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The measure of ∠BCD = (68.5)°

Step-by-step explanation:

Here, given:

AB II CD, BC II AE,  ∠ABD = (3x +4)°,   ∠BCD = (6x -8)° and   ∠EDF = (7x-20)°

Now, as given

∠ABD = (3x +4)°

⇒∠BDC = (3x +4)°  ( as AB II CD,  PAIR OF ALTERNATE ANGLES)   ... (1)

and , ∠EDF = (7x-20)°

⇒∠ADB = (7x-20)°  (PAIR OF VERTICALLY OPPOSITE ANGLES)

⇒∠DBC = (7x-20)°  ( as BC II AE,  PAIR OF ALTERNATE ANGLES)   ... (2)

Now, in Δ DBC:

∠BDC + ∠DBC + ∠BCD   = 180°  ( ANGLE SUM PROPERTY of a Δ)

⇒ (3x +4)° +  (7x-20)° + (6x -8)°  = 180

or, 16 x  - 24 = 180

or, 16 x = 204

x = 204/16 =  12.75 , or x = 12.75

⇒ ∠BCD = (6x -8)° = 6(12.75) - 8 =  68.5

Hence, the measure of ∠BCD = (68.5)°

68.5°

Step-by-step explanation:

ABD + BCD+ EDF = 180

3x + 4 + 6x - 8 + 7x - 20 = 180

16x = 204

x = 51/4

6x - 8

6(51/4) - 8

68.5

a) t'(-3, 6) and v'(0, 3)  

step-by-step explanation:

your answer is -9

step-by-step explanation: