Question 1184068
Here's how to solve this problem:

**Let:**

*  'x' be the amount invested in the savings account at 4% interest.
*  'y' be the amount invested in the mini-mall development at 10% interest.

**Set up the equations:**

* **Equation 1 (Total investment):** x + y = 5650
* **Equation 2 (Total interest):** 0.04x + 0.10y = 445

**Solve the equations:**

One way to solve this is using substitution:

1. **Solve Equation 1 for x:**  x = 5650 - y

2. **Substitute this value of x into Equation 2:**
   0.04(5650 - y) + 0.10y = 445

3. **Simplify and solve for y:**
   226 - 0.04y + 0.10y = 445
   0.06y = 219
   y = 3650

4. **Substitute the value of y back into Equation 1 to find x:**
   x + 3650 = 5650
   x = 2000

**Answer:**

* $2000 was invested in the savings account at 4% interest.
* $3650 was invested in the mini-mall development at 10% interest.