a varies directly as the square of b and inversely as the cube of c if a = 195 when b=2 and c=4 find a if b = 6 and c=9. round your answer to two decimal places if necessary
To solve this problem, we start by setting up the direct and inverse variation equation:
a = k * (b^2) / (c^3)
We can solve for k by substituting the given values of a, b, and c:
195 = k * (2^2) / (4^3)
195 = k * 4 / 64
195 = k / 16
k = 195 * 16
k = 3120
Now that we have the value of k, we can find a when b = 6 and c = 9:
a = 3120 * (6^2) / (9^3)
a = 3120 * 36 / 729
a = 153120 / 729
a ≈ 210.14
Therefore, when b = 6 and c = 9, a is approximately 210.14 (rounded to two decimal places).
To find the value of a when b = 6 and c = 9, we can use the direct variation relationship between a and the square of b and the inverse variation relationship between a and the cube of c.
Step 1: Write the direct variation equation: a = k * b^2, where k is the constant of variation.
Step 2: Write the inverse variation equation: a = k / c^3.
To solve for the constant of variation (k), we can substitute the given values of a, b, and c from the first scenario into either equation.
Using the given values a = 195, b = 2, and c = 4 in the direct variation equation:
195 = k * (2^2),
195 = k * 4,
k = 195 / 4 = 48.75.
Now we can use this value of k in the inverse variation equation to find the value of a when b = 6 and c = 9.
a = (48.75) / (9^3),
a = 48.75 / 729,
a ≈ 0.067 (rounded to two decimal places).
Therefore, a is approximately 0.067.