Question

suppose that y is proportional to the cube of x, and that y = 16 when x = 6. what is y when x = 14? round your answer to two decimal places if necessary.

Explanation:

Step1: Establish the proportional - relationship equation

Since $y$ is proportional to the cube of $x$, the equation is $y = kx^{3}$, where $k$ is the constant of proportionality. When $y = 16$ and $x = 6$, we substitute these values into the equation: $16=k\times6^{3}$.

Step2: Solve for the constant $k$

First, calculate $6^{3}=216$. Then, from $16 = k\times216$, we can solve for $k$ as $k=\frac{16}{216}=\frac{2}{27}$.

Step3: Find $y$ when $x = 14$

Substitute $k=\frac{2}{27}$ and $x = 14$ into the equation $y = kx^{3}$. We have $y=\frac{2}{27}\times14^{3}$. Calculate $14^{3}=14\times14\times14 = 2744$. Then $y=\frac{2\times2744}{27}=\frac{5488}{27}\approx203.26$.

Answer:

$203.26$