Can someone solve this?

Step-by-step explanation:

1A. Opposite angles of an inscribed quadrilateral are supplementary.

  ∠K = 180° -∠M

  7x = 180 -131 = 49

  x = 7 . . . . . divide by 7

  ∠L = 180 -∠J

  11x +y = 180 -99 . . . fill in values

  11(7) +y = 81 . . . . . .  fill in x

  y = 4 . . . . . subtract 77

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1B. The arc measure is twice the inscribed angle measure:

  arc JKL = 2∠M = 2(131°)

  arc JKL = 262°

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2. Tangents from the same point are congruent.

  AM = AL = 3

  CN = CM = 10 -AM = 7

  BL = BN = 9 -CN

  BL = 2

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3. Linear angles are supplementary.

  ∠2 = 180° -∠1 = 180° -110°

  ∠2 = 70°

Angles where chords cross are the average of the subtended arcs.

  ∠3 = (100° +60°)/2

  ∠3 = 80°

  ∠4 = 180° -∠3 = 100°

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4. The product of the segments from the crossing point to the circle intersection points is a constant.

  12x = 8(x+12)

  4x = 96 . . . . subtract 8x

  x = 24

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5. Arc length is given by the formula ...

  s = rθ . . . . θ in radians

  s = (7 cm)(100°(π/180°)) = (7 cm)(5π/9) ≈ 12.22 cm

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step-by-step explanation: