Suppose that $$ y $$ varies directly with $$ x $$, and $$ y=15 $$ when $$ x=6 $$. (a) Write a direct variation equation that relates $$ x $$ and $$ y $$. Equation:

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y = 5/2 * x Since $$ y $$ varies directly with $$ x $$, we can express this relationship with the equation: $$ y = kx $$ where $$ k $$ is the constant of variation. To find $$ k $$, we can use the given values $$ y = 15 $$ when $$ x = 6 $$. Substituting these values into the equation: $$ 15 = k \cdot 6 $$ To solve for $$ k $$, divide both sides by 6: $$ k = \frac{15}{6} = \frac{5}{2} $$ Now that we have the value of $$ k $$, we can write the direct variation equation: $$ y = \frac{5}{2}x $$ So the direct variation equation that relates $$ x $$ and $$ y $$ is: $$ \boxed{y = \frac{5}{2}x} $$

Suppose that \( y \) varies directly with \( x \), and \( y=15 \) when \( x=6 \). (a) Write a direct variation equation that relates \( x \) and \( y \). Equation:

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Suppose that \( y \) varies directly with \( x \), and \( y=15 \) when \( x=6 \). (a) Write a direct variation equation that relates \( x \) and \( y \). Equation:

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