Study the given ray. State the ray that is equal to IM. Name the triangle by its side property. What is the sum of the interior angles of the shape below? How many different quadrilaterals can be obtained from the hexagon?

The diagram shows a ray starting at point J and extending through points K, L, and M to the right. This ray can be named $\vec{JK}$, $\vec{JL}$, or $\vec{JM}$. The question asks to "State the ray that is equal to $\vec{M}$". The notation $\vec{M}$ for a ray is non-standard. However, in basic geometry, if a single letter with an arrow is used, it often refers to a ray starting at that point. Given the context, it is most likely asking for a ray that starts at point M and extends in the opposite direction of the main ray shown (which goes to the right). A ray starting at M and extending to the left would pass through L, K, and J.

The ray starting at M and extending to the left can be named $\vec{ML}$, $\vec{MK}$, or $\vec{MJ}$. We will use $\vec{ML}$.
The answer is $\boxed{\vec{ML}}$.

The triangle shown has two sides marked with a single tick mark, indicating that these two sides are equal in length. The third side has two tick marks, indicating it has a different length.
A triangle with exactly two sides of equal length is classified as an isosceles triangle.
The answer is $\boxed{\text{Isosceles triangle}}$.

The shape shown is a polygon. By counting its sides, we find that it has 5 sides. A polygon with 5 sides is called a pentagon.
The formula for the sum of the interior angles of a polygon with $n$ sides is given by $(n-2) \times 180^\circ$.
For a pentagon, $n=5$.

The answer is $\boxed{540^\circ}$.

The question asks how many different quadrilaterals can be obtained from a hexagon.
A hexagon has 6 vertices. A quadrilateral is a polygon with 4 vertices.
To form a quadrilateral from a hexagon, we need to choose 4 vertices out of the 6 available vertices. The order of choosing the vertices does not matter, so this is a combination problem.
The number of combinations of choosing $k$ items from a set of $n$ items is given by the formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.
Here, $n=6$ (total vertices in a hexagon) and $k=4$ (vertices for a quadrilateral).

The answer is $\boxed{15}$.