If the quantity P varies as the square of Q and inversely as the square root R. If P=12,Q=3 and R=81,find P if Q=1 and R°25
P = kQ^2/√R
so, P√R/Q^2 = k, which is constant.
So, you want to find P such that
P√25/1^2 = 12√81/3^2
now just simplify and solve for P
To solve this problem, we need to understand how P, Q, and R are related. According to the given information, the quantity P varies directly with the square of Q and inversely with the square root of R.
We can express this relationship in an equation as follows:
P ∝ Q^2 / √R
Here, the symbol "∝" represents proportionality. To find the constant of proportionality, we need to find the value of P when Q = 3 and R = 81.
Given:
P = 12 (when Q = 3 and R = 81)
Q = 1
R = 25
To find the constant of proportionality, we can set up a proportion using the initial and new values of Q and R:
(P1 / P2) = [(Q1^2 / √R1) / (Q2^2 / √R2)]
Plugging in the values we know, we get:
12 / P = [(3^2 / √81) / (1^2 / √25)]
Simplifying further:
12 / P = [(9 / 9) / (1 / 5)]
We can calculate and simplify the right side:
12 / P = 5
To isolate P, we multiply both sides by P:
12 = 5P
Divide both sides by 5:
P = 12 / 5
P = 2.4
Therefore, when Q = 1 and R = 25, the value of P is 2.4.