Is there a photo or something?
<span>The diagonals of a parallelogram bisect each other.
So we just equate the two equations BE and DE
2x^2 - x = x^2 + 6
2x^2 - x - x^2 - 6 = 0
x^2 - x - 6 = 0
This is quadratic eqn. So we obtain.
(x - 3) ( x + 2) = 0
Setting each to 0 and solving for x, we have that x = 3 and x = -2
So we have two possible values for BD.
But Since BE = DE then we can simply double each
If x = 3 BD = 2 (2(3)^2 - 3) = 2 ( 18 - 3) = 30 units
If x = - 2 BD = 2((-2)^2 -3 ) = 2 (8 - 3) = 10 units</span>
<h3>
Answer:</h3><h3>
-3.5,-0.5</h3>
Step-by-step explanation:
Answer:
cute scalene triangle.
Sides: a = 1 b = 2 c = 1.906
Area: T = 0.94
Perimeter: p = 4.906
Semiperimeter: s = 2.453
Angle ∠ A = α = 29.543° = 29°32'36″ = 0.516 rad
Angle ∠ B = β = 80.457° = 80°27'24″ = 1.404 rad
Angle ∠ C = γ = 70° = 1.222 rad
Height: ha = 1.879
Height: hb = 0.94
Height: hc = 0.986
Median: ma = 1.888
Median: mb = 1.147
Median: mc = 1.262
Inradius: r = 0.383
Circumradius: R = 1.014
Vertex coordinates: A[1.906; 0] B[0; 0] C[0.166; 0.986]
Centroid: CG[0.691; 0.329]
Coordinates of the circumscribed circle: U[0.953; 0.347]
Coordinates of the inscribed circle: I[0.453; 0.383]
Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 150.457° = 150°27'24″ = 0.516 rad
∠ B' = β' = 99.543° = 99°32'36″ = 1.404 rad
∠ C' = γ' = 110° = 1.222 rad
Step-by-step explanation:
