the weight of a symetry ball varies directly with the cube of it's radius a ball with the radius of 2 inches weighs 5.60. find the weight of a ball with a 3 inch radius
weight = k r^3 , where k is a constant
when r = 2, weight = 5.6
5.6 = k(2^3)
k = 5.6/8 = .7
so weight = .7r^3
when r = 3
weight = .7(27) = 18.9
or ... just use a ratio
w/5.6 = 3^3/2^3
w = 5.6(27/8) = 18.9
To find the weight of a ball with a 3 inch radius, we can use the concept of direct variation provided in the question.
Direct variation means that two quantities are related in such a way that they always have the same ratio. In this case, the weight (W) of the ball is directly proportional to the cube of its radius (r).
Mathematically, we can express this relationship as:
W = k * r^3,
where W represents the weight of the ball, r represents the radius of the ball, and k is the constant of variation.
To find the value of k, we can use the given information. When the radius is 2 inches, the weight is 5.60. Plugging these values into the equation, we get:
5.60 = k * 2^3,
5.60 = k * 8.
To find the value of k, divide both sides of the equation by 8:
k = 5.60 / 8,
k = 0.70.
Now that we have the value of k, we can find the weight of the ball with a 3 inch radius:
W = 0.70 * 3^3,
W = 0.70 * 27,
W ≈ 18.90.
Therefore, a ball with a 3 inch radius would weigh approximately 18.90 units.