A racecar driver is doing a time trial of a new car on a straight back. The car's distance from the timekeeper is represented by y=293x+50 where x is time in seconds and y is the distance from the timekeeper's position. How many feet from the timekeeper is the race car at the beginning?

50 feet

Step-by-step explanation:

The car's distance from the timekeeper us given by :

y = 293x+50

Where x is time in seconds and y is the distance from the timekeeper's position.

The general equation of a line is given by :

y = mx +c

On comparing, we get : c = 50

It means the timekeeper has covered 50 feets at the beginning.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°