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Evaluate (x^-2y^0)/(x^3y^-2) when x = 2 and y = 5. a. 25/2 b. 25/32 c. 50

User PicxyB
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2 Answers

1 vote
The answer is B. 25/32
3 votes

Answer:

b. 25/32

Explanation:

To evaluate the function


f(x,y) = (x^(-2)y^(0))/(x^(3)y^(-2))

When
x = 2, y = 5 we replace x by 2 and y by 5 in the function. So:


f(x,y) = (x^(-2)y^(0))/(x^(3)y^(-2))


f(2,5) = (2^(-2)5^(0))/(2^(3)5^(-2))

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))


f(2,5) = (5^(2)*1)/(8*4) = (25)/(32)

The correct answer is

b. 25/32

User Ameer Tamboli
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6.2k points