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.