Question

(x^-4y^0z^2)^-2
a) x^8/y^2z^4
b) 1/x^8z^4
c) x^8/z^4
d) x^6y/z^4

Answer 1

I solved the equation to
c) x^8/z^4
Is this correct?

Answer 2

oh wait.....by using exponent rule for negative exponents...If "a" is a real number other than "0" and "n" is an integer, then, a^-n=1/a^n and 1/a^-n=a^n. I attempted the equation again and solved to
b) 1/x^8z^4

Answer 3

x^-4 = 1/x^4
y^0 =1
z^2 =z^2
so than your exercise will be ( (z^2)/(x^4) )^-2

so than

1 1 x^8
--------------- = ----------- = ------
((z^2)/(x^4))^2 (z^4)/(x^8) z^4

hope so much that is understandably

good luck

bye

Answer 4

c. is right sure

Answer 5

Thank you!

Answer 6

yw

good luck

bye