Question

suppose that y is inversely proportional to the cube of x, and that y = 6 when x = 11. what is y when x = 6? round your answer to two decimal places if necessary.

Explanation:

Step1: Write the inverse - proportion equation

Since $y$ is inversely proportional to the cube of $x$, the equation is $y=\frac{k}{x^{3}}$, where $k$ is the constant of proportionality.

Step2: Find the value of $k$

We know that $y = 6$ when $x = 11$. Substitute these values into the equation $y=\frac{k}{x^{3}}$:
$6=\frac{k}{11^{3}}$
$k=6\times11^{3}=6\times1331 = 7986$.

Step3: Find $y$ when $x = 6$

Substitute $k = 7986$ and $x = 6$ into the equation $y=\frac{k}{x^{3}}$:
$y=\frac{7986}{6^{3}}=\frac{7986}{216}\approx36.97$.

Answer:

$36.97$