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professor190 [17]
3 years ago
15

A student on the cross- county team runs 30 minutes a day as a part of her training. Write an equation to describe the relations

hip between the distance she runs in miles, D, andher running speed,in miles per hour,when she runs at a constant speed of 5.4miles per hour for m minutes,and then at be miles per hour for n minutes
Mathematics
1 answer:
KengaRu [80]3 years ago
6 0

Answer:

The answer is below

Step-by-step explanation:

A student on the cross- county team runs 30 minutes a day as a part of her training. Write an equation to describe the relationship between the distance she runs in miles, D, and her running speed, in miles per hour, when she runs at a constant speed of 5.4miles per hour for m minutes, and then at b miles per hour for n minutes

Solution:

Given that the student runs for a total of 30 minutes per day. He runs at 5.4 miles per hour for m minutes, and then at b miles per hour for n minutes.

Hence:

m + n = 30

60 minute = 1 hour. Therefore, m minute = m/60 hour, n minute = n/60 hour.

The distance traveled is the product the speed in miles per hour and the time taken to cover the distance. Distance = speed * time

The total distance ran (D) is:

D = 5.4(m/60) + b(n/60)

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
A train leaves Philadelphia at 2:00 PM. A second train leaves the same city in the same direction at 4:00 PM. The second train t
polet [3.4K]

Answer: First train speed shall be 42 mi/hr

Second train speed shall be 84 mi/hr

Yes it is the valid answer, it gets annoying when almost everyone enters the wrong answer for a good laugh. Either way, hope this helps! Have a spectacular day.

Step-by-step explanation:

What is the 1st train's head start in miles?

Let +s+ = the speed of the 1st train in mi/hr

From 2PM to 4PM is 2 hrs

+d%5B1%5D+=+s%2A2+

---------------------

Let +d+ = distance the 2nd train travels until

it overtakes the 1st train

From 4PM to 6PM is 2 hrs

Start time when the 2nd train leaves

---------------------

Equation for 1st train:

(1) +d+-+2s+=+s%2A2+

Equation for 2nd train:

(2) +d+=+%28+s%2B42+%29%2A2+

--------------------

Substitute (2) into (1)

(1) +%28+s%2B42+%29%2A2+-+2s+=+s%2A2+

(1) +2s+%2B+84+-+2s+=+2s+

(1) +2s+=+84+

(1) +s+=+42+

and

+s%2B+42+=+42+%2B+42+

+s+%2B+42+=+84+

------------------

The 1st train's speed is 42 mi/hr

The 2nd train's speed is 84 mi/hr

---------------------------

check:

(2) +d+=+%28+s%2B42+%29%2A2+

(2) +d+=84%2A2+

(2) +d+=+168+

and

(1) +d+-+2s+=+s%2A2+

(1) +d+-+2%2A42+=+42%2A2+

(1) +d+=+84+%2B+84+

(1) +d+=+168+

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x2=400

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