1. The problem asks for the distance between the points $(-3, -4)$ and $(2, 6)$. 2. We use the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$: $$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ 3. For the first problem, substitute the coordinates: $$ d = \sqrt{(2 - (-3))^2 + (6 - (-4))^2} = \sqrt{(2 + 3)^2 + (6 + 4)^2} = \sqrt{5^2 + 10^2} = \sqrt{25 + 100} = \sqrt{125} $$ 4. Simplify the square root: $$ \sqrt{125} = 5\sqrt{5} $$ 5. Thus, the distance between the points $(-3, -4)$ and $(2, 6)$ is $5\sqrt{5}$ units. 6. For the second problem, given points $(x, 6)$ and $(28, 13)$ and distance $25$, use the distance formula and set equal to $25$: $$ 25 = \sqrt{(28 - x)^2 + (13 - 6)^2} = \sqrt{(28 - x)^2 + 7^2} = \sqrt{(28 - x)^2 + 49} $$ 7. Square both sides to eliminate the square root: $$ 25^2 = (28 - x)^2 + 49 $$ 8. Calculate $25^2$: $$ 625 = (28 - x)^2 + 49 $$ 9. Subtract $49$ from both sides: $$ 625 - 49 = (28 - x)^2 $$ $$ 576 = (28 - x)^2 $$ 10. Take the square root of both sides: $$ \sqrt{576} = |28 - x| $$ $$ 24 = |28 - x| $$ 11. This gives two cases: - $28 - x = 24$ which gives $x = 4$ - $28 - x = -24$ which gives $x = 52$ 12. Therefore, the values of $x$ that make the distance 25 units are $4$ and $52$.