A sandcastle mold is in the shape of a cone, with a diameter of 4 meters and a height of 6 meters. What volume of sand is needed to fill the mold? Use 3.14 for π .
To find the volume of a cone, we need to use the formula: V = (1/3)πr^2h, where V is the volume, r is the radius, and h is the height.
First, we need to find the radius of the cone. Since the diameter is given, we can divide it by 2 to find the radius: r = 4 / 2 = 2 meters.
Now, we can plug the values into the formula: V = (1/3)(3.14)(2^2)(6) = (1/3)(3.14)(4)(6) = (1/3)(3.14)(24) = 3.14(8) = 25.12 cubic meters.
Therefore, approximately 25.12 cubic meters of sand is needed to fill the mold.
A hanging flower vase is in the shape of a cone with a radius of 5 inches and a height of 15 inches. What volume of water can the flower vase hold? Use 3.14 for π .
To find the volume of the cone-shaped flower vase, we can use the formula: V = (1/3)πr^2h, where V is the volume, r is the radius, and h is the height.
Given that the radius is 5 inches and the height is 15 inches, we can substitute these values into the formula: V = (1/3)(3.14)(5^2)(15) = (1/3)(3.14)(25)(15) = (1/3)(3.14)(375) = 392.5 cubic inches.
Therefore, the flower vase can hold approximately 392.5 cubic inches of water.
To find the volume of the sandcastle mold, we can use the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
where V is the volume, π is a constant approximately equal to 3.14, r is the radius of the base (which is half of the diameter), and h is the height of the cone.
Let's break down the problem step by step:
1. Find the radius of the base:
The diameter of the cone is given as 4 meters, so the radius (r) is half of the diameter:
r = 4 / 2 = 2 meters
2. Calculate the volume of the cone:
Now we can substitute the values into the formula:
V = (1/3) * 3.14 * 2^2 * 6
Simplifying the equation:
V = (1/3) * 3.14 * 4 * 6
V = (1/3) * 3.14 * 24
V = 8.3733333 cubic meters (approximately, depending on rounding)
Therefore, approximately 8.37 cubic meters of sand is needed to fill the mold.