What is #(-19y^0 z^4)/(-3z^16)#?

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Note: The numbers after the ^ sign represent exponents

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Answer 1 · 2016-01-05T17:30:02

#(-19y^0z^4)/(-3z^16)=19/(3z^12)#

#(-19y^0z^4)/(-3z^16)#

Dividing two negative numbers equals a positive number.

#(19y^0z^4)/(3z^16)#

Apply exponent rule #a^0=1 rArr y^0=1#.

#(19xx1xxz^4)/(3z^16)#

Simplify.

#(19z^4)/(3z^16)#

Apply exponent rule #a^m/a^n=a^((m-n))#+

#(19z^((4-16)))/(3)#

Simplify.

#(19z^-12)/3#

Apply negative exponent rule #a^(-m)=1/a^m#.

#19/(3z^12)#