Answer:
We conclude that:
∠A = ∠ FEC
Step-by-step explanation:
Given
△ABD ≅ △EFC
To determine:
∠A =
As
△ABD ≅ △EFC
So the triangles △ABD and △EFC are congruent to each other.
Please check the attached graph.
From the graph, it is clear that ∠A is correspondent to ∠E.
∠A = ∠E
From the attached figure, it is clear that:
∠F can also be denoted by ∠ FEC
as
∠A = ∠E
so
∠A = ∠ FEC
Therefore, we conclude that:
∠A = ∠ FEC
The best way to do this is to draw a picture of ΔFKL and include line segment KM that is perpendicular to FL. This creates ΔFKM which is a 45°-45°-90° triangle and ΔLKM which is a 30°-60°-90° triangle.
Find the lengths of FM and ML. Then, FM + ML = FL
<u>FM</u>
ΔFKM (45°-45°-90°): FK is the hypotenuse so FM =
<u>ML</u>
ΔLKM (30°-60°-90°): from ΔFKM, we know that KM = , so KL =
<u>FM + ML = FL</u>
=
Answer:
Step-by-step explanation:
slope-intercept form: y = mx + b
y = (x , y)
x = (x , y)
m = slope = 1/2
b = y-intercept = 89
Plug in the corresponding numbers & variables to the corresponding variables:
y = (1/2)x + 89
is your answer.
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