Suppose T varies jointly with r and the square of n. When r=5 and n=3, then T=180. Find T if r=9 and n=4.
T = krn^2
so, T/(rn^2) is constant. You want T such that
T/(9*16) = 180/(5*9)
Please explain the answer
To solve this problem, let's write down the given information:
T varies jointly with r and the square of n.
When r = 5 and n = 3, then T = 180.
Let's form the equation using the given information:
T = k * r * n^2
where k is the constant of variation.
Now, let's find the value of k using the first set of values:
180 = k * 5 * 3^2
180 = k * 5 * 9
180 = 45k
k = 180/45
k = 4
Now that we have the value of k, we can find T when r = 9 and n = 4:
T = 4 * 9 * 4^2
T = 4 * 9 * 16
T = 576
Therefore, when r = 9 and n = 4, T = 576.
To solve this problem, we need to use the concept of joint variation and solve for the constant of variation.
First, let's identify the joint variation equation:
T = k * r * n^2
Given that T = 180 when r = 5 and n = 3, we can insert these values into the equation:
180 = k * 5 * 3^2
Simplifying further:
180 = 45k
Divide both sides by 45 to solve for k:
k = 180/45
k = 4
Now that we have the value of k, we can substitute it into the joint variation equation and solve for T when r = 9 and n = 4:
T = 4 * 9 * 4^2
T = 4 * 9 * 16
T = 576
Therefore, when r = 9 and n = 4, T = 576.