Question

Can someone help with 16 and 18

Answer 1

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Answer:

  16. k^5/j^3

  18. 81y^20

Step-by-step 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.

  $\dfrac{k^6j^2}{kj^5}=k^{6-1}j^{2-5}=k^5j^{-3}=\boxed{\dfrac{k^5}{j^3}}$

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18.

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