can i get a picture of the graph please?
Y-70
X-30
Z-80
You’re welcome
-6(12)= -36
(13)(-3)= -39
(-5)(16)= -80
Answer:
<u><em>ΔEAB ≅ΔFCD</em></u>
Step-by-step explanation:
As ABCD is a parallelogram opposite sides are equal s AE ≅CF ,also AB≅ DC . Both E and F bisect the opposite lines in equal halves.
The parallelogram has a property that its opposite angles are equal . Therefore ∠c≅∠a, ∠d≅∠b
Using the SAA≅SAA postulate we conclude that ΔEAB ≅ΔFCD
Answer:
<h2>The central angle of the sector is 126°.</h2>
Step-by-step explanation:
Givens
![A=140 \pi \ cm^{2} \\r=20 \cm](https://tex.z-dn.net/?f=A%3D140%20%5Cpi%20%5C%20cm%5E%7B2%7D%20%5C%5Cr%3D20%20%5Ccm)
The area of a circular sector is defined as
![A=\frac{\pi r^{2} \theta }{360\°}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%20r%5E%7B2%7D%20%5Ctheta%20%7D%7B360%5C%C2%B0%7D)
Replacing given values, and solving for ![\theta](https://tex.z-dn.net/?f=%5Ctheta)
![140 \pi =\frac{\pi 20^{2} \theta }{360\°}\\ 400 \theta = 50,400\\\theta = \frac{50,400}{400} \approx 126 \°](https://tex.z-dn.net/?f=140%20%5Cpi%20%3D%5Cfrac%7B%5Cpi%2020%5E%7B2%7D%20%5Ctheta%20%7D%7B360%5C%C2%B0%7D%5C%5C%20400%20%5Ctheta%20%3D%2050%2C400%5C%5C%5Ctheta%20%3D%20%5Cfrac%7B50%2C400%7D%7B400%7D%20%5Capprox%20126%20%5C%C2%B0)
Therefore, the central angle of the sector is 126°.