Points equidistant from DE EF are in the bisector of angle DEF
points equidistant from EF DF are in the bisector of angle EFD
the sought after point is the intersection of bisectricess of triangle
<span>(x – h)^2 + (y – k)^2 = r<span>^2
this equation is a derivative of the equation of a circle
x^2 + y^2 = r^2
This is from the origin. If we move the in x or y then the radius will change positions in x or y
with h = -3 and k = 1
we can plug in each set of numbers and solve.
we find Z to be on the circle edge!</span></span>
Answer:

Step-by-step explanation:
step 1
Find the radius of the circle
we know that
The circumference of a circle is equal to

we have


substitute the values



step 2
Find the area of the circle
The area of the circle is equal to

substitute the values


Answer: ![(-\infty, -4]](https://tex.z-dn.net/?f=%28-%5Cinfty%2C%20-4%5D)
Curved parenthesis at negative infinity
Square bracket at -4
====================================================
Work Shown:

The last inequality shown above is the same as saying 
Converting this to interval notation leads to the final answer of ![(-\infty , -4]](https://tex.z-dn.net/?f=%28-%5Cinfty%20%2C%20-4%5D)
Note the use of a square bracket at -4 to include this endpoint. We can never include either infinity, so we always use a parenthesis for either infinity.
Answer:
<A = 16.82 degrees
Step-by-step explanation:
Tan^-1 = opposite/adjacent
Tan^-1 (13/43) = 16.82 degrees