Answer:
x° = ∠OBR = ∠ABC (base angles of a cyclic isosceles trapezoid)
Step-by-step explanation:
APRB form a cyclic trapezoid
∠APO = x° (Base angle of an isosceles triangle)
∠OPR = ∠ORP (Base angle of an isosceles triangle)
∠ORB = ∠OBR (Base angle of an isosceles triangle)
∠APO + ∠OPR + ∠OBR = 180° (Sum of opposite angles in a cyclic quadrilateral)
Similarly;
∠ORB + ∠ORP + x° = 180°
Since ∠APO = x° ∠ORB = ∠OBR and ∠OPR = ∠ORP we put
We also have;
∠OPR = ∠AOP = ∠BOR (Alternate interior angles of parallel lines)
Hence 2·x° + ∠AOP = 180° (Sum of angles in a triangle) = 2·∠OBR + ∠BOR
Therefore, 2·x° = 2·∠OBR, x° = ∠OBR = ∠ABC.
Answer:
92832-31504=61328...........
If you add a constant to the original equation, you are actually shifting up or down
Since the constant is positive, you are shifting the graph 20 units up
y=-2x+7
Step-by-step explanation:
line segment xy has endpoints x(5 7) and y(-3 3)
for the equation of the perpendicular bisector of line segment xy
slope of line segment xy = 7-3 / 5+3
= 4/8
=1/2
so slope of perpendicular bisector is -2
as m1m2= -1 or m1= -1/m2
As perpendicular bisector goes through midpoint of xy , let's find midpoint of xy
midpoint(x,y) = (5-3 / 2 , 7+3 / 2)
=(2/2, 10/2)
=(1,5)
find the equation of line(perpendicular bisector) passing through (1,5) and the slope -2
y-5 = -2(x-1)
y-5 =-2x+2
y=-2x+7
