Question

HELP ASAP I WILL GIVE U BRAINLIEST

Answer 1

Answer: the answer is probably b theoreticaly

Step-by-step explanation:

Answer 2

Answer:

The final answer is $\frac{8}{27}$ .

Step-by-step explanation:

Solve the following equation:

$(2^{2} \cdot 3^{-1} \cdot 4^{-1})^{-1} \over 2^{-3} \cdot 3^{4}$

-Simplify $2^{2}$ :

$(2^{2} \cdot 3^{-1} \cdot 4^{-1})^{-1} \over 2^{-3} \cdot 3^{4}$

$(4 \times 3^{-1} \times 4^{-1})^{-1} \over 2^{-3} \times 3^{4}$

-Simplify $3^{-1}$ :

$(4 \times 3^{-1} \times 4^{-1})^{-1} \over 2^{-3} \times 3^{4}$

$(4 \times( \frac{1}{3}) \times 4^{-1})^{-1} \over 2^{-3} \times 3^{4}$

-Multiply both $4$ and $\frac{1}{3}$ :

$(4 \times( \frac{1}{3}) \times 4^{-1})^{-1} \over 2^{-3} \times 3^{4}$

$(\frac{4}{3} \times 4^{-1})^{-1} \over 2^{-3} \times 3^{4}$

-Simplify $4^{-1}$ :

$(\frac{4}{3} \times 4^{-1})^{-1} \over 2^{-3} \times 3^{4}$

$(\frac{4}{3} \times (\frac{1}{4}) )^{-1} \over 2^{-3} \times 3^{4}$

-Multiply both $\frac{4}{3}$ and $\frac{1}{4}$ together:

$(\frac{4}{3} \times (\frac{1}{4}) )^{-1} \over 2^{-3} \times 3^{4}$

$(\frac{1}{3})^{-1} \over 2^{-3} \times 3^{4}$

-Simplify $\frac{1}{3}^{-1}$ :

$(\frac{1}{3})^{-1} \over 2^{-3} \times 3^{4}$

$3 \over 2^{-3} \times 3^{4}$

-Then, on the denominator,  simplify $2^{-3}$ :

$3 \over 2^{-3} \times 3^{4}$

$3 \over \frac{1}{8} \times 3^{4}$

Simplify $3^{4}$ :

$3 \over \frac{1}{8} \times 3^{4}$

$3 \over \frac{1}{8} \times 81$

Multiply both $\frac{1}{8}$ and $81$ together:

$3 \over \frac{1}{8} \times 81$

$3 \over \frac{81}{8}$

Divide $3$ and $\frac{81}{8}$ by multiplying $3$ by the reciprocal of $\frac{81}{8}$ :

$3 \over \frac{81}{8}$

$3 \times (\frac{81}{8})$

-And you multiply $3$ and $\frac{81}{8}$ together:

$3 \times (\frac{81}{8})$

$\frac{8}{27}$

So, the final answer would be $\frac{8}{27}$ .