Answer:
A non-equilateral rhombus.
Step-by-step explanation:
We can solve this graphically.
We start with square:
ABCD
with:
A = (11, - 7)
B = (9, - 4)
C = (11, - 1)
D = (13, - 4)
Only with the vertices, we can see that ABCD is equilateral, as the length of each side is:
AB = √( (11 - 9)^2 + (-7 -(-4))^2) = √( (2)^2 + (3)^2) = √(4 + 9) = √13
BC = √( (11 - 9)^2 + (-1 -(-4))^2) = √13
CD = √( (11 - 13)^2 + (-1 -(-4))^2) = √13
DA = √( (11 - 13)^2 + (-7 -(-4))^2) = √13
And we change C by C' = (11, 1)
In the image you can see the 5 points and the figure that they make:
The figure ABCD is a rhombus, and ABC'D is also a rhombus, the only difference between the figures is that ABCD is equilateral while ABC'D is not equilateral.
I think its -40 because it looks like you have negative 40 then ur adding 10
(X+3)(y-2)
F(x-3)-2
This cheat sheet may help
Answer:
AN = 16 units
Step-by-step explanation:
So we can use the fact that 
- Since this is an isosceles triangle, segments FO and FR are both equal to 7
Manipulating the equation above to solve for AN we have 
- Substituting in the known values we have

<span>17x=18(1/3)
17x = 6
x = 6/17
hope it helps</span>