16. The region bounded by $$ y=4-2 x $$, the $$ x $$-axis, and the $$ y $$-axis.

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1. **Determine the vertices of the region:**

- The line is given by $$ y = 4 - 2x $$.
   - The intersection with the $$ x $$-axis occurs when $$ y = 0 $$:
     $$
     4 - 2x = 0 \quad \Longrightarrow \quad x = 2.
     $$
     So, the point is $$ (2, 0) $$.
   - The intersection with the $$ y $$-axis occurs when $$ x = 0 $$:
     $$
     y = 4 - 2(0) = 4.
     $$
     So, the point is $$ (0, 4) $$.
   - The origin $$ (0, 0) $$ is the intersection of the $$ x $$-axis and $$ y $$-axis.

Thus, the vertices of the triangle are $$ (0, 0) $$, $$ (2, 0) $$, and $$ (0, 4) $$.

2. **Calculate the area of the triangle:**

The area $$ A $$ of a right triangle with legs of lengths $$ b $$ and $$ h $$ is given by:
   $$
   A = \frac{1}{2} b h.
   $$
   
   - The base $$ b $$ is the distance along the $$ x $$-axis from $$ (0,0) $$ to $$ (2,0) $$, so $$ b = 2 $$.
   - The height $$ h $$ is the distance along the $$ y $$-axis from $$ (0,0) $$ to $$ (0,4) $$, so $$ h = 4 $$.

Substitute the values:
   $$
   A = \frac{1}{2} \times 2 \times 4 = 4.
   $$

3. **Final Answer:**

The area of the region bounded by $$ y = 4 - 2x $$, the $$ x $$-axis, and the $$ y $$-axis is $$ 4 $$.
The area of the region is 4 square units.

The region bounded by , the -axis, and the -axis.

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