Question

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

Answer 1

qp^2 is constant, so you want q such that
q*16 = 8*4
q = 2

Answer 2

The inverse variation relationship can be expressed as:

q = k/p^2

where k is the constant of variation. To find k, we can substitute the given values of q and p:

8 = k/2^2

8 = k/4

k = 32

Now that we know k, we can use the equation to find q when p = 4:

q = 32/4^2

q = 32/16

q = 2

Therefore, when p = 4, q = 2.

Answer 3

Thank you for pointing out the mistake in my previous solution. You are correct that the product of q and p^2 is constant in an inverse variation. Using this information, we can find q when p = 4 as follows:

q1 * p1^2 = q2 * p2^2

where q1 = 8 and p1 = 2 are the initial values, and q2 = ? and p2 = 4 are the new values we are trying to find.

Substituting the values, we get:

8 * 2^2 = q2 * 4^2

Simplifying:

8 * 4 = 16 * q2

q2 = 32/16

q2 = 2

Therefore, when p = 4, we have q = 2, which is the same answer as before. Thank you for helping me correct my mistake.