In the diagram A, B, and C are the points on the circle. Use the diagram to prove the theorem which states that: if O is the center of the circle then AÔB = 2Ĉ.
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Explanation:

Here's one way to go at it.

Draw segments AB and CO. Define angles as follows. (The triangles with sides that are radii are all isosceles, so their base angles are congruent.)

  x = angle OAB = angle OBA

  y = angle OAC = angle OCA

  z = angle OBC = angle OCB

Consider the angles at each of the points A, B, C.

At A, we have ...

  angle CAB = x + y

At B, we have ...

  angle CBA = x + z

At C, we have ...

  angle ACB = y + z

The sum of the angles of triangle ABC is 180°, as is the sum of angles in triangle ABO. This gives ...

  x + x + ∠AOB = (x+y) +(x+z) +(y+z)

  ∠AOB = 2(y+z) = 2∠ACB

This shows ∠AOB = 2×∠C, as required.

we have the parent function given by,

1. the function is translated 5 units to the left.

so, we get,   becomes

2. the function is translated 5 units to the right.

so, we get,   becomes

3. the function is translated 5 units upwards.

so, we get,   becomes

4. the function is translated 5 units downwards.

so, we get,   becomes

5. the function is reflected across x-axis.

so, we get,   becomes

6. the function is reflected across y-axis.

so, we get,   becomes .

step-by-step explanation:

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