Question

Apply the properties of integer exponents to generate an equvalint algebraic expression to (4x^4 y^-4)^3/(x^7y^3)^2

Answer 1

what is the answer

Answer 2

can you write it normally

Answer 3

what is that i don't understand write regularly

Answer 4

To simplify the expression \((4x^4 y^{-4})^3/(x^7y^3)^2\) using properties of integer exponents, we can proceed step by step.

1. **Simplify the numerator**:
\[
(4x^4 y^{-4})^3 = 4^3 (x^4)^3 (y^{-4})^3
\]
Now calculate each part:
\[
4^3 = 64, \quad (x^4)^3 = x^{12}, \quad (y^{-4})^3 = y^{-12}
\]
Therefore:
\[
(4x^4 y^{-4})^3 = 64x^{12}y^{-12}
\]

2. **Simplify the denominator**:
\[
(x^7 y^3)^2 = (x^7)^2 (y^3)^2
\]
Now calculate each part:
\[
(x^7)^2 = x^{14}, \quad (y^3)^2 = y^{6}
\]
Therefore:
\[
(x^7 y^3)^2 = x^{14} y^{6}
\]

3. **Combine the results**:
Now combine the results from the numerator and denominator:
\[
\frac{64x^{12}y^{-12}}{x^{14}y^{6}}
\]

4. **Apply the properties of exponents**:
Using the property of exponents \(\frac{a^m}{a^n} = a^{m-n}\), we can simplify:
\[
= 64 \cdot \frac{x^{12}}{x^{14}} \cdot \frac{y^{-12}}{y^{6}}
\]
Thus:
\[
= 64 \cdot x^{12 - 14} \cdot y^{-12 - 6}
\]
Which simplifies to:
\[
= 64 \cdot x^{-2} \cdot y^{-18}
\]

5. **Final expression**:
Since \(x^{-2} = \frac{1}{x^2}\) and \(y^{-18} = \frac{1}{y^{18}}\), we can write the final expression as:
\[
= \frac{64}{x^2 y^{18}}
\]

Thus, the equivalent algebraic expression is:

\[
\frac{64}{x^2 y^{18}}
\]

Answer 5

The answer to the expression \((4x^4 y^{-4})^3/(x^7y^3)^2\) simplified using the properties of integer exponents is:

\[
\frac{64}{x^2 y^{18}}
\]

Answer 6

Certainly! The simplified expression is:

**64 / (x² * y¹⁸)**.

Answer 7

I apologize for any confusion. Here's the answer written in a straightforward way:

**64 divided by (x squared times y raised to the 18th power)**.

In mathematical notation, that looks like:

**64 / (x² * y¹⁸)**.

If you have any further questions or need additional explanations, feel free to ask!