X is less than or equal to 3
X is less than 9
X is less than -5
X is greater than or equal to 13
Answer:
10-5
Step-by-step explanation:
As per the attached figure, right angled
has an inscribed circle whose center is
.
We have joined the incenter
to the vertices of the
.
Sides MD and DL are equal because we are given that 
Formula for <em>area</em> of a
As per the figure attached, we are given that side <em>a = 10.</em>
Using pythagoras theorem, we can easily calculate that side ML = 10
Points P,Q and R are at
on the sides ML, MD and DL respectively so IQ, IR and IP are heights of
MIL,
MID and
DIL.
Also,


So, radius of circle = 
C^2-64 is a square roots factoring one. you take the square root of c^2 which is c and then the square root of 64 which is 8 or -8 so the answer is (c-8)(c+8). and it is a special product