Question

Can someone help with 16 and 18

Answer 1

9514 1404 393

Answer:

16. k^5/j^3

18. 81y^20

Explanation:

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

(a^b)/(a^c) = a^(b-c) . . . . where b may be 0

(a^b)^c = a^(bc)

(ab)^c = (a^c)(b^c)

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We can simplify the given expressions as follows.

16.

`(k^6j^2)/(kj^5)=k^(6-1)j^(2-5)=k^5j^(-3)=\boxed{(k^5)/(j^3)}`

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18.

`(x^(-2))^2(3xy^5)^4=(x^(-2\cdot2))(3^4x^4y^(5\cdot4))=81x^(-4+4)y^(20)=\boxed{81y^(20)}`