To solve this inequality, we can examine the given expression step-by-step to confirm if the implication \( |x + 2| < 0.3 \Rightarrow |4x + 8| < 1.2 \) holds.

### Step 1: Analyze the inequality \( |x + 2| < 0.3 \)

This inequality implies:
\[
-0.3 < x + 2 < 0.3
\]
Subtracting 2 from all parts:
\[
-2.3 < x < -1.7
\]

Thus, the solution for \( |x + 2| < 0.3 \) is:
\[
x \in (-2.3, -1.7)
\]

### Step 2: Test the implication \( |4x + 8| < 1.2 \)

We want to see if \( x \in (-2.3, -1.7) \) guarantees that \( |4x + 8| < 1.2 \).

Rewrite \( |4x + 8| \) using the fact that \( 4x + 8 = 4(x + 2) \):
\[
|4x + 8| = |4(x + 2)| = 4 |x + 2|
\]

Since \( |x + 2| < 0.3 \), multiplying by 4 gives:
\[
|4x + 8| < 4 \cdot 0.3 = 1.2
\]

### Conclusion
Since \( |4x + 8| < 1.2 \) directly follows from \( |x + 2| < 0.3 \), the implication \( |x + 2| < 0.3 \Rightarrow |4x + 8| < 1.2 \) is **true**.

Would you like further clarification or have any questions?

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Here are some related questions you might find interesting:
1. What happens if we change the range in the initial inequality, say to \( |x+2| < 0.5 \)?
2. How would the result change if we had \( |x + 2| \leq 0.3 \) instead?
3. Can this reasoning apply to any expression where a variable term is scaled?
4. What if \( |x + 2| > 0.3 \)? Would the implication still hold?
5. How would the problem change if we had \( |4x + 8| \leq 1.2 \)?

**Tip:** Inequalities often maintain their direction when multiplying by positive constants, but not when dealing with negatives—always verify signs when solving.

Explore the solution to the inequality |x+2| < 0.3 and the resulting implication |4x + 8| < 1.2. This problem highlights absolute value inequalities and the implications of scaling values, suitable for Grades 10-12.

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