Problem: yyy varies inversely as the square of xxx. When y=16,x=1y=16, x=1y=16,x=1. When x=8,yx=8, yx=8,y equals
Answer Choices:
A. 222
B. 128128128
C. 646464
D. 1/41 / 41/4
E. 102410241024 Solution:
y=k/x2,16=k/1,k=16;∴y=16/x2,y=16/8⋅8=1/4;y=k / x^{2}, \quad 16=k / 1, \quad k=16 ; \quad \therefore y=16 / x^{2}, \quad y=16 / 8 \cdot 8=1 / 4; y=k/x2,16=k/1,k=16;∴y=16/x2,y=16/8⋅8=1/4;
or
y1/y2=x22/x12,16/y2=82/12;∴y2=1/4y_{1} / y_{2}=x_{2}^{2} / x_{1}^{2}, \quad 16 / y_{2}=8^{2} / 1^{2} ; \quad \therefore y_{2}=1 / 4 y1​/y2​=x22​/x12​,16/y2​=82/12;∴y2​=1/4