math 20-3 unit 3b: volume & capacity name maisy 2. a box with a length of 4 m and a width of 6 m has a volume of 84m³. what is the height of the box? 4×6×? 3. lhorine is packing boxes into a room with a volume of 40m³. if each box is 0.2 m by 0.4 m by 0.6 m, how many boxes can fit in the room? 4. a can of soup has a radius of 3.9 cm and a height of 10.4 cm a. what is the volume of the can? πr²h π×3.9²×10.4 = 496.9 v=496.9 b. how much soup (in litres) does the can hold? 5. a cylinder has a capacity of 2.4 l if it has a diameter of 12 cm, what is the height of the cylinder? 2.4l ×1000 v=2400 v/(πr²) 22.5

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Accepted Answer

### Problem 2:
# Explanation:
## Step1: Recall volume formula for rectangular prism
The volume $ V $ of a rectangular prism is given by $ V = l \times w \times h $, where $ l $ is length, $ w $ is width, and $ h $ is height. We know $ V = 84 \, \text{m}^3 $, $ l = 4 \, \text{m} $, $ w = 6 \, \text{m} $. We need to solve for $ h $.
## Step2: Rearrange formula to solve for $ h $
From $ V = l \times w \times h $, we get $ h=\frac{V}{l \times w} $.
## Step3: Substitute values
Substitute $ V = 84 $, $ l = 4 $, $ w = 6 $ into the formula: $ h=\frac{84}{4\times6}=\frac{84}{24}=3.5 $.
# Answer:
The height of the box is $ 3.5 \, \text{m} $.

### Problem 3:
# Explanation:
## Step1: Find volume of one box
The volume $ V_{box} $ of a rectangular box is $ V_{box}=l\times w\times h $. Given $ l = 0.2 \, \text{m} $, $ w = 0.4 \, \text{m} $, $ h = 0.6 \, \text{m} $, so $ V_{box}=0.2\times0.4\times0.6 = 0.048 \, \text{m}^3 $.
## Step2: Find number of boxes
The number of boxes $ n $ that fit in the room is the volume of the room $ V_{room} $ divided by the volume of one box $ V_{box} $. Given $ V_{room}=40 \, \text{m}^3 $, so $ n=\frac{V_{room}}{V_{box}}=\frac{40}{0.048}\approx833.33 $. Since we can't have a fraction of a box, we take the whole number part, so $ n = 833 $ (we round down because we can't fit a partial box).
# Answer:
Approximately $ 833 $ boxes can fit in the room.

### Problem 4a:
# Explanation:
## Step1: Recall volume formula for cylinder
The volume $ V $ of a cylinder is $ V=\pi r^2h $, where $ r $ is radius and $ h $ is height. Given $ r = 3.9 \, \text{cm} $, $ h = 10.4 \, \text{cm} $.
## Step2: Substitute values into formula
$ V=\pi\times(3.9)^2\times10.4 $. Calculate $ (3.9)^2 = 15.21 $, then $ 15.21\times10.4 = 158.184 $, then $ V=\pi\times158.184\approx3.1416\times158.184\approx496.9 \, \text{cm}^3 $.
# Answer:
The volume of the can is approximately $ 496.9 \, \text{cm}^3 $.

### Problem 4b:
# Explanation:
## Step1: Convert volume to litres
We know that $ 1 \, \text{L} = 1000 \, \text{cm}^3 $. So to convert $ 496.9 \, \text{cm}^3 $ to litres, we divide by $ 1000 $.
## Step2: Calculate the volume in litres
$ V_{litres}=\frac{496.9}{1000}=0.4969 \, \text{L} \approx0.5 \, \text{L} $.
# Answer:
The can holds approximately $ 0.5 \, \text{L} $ (or $ 0.4969 \, \text{L} $) of soup.

### Problem 5:
# Explanation:
## Step1: Convert capacity to cubic centimetres
We know that $ 1 \, \text{L} = 1000 \, \text{cm}^3 $, so $ 2.4 \, \text{L} = 2.4\times1000 = 2400 \, \text{cm}^3 $. The diameter $ d = 12 \, \text{cm} $, so the radius $ r=\frac{d}{2}=\frac{12}{2}=6 \, \text{cm} $.
## Step2: Recall volume formula for cylinder and solve for $ h $
The volume of a cylinder is $ V=\pi r^2h $, so we can solve for $ h $: $ h=\frac{V}{\pi r^2} $.
## Step3: Substitute values
Substitute $ V = 2400 $, $ r = 6 $ into the formula: $ h=\frac{2400}{\pi\times(6)^2}=\frac{2400}{\pi\times36}\approx\frac{2400}{113.097}\approx21.22 \, \text{cm} $ (if we use $ \pi\approx3.1416 $, $ \frac{2400}{3.1416\times36}=\frac{2400}{113.0976}\approx21.22 $, or if we use the approximate value from the sketch (22.5) there might be a different approximation of $ \pi $, but with $ \pi\approx3.14 $, $ \frac{2400}{3.14\times36}=\frac{2400}{113.04}\approx21.23 $).
# Answer:
The height of the cylinder is approximately $ 21.2 \, \text{cm} $ (or around $ 22.5 \, \text{cm} $ if a different $ \pi $ approximation was used).

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QUESTION IMAGE

Question

Explanation:

Problem 2:

Step1: Recall volume formula for rectangular prism

Step2: Rearrange formula to solve for \( h \)

Step3: Substitute values

Step1: Find volume of one box

Step2: Find number of boxes

Step1: Recall volume formula for cylinder

Step2: Substitute values into formula

Answer:

Problem 3: