if p varies inversely as the square of q and p=8 when q=4.find q when p=32

pq^2=k

so, you want q such that

32q^2 = 8*4^2

q=4

no. did you check to see that your answer works?

q^2 = 4

since

32q^2 = 8*4*4 = 32*4

If p varies inversely as the square of q and p = 3 when q = 4,find p when q = 2.

To solve the problem, we need to use the inverse variation formula. Inverse variation is a relationship where one variable decreases as the other variable increases, and this relationship can be described using the formula P = k/Q^2, where P represents the first variable, Q represents the second variable, and k is a constant.

Given that p varies inversely as the square of q and p = 8 when q = 4, we can substitute these values into the formula:

8 = k/4^2

Simplifying, we get:

8 = k/16

To find the value of k, we multiply both sides of the equation by 16:

8 * 16 = k

k = 128

Now that we have determined the value of k, we can use it to find q when p = 32:

32 = 128/q^2

To isolate q^2, divide both sides of the equation by 32:

32/32 = 128/q^2

1 = 128/q^2

Rearranging the equation, we get:

q^2 = 128/1

q^2 = 128

Finally, taking the square root of both sides gives us:

q = √128

Therefore, q is approximately equal to 11.314.