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The problem describes a series of translations for triangle \(ABC\) with vertices \(A(-2, -2)\), \(B(-3, 1)\), and \(C(1, 1)\). The translations occur in two steps:

1. The first translation moves the points by \((x, y) \to (x - 1, y + 3)\), producing triangle \(A'B'C'\).
2. The second translation moves triangle \(A'B'C'\) by \((x, y) \to (x + 4, y - 1)\), producing triangle \(A''B''C''\).

### Part (a): Find the coordinates for the vertices of triangle \(A''B''C''\).

1. **First translation** (for triangle \(A'B'C'\)):
   - For vertex \(A(-2, -2)\): 
     \[
     (-2, -2) \to (-2 - 1, -2 + 3) = (-3, 1)
     \]
   - For vertex \(B(-3, 1)\): 
     \[
     (-3, 1) \to (-3 - 1, 1 + 3) = (-4, 4)
     \]
   - For vertex \(C(1, 1)\): 
     \[
     (1, 1) \to (1 - 1, 1 + 3) = (0, 4)
     \]

   So, after the first translation, the vertices of triangle \(A'B'C'\) are:
   \[
   A'(-3, 1), B'(-4, 4), C'(0, 4)
   \]

2. **Second translation** (for triangle \(A''B''C''\)):
   - For vertex \(A'(-3, 1)\): 
     \[
     (-3, 1) \to (-3 + 4, 1 - 1) = (1, 0)
     \]
   - For vertex \(B'(-4, 4)\): 
     \[
     (-4, 4) \to (-4 + 4, 4 - 1) = (0, 3)
     \]
   - For vertex \(C'(0, 4)\): 
     \[
     (0, 4) \to (0 + 4, 4 - 1) = (4, 3)
     \]

   So, the vertices of triangle \(A''B''C''\) are:
   \[
   A''(1, 0), B''(0, 3), C''(4, 3)
   \]

### Part (b): Write a rule for one translation that maps triangle \(ABC\) to triangle \(A''B''C''\).

Instead of performing two translations separately, we can combine them into one:

1. The first translation is \((x, y) \to (x - 1, y + 3)\).
2. The second translation is \((x, y) \to (x + 4, y - 1)\).

We can combine these transformations as follows:
\[
(x, y) \to ((x - 1) + 4, (y + 3) - 1) = (x + 3, y + 2)
\]

So, the combined translation that maps triangle \(ABC\) directly to triangle \(A''B''C''\) is:
\[
(x, y) \to (x + 3, y + 2)
\]

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Do you want more details or have any other questions? Here are some related questions:

1. What would happen if you only apply the second translation to triangle \(ABC\)?
2. Can you reverse the process to find the original triangle from \(A''B''C''\)?
3. How can you describe the translations in vector form?
4. What would the vertices of triangle \(A'B'C'\) be if the first translation was different?
5. How do these translations affect the area and orientation of the triangle?

**Tip:** When combining multiple transformations, always apply one transformation at a time and simplify the overall movement into a single translation if possible.

Triangle ABC with vertices A(-2, -2), B(-3, 1), and C(1, 1) is translated by (x, y) → (x - 1, y + 3). Then the image, triangle A'B'C', is translated by (x, y) → (x + 4, y - 1), resulting in triangle A''B''C''. Find the coordinates for the vertices of triangle A''B''C'' and write a rule for the translation that maps triangle ABC to triangle A''B''C''.

Solve the problem of translating a triangle in the coordinate plane with vertices A(-2, -2), B(-3, 1), and C(1, 1). The triangle is first translated by (x, y) → (x - 1, y + 3) and then by (x, y) → (x + 4, y - 1), resulting in triangle A''B''C''. Learn how to find the coordinates of the new triangle and combine the transformations.

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