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Sagot :
Answer:
1. 4π/3 in³
2. 2.6 ounces
3. 1081.5 ft³
Step-by-step explanation:
1. A soap manufacturing company changes the design of their soap from a cone to a sphere shape. The cone had a height of 4 inches and radius of 3 inches. The sphere has a diameter of 4 inches. In cubic inches, how much less soap is in the new design than the old design? Leave answer in terms of π and as a simplified fraction.
We find the volume of the cone and sphere to know the quantity of soap in each of them.
Volume of cone of soap, V = πr²h/3 where r = radius of cone = 3 inches and h = height of cone = 4 inches
So, V = πr²h/3
= π(3 in)² × 4 in/3
= 9π in² × 4 in/3
= 3π in² × 4 in
= 12π in³
We now find the volume of the sphere, V' = 4πR³/3 where R = radius of sphere = d/2 where d = diameter of sphere = 4 inches. So. R = d/2 = 4 in/2 = 2 in.
So, V = 4πR³/3
= 4π(2 in)³/3
= 4π × 8 in³/3
= 32π in³/3
So the amount of soap less used in the new design than the old design is the old volume - new volume = V - V' = 12π in³ - 32π in³/3 = (36π in³ - 32π in³)/3 = 4π/3 in³
2. Austin uses a mold to make cone shaped cupcakes. The diameter of the mold is 3 inches, and the height of the mold is 2 inches. If 1 cubic inch is about 0.55 ounce, how many ounces will ten cupcakes weigh? Use 3.14 for pi. Round to the nearest tenth of an ounce.
Since the cake is a cone, we find its volume to know its quantity.
So, Volume of cone of cupcakes, V = πr²h/3 where r = radius of cone = d/2 where d = diameter = 3 inches. So r = d/2 = 3/2 = 1.5 in and h = height of cone = 2 inches
So, V = πr²h/3
= π(1.5 in)² × 2 in/3
= 2.25π in² × 2 in/3
= 4.5π in³/3
= 4.5(3.14) in³/3
= 14.13 in³/3
= 4.71 in³
Now since 1 cubic inch = 0.55 ounces,
4.71 in³ = 4.71 × 1 in³
= 4.71 × 0.55 ounces
= 2.591 ounces
≅ 2.6 ounces to the nearest tenth of an ounce.
3. Walden is planning to install an above ground pool in he backyard. The pool has a diameter of 18 feet and depth of 5 feet. He is going to fill it to 85% capacity. How much water will be in the pool when he is finished?
We need to find the volume of the empty pool, V.
Since the pool has a diameter, d = 18 feet, it has a circular cross-section whose area, A = πd²/4.
Since the depth of the pool, h = 5 feet, the volume of the pool is V = Ah = πd²h/4.
So, V = πd²h/4
= π(18 ft)² × 5 ft/4
= 324π ft² × 5 ft/4
= 81π ft² × 5 ft
= 405π ft³
= 1272.35 ft³
Since Walden fills the pool to 85 % capacity, the volume of water would thus be V' = 85% of V
= 0.85V
= 0.85 × 1272.35 ft³
= 1081.5 ft³
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