Suppose x varies inversely as y, y varies directly as the square of t and x = 1, when t = 3. Find x when t = \(\frac{1}{3}\).
The correct answer is A. 81
Since x varies inversely to y, we can write their relationship as x = k/y, where k is the constant of proportionality.
Since y varies directly as the square of t, we can write their relationship as y = kt^2, where k is the constant of proportionality.
Substituting the second equation into the first equation, we get x = k/(kt^2), which simplifies to x = 1/t^2.
When t = 3, x = 1, so we can substitute these values into the equation x = 1/t^2 to find the value of k:
1 = 1/3^2
k = 9
Now that we know the value of k, we can use the equation x = k/t^2 to find the value of x when t = 1/3:
x = 9/(1/3)^2
x = 9 * 9
x = 81
So, when t = 1/3, x is 81.
Previous question Next question