Questions: If a varies directly as m and n^2 and inversely as y^3, and a=9 when m=4, n=9, and y=3, find a if m=7, n=2, and y=5. a=0 (Type an integer or a simplified fraction.)

$If a varies directly as m and n^2 and inversely as y^3, and a=9 when m=4, n=9, and y=3, find a if m=7, n=2, and y=5. a=0 (Type an integer or a simplified fraction.)$

Transcript text: If a varies directly as $m$ and $n^{2}$ and inversely as $y^{3}$, and $a=9$ when $m=4, n=9$, and $y=3$, find a if $m=7, n=2$, and $y=5$. $a=0$ (Type an integer or a simplified fraction.)

Solution

Solution Steps

Step 1: Determine the constant of proportionality ($k$)

To find the constant of proportionality $k$, we use the initial conditions given and the formula $a = k \cdot \frac{m \cdot n^2}{y^3}$. Substituting the initial values into the formula, we get $k = \frac{a_{initial} \cdot y_{initial}^3}{m_{initial} \cdot n_{initial}^2} = 0.75$.

Step 2: Find the new value of $a$

Using the previously found $k$ and the new values of $m$, $n$, and $y$, we substitute them into the formula to find the new value of $a$: $a = k \cdot \frac{m_{new} \cdot n_{new}^2}{y_{new}^3} = 0$.

Final Answer:

The new value of $a$, rounded to 0 decimal places, is 0.

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