Answer:
b. 25/32
Explanation:
To evaluate the function
![f(x,y) = (x^(-2)y^(0))/(x^(3)y^(-2))](https://img.qammunity.org/2019/formulas/mathematics/college/277f75mkb7q2morikrk0be4n27h1lpjxxc.png)
When
we replace x by 2 and y by 5 in the function. So:
![f(x,y) = (x^(-2)y^(0))/(x^(3)y^(-2))](https://img.qammunity.org/2019/formulas/mathematics/college/277f75mkb7q2morikrk0be4n27h1lpjxxc.png)
![f(2,5) = (2^(-2)5^(0))/(2^(3)5^(-2))](https://img.qammunity.org/2019/formulas/mathematics/college/1owxiiv2z9imgyknkvdyw0xkwqd64s68az.png)
Expressions wth negative exponents in the numerator go to the denominator with positive exponent. On the denominator goes to the numerator with positive exponent. So:
![f(2,5) = (5^(2)5^(0))/(2^(3)2^(2))](https://img.qammunity.org/2019/formulas/mathematics/college/dr8kgim0j13u33cy3d2gyl73rjq3xx0ayl.png)
![f(2,5) = (5^(2)*1)/(8*4) = (25)/(32)](https://img.qammunity.org/2019/formulas/mathematics/college/85kzth2lundbk7rfd1zyfzn427005ki226.png)
The correct answer is
b. 25/32