Suppose there is a 3-degree drop in temperature for every thousand feet that an airplane climbs into the sky. If the temperature on the ground is 51 degrees what would be the temperature when the plane reaches an altitude of 21,000 1?

The temperature would be degree(s).

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°

c.

step-by-step explanation:

because first you have to add the 4x, 9x, and the 7x together and you get 20x and then you add the last numbers up and you get 5and then you put the subtraction sign in there and that's how you get the answer.