Answer:
See Below.
Step-by-step explanation:
In the given figure, O is the center of the circle. Two equal chords AB and CD intersect each other at E.
We want to prove that I) AE = CE and II) BE = DE
First, we will construct two triangles by constructing segments AD and CB. This is shown in Figure 1.
Recall that congruent chords have congruent arcs. Since chords AB ≅ CD, their respective arcs are also congruent:

Arc AB is the sum of Arcs AD and DB:

Likewise, Arc CD is the sum of Arcs CB and DB. So:

Since Arc AB ≅ Arc CD:

Solve:

The converse tells us that congruent arcs have congruent chords. Thus:

Note that both ∠ADC and ∠CBA intercept the same arc Arc AC. Therefore:

Additionally:

Since they are vertical angles.
Thus:

By AAS.
Then by CPCTC:

Please equate 2x^2 -8x -90 to zero: <span>2x^2 -8x -90=0
Next, reduce </span>2x^2 -8x -90=0 by div. all terms by 2: x^2 -4x -45=0
Next, factor the poly: (x-9)(x+5) = 0
Set each factor = to 0 and solve for x:
x-9=0 => x=9
x+5=0 => x= -5
Answer:
If both lines highlighted gone in both directions forever they will never ever meet.
Step-by-step explanation:
hope this helps.
Answer:
2x³+3x²-8x+3
Step-by-step explanation:
So we know x=½,x=1 and x=-3
Then (2x-1)(x-1)(x+3) are the factors
(2x-1)[(x²+3x-x-3)]
(2x-1)[(x²+2x-3)]
2x³+4x²-6x-x²-2x+3
2x³+3x²-8x+3
The answer to this equation is y=-8x-19
I hope this helps!!!!!!!!