The theorem is used to prove △ABC≅△DEC is Side-Angle-Side(SAS) theorem
The two triangles are
△ABC and △DEC
Here we have to prove that the △ABC and △DEC are congruent
It is given that
The length of AC = The length of CD
The angle ACB = The angle ECD
The point C is the midpoint of EB
Therefore,
The length of BC = The length of EC
Therefore, the two sides and one angle of triangle ABC is equal to the two sides and one angle of triangle DEC, so the both triangles are congruent by SAS rule
Hence, the theorem is used to prove △ABC≅△DEC is Side-Angle-Side(SAS) theorem.
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Answer:
y = x/16 - 0
Step-by-step explanation:
Using a point slope form equation
y - y1 = m(x - x1)
Given
( 16 , 1)
Slope = 1/16
x1 = 16
y1 = 1
Insert the values
y - y1 = m(x - x1)
y - 1 = 1/16( x - 16)
Using a slope intercept form equation
y = mx + c
Open the bracket
y - 1 = 1/16(x - 16)
y - 1 = (x - 16)/16
y = (x - 16)/16 + 1
LCM = 16
y = (x - 16 + 16)/16
y = (x - 0)/16
Following the slope intercept form equation
y = x/16 - 0/16
y = x/16 - 0
The equation is y = x/16 - 0
Answer:
3
Step-by-step explanation:
If you look on the graph, you can see that the y intercept is at -4. This easily takes out answers 1 and 2 because they have a y intercept a +3. Next you notice that any point with a y value must be equal to or less than 4/3x-4. Since every point below the line has a y value less than any point on the line, it must mean that y< or =4/3x-4