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.
Steps to solve:
5(x + 1) = 20
~Distributive property
5x + 5 = 20
~Subtract 5 to both sides
5x + 5 - 5 = 20 - 5
~Simplify
5x = 15
~Divide 5 to both sides
5x/5 = 15/5
~Simplify
x = 3
Best of Luck!
The slope of a line is constant because the acceleration is equal to zero.
The slope of a line is always just (y2-y1)/(x2-x1), so in this case:
m=(-7-6)/(4--6)
m=-13/10
m=-1.3
Number one: if line segments SR & RT are perpendicular line segment Tu and US are perpendicular and angle STR is congruent to angle TSU, then triangle TRS is congruent to triangle SUT
Number two: if line segment AC is congruent to line segment CB and line segment CB bisects line segment AB, then < a is congruent to < B