If x = -2 and y = 1, then what is the value of $(\frac{4}{x}+y)(\frac{16}{x^2}+y^2 -\frac{4y}{x})$?
Answer & explanation
Correct answer: option 3
x = -2
y = 1
Then what is the value of $(\frac{4}{x}+y)(\frac{16}{x^2}+y^2 -\frac{4y}{x})$
= $(\frac{4}{-2}+1)(\frac{16}{-2^2}+1^2 -\frac{4×1}{-2})$
= (-1)(4+1+2)
= -7