If q varies inversely as the square of p and if q=8 and p= 2, find a when p= 4

POS doesn't answer the question. Bullsh@t

To find the value of "a" when p=4, we need to use the inverse variation formula.

The inverse variation equation is q = k/p^2, where k is the constant of variation.

Given that q = 8 when p = 2, we can substitute these values into the equation to solve for k.

8 = k/2^2
8 = k/4

To find the value of k, we can multiply both sides of the equation by 4.

8 * 4 = k
32 = k

Now we have the constant of variation, k = 32.

To find the value of q when p = 4, we substitute the known values into the inverse variation equation.

q = k/p^2
q = 32/4^2
q = 32/16
q = 2

Therefore, when p = 4, q = 2.

To find the value of 'a' when p = 4, we can use the inverse variation equation:

q = k / p^2

where q is the dependent variable, p is the independent variable, and k is the constant of variation.

Given that q = 8 when p = 2, we can plug these values into the equation to solve for k:

8 = k / 2^2
8 = k / 4
k = 8 * 4
k = 32

Now, we can substitute the value of k into the equation to find the value of q when p = 4:

q = 32 / 4^2
q = 32 / 16
q = 2

Therefore, when p = 4, q = 2.

q varies inversely as the square of p :

q = k*1/p^2
when q=8,p=2
8 = k(1/4)
k = 32

q = 32/p^2
so when p = 4, q = .... ?