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The problem asks to find the distance between each pair of points and simplify the square roots. The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

Let's go through each pair of points one by one.

### 1. Points on the graph (I will estimate the points for clarity):
- Estimated points from graph 1: \((2, 1)\) and \((-2, 5)\)

Using the distance formula:
\[
d = \sqrt{((-2 - 2)^2 + (5 - 1)^2)} = \sqrt{(-4)^2 + (4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
\]

### 2. Points from graph 2:
- Estimated points from graph 2: \((3, 4)\) and \((-2, -2)\)

Using the distance formula:
\[
d = \sqrt{((-2 - 3)^2 + (-2 - 4)^2)} = \sqrt{(-5)^2 + (-6)^2} = \sqrt{25 + 36} = \sqrt{61}
\]

### 3. Points \((5, 8)\) and \((2, 5)\):

\[
d = \sqrt{((2 - 5)^2 + (5 - 8)^2)} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}
\]

### 4. Points \((7, 8)\) and \((-5, -1)\):

\[
d = \sqrt{((-5 - 7)^2 + (-1 - 8)^2)} = \sqrt{(-12)^2 + (-9)^2} = \sqrt{144 + 81} = \sqrt{225} = 15
\]

### 5. Points \((-1, -6)\) and \((4, -5)\):

\[
d = \sqrt{((4 + 1)^2 + (-5 + 6)^2)} = \sqrt{(5)^2 + (1)^2} = \sqrt{25 + 1} = \sqrt{26}
\]

### 6. Points \((6, 2)\) and \((4, -1)\):

\[
d = \sqrt{((4 - 6)^2 + (-1 - 2)^2)} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}
\]

### 7. Points \((-6, 1)\) and \((3, -5)\):

\[
d = \sqrt{((3 + 6)^2 + (-5 - 1)^2)} = \sqrt{(9)^2 + (-6)^2} = \sqrt{81 + 36} = \sqrt{117} = 3\sqrt{13}
\]

### 8. Points \((2, -2)\) and \((7, 8)\):

\[
d = \sqrt{((7 - 2)^2 + (8 + 2)^2)} = \sqrt{(5)^2 + (10)^2} = \sqrt{25 + 100} = \sqrt{125} = 5\sqrt{5}
\]

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Would you like further details on any of these solutions or clarifications?

### 5 related questions:
1. How do you derive the distance formula from the Pythagorean theorem?
2. What is the significance of simplifying square roots in geometry?
3. How does the distance formula relate to vectors and magnitude in physics?
4. Can the distance formula be applied in three-dimensional space?
5. How can the concept of distance between points be extended to non-Euclidean geometries?

### Tip:
Always double-check your arithmetic calculations when simplifying square roots, as small errors can lead to incorrect results.

Find the distance between each pair of points. Simplify all roots.

Explore how to find the distance between pairs of points using the distance formula, derived from the Pythagorean theorem. This lesson includes step-by-step solutions to calculate distances on a coordinate plane, suitable for students in Grades 9-12.

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