Question

4^−8 : (4^−5)^2 = 16

Answer 1

Ответ:

хочу халявнве баллв, поэтому....

Answer 2

Let's break it down! The expression is $\frac{4^{-8}}{(4^{-5})^2}$.

First, remember that $a^{-n}$ is the same as $\frac{1}{a^n}$. So, we can rewrite the expression:

$\frac{1}{4^{8}} \div \frac{1}{(4^{5})^2}$

Now, when you divide by a fraction, it's the same as multiplying by its reciprocal:

$\frac{1}{4^{8}} \times \frac{(4^{5})^2}{1}$

Next, use the rule $a^{mn} = (a^m)^n$:

$\frac{1}{4^{8}} \times \frac{4^{10}}{1}$

Now, subtract the exponents when you multiply with the same base:

$\frac{1 \times 4^{10}}{4^{8} \times 1}$

Finally, subtract the exponents in the denominator:

$\frac{1 \times 4^{10}}{4^{8}}$

To simplify further, subtract the exponents in the numerator:

$\frac{4^{10-8}}{1}$

$\frac{4^2}{1}$

$16$

So, indeed, the expression is equal to 16!