) 2/pq
b) p/2q
c) 8pq^3
d) 2q^3/p

Thank you for your help

Answer 1

Hi,
1. (3x^-2y)^-1/(2xy^-2)^0

a) 1/3x^2y
b) 3xy
c) 3y/x^2
d) x^2/3y <== ANSWER

(3x^-2y)^-1/(2xy^-2)^0
Remember first that anything to the zero power = 1.
(3x^-2y)^-1/(2xy^-2)^0 = (3x^-2y)^-1/1 = (3x^-2y)^-1

(3x^-2y)^-1 = 3^(-1)x^(2)y^(-1) =

..x²
----- <== ANSWER
3y








2) [2pq^-1/4q^2]^-1

a) 2/pq
b) p/2q
c) 8pq^3
d) 2q^3/p <== ANSWER

[2pq^-1/4q^2]^-1 =

[2pq^-1]^-1
---------------- =
[4q^2]^-1

2^(-1)p^(-1) q^1
---------------------- =
4^(-1)q^(-2)

4^(1)q^(2) q^1
---------------------- =
2^(1)p^(1)


4q³
----- =
2p

2q³
----- <== ANSWER
..p

I hope that helps!! :-)

Answer 2

1. (3x^-2y)^-1/(2xy^-2)^0
= (3x^-2y)^-1/1 [anything to zero power is 1]
=(3y/x^2)^-1
= x^2/3y [just take reciprocal]

2) [2pq^-1/4q^2]^-1
= [2p/q / 4q^2]^-1
=[p/2q^3]^-1 = 2q^3/p

Answer 3

what exactly are we solving for now?

Answer 4

1/{3x-2y} !!!!!!!!!