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.

Question

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.

Knowee AI · Accepted Answer

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.

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