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zero exponent property (2x^2y^-3/-12z^-1)^0
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Know that a^0 = 1
so all equals 1?
(2x^2y^-3/-12z^-1)^0 = 1
can i show u another problem?
yep, bearsams
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no problem
(-9x^-3z^4/6y^6)^0 (-2x^2)^3
Anything to the zero power is just one. It's the Law — the Laws of Exponents
Are they two different terms?
yah but 1 problem
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Then (-9x^-3z^4/6y^6)^0 (-2x^2)^3 = 1{(-2 x -2 x -2)(x^2)(x^2)(x^2)} = -8x^6
Are you okay with that problem?
im sorry but i dont know how to do that
In indices, \[a^m \times a^n= a^\left( m+n \right)\]
This implies \[(-2)^3 = -2\times-2\times-2=-8\]
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Also, \[x^2\times x^2\times x^2 = x^\left( 2+2+2 \right)\] which equals \[x^6\]
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