Which expression is equivalent to `((2xy)^(-2))/(4y^(-5))`?

Options

A. `-(y^(3))/(x^(2))`
B. `-(y^(3))/(16x^(2))`
C. `(y^(3))/(x^(2))`
D. `(y^(3))/(16x^(2))`

Answer

To simplify the expression \(\frac{(2xy)^{-2}}{4y^{-5}}\), we will follow these steps:

### Step 1: Rewrite the expression
Start with the original expression:
\[ \frac{(2xy)^{-2}}{4y^{-5}} \]

### Step 2: Apply the negative exponent
Using the property \(a^{-n} = \frac{1}{a^n}\), rewrite the numerator:
\[ (2xy)^{-2} = \frac{1}{(2xy)^2} \]
Thus, the expression becomes:
\[ \frac{1}{(2xy)^2 \cdot 4y^{-5}} \]

### Step 3: Expand the numerator
Now, expand \((2xy)^2\):
\[ (2xy)^2 = 2^2 \cdot x^2 \cdot y^2 = 4x^2y^2 \]
So the expression now looks like:
\[ \frac{1}{4x^2y^2 \cdot 4y^{-5}} = \frac{1}{4x^2y^2} \cdot \frac{1}{4y^{-5}} \]

### Step 4: Simplify the denominator
Now, simplify \(4y^{-5}\):
\[ 4y^{-5} = 4 \cdot \frac{1}{y^5} = \frac{4}{y^5} \]
Thus, we can rewrite the expression as:
\[ \frac{1}{4x^2y^2} \cdot \frac{y^5}{4} = \frac{y^5}{16x^2y^2} \]

### Step 5: Combine the terms
Now, combine the terms in the fraction:
\[ \frac{y^5}{16x^2y^2} = \frac{y^{5-2}}{16x^2} = \frac{y^3}{16x^2} \]

### Final Result
The final simplified expression is:
\[ \frac{y^3}{16x^2} \]

### Conclusion
Thus, the expression equivalent to \(\frac{(2xy)^{-2}}{4y^{-5}}\) is:
\[ \frac{y^3}{16x^2} \]