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 = .... ?