Respuesta :
Answer:
b. 25/32
Step-by-step explanation:
To evaluate the function
[tex]f(x,y) = \frac{x^{-2}y^{0}}{x^{3}y^{-2}}[/tex]
When [tex]x = 2, y = 5[/tex] we replace x by 2 and y by 5 in the function. So:
[tex]f(x,y) = \frac{x^{-2}y^{0}}{x^{3}y^{-2}}[/tex]
[tex]f(2,5) = \frac{2^{-2}5^{0}}{2^{3}5^{-2}}[/tex]
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:
[tex]f(2,5) = \frac{5^{2}5^{0}}{2^{3}2^{2}}[/tex]
[tex]f(2,5) = \frac{5^{2}*1}{8*4} = \frac{25}{32}[/tex]
The correct answer is
b. 25/32