There are 8! Eight sides on the shape, and all sides are the same shape.
Answer:
9
Step-by-step explanation:
2*2=4*2=8+2=10-1=9
The midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9) is 
<u>Solution:</u>
Given, two points are (-6, 6) and (-3, -9)
We have to find the midpoint of the segment formed by the given points.
The midpoint of a segment formed by
is given by:


Plugging in the values in formula, we get,

Hence, the midpoint of the segment is 
Step-by-step explanation:
is this the question you asked
Answer:

Step-by-step explanation:
We need to find m ∠EGD.
From the given figure, we can see that, ∠BGA & ∠EGD are vertically opposite angles (opposite angles that share the same vertex ⟶ G).
Also, vertically opposite angles are equal to each other.
Given, ∠BGA = 30°
So, ∠EGD = ∠BGA = 30°.
