Questions: Problem 20: (first taught in lesson 1) Consider the four relations: I. (x, y) x+y=8 II. (x, y) x^2+y^2=8 III. (x, y) x^2+y=8 IV. (x, y) x+y^2=8 Of the four relations, which are also functions of y in terms of x ? After you pick your answer press GO. A. I and II B. I and III C. I and IV D. III and II E. III and IV

Solution Steps

A relation is a function if each input \( x \) corresponds to exactly one output \( y \). In other words, for a relation to be a function, no vertical line should intersect its graph more than once.

Relation I is given by \( \{(x, y) \mid x + y = 8\} \). This can be rewritten as \( y = 8 - x \). For every value of \( x \), there is exactly one corresponding value of \( y \). Therefore, Relation I is a function.

Relation II is given by \( \{(x, y) \mid x^{2} + y^{2} = 8\} \). This represents a circle centered at the origin with radius \( \sqrt{8} \). For some values of \( x \), there are two corresponding values of \( y \) (one positive and one negative). Therefore, Relation II is not a function.

Relation III is given by \( \{(x, y) \mid x^{2} + y = 8\} \). This can be rewritten as \( y = 8 - x^{2} \). For every value of \( x \), there is exactly one corresponding value of \( y \). Therefore, Relation III is a function.

Final Answer