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:

Answer:
b 12
Step-by-step explanation:
Answer:
195.25
Step-by-step explanation:
Consider geometric series S(n) where initial term is a
So S(n)=a+ar^1+...ar^n
Factor out a
S(n)=a(1+r+r^2...+r^n)
Multiply by r
S(n)r=a(r+r^2+r^3...+r^n+r^n+1)
Subtract S(n) from S(n)r
Note that only 1 and rn^1 remain.
S(n)r-S(n)=a(r^n+1 -1)
Factor out S(n)
S(n)(r-1)=a(r^n+1 -1)
The formula now shows S(n)=a(r^n+1 -1)/(r-1)
Now use the formula for the problem
Expand
Simplify 0.9x + 1.26 - 2.3 + 0.1 * x to x - 1.04
add 1.04 to both sides
simplify 1.6 + 1.04 to 2.64
Answer: x = 2.64
Answer:
7-3√2/31
Step-by-step explanation:
Rationalise it and you'll get your answer.