Suppose x varies inversely as y, y varies directly as the square of t and x = 1, when t = 3. Find x when t = \(\frac{1}{3}\).
 

  • A 81
  • B 27
  • C \(\frac{1}{9}\)
  • D \(\frac{1}{27}\)
  • E \(\frac{1}{81}\)

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.

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