Let $V$ be the region in the Cartesian plane consisting of all points $(x,y)$ satisfying the simultaneous conditions
$$|x|\le y\le|x|+3\text{ and }y\le4.$$Find the centroid of $V$.
To find the centroid of the region \( V \) defined by the inequalities \( |x| \le y \le |x| + 3 \) and \( y \le 4 \), we need to follow these steps:

1. **Graph the Region**:
   - The inequality \( |x| \le y \) represents the region above the lines \( y = x \) and \( y = -x \).
   - The inequality \( y \le |x| + 3 \) represents the region below the lines \( y = x + 3 \) and \( y = -x + 3 \).
   - The inequality \( y \le 4 \) represents the region below the line \( y = 4 \).

   Combining these, we get the region bounded by the lines \( y = x \), \( y = -x \), \( y = x + 3 \), \( y = -x + 3 \), and \( y = 4 \).

2. **Determine the Intersection Points**:
   - The lines \( y = x + 3 \) and \( y = 4 \) intersect at \( x + 3 = 4 \Rightarrow x = 1 \). So, the point is \( (1, 4) \).
   - The lines \( y = -x + 3 \) and \( y = 4 \) intersect at \( -x + 3 = 4 \Rightarrow x = -1 \). So, the point is \( (-1, 4) \).
   - The lines \( y = x \) and \( y = x + 3 \) intersect at \( x = 0 \). So, the point is \( (0, 3) \).
   - The lines \( y = -x \) and \( y = -x + 3 \) intersect at \( x = 0 \). So, the point is \( (0, 3) \).

3. **Set Up the Integral for the Centroid**:
   The centroid \((\bar{x}, \bar{y})\) of a region \( V \) is given by:
   \[
   \bar{x} = \frac{1}{A} \iint_V x \, dA, \quad \bar{y} = \frac{1}{A} \iint_V y \, dA
   \]
   where \( A \) is the area of the region \( V \).

4. **Calculate the Area \( A \)**:
   The region is symmetric about the y-axis, so we can calculate the area of one half and double it.
   \[
   A = 2 \int_{0}^{1} \int_{x}^{x+3} dy \, dx + 2 \int_{1}^{3} \int_{x}^{4} dy \, dx
   \]
   \[
   A = 2 \left( \int_{0}^{1} (x+3 - x) \, dx + \int_{1}^{3} (4 - x) \, dx \right)
   \]
   \[
   A = 2 \left( \int_{0}^{1} 3 \, dx + \int_{1}^{3} (4 - x) \, dx \right)
   \]
   \[
   A = 2 \left( 3 \left[ x \right]_{0}^{1} + \left[ 4x - \frac{x^2}{2} \right]_{1}^{3} \right)
   \]
   \[
   A = 2 \left( 3(1 - 0) + \left( 4(3) - \frac{3^2}{2} - (4(1) - \frac{1^2}{2}) \right) \right)
   \]
   \[
   A = 2 \left( 3 + \left( 12 - \frac{9}{2} - 4 + \frac{1}{2} \right) \right)
   \]
   \[
   A = 2 \left( 3 + \left( 12 - 4.5 - 4 + 0.5 \right) \right)
   \]
   \[
   A = 2 \left( 3 + 4 \right) = 2 \times 7 = 14
   \]

5. **Calculate \(\bar{x}\)**:
   Since the region is symmetric about the y-axis, \(\bar{x} = 0\).

6. **Calculate \(\bar{y}\)**:
   \[
   \bar{y} = \frac{1}{A} \iint_V y \, dA
   \]
   \[
   \bar{y} = \frac{1}{14} \left( 2 \int_{0}^{1} \int_{x}^{x+3} y \, dy \, dx + 2 \int_{1}^{3} \int_{x}^{4} y \, dy \, dx \right)
   \]
   \[
   \bar{y} = \frac{2}{14} \left( \int_{0}^{1} \left[ \frac{y^2}{2} \right]_{x}^{x+3} \, dx + \int_{1}^{3} \left[ \frac{y^2}{2} \right]_{x}^{4} \, dx \right)
   \]
   \[
   \bar{y} = \frac{1}{7} \left( \int_{0}^{1} \left( \frac{(x+3)^2}{2} - \frac{x^2}{2} \right) \, dx + \int_{1}^{3} \left( \frac{16}{2} - \frac{x^2}{2} \right) \, dx \right)
   \]
   \[
   \bar{y} = \frac{1}{7} \left( \int_{0}^{1} \left( \frac{x^2 + 6x + 9 - x^2}{2} \right) \, dx + \int_{1}^{3} \left( 8 - \frac{x^2}{2} \right) \, dx \right)
   \]
   \[
   \bar{y} = \frac{1}{7} \left( \int_{0}^{1} \left( 3 + 3x \right) \, dx + \int_{1}^{3} \left( 8 - \frac{x^2}{2} \right) \, dx \right)
   \]
   \[
   \bar{y} = \frac{1}{7} \left( \left[ 3x + \frac{3x^2}{2} \right]_{0}^{1} + \left[ 8x - \frac{x^3}{6} \right]_{1}^{3} \right)
   \]
   \[
   \bar{y} = \frac{1}{7} \left( \left( 3(1) + \frac{3(1)^2}{2} - 0 \right) + \left( 8(3) - \frac{3^3}{6} - (8(1) - \frac{1^3}{6}) \right) \right)
   \]
   \[
   \bar{y} = \frac{1}{7} \left( 3 + \frac{3}{2} + 24 - \frac{27}{6} - 8 + \frac{1}{6} \right)
   \]
   \[
   \bar{y} = \frac{1}{7} \left( 3 + 1.5 + 24 - 4.5 - 8 + 0.1667 \right)
   \]
   \[
   \bar{y} = \frac{1}{7} \left( 16.1667 \right) \approx 2.31
   \]

The final answer is \( \boxed{ (0, 2.31) } \).