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The problem displayed on the screen involves setting up a one-step unit conversion, specifically solving for a variable \( T \) in an equation involving scientific notation and conversion factors. Here’s how to interpret and solve it.

The equation shown is:
\[
(-3.1 \times 10^6 \, \text{cm}^3) \times 10^{-6} = T \, \text{m}^3
\]

### Step-by-Step Solution

1. **Interpret Units**: The initial value is in \( \text{cm}^3 \) (cubic centimeters) and needs to be converted to \( \text{m}^3 \) (cubic meters).

2. **Conversion Factor**: 
   To convert from \( \text{cm}^3 \) to \( \text{m}^3 \), recall that \( 1 \, \text{m} = 100 \, \text{cm} \), so:
   \[
   1 \, \text{m}^3 = (100 \, \text{cm})^3 = 10^6 \, \text{cm}^3
   \]
   Thus, \( 1 \, \text{cm}^3 = 10^{-6} \, \text{m}^3 \).

3. **Calculate \( T \)**:
   Multiply the given volume in \( \text{cm}^3 \) by \( 10^{-6} \) to convert it to \( \text{m}^3 \):
   \[
   T = (-3.1 \times 10^6) \times 10^{-6} = -3.1 \, \text{m}^3
   \]

Therefore, the missing part of the equation is:
\[
T = -3.1 \, \text{m}^3
\]

Would you like further clarification on this solution? 

Here are five related questions for further practice:
1. How would you convert \( 5.2 \times 10^4 \, \text{cm}^3 \) to \( \text{m}^3 \)?
2. What is the result of converting \( -1.5 \times 10^7 \, \text{cm}^3 \) to \( \text{m}^3 \)?
3. If \( T = 2.8 \, \text{m}^3 \), what is its equivalent in \( \text{cm}^3 \)?
4. What conversion factor should be used to convert \( \text{cm}^2 \) to \( \text{m}^2 \)?
5. How does scientific notation help in performing large unit conversions?

**Tip**: Remember that unit conversion in cubic measurements requires cubing the conversion factor for linear measurements!

A student sets up the following equation to convert a measurement: (-3.1 x 10^6 cm^3) x 10^-6 = T m^3. Fill in the missing part of this equation.

Learn how to convert -3.1 x 10^6 cm^3 to m^3 using scientific notation in this detailed solution. This problem covers unit conversion and scientific notation concepts, ideal for high school students.

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