If q varies inversely as the square of p and if q=8 and p= 2, find a when p= 4
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.
qp^2 is constant, so you want q such that
q*16 = 8*4
q = 2
To find the value of a when p = 4, you can use the inverse variation equation:
q = k/p^2
First, let's find the value of k using the given values of q and p.
q = 8 and p = 2
8 = k/2^2
8 = k/4
To solve for k, multiply both sides of the equation by 4:
4 * 8 = k
32 = k
Now that we have the value of k, we can find the value of a when p = 4:
q = k/p^2
Substitute the known values:
8 = 32/4^2
8 = 32/16
8 = 2
So, when p = 4, q = 2. Hence, the value of a is 2.
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.