52-46=6 drop down the 1 = 61- 46=15 drop down the 0 =150 46x3= 138 so 150-138= 12
Answer:
the coordinates of C' = (2,1)
Step-by-step explanation:
The coordinates of point C can be found by looking at the graph.
Coordinates of C are C= (6,3)
If ABCD is dilated by a factor of 1/3 then the coordinates of C' can be found by multiplying the coordinates of C by 1/3
C = (6,3)
C' =(1/3*6,1/3*3)
C' = (2,1)
So, the coordinates of C' = (2,1)
Well you see you take the sector and tie it 3 times the radiants which will then answer the question that you seek from within the depths of bones and minds.
AE = AC = 4
m<CAB = 60 (equilateral triangle)
m<CAE = 90 (square)
m<BAE = 150 (= 60 + 90)
Triangle BAE is isosceles since AB = AE;
therefore, m<AEB = m<ABE.
m<AEB + m<ABE + m<BAE = 180
m<AEB + m< ABE + 150 = 180
m<AEB + m<AEB = 30
m<AEB = 15
In triangle ABE, we know AE = AB = 4;
we also know m<BAE = 150, and m<AEB = 15.
We can use the law of sines to find BE.
BE/(sin 150) = 4/(sin 15)
BE = (4 sin 150)/(sin 15)
BE = 7.727
765 divides by 5 = 153.
Hope that helped! :)
