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The problem asks for the distance between the two points \((-3, 4)\) and \((-8, -8)\).

We use the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
where \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points. 

In this case:
\[
x_1 = -3, \quad y_1 = 4, \quad x_2 = -8, \quad y_2 = -8
\]

Substituting into the formula:
\[
d = \sqrt{(-8 - (-3))^2 + (-8 - 4)^2}
\]
\[
d = \sqrt{(-8 + 3)^2 + (-12)^2}
\]
\[
d = \sqrt{(-5)^2 + (-12)^2}
\]
\[
d = \sqrt{25 + 144}
\]
\[
d = \sqrt{169}
\]
\[
d = 13
\]

So, the distance between the two points is **13 units**.

Do you need further details or have any questions?

Here are some related questions for practice:
1. What is the distance between the points \((2, 5)\) and \((-1, -3)\)?
2. Can the distance between two points ever be negative?
3. How would the distance formula change in three-dimensional space?
4. What is the midpoint between the points \((-3, 4)\) and \((-8, -8)\)?
5. How is the distance formula related to the Pythagorean theorem?

**Tip:** The distance formula is derived directly from the Pythagorean theorem, considering the horizontal and vertical differences between two points as the legs of a right triangle.

Find the distance between the two points (-3, 4) and (-8, -8), rounding to the nearest unit if necessary.

Calculate the distance between the two points (-3, 4) and (-8, -8) using the distance formula. Learn how to apply the Pythagorean theorem in coordinate geometry. This problem is perfect for students in Grades 8-10.

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