If A varies inversely as the square of B and A=25 when B=1/5.Find B when A=49
If A varies inversely with the square of B,
A*B^2 = constant.
In this case, the constant is
A/B^2 = 25*(1/5)^2 = 1
When A = 49, B = 1/7
To solve this problem, we can use the inverse variation equation:
A = k/B^2,
where k is the constant of variation.
We know that A is 25 when B is 1/5, so we can substitute these values into the equation:
25 = k/(1/5)^2.
To simplify, let's square 1/5:
25 = k/(1/25).
To get rid of the fraction on the right side, we can multiply both sides of the equation by 25:
25 * 25 = k.
625 = k.
Now we have the value of k, which is 625.
To find B when A is 49, we can use the inverse variation equation again, plugging in the known values:
49 = 625/B^2.
To isolate B^2, we can divide both sides of the equation by 49:
49/49 = 625/B^2 / 49.
1 = 625/(B^2 * 49).
Now, we can multiply both sides by B^2 * 49 to solve for B^2:
B^2 * 49 = 625.
We can divide both sides by 49 to isolate B^2:
B^2 = 625/49.
Finally, we can take the square root of both sides to find B:
B = √(625/49).
B ≈ 2.5.
Therefore, when A is 49, B is approximately 2.5.