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\[\huge \frac{ x }{ (2x^0)^2 }\]
\[\large \frac{ x }{ 4 }\]
@Nnesha
what happened to x^0 :o ???
\[\large (2x^0)^2 = 2^2 \times (x^0)^2 = 4 \times x^0 = 4 \times 1 = 4\] would the x^0 be considered as 1 ?
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PERFECT!yes (anything)^0 = !
1**
So my answer is correct?
yes!
Thank you!
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(anything except 0)^0 = 1
np :=)
ye (hartnn)^0= 1
yay! i am non-0 :P :D
ohh wait i just noticed double nn's at the end of ur username!! :o
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and at the beginning of yours :)
yeah hahah
hartnnesha
LOL!
heheh ^.*
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