The circumscribed circle is a circle which passes through all the vertices of a triangle.
This means the center of the circle must be at the same distance from all the vertices.
A point equidistant from the end point of a line segment always lies on the perpendicular bisector.
Also the point of concurrence of the perpendicular bisectors of the sides of the triangle is the circum center which is the center of circum cirle.
Hence we must draw the perpendicular bisector of any one side.
So Patrick's first step should be to construct the perpendicular bisector of ST.
Option A is the right answer.
The answer is 3. Point and line symmetry
If you mean graphically x^2+500 would work on a graph
Answer:
y + 4/7 6x
Step-by-step explanation:
Answer:
√73 or 8.54 units to the nearest hundredth.
Step-by-step explanation:
Use the distance formula with points (8, 3) and (0, 0).
D = √ [(8 - 0)^2 + (3 - 0)^2]
= √(64 + 9)
= √73 units.