Two trains, old steamy and chug-a-lug, are 290 miles apart from each other and headed for the same station. they started toward the station at 8: 00 a. m. if they are both set to arrive at 10: 30 a. m. and old steamy is going 6.14 mph, how fast must chug-a-lug be going? how far does chug-a-lug have to travel?

8 am to 10:30 am is 2.50 hours.

Old Steamy is going 6.14 mph for 2.5 hours so travels : 6.14 * 2.5 = 15.35 miles.

This means Chug a lug has to travel 290 - 15.35 = 274.65 miles in 2.5 hours.

The speed would be 274.65 / 2.5 = 109.86 miles per hour.

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°

answer:   rotate 270° clockwise about (7, 1) and reflect across the line x = 7

step-by-step explanation:

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