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KengaRu [80]
3 years ago
5

The tail of a 1-mile long train exits a tunnel exactly 3 minutes after the front of the train entered the tunnel. If the train i

s moving 60 miles per hour, how many miles long is the tunnel?
Mathematics
2 answers:
kow [346]3 years ago
7 0
60 mph is One mile a minute. So the train will start to exit the tunnel after 7 minutes. It will finish exiting the tunnel 1.5 mins later.
igomit [66]3 years ago
4 0

The tail of the train starts one mile from the mouth of the tunnel and it must travel that distance plus the length of the tunnel in 3 minutes.

 

Let t = tunnel length

 

t + 1  = distance the tail must travel in 3 minutes.

3 minutes = 1/20 hr

 

 (t+1) miles  /  60 m/hr  = 1/20 hr

t + 1 = 60/20

t+1 = 3

Subtract 1 from each side:

t = 2

The tunnel is 2 miles long.

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We have been given that a person places $6340 in an investment account earning an annual rate of 8.4%, compounded continuously. We are asked to find amount of money in the account after 2 years.

We will use continuous compounding formula to solve our given problem as:

, where

A = Final amount after t years,

P = Principal initially invested,

e = base of a natural logarithm,

r = Rate of interest in decimal form.  

Upon substituting our given values in above formula, we will get:

Upon rounding to nearest cent, we will get:

Therefore, an amount of $7499.82 will be in account after 2 years.

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3 years ago
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On Thursday afternoon at camp Alice played basketball and went swimming before dinner she spent 45 minutes playing basketball an
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Answer:

3:45 PM.

Step-by-step explanation:

We have been given that on Thursday afternoon at camp Alice played basketball and went swimming before dinner she spent 45 minutes playing basketball and one hour swimming dinner lasted for an hour. The dinner ended at 5:30 PM.

Since she spent 1 hour for swimming, so 1 hour before 5:30 PM would be 4:30 PM.

To find the time, when Alice started playing basketball, we need to find 45 minutes before 4:30 PM.

30 minutes before 4:30 PM would be 4:00 PM and 15 minutes before 4:00 PM would be 3:45 PM.

Therefore, Alice started playing basket ball at 3:45 PM.  

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3 years ago
3x-4y=16 and 5x+2y=44 using elimination strategy and comparing strategy
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NO LINKS!!! Part 2: Find the Lateral Area, Total Surface Area, and Volume. Round your answer to two decimal places.​
slava [35]

Answer:

<h3><u>Question 7</u></h3>

<u>Lateral Surface Area</u>

The bases of a triangular prism are the triangles.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the triangles (bases).

\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2

<u>Total Surface Area</u>

Area of the isosceles triangle:

\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2

Total surface area:

\implies \sf T.A.=2\:bases+L.A.=2\left(\dfrac{3\sqrt{391}}{4}\right)+138=167.66\:\:m^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)

<h3><u>Question 8</u></h3>

<u>Lateral Surface Area</u>

The bases of a hexagonal prism are the pentagons.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the pentagons (bases).

\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2

<u>Total Surface Area</u>

Area of a pentagon:

\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2

where a is the side length.

Therefore:

\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)

Total surface area:

\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)

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