Find each measurement. (The figure is not drawn to scale.)

a. m∠Z =  62

b. = 118

c.  m∠W = 62

d. = 122°

Step-by-step explanation:

a. The given parameters are;

m∠X = 118

m∠Y = 120

m∠X + m∠Z = 180  Angles in opposite segment are supplementary

m∠Z = 180 - m∠X  = 180 - 118 = 62

m∠Z =  62

b. Given line drawn from W to Y forms isosceles triangles WZY, with base angles ∠WYZ and ∠YWZ equal (Base angles of an isosceles triangle)

Therefore

∠WYZ + ∠YWZ + m∠Z = 180 (Angle sum theorem)

∠WYZ = ∠YWZ (Substitution property of equality)

∠WYZ + ∠YWZ + m∠Z = ∠WYZ + ∠WYZ + m∠Z =180

2×∠WYZ + 62 =180

2×∠WYZ = 180 -62 = 118°

∠WYZ =  118°/2 =59

∠WYZ = ∠YWZ = 59

subtends chord WZ at the center  =  ∠WYZ  subtends chord WZ at the circumference

∴ 2×∠WYZ =

 = 2×59 = 118

= 118

c.  m∠X + m∠Y + m∠Z + m∠W = 360 (Sum of angles in a quadrilateral)

m∠W = 360 - (m∠X + m∠Y + m∠Z) = 360 - (118 + 120 + 60) = 62

m∠W = 62

d. + =  (Angle addition postulate)

= 2 × ∠Y (Angle subtended at the center = 2 × Angle subtended at the circumference

∴ = 2 × 120 = 240

= -

= 240 - 118 = 122°

= 122°.

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.