W varies inversely as the square of t.  If W = 12 when t = 2. Find t when W = 27.Options :27t = 2427t2=48108 = 12t212t=54

The problem states that "W varies inversely as the square of t". This can be written as W = k/(t^2), where k is the constant of variation.

Given that W = 12 when t = 2, we can substitute these values into the equation to find k:

12 = k/(2^2) 12 = k/4 k = 12 * 4 k = 48

Now that we have the constant of variation, we can find t when W = 27 by substituting these values into the equation:

27 = 48/(t^2) 27t^2 = 48 t^2 = 48/27 t^2 = 1.7778 t = sqrt(1.7778) t = 1.33 (rounded to two decimal places)

So, the value of t when W = 27 is approximately 1.33.

The problem states that "W varies inversely as the square of t". This can be written as W = k/(t^2), where k is the constant of variation.

Given that W = 12 when t = 2, we can substitute these values into the equation to find k:

12 = k/(2^2) 12 = k/4 k = 12 * 4 k = 48

Now that we have the constant of variation, we can find t when W = 27 by substituting these values into the equation:

27 = 48/(t^2) 27t^2 = 48 t^2 = 48/27 t^2 = 1.7778 t = sqrt(1.7778) t = 1.33 (rounded to two decimal places)

So, the value of t when W = 27 is approximately 1.33.