PreCal
posted by Anonymous .
Solve for x.
3x^2+8=4

PreCal 
Steve
3x^2 = 12
x^2 = 4
x = ±2 
PreCal 
Anonymous
thanks and what about this one ?
Simplify the expression
(3x^1y^2)^3(5x^2y)0
I'm getting confused because of the negative exponent 
PreCal 
Steve
anything^0 = 1, so we just have
(3x^1y^2)^3 = 27 x^3 y^6 = 27y^6/x^3
negative exponents indicate a swap between numerator and denominator
x^3 = 1/x^3
1/x^3 = x^3
recall that x^a/x^b = x^(ab)
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