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dannyboy938
19.03.2020 •
Mathematics
Quadrilateral ABCD has vertices at A(3,−2), B(4,3), C(−2,0), and D(−3,−5). Quadrilateral A B C D in the plane with the given coordinates. © 2019 StrongMind. Created using GeoGebra. Which answers prove that quadrilateral ABCD is a parallelogram? Select all that apply. slope of AB¯¯¯¯¯¯¯¯=slope of DC¯¯¯¯¯¯¯¯; AD¯¯¯¯¯¯¯¯≅BC¯¯¯¯¯¯¯¯ AB¯¯¯¯¯¯¯¯∥DC¯¯¯¯¯¯¯¯ and AD¯¯¯¯¯¯¯¯∥BC¯¯¯¯¯¯¯¯ ∠A is supplementary to ∠B; ∠A≅∠C AD¯¯¯¯¯¯¯¯≅BC¯¯¯¯¯¯¯¯≅CD¯¯¯¯¯¯¯¯ slope of AD¯¯¯¯¯¯¯¯=slope of BC¯¯¯¯¯¯¯¯; AD¯¯¯¯¯¯¯¯≅BC¯¯¯¯¯¯¯¯
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Ответ:
The quadrilateral ABCD is a parallelogram as, slope of AB = slope of DC = 5, and slope of AD = slope of BC = 0.5.
Step-by-step explanation:
Quadrilateral ABCD has vertices at A(3,−2), B(4,3), C(−2,0), and D(−3,−5).
So, the slope of AB =![\frac{3 + 2}{4 - 3} = 5](/tpl/images/0553/8114/b5922.png)
Now, the slope of BC =![\frac{0 - 3}{- 2 - 4} = \frac{1}{2}](/tpl/images/0553/8114/ffd95.png)
The slope of CD =![\frac{- 5 - 0}{- 3 + 2} = 5](/tpl/images/0553/8114/ae75f.png)
Finally, the slope of DA =![\frac{- 2 + 5}{3 + 3} = \frac{1}{2}](/tpl/images/0553/8114/a84e2.png)
Therefore, the quadrilateral ABCD is a parallelogram as, slope of AB = slope of DC = 5, and slope of AD = slope of BC = 0.5. (Answer)
Note: A quadrilateral to be a parallelogram the slope of each pair of opposite sides must be the same.
Ответ:
The formula for circumference is pi x diameter.
C= pi x d
C= 3.14 x 3000
C= 9420 feet