What is [(3/4+1/3-1/6+1/3)^{-1}]^{-2} ?

The solution to [(3/4+1/3-1/6+1/3)^{-1}]^{-2} is 25/16

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simplify [(3/4+1/3-1/6+1/3)^{-1}]^{-2}

simplify

[(\frac{3}{4} + \frac{1}{3} - \frac{1}{6} + \frac{1}{3})^{- 1}]^{- 2}

\frac{25}{16}

Solution steps

((\frac{3}{4} + \frac{1}{3} - \frac{1}{6} + \frac{1}{3})^{- 1})^{- 2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}, a \geq 0

= (\frac{3}{4} + \frac{1}{3} - \frac{1}{6} + \frac{1}{3})^{- 1 \cdot (- 2)}

- 1 \cdot (- 2) = 2

= (\frac{3}{4} + \frac{1}{3} - \frac{1}{6} + \frac{1}{3})^{2}

\frac{3}{4} + \frac{1}{3} - \frac{1}{6} + \frac{1}{3} = \frac{5}{4}

= (\frac{5}{4})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{5 ^{2}}{4 ^{2}}

5^{2} = 25

= \frac{25}{4 ^{2}}

4^{2} = 16

= \frac{25}{16}

x + 7 > 2

a d + 3 a - d^{2} - 3 d

ab + a + b + 1

3 x + 10 = 7 x - 6

∣ 2 x + 1 ∣ + 4 \leq 13

What is [(3/4+1/3-1/6+1/3)^{-1}]^{-2} ?
The solution to [(3/4+1/3-1/6+1/3)^{-1}]^{-2} is 25/16