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Mumz [18]
4 years ago
7

Larry and Paul start out running at a rate of 5 mph. Paul speeds up his pace after 5 miles to 10 mph but Larry continues the sam

e pace. How long after they start will they be 10 miles apart?
Mathematics
1 answer:
klasskru [66]4 years ago
3 0
<h2>Hello!</h2>

The answer is:

They will be 10 miles apart after 3 hours.

<h2>Why?</h2>

To calculate how long after they start will they be 10 miles apart, we need to assume that after 1 one hour, they were at the same distance (5 miles), then, calculate the time when they are 10 miles apart, knowing that Paul increased its speed two times, running first at 5mph and then, at 10 mph.

The time that will pass to be 10 miles apart can be calculated using the following equation:

TotalTime=TimeToReach5miles+TimeToBe10milesApart

Calculating the time to reach 5 miles for both Larry and Paul, at a speed of 5 mph, we have:

x=xo+v*t\\\\5miles=0+5mph*t\\\\t=\frac{5miles}{5mph}=1hour

We have that to reach a distance of 5 miles, they needed 1 hour. We need to remember that at this time, they were at the same distance.

If we want to know how many time will it take for them to be 10 miles apart with Paul increasing its speed to 10mph, we need to assume that after that time, the distance reached by Paul will be the distance reached by Larry plus 10 miles.

So, for the second moment (Paul increasing his speed) we have:

For Larry:

x_{L}=5miles+5mph*t

Therefore, the distance of Paul will be equal to the distance of Larry plus 10 miles.

For Paul:

x{L}+10miles=xo+10mph*t\\\\5miles+5mph*t+10miles=5miles+10mph*t\\\\5miles+10miles-5miles=10mph*t-5mph*t\\\\10miles=5mph*t\\\\t=\frac{10miles}{5mph}=2hours

Then, there will take 2 hours to Paul to be 10 miles apart from Larry after both were at 5 miles and Paul increased his speed to 10 mph.

Hence, calculating the total time, we have:

TotalTime=TimeToReach5miles+TimeToBe10milesApart

TotalTime=1hour+2hours=3hours

Have a nice day!

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Given:

The figure of rectangle.

To find:

a. The diagonal of the rectangle.

b. The area of the rectangle.

c. perimeter of the rectangle.

Solution:

(a)

In a right angle triangle,

\sin \theta=\dfrac{Perpendicular}{Hypotenuse}

\sin 30=\dfrac{12}{Hypotenuse}

\dfrac{1}{2}=\dfrac{12}{Hypotenuse}

Hypotenuse=12\times 2

Hypotenuse=24

So, the diagonal of the of the rectangle is 24 units.

(b)

In a right angle triangle,

\tan \theta=\dfrac{Perpendicular}{Base}

\tan 30=\dfrac{12}{Base}

\dfrac{1}{\sqrt{3}}=\dfrac{12}{Base}

Base=12\sqrt{3}

Length of the rectangle is 12 and width of the rectangle is 12\sqrt{3}. So, the area of the rectangle is:

Area=length \times width

Area=12 \times 12\sqrt{3}

Area=144\sqrt{3}

So, the area of the rectangle is 144\sqrt{3} sq. units.

(c)

Perimeter of the rectangle is:

P=2(length+width)

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P=24+24\sqrt{3}

P\approx 65.57

Therefore, the perimeter of the rectangle is about 65.57 units.

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