If t varied jointly with u and the square of v, and t is 1152 when u is 8 and v is 4, find t when v is 5 and u is 2.
Answer choices:
A) 144
B) 785
C) 450
D)225
To solve this problem, we need to use the concept of joint variation. When t varies jointly with u and the square of v, it means that t can be expressed as a product of u and v squared, multiplied by some constant.
Let's denote the constant of variation as k. Therefore, we can set up the equation as follows:
t = k * u * v^2
We have the following information:
t = 1152
u = 8
v = 4
Plugging in these values into the equation, we can solve for k:
1152 = k * 8 * 4^2
1152 = 32k
Dividing both sides by 32:
k = 36
Now that we have determined the constant of variation, we can use it to find t when v is 5 and u is 2:
t = k * u * v^2
t = 36 * 2 * 5^2
t = 36 * 2 * 25
t = 1800
Therefore, t is equal to 1800.
None of the answer choices provided matches this value of t = 1800.
To solve this problem, we need to use the concept of direct and joint variation.
Given that t varies jointly with u and the square of v, we can write the equation as:
t = k * u * v^2
where k is the constant of variation.
We are given that t is 1152 when u is 8 and v is 4. Let's plug these values into the equation to find the value of k:
1152 = k * 8 * 4^2
1152 = k * 8 * 16
1152 = 128k
k = 9
Now, we can use this value of k to find t when u is 2 and v is 5:
t = 9 * 2 * 5^2
t = 9 * 2 * 25
t = 450
Therefore, the value of t when u is 2 and v is 5 is 450.
The correct answer choice is C) 450.
t = kuv^2
So, t/(uv^2) = k, a constant
You want t such that
t/(2*5^2) = 1152/(8*4^2)