Consider the figure below. Line FG is parallel to line JK. Part A:Solve for a justify your answer
Part B:Determine m< FNT
Part C:Determine m

A. a = 10

B. m<FNT = 120°

C. m<KTU = 60°

Step-by-step explanation:

A. (7a + 50)° and (14a - 20)° are corresponding angles. Therefore:

(7a + 50)° = (14a - 20)°

Use this equation to find the value of a

7a + 50 = 14a - 20

Combine like terms

7a - 14a = - 50 - 20

-7a = -70

Divide both sides by -7

-7a/-7 = -70/-7

a = 10

B. m<FNT = (14a - 20)° (alternate interior angles are congruent)

Plug in the value of a

m<FNT = 14(10) - 20

m<FNT = 140 - 20

m<FNT = 120°

C. m<KTU + (14a - 20)° = 180° (linear pair)

Plug in the value of a

m<KTU + 14(10) - 20 = 180

m<KTU + 120 = 180

m<KTU + 120 - 120 = 180 - 120

m<KTU = 60°

whats the

the rocket cannot penetrate the ground, in a contextual sense, therefore the roots of this equation should be the domain of this function. i'll let you find the roots of this quadratic equation (hint: use the quadratic formula:  

1- c

2-a

step-by-step explanation:

mark brainliest!