Answer:
Since the pentagon is regular, the line connecting the centre of the circle to the verticles (A,B,C,D,E) should inscribe an equal angle in them
You can see what I mean in the attachment
So to mark other points we follow the following steps
- Connect OA
- With OA as base, make angle of 72°. Say th new line is OB.
- Similarly, do for the rest of them
- This is a cyclic pentagon
Step-by-step explanation:
Hope this helps... Pls vote as brainliest
The correct answer is (0, 3)
<em>Answer:</em>
<em>∠x = 70°</em>
<em>Step-by-step explanation:</em>
<em>Given that PQ||BR</em>
<em>Line PQ and BR with transversal AC,</em>
<em>∠x + ∠ACR = 180° {Co-interior angles}</em>
<em>⇒∠x + 110° = 180° {From what I can see it looks like 110.}</em>
<em>⇒∠x = 180° - 110°</em>
<em>⇒</em><em>∠x = 70°</em>
Answer:
y = 7/2x + 2
Step-by-step explanation:
To put the equation in slope intercept form [ y = mx + b ], we need to solve for y.
14x = 6y - 12
~Subtract 6y to both sides
14x - 6y = -12
~Subtract 14x to both sides
-6y = -12 - 14x
~Divide -6 to everything
y = 2 + 7/3x
~Reorder
y = 7/2x + 2
Best of Luck!
I believe there is only 1 real zero
hope this helps !!
