Can anyone tell me how to do this? (Steps) and can you answer these two questions for me plz
1 answer:
Ax²+bx+c=0
The formula to solve for x is:
So for 3x²+13x+10=0, plug in the data
x=-13±√13²-4(3)(10) / 2(3)
x=-13±√169-120 / 6
x=-13±√49 /6
x=-13<span>±7 / 6
x=-13+7/6
x=-6/6
x=-1
x=-13-7/ 6
x=-20/6 or 3.3repeating
3x</span>²+14x+15=0
x=-14±√14²-4(3)(15) / 2(3)
x=-14±√196-180 /6
x=-14<span>±</span>√16 / 6
x=-14+4/6
x=-10/6 or 1.6repeating
x=-14-4/6
x=-18/6
x=-3
The formula to solve for x is:
So for 3x²+13x+10=0, plug in the data
x=-13±√13²-4(3)(10) / 2(3)
x=-13±√169-120 / 6
x=-13±√49 /6
x=-13<span>±7 / 6
x=-13+7/6
x=-6/6
x=-1
x=-13-7/ 6
x=-20/6 or 3.3repeating
3x</span>²+14x+15=0
x=-14±√14²-4(3)(15) / 2(3)
x=-14±√196-180 /6
x=-14<span>±</span>√16 / 6
x=-14+4/6
x=-10/6 or 1.6repeating
x=-14-4/6
x=-18/6
x=-3
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Answer:
(the statement does not appear to be true)
Step-by-step explanation:
I don't think the statement is true, but you CAN compute the intercepted arc from the angle.
Note that BFDG is a convex quadrilateral, so its angles sum to 360. Since we know the inscribed circle touches the angle tangentially, angles BFD and BGD are both right angles, with a measure of 90 degrees.
Therefore, adding the angles together, we have:
alpha + 90 + 90 + <FDG = 360
Therefore, <FDG, the inscribed angle, is 180-alpha (ie, supplementary to alpha)
