How do you divide #(4p ^ { 3} q ^ { 5} ) ^ { 0} \div ( 6p ^ { 2} ) ^ { - 1}#?

2 Answers
Ellie
Mar 18, 2017

#6p^2#

Explanation:

Any number to the 0 power is equal to 1

#(4p^3q^5)^0# = 1

Distribute the -1 to 6 and #p^2#

#1/((6^-1)(p^(2(-1))))# = #1/((6^-1)(p^-2))#

Any number to a negative exponent is equal to its reciprocal
For example: #a^-2# = #1/(a^2)#

#1/((6^-1)(p^-2))#= #6##p^2#

ANSWER: #6##p^2#

Tony B
Mar 18, 2017

#6p^2#

Explanation:

I have used substitutions a and b to demonstrate what is happening to groups. I do this in an attempt reduce confusion about the action of mathematical processes

Set #a=4p^3q^5# so that we have #a^0#

But #a^0=1# so #(4p^3q^5)^0=1#

Now we have: #1-:(6p^2)^(-1)#

Set #b=(6p^2)^(-1)#

#1-:b" "=" " 1xx1/b#

#=1xx1/((6p^2)^(-1)#

#=1xx6p^2#

#=6p^2#

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