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Vera_Pavlovna [14]
2 years ago
5

Duane rode his bike at 14 miles per hour for 3 hours, then put the bike down and began walking at miles 2 per hour. Elizabeth st

arted jogging at 5 miles per hour along the same path at the same time that Duane started biking. When she reached the bike, she began riding it at 10 miles per hour. How many hours will it have been since they started when Elizabeth catches up to Duane
Mathematics
1 answer:
Colt1911 [192]2 years ago
4 0

Elizabeth will take <u>9 hours and 45 minutes</u> to catch up with Duane using the given speed-distance-time values.

The speed-distance-time relations are:

Speed = Distance/Time, Distance = Speed*Time, and Time = Distance/Speed.

The distance traveled by Duane on the bike at 14 miles per hour for 3 hours = 14*3 miles = 42 miles.

The time that Elizabeth takes to cover 42 miles jogging at 5 miles per hour = 42/5 hours = 8.4 hours.

The distance traveled by Duane in the extra time which Elizabeth took to reach his bike (8.4 - 3 = 5.4 hours) walking at 2 miles per hour = 2*5.4 = 10.8 miles.

Now, the distance between Elizabeth and Duane is 10.8 miles, before Elizabeth starts riding the bike.

Keeping the distance constant, we calculate the relative speed to calculate the time to cover this distance.

Relative speed = Elizabeth's speed - Duane's speed {As Elizabeth is moving towards Duane, that is, covering the distance but Duane is moving away from Elizabeth, that is, increasing the distance}.

or, Relative speed = 10 - 2 = 8 miles per hour.

Time taken to cover the distance = 10.8/8 = 1.35 hours.

Therefore, the total time taken by Elizabeth to catch up with Duane = 8.4 + 1.35 hours = 9.75 hours or 9 hours and 45 minutes.

Learn more about the speed-distance-time relations at

brainly.com/question/4931057

#SPJ4

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<h3>Which method can be used to find the measure of angle m<D?</h3>

In triangle ∆MDK, we have;

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Using cosine rule, we get;

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Therefore;

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D = arccos \left( \frac{ {m}^{2} +  {k}^{2}  -  {d}^{2}   }{2 \cdot m \cdot k}  \right)

Which gives;

D = arccos \left( \frac{ {16.4}^{2} +  {19}^{2}  -  {22}^{2}   }{2  \times  16.4  \times  19}  \right)  \approx  {76}^{ \circ}

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The correct option is therefore;

  • c) 76°

Learn more about the rule of cosines here:

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