Yes the diagonals of a parallelogram have the same midpoint since they ... of the intersection of the diagonals of parallelogram AB CD given the vertex points ... If a parallelogram is a rhombus then its diagonals are? , statement 2 is the answer
Answer:
regular pentadecagon
Step-by-step explanation:
it has 15 sides
that is what matches the exterior angled sum property - 360degree
let me explain it
1 side = 180 - 156 = 24
360/24 = 15
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Answer:
3/10
Step-by-step explanation:
i answered it already
Answer:
the answer is c!
Step-by-step explanation:
Answer:
Step-by-step explanation:
To evaluate for such, the following comprehension is required,
Equation Required: Distance Formula: d(P, Q) = √ (x2 − x1)^2 + (y2 − y1)^2
Denote the configurations as the following,
(5, -1). (5, -4)
X1 Y1. X2. Y2
D(P, Q) = √(5 - 5)^2 + (-4 +1)^2. <== Since the double negative is present, the operation is acknowledged as positive.
D(P, Q) = √(0)^2 + (-3)^2
D(P, Q) = √9 = 3
Thus, the agglomerate distance between the points situated in the Cartesian plane is disclosed, and is, henceforth, disseminated as 03 units.
