Respuesta :
[tex] (-\dfrac{3^2}{3^3} )^2 = (-\dfrac{1}{3} )^2 = \dfrac{1}{9} [/tex]
OR (in more details):
[tex](-\dfrac{3^2}{3^3} )^2 = (-3^{2-3})^2 = (-3^{-1})^2 = (-\dfrac{1}{3} )^2 = \dfrac{1}{9} [/tex]
So either you meant [tex] (\frac{-3^2}{-3^3} )*\frac{2}{1} [/tex], or you actually meant [tex] (\frac{-3^2}{-3^3} )^2 [/tex] . I'll solve both.
[tex] (\frac{-3^2}{-3^3} )*\frac{2}{1} [/tex]
Firstly, do the division. Since the numerator and the denominator have the same base, you can write it as -3^-1, which can be rewritten as -1/3. [tex] -\frac{1}{3}*\frac{2}{1} [/tex]
Next, multiply the fractions, and your answer should be [tex] -\frac{2}{3} [/tex]
[tex] (\frac{-3^2}{-3^3} )^2 [/tex]
Firstly, multiply square with the powers in the fraction to get [tex] \frac{-3^4}{-3^6} [/tex]
Next, divide. Like before, they have the same base, so the answer can be written as -3^-2, which can be rewritten as 1/3^2 = 1/9.