How do you simplify #5( x ^ { 2} y ^ { 5} ) ^ { 0}#?

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Answer 1 · 2017-03-18T11:53:15

5

Any positive number with the index 0 will result in 1, no matter it being 0.0000...0001 or #oo#.

Hence,

#5(x^2y^5)^0=5(1)=5#

Answer 2 · 2017-03-18T12:02:51

5

#color(blue)("Preamble")#

Consider the following explanation by example.

It is known that #8-:2=4#

But #8-:2 = 8/2=(2^2xx2)/2 = 2^2=4#

Now look at what we can do with the indices (powers).

#8^1/2^1=2^3/2^1=2^(3-1)=2^2# So we can subtract indices in this context.

Now consider #2/2=1=2^1/2^1=2^(1-1)=2^0=1#

So raised to a power of 0 means that the whole becomes 1
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#color(blue)("Answering the question")#

Given:#" "5(x^2y^5)^0#

This means that whatever value is inside the brackets it end up being raised to the power of 0. Thus, eventually it has to become the value of 1

So #5(x^2y^5)^0" is the same as "5xx1=5#

Answer 3 · 2017-03-18T13:31:37

#5#

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

#:.=5(x^(2 xx 0)y^(5 xx 0))#

#a^0=1#

#:.=5( x^0y^0)#

#:.=5(1 xx 1))#

#:.=5#