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what is the value of 3x2y3c when x=3 y=-2 and c= 1/3?
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3*(3^2)*(-2^3)*(1/3)
Solution.. \[(3x^{2})(y ^{3})c\] \[(3(3)^{2})(-2^{2})(1/3)\]\[(3(9))(4)(1/3)\]\[(27)(4)(1/3)\]\[108/3\]\[36\] This is the answer...
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