Answer:
No, it is a reflection and a translation
Step-by-step explanation:
It reflects off of A and translates to the right over B
Answer:
K(x) = ![\frac{-10}{[1 + (-10x)^2]^{\frac{3}{2} } }](https://tex.z-dn.net/?f=%5Cfrac%7B-10%7D%7B%5B1%20%2B%20%28-10x%29%5E2%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%7D)
( curvature function)
Step-by-step explanation:
considering the Given function
F(x) = 
first Determine the value of F'(x)
F'(x) = 
F'(x) = -10x
next we Determine the value of F"(x)
F"(x) = 
F" (x) = -10
To find the curvature function we have to insert the values above into the given formula
K(x) ![= \frac{|f"(x)|}{[1 +( f'(x)^2)]^{\frac{3}{2} } }](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%7Cf%22%28x%29%7C%7D%7B%5B1%20%2B%28%20f%27%28x%29%5E2%29%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%7D)
K(x) = ( curvature function)
Answer:
c bro okkkkkkkkkkkkkkmmmmmmmkkkkkkkkkbey
The correct answer I believe in b
Answer:
Table :
Statements Reasons
1)ΔEGC≅ΔEGB Given
2)EB = EC Corresponding Parts of Congruent
Triangles are Congruent
3) BC = BE Given
4) By 2 and 3 EB≅EC≅BC
5)∠BEC = 60° ΔBEC is an equilateral triangle
Step-by-step explanation:
We are given that BE = BC ---1
We are also given that ΔEGC≅ΔEGB
So, BE = CE(Corresponding Parts of Congruent Triangles are Congruent) ---2
Now By 1 and 2
BE=BC=CE
In ΔBEC
BE=BC=CE
Since all sides are equal . So, it is an equilateral triangle
All angles of equilateral triangle are also equal i.e. 60°
So, ∠BEC = 60°
Hence proved
Table :
Statements Reasons
1)ΔEGC≅ΔEGB Given
2)EB = EC Corresponding Parts of Congruent
Triangles are Congruent
3) BC = BE Given
4) By 2 and 3 EB≅EC≅BC
5)∠BEC = 60° ΔBEC is an equilateral triangle
