How do you multiply #(2y ^ { 0} ) ^ { 4} \cdot 2x ^ { 5}#?

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Answer 1 · 2018-01-18T04:21:59

#color(blue)((2x)^5)#

Anything with power 0 is 1.

In our case #y^0 = 1#

Hence the first term of the sum is #(2 * y^0)^4 = (2* 1)^4 = 2^4#

Rewriting the sum,

#= 2^4 * 2 * x^5 = 2^(4+1) * x^5 = 2^5 * x^5 = color(blue)((2x)^5)#

Answer 2 · 2018-01-18T04:38:16

#(2y^0)^4*2x^5=color(blue)(32x^5#

Multiply:

#(2y^0)^4*2x^5#

Any number raised to the power of #0# is #1#. #n^0=1#

Simplify.

#(2*1)^4*2x^5#

Simplify #2*1# to #2#.

#2^4*2x^5#

Simplify #2^4# to #16#.

#16*2x^5#

Simplify.

#32x^5#