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Sagot :
Let's analyze the given problem step by step.
### 1. Identify the given erosion rate:
The beach is eroding at a rate of 4 centimeters per year.
### 2. Convert the erosion rate to millimeters:
We know that 1 centimeter is equivalent to 10 millimeters. Therefore:
[tex]$ 4 \text{ cm/year} \times 10 \frac{\text{mm}}{\text{cm}} $[/tex]
### 3. Convert the time period from years to days:
There are 365 days in a year. Therefore:
[tex]$ 1 \text{ year} = 365 \text{ days} $[/tex]
### 4. Combine these conversions:
We want the erosion rate in millimeters per day. To obtain this, we need to convert the period from years to days while the length from centimeters to millimeters:
[tex]$ 4 \text{ cm/year} \times 10 \frac{\text{ mm}}{\text{ cm}} \times \frac{1 \text{ year}}{365 \text{ days}} $[/tex]
Now, let's refer to the given expressions and evaluate them:
- Expression 1:
[tex]$ \frac{4 \text{ cm}}{1 \text{ year}} \times \frac{10 \text{ mm}}{1 \text{ cm}} \times \frac{1 \text{ year}}{365 \text{ days}} $[/tex]
Cancel out the 'cm' and 'year' units:
[tex]$ \frac{4 \times 10 \text{ mm}}{365 \text{ days}} $[/tex]
This simplifies to:
[tex]$ \frac{40 \text{ mm}}{365 \text{ days}} \approx 0.1095890410958904 \text{ mm/day} $[/tex]
- Expression 2:
[tex]$ \frac{4 \text{ cm}}{1 \text{ year}} \times \frac{1 \text{ cm}}{10 \text{ mm}} \times \frac{365 \text{ days}}{1 \text{ year}} $[/tex]
Cancel out the 'cm' and 'year' units:
[tex]$ \frac{4 \times 365 \text{ days}}{10 \text{ mm} \times 1 \text{ year}} $[/tex]
This simplifies to:
[tex]$ \frac{1460 \text{ days/year}}{10 \text{ mm/year}} = 146 \text{ days/mm} $[/tex]
However, this simplifies to units of days per millimeter, which is incorrect as we need mm/day.
### Conclusion:
Expression 1 results in the correct units (millimeters per day) and numerical value:
[tex]$ \frac{40 \text{ mm}}{365 \text{ days}} \approx 0.1095890410958904 \text{ mm/day} $[/tex]
Therefore, the correct expression is:
[tex]$ \frac{4 \text{ cm}}{1 \text{ year}} \times \frac{10 \text{ mm}}{1 \text{ cm}} \times \frac{1 \text{ year}}{365 \text{ days}} $[/tex]
The erosion rate of the beach is approximately [tex]\( 0.1095890410958904 \text{ mm/day} \)[/tex].
### 1. Identify the given erosion rate:
The beach is eroding at a rate of 4 centimeters per year.
### 2. Convert the erosion rate to millimeters:
We know that 1 centimeter is equivalent to 10 millimeters. Therefore:
[tex]$ 4 \text{ cm/year} \times 10 \frac{\text{mm}}{\text{cm}} $[/tex]
### 3. Convert the time period from years to days:
There are 365 days in a year. Therefore:
[tex]$ 1 \text{ year} = 365 \text{ days} $[/tex]
### 4. Combine these conversions:
We want the erosion rate in millimeters per day. To obtain this, we need to convert the period from years to days while the length from centimeters to millimeters:
[tex]$ 4 \text{ cm/year} \times 10 \frac{\text{ mm}}{\text{ cm}} \times \frac{1 \text{ year}}{365 \text{ days}} $[/tex]
Now, let's refer to the given expressions and evaluate them:
- Expression 1:
[tex]$ \frac{4 \text{ cm}}{1 \text{ year}} \times \frac{10 \text{ mm}}{1 \text{ cm}} \times \frac{1 \text{ year}}{365 \text{ days}} $[/tex]
Cancel out the 'cm' and 'year' units:
[tex]$ \frac{4 \times 10 \text{ mm}}{365 \text{ days}} $[/tex]
This simplifies to:
[tex]$ \frac{40 \text{ mm}}{365 \text{ days}} \approx 0.1095890410958904 \text{ mm/day} $[/tex]
- Expression 2:
[tex]$ \frac{4 \text{ cm}}{1 \text{ year}} \times \frac{1 \text{ cm}}{10 \text{ mm}} \times \frac{365 \text{ days}}{1 \text{ year}} $[/tex]
Cancel out the 'cm' and 'year' units:
[tex]$ \frac{4 \times 365 \text{ days}}{10 \text{ mm} \times 1 \text{ year}} $[/tex]
This simplifies to:
[tex]$ \frac{1460 \text{ days/year}}{10 \text{ mm/year}} = 146 \text{ days/mm} $[/tex]
However, this simplifies to units of days per millimeter, which is incorrect as we need mm/day.
### Conclusion:
Expression 1 results in the correct units (millimeters per day) and numerical value:
[tex]$ \frac{40 \text{ mm}}{365 \text{ days}} \approx 0.1095890410958904 \text{ mm/day} $[/tex]
Therefore, the correct expression is:
[tex]$ \frac{4 \text{ cm}}{1 \text{ year}} \times \frac{10 \text{ mm}}{1 \text{ cm}} \times \frac{1 \text{ year}}{365 \text{ days}} $[/tex]
The erosion rate of the beach is approximately [tex]\( 0.1095890410958904 \text{ mm/day} \)[/tex].
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