https://file.aiquickdraw.com/mathbot/file/1725855871771rji2joid.jpg

The expression provided in the image is as follows:

\[
\frac{\left(2 \times 10^2 \, \text{g} + 8.123 \, \text{g}\right)}{(0.091 \, \text{cm}^3 - 0.079 \, \text{cm}^3)} - \square \times \frac{5}{7} \, \text{cm}
\]

To solve this expression step-by-step, we can proceed as follows:

### Step 1: Simplify the numerator
The numerator is:
\[
2 \times 10^2 \, \text{g} + 8.123 \, \text{g}
\]

\[
2 \times 10^2 = 200 \, \text{g}
\]
So:
\[
200 \, \text{g} + 8.123 \, \text{g} = 208.123 \, \text{g}
\]

### Step 2: Simplify the denominator
The denominator is:
\[
0.091 \, \text{cm}^3 - 0.079 \, \text{cm}^3
\]
\[
0.091 \, \text{cm}^3 - 0.079 \, \text{cm}^3 = 0.012 \, \text{cm}^3
\]

### Step 3: Divide the numerator by the denominator
\[
\frac{208.123 \, \text{g}}{0.012 \, \text{cm}^3} = 17343.5833 \, \text{g/cm}^3
\]

### Step 4: Subtract the expression on the right
The expression now is:
\[
17343.5833 - \square \times \frac{5}{7} \, \text{cm}
\]

Since the square is a placeholder, you need to determine the value or leave it as is. Therefore, the expression can be simplified as:

\[
17343.5833 - \square \times \frac{5}{7} \, \text{cm}
\]

Would you like to proceed with more details on this calculation, or do you have any other questions?

### Related Questions:
1. What happens if the value in the square is zero?
2. How do unit conversions impact the final expression?
3. How can this expression be applied to real-world physics problems?
4. What if the numerator was a different number?
5. How would this expression change if the denominator values were negative?

### Tip:
Always ensure units are consistent throughout the calculation to avoid errors, especially when dealing with physical quantities like mass, volume, and length.

Simplify the expression: (2 × 10^2 g + 8.123 g) / (0.091 cm^3 - 0.079 cm^3) - □ × (5/7) cm.

Step-by-step guide to simplifying the expression (2 × 10^2 g + 8.123 g) / (0.091 cm^3 - 0.079 cm^3) - □ × (5/7) cm. Learn how to handle arithmetic operations and unit conversions, suitable for high school students in Grades 10-12.

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