Mathematics, 04.09.2019 16:30 segobins
X3 (x 2 β5)=β4xspace, x, start superscript, 3, end superscript, left parenthesis, x, start superscript, 2, end superscript, minus, 5, right parenthesis, equals, minus, 4, x if x > 0x> 0x, is greater than, 0, what is one possible solution to the equation above?
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X3 (x 2 β5)=β4xspace, x, start superscript, 3, end superscript, left parenthesis, x, start superscri...
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